Unit 7 — End of Topic Tests (reviewer copy: all keys and model answers visible)

End of Topic Test: Counting by weighing and the mole & Molar mass & Mass and moles (u07-t1 · 5 forms × 23 items · 115 total)

David's notes for James:
full notes file
# u07-t1 — notes/queries from David's side (U7-G1)

1. CALC-HEAVY FRQ SPLIT - frq_per_form = 3, OVERRIDING your gate data's frq_per_form=1 baseline (confirm the composition): QB-v2 sec.4 places 'mole conversions' BY NAME in the 2-3-FRQ class, and this scope carries 7 atomKind=quant lessons + 2 more with numeric gate seeds (9 of 21 lessons calculation-bearing, 15 of 63 pool seeds numeric) and THREE distinct multi-step procedure families. Three FRQs = exactly one per procedure family AND one per topic: FRQ-1 (seq 21, MOL-002) = counting-by-weighing division with the error-diagnosis part, FRQ-2 (seq 22, MOL-013) = molar-mass assembly incl. the PARENTHESES formula, FRQ-3 (seq 23, MOL-021, terminal capstone) = the two-stage mass<->moles chain both directions. Two FRQs would leave a full calculation family MCQ/numeric-only - what QB sec.4 says multiple choice measures worst. Totals stay inside the mandated window: 20 objective + 3 FRQ = 23 items/form. D5: min(20, 6 x 21 lessons = 126) -> the cap binds at 20, no scaling. Full adjudication text in blueprint.composition.frq_split.
2. 21 LESSONS vs 20 OBJECTIVE SLOTS (confirm - U3-G3 precedent): MOL-021 (the capstone) is covered by the seq-23 FRQ ALONE; MOL-002 and MOL-013 each hold an objective row AND their FRQ home. Every other scope lesson holds exactly one objective row.
3. INTERACTION FORMATS (confirm): per form 13 mcq / 7 numeric ENTRY / 3 FRQ. The HS numeric-entry rule (never numeric-answer mcq) applied to every quant lesson with a typeable plain-number answer; MOL-019 stays mcq (ranking has no numeric-entry shape; all 3 of its gate seeds are mcq - format parity); MOL-016 is the equation FACT atom - mcq. MOL-005's numeric gate seed displays 6.022 x 10^23, which the shipped V-T6 contract cannot pass (plain-numeral display regex) - its row stays mcq and the ADDITIVE V-T6 e-notation display lane (parse 'coefficient x 10^exp', count sigFigs on the coefficient, + a make_blind_packet numeric echo) is PROPOSED-NOT-APPLIED for _shared (read-only this wave; full spec in gate_config.adjudication_notes). Format parity holds meanwhile (2 of MOL-005's 3 gate seeds are mcq). No scientific-notation entry slots exist this gate.
4. ADJUDICATED INSTANCE FAMILIES (recorded in gate_config.json, enforced by the gate-local validate_with_adjudications.py over the UNTOUCHED shared validator - U5-G2/U3-G3 wrapper protocol, extended so the wrapper RE-DERIVES the full colliding-gram sets and no hidden '(+N more)' gram is excused unseen): (1) avogadro-rule-content - 6.022 x 10^23 is the TAUGHT FACT of MOL-005/MOL-010, never freshened (132 fail lines; seq 4/17 option content + B2/E2/B9/E9); (2) canonical-element-molar-masses - QB rule 8 MANDATES real 1-decimal element M values, supplied in stems per the course's own '(molar masses: ...)' shell (347 lines); (3) canonical-compound-molar-masses - real M of named compounds: N2 28.0, ethane 30.0, acetylene 26.0, urea 60.0 + ammonium nitrate 80.0 at the kept seed pair 18A (6 lines); (4) constants-shell-grams - grams built ONLY of shell words + element symbols + canonical constants, token-by-token verified (61 lines). Every sample mass, mole amount, answer, and substance outside the families stays fresh and fully scanned. 546/546 V-T7 fail lines covered, 0 uncovered, 11/12 shared validators PASS -> wrapper exit 0.
5. DETECTION EXTRAS NOT ADOPTED (flagged for you + David): stage-1 candidate extra_materials (potassium/barium/ammonium/nitrate...) were NOT added to the validator's detection lexicon - the authoring ledger legitimately places those words on two same-form slots (recorded repeats), so adding them would manufacture cross-form fails the ledger already governs; word-level exclusivity for non-lexicon words stays enforced by the ledger + blind review. The pool-60 'nitrate' overlap with MOL-021's own Ca(NO3)2 fluency is recorded so any future adoption is made knowingly.
6. SIG-FIG / PRECISION CONVENTIONS (confirm - the U7-G2 sol canon applied under sol's authoritative loop-1 ruling, all content made propagation-correct): seq-11 displays report the 3 sf their least-precise given supports (124 / 257) with half-unit tolerance; seq-16 given amounts re-valued to 4 sf (4.000/2.000/1.250 mol) so the 1-decimal keyed products are exactly correct; FRQ23(b) amounts 4 sf across stem/prompt/rubric/model; FRQ23(c) carries a '3 significant figures' ask on every form; mole-entry slots carry NO precision ask and keep minimal exact displays (value-compare grading cannot assess trailing zeros); FRQ finals ANNOUNCE precision in the stem. Every numeric item exports precisionAssessed + gradingNote + wrongValueDiagnostics.
7. NATURE-FIXED CASE SPACE AT SEQ 11 (MOL-011, confirm): real molecular elements form a closed set (H2 N2 O2 O3 F2 Cl2 Br2 I2 P4 S8) largely consumed by the lesson's own teaching text - five parallel forms cannot all be corpus-fresh. Resolution: P4 (your seed) / I2 / S8 / O3 / S2 with figure-borne printed values (the pt-square reference-figure blind family) where a stem-borne constant would collide; the A5 same-seq sulfur share is recorded and was routed to blind review (no telegraphing found).
8. SEED CENSUS (SEED RULING - confirm): 51 of your 63 pool seeds placed (stage-1 A=22 incl. 2 FRQ part seeds / B=16 / C=13; D and E fully authored), 12 spares each with a recorded reason; 22 placements flagged must_freshen_instance and freshened in CONTENT (no validator relief); 0 NOTA/AOTA in the pool (replace_distractor unused). FRQ-3 is demo-seeded: your curated pool 60/61 became form-A parts (a)/(b) near-verbatim; pool 62 was RESTRUCTURED into part (c) with its NaOH freshened (it is MOL-021's own worked-example instance - memory-answerable). Freshening a curated seed is flagged for your confirmation.
9. FRESHNESS DEVIATIONS FROM THE AUTHORING PLAN (all recorded in item source strings + fix_log): 13A K2SO4->BaSO4 (adjacent-potassium repeat), 15C SO2->benzene + 19C CaS->SO2 ('calcium'/cross-form sulfur), 16C NiO->GeO2, 16D CdS->Ga2O3, 19A LiCl+helium->SrCl2+formula-only H2 ('lithium'/'helium' are MOL-019-OWN corpus words), 19B NaBr leg re-valued to 51.45 g + formula-only ('sodium' MOL-019-own), 22C Ca3(PO4)2->Mn2(SO4)3 + 22D Ba(NO3)2->Bi(NO3)3 ('calcium'/'barium' MOL-013-OWN), 20A glucose->unnamed compound (180.0 corpus-consumed), float-exactness re-values 23B(a) 470.1->626.8 / 15D 276.3->184.2 / 16A/B/D moles, arithmetic corrections to canonical 1-decimal sums (Sr(NO3)2 211.6, CoCl2 129.9, Zn3(PO4)2 386.2), and A#6 seed per-bearing mass 4.5->5.4 g (broke a same-form same-value share with the A#21 FRQ seed - blind round-2 catch).
10. FRQ TYPED-WORK CONVENTIONS (confirm): model answers/rubrics show the full working but avoid literal scaffold-marker strings (V-T9 boundary); nothing typed requires sub/superscripts (V-T12); FRQ21 stems state 'Treat the given masses as exact.' and the tare-ambiguity phrasing 'the objects alone have a total mass' (blind round-1 sweep); seq-16 and FRQ23(b) ask for '1 decimal place' (course convention: taught molar masses and given amounts are exact bookkeeping values); FRQ21(c) diagnose-the-error part mirrors MOL-002's taught error-diagnosis (your demo pool design).
11. ACCEPTED ADVISORY CLASSES (recorded, David's call to reopen): (a) cross-item procedural cueing (seq6<->seq21, seq16<->seq23(b), seq-3 equation row -> later numerics) - blueprint-inherent: recognition and execution of the same taught route are distinct blueprint rows and the FRQ stems themselves mandate the worked stages (U7-G2 precedent); (b) row-5 residual purism - MOL-001's own rule lines ('No instrument you can buy reads out a particle count'; 'worked out from the mass') are course-local authority; (c) house longest-option / taught-wording-echo style - V-T4 quantified balance PASS, 148/148 blind key picks, 0 defensible-distractor fails; (d) 'engineered exact divisions' (D#23) - the V-T6 exact-division law; (e) sol PASS-MINORS left as recorded: A#12 uniform 1-dp option style, B#17 whole-number foil visibly contradicted by the stem's 50.9, E#17 d1 loose 'g/mol times more' wording.
12. BINDING INSTANCE RULE (course-wide): no fissile/weapons/dose framings anywhere - 0 hits across all 115 items at authoring and re-swept at packaging on forms_final + the export body (lanthanide/actinide line respected: cerium/lanthanum used, no actinides); the palette is classroom-safe throughout.
13. FRQ EXPORT LAYOUT (packaging patch, confirm the seq-23 rubric attachment): the 15 FRQs were authored with part markers inline; the exporter's lossless-split contract (U01-G2 canon) required blank-line-separated prompt parts + literal '(part x)' rubric tags, so packaging patch 1 inserted the whitespace breaks (7 per form, zero token change) and the tags (9 per form; leading '(a)' -> '(part a)' etc.). On seq 23 the two 35-point stage-criteria each grade work spanning parts (a) AND (b) ('Stage-1 molar masses assembled ... in both (a) and (b)'; 'Stage-2 equation work: (a) ... (b) ...') and carry the verbatim tag '(parts a and b)' - per the U01-G2 rule they attach to the LAST part named, so in the export part (a) shows 0 standalone rubric rows and part (b) carries 70 points; part (a)'s observables live inside those two spanning criteria and points are counted once (sums stay 100). Confirm this reads correctly in your renderer.
14. CRASH-RESUME PROVENANCE (disclosure, no action needed): the authoring fleet for this gate was killed THREE times mid-run by laptop restarts - mid stage-2 authoring (resumed: 13 pre-crash JSONs audited, one schema-drift row quarantined verbatim at _crash_quarantine/ and carried content-unchanged), before the stage-4-7 QC session (resumed: every JSON json.loads-verified, adjudicated validator re-run byte-consistent, no pre-crash blind/sol artifacts existed so stages 4-7 ran in full), and after the stage-7 finalize wrote forms_final.json + validator_run_final.txt (resumed at packaging, this session). Each resume re-audited the full disk state (json.loads on every artifact, 5x23 form census, forms_final payload re-asserted identical to forms_v1) before running ONLY the missing stages; no truncated artifact was found.

Form A

MOL-004 · easy · recall · seq 1 · seed:U7-G1/pool[9] (= MOL-004 gateItems[0]) — FRESHENED per rows/_AUTHORING_PLAN.md row 1: methane -> propane (cross-scope word), factor 0.5 -> 5 (0.5 corpus-consumed); whole-number-only foil replaced by the needs-molar-mass foil; inverted + elastic trap classes kept · mcq4
How does the number of molecules in 5 mol of propane, C₃H₈, compare with the number in 1 mol of propane?
Five times as many. KEY
One-fifth as many.
The same number.
It cannot be determined without propane's molar mass.
MOL-007 · easy · recall · seq 2 · seed:U7-G1/pool[20] (= MOL-007 gateItems[2]) — verbatim (field mapping only) · mcq4
What does a substance's molar mass state?
The mass in grams of one mole of the substance. KEY
The mass in grams of one particle of the substance.
The number of particles in one mole of the substance.
The number of moles in one gram of the substance.
MOL-016 · easy · recall · seq 3 · seed:U7-G1/pool[45] (= MOL-016 gateItems[0], James) — verbatim (field mapping only) · mcq4
In the equation n = m/M, which symbol stands for the quantity an electronic balance measures?
m — the mass of the sample KEY
n — the amount of substance in moles
M — the molar mass
None of them — a balance measures moles directly.
MOL-005 · easy · recall · seq 4 · seed:U7-G1/pool[14] (= MOL-005 gateItems[2], James) — verbatim (field mapping only; HTML entities/tags → Unicode) · mcq4
Which statement describes Avogadro's number?
The number of particles in one mole of any substance: 6.022 × 10²³. KEY
The number of particles in one gram of any substance.
The largest number of particles a balance can detect.
The number of moles in any laboratory sample.
MOL-001 · medium · application · seq 5 · seed:U7-G1.pool[0] (MOL-001 gateItems[0], silver-ring sensitive-balance claim) — kept verbatim (silver = cross-scope WARN-class single use, recorded silver(5A) in the _AUTHORING_PLAN ledger; no numbers) · mcq4
A student says a laboratory's most sensitive balance can display the number of atoms in a silver ring. What is wrong with the claim?
A balance of any sensitivity measures mass only — the atom count must be worked out from that mass. KEY
Nothing is wrong — top laboratory balances switch to a count display.
The claim is right for a pure element like silver — a count display only fails for mixtures.
The claim fails only because silver atoms happen to be unusually heavy.
MOL-002 · medium · application · seq 6 · seed:U7-G1/pool[3] (= MOL-002 gateItems[0]) — stem, worded 5-line scaffold, and inverted/product diagnostics carried near-verbatim; tolerance tightened 0.01 -> 0 with integer:true per _AUTHORING_PLAN row 6 (exact integer counts, RECORDED); off-by-decimal diagnostic added per the plan's fixed set; inverted-quotient diagnostic value rounded from the seed's printed 0.009259259259259259 | BLIND WAVE-2 FRESHEN: the seed's 4.5 g per-bearing mass duplicated A#21's seed pin mass on the same form (r3 cross-item cue) — re-valued to 5.4 g -> 90 bearings (exact division), diagnostics re-computed; SEED RULING (no freshness exemption), flagged for David/James. · numeric
A jar holds identical steel ball bearings. The bearings alone have a total mass of 486.0 g, and one bearing has a mass of 5.4 g. How many bearings are in the jar?
Answer: 90 bearings KEY
also accepted: 90 bearings
wrongValueDiagnostics
[
 {
  "value": 0.0111,
  "misconception": "you divided the mass of one bearing by the total mass — the total goes on top",
  "feedback": "This answer is wrong — you divided the mass of one bearing by the total mass — the total goes on top"
 },
 {
  "value": 2624.4,
  "misconception": "you multiplied the two masses instead of dividing",
  "feedback": "This answer is wrong — you multiplied the two masses instead of dividing"
 },
 {
  "value": 900,
  "misconception": "you slipped a decimal place in the division — check by multiplying back: this count times the mass of one bearing comes to ten times the jar's total mass",
  "feedback": "This answer is wrong — you slipped a decimal place in the division — check by multiplying back: this count times the mass of one bearing comes to ten times the jar's total mass"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\ntotal mass = 486.0 g\nmass of one bearing = 5.4 g\nWrite down the equation:\nnumber of bearings = total mass / mass of one bearing\nSubstitute in the values, and calculate:\nnumber of bearings = 486.0 / 5.4\nnumber of bearings = 90",
 "anticipatedWrong": [
  {
   "value": 0.0111,
   "misconception": "you divided the mass of one bearing by the total mass — the total goes on top",
   "feedback": "This answer is wrong — you divided the mass of one bearing by the total mass — the total goes on top"
  },
  {
   "value": 2624.4,
   "misconception": "you multiplied the two masses instead of dividing",
   "feedback": "This answer is wrong — you multiplied the two masses instead of dividing"
  },
  {
   "value": 900,
   "misconception": "you slipped a decimal place in the division — check by multiplying back: this count times the mass of one bearing comes to ten times the jar's total mass",
   "feedback": "This answer is wrong — you slipped a decimal place in the division — check by multiplying back: this count times the mass of one bearing comes to ten times the jar's total mass"
  }
 ]
}
MOL-003 · medium · application · seq 7 · seed:U7-G1/pool[6] (= MOL-003 gateItems[0], N2 doubter-reply) — carried verbatim (HTML subscripts -> Unicode N₂ only); nitrogen = cross-scope WARN-class single use, plan-ledger record nitrogen(7A+19A same-form-A non-adjacent); key = the taught mechanism sentence class (A3) · mcq4
A tank holds pure nitrogen gas, N₂. A student doubts that weighing can count its molecules, since nobody has measured each molecule separately. Which reply is correct?
N₂ molecules share one steady average mass, so the tank's mass counts the molecules. KEY
The student is right — counting requires each particle to be measured once.
Weighing counts molecules only in solids, where the particles hold still.
Counting is possible because nitrogen molecules all move at the same speed.
MOL-006 · medium · application · seq 8 · seed:U7-G1/pool[15] (= MOL-006 gateItems[0]) — verbatim (field mapping only; magnesium = cross-scope WARN-class single use, label-only usage, recorded magnesium(8A) in the _AUTHORING_PLAN ledger) · mcq4
Which particle does a mole of magnesium oxide, MgO, count?
Formula units KEY
Molecules
Atoms
Ions
MOL-009 · medium · application · seq 9 · seed:U7-G1/pool[26] (= MOL-009 gateItems[2]) — verbatim (field mapping only); silicon + 28.1 corpus-clean (stage-1 scan: silicon appears only in MOL-009's own gate, not the corpus) · mcq4
Silicon's periodic-table square prints 28.1. Which pair of statements is right?
One silicon atom: 28.1 amu. 1 mol of silicon atoms: 28.1 g. KEY
One silicon atom: 28.1 g. 1 mol of silicon atoms: 28.1 amu.
One silicon atom: 28.1 amu. 1 mol of silicon atoms: 28.1 amu.
One silicon atom: 28.1 g. 1 mol of silicon atoms: 28.1 g.
MOL-008 · medium · application · seq 10 · seed:U7-G1/pool[21] (= MOL-008 gateItems[0]) — carried near-verbatim; printed square moved from stem prose into the pt-square reference-table figure payload (blind-family requirement, blueprint row 10); doubled-value diagnostic added as the plan's third class · numeric
The periodic-table square for copper is shown. State copper's molar mass, in g/mol. Give your answer to 3 significant figures.
FIGURE (pt-square-reference-table · stimulus) — title: "periodic-table square: copper" · rows: [["atomic number", "29"], ["symbol", "Cu"], ["average atomic mass", "63.5"]] · style: "single periodic-table element square, exactly as printed on the classroom periodic table"
raw spec
{
 "figureRequired": true,
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "pt-square-reference-table",
  "title": "periodic-table square: copper",
  "rows": [
   [
    "atomic number",
    "29"
   ],
   [
    "symbol",
    "Cu"
   ],
   [
    "average atomic mass",
    "63.5"
   ]
  ],
  "style": "single periodic-table element square, exactly as printed on the classroom periodic table"
 }
}
Answer: 63.5 g/mol KEY
also accepted: 63.5 g/mol
wrongValueDiagnostics
[
 {
  "value": 29,
  "misconception": "you read the atomic number — the smaller whole number counts protons and is not a mass",
  "feedback": "This answer is wrong — you read the atomic number — the smaller whole number counts protons and is not a mass"
 },
 {
  "value": 34.5,
  "misconception": "you subtracted the atomic number from the atomic mass — that neutron-count move belongs to isotope work, not molar mass",
  "feedback": "This answer is wrong — you subtracted the atomic number from the atomic mass — that neutron-count move belongs to isotope work, not molar mass"
 },
 {
  "value": 127.0,
  "misconception": "you doubled the printed value — the square's average atomic mass is already the whole molar mass for an element of single atoms",
  "feedback": "This answer is wrong — you doubled the printed value — the square's average atomic mass is already the whole molar mass for an element of single atoms"
 }
]
feedback
{
 "keyExplanation": "The square prints two numbers: the atomic number 29 and the average atomic mass 63.5. The average atomic mass, read in grams per mole, is the molar mass: copper's is 63.5 g/mol. The 29 counts protons and is never a mass.",
 "anticipatedWrong": [
  {
   "value": 29,
   "misconception": "you read the atomic number — the smaller whole number counts protons and is not a mass",
   "feedback": "This answer is wrong — you read the atomic number — the smaller whole number counts protons and is not a mass"
  },
  {
   "value": 34.5,
   "misconception": "you subtracted the atomic number from the atomic mass — that neutron-count move belongs to isotope work, not molar mass",
   "feedback": "This answer is wrong — you subtracted the atomic number from the atomic mass — that neutron-count move belongs to isotope work, not molar mass"
  },
  {
   "value": 127.0,
   "misconception": "you doubled the printed value — the square's average atomic mass is already the whole molar mass for an element of single atoms",
   "feedback": "This answer is wrong — you doubled the printed value — the square's average atomic mass is already the whole molar mass for an element of single atoms"
  }
 ]
}
MOL-011 · medium · application · seq 11 · seed:U7-G1/pool[31] (= MOL-011 gateItems[1]) — carried near-verbatim (placement scan clean; 31.0 reserved for this slot per stage-1 flag 9); divided-by-subscript diagnostic added as the third class · numeric
White phosphorus exists as P₄ molecules. Phosphorus's periodic-table square prints 31.0. Calculate the molar mass of P₄, in g/mol. Give your answer to 3 significant figures.
Answer: 124 g/mol KEY
also accepted: 124 g/mol
wrongValueDiagnostics
[
 {
  "value": 31.0,
  "misconception": "you ignored the subscript — each P₄ molecule holds four atoms, so the atom's value quadruples",
  "feedback": "This answer is wrong — you ignored the subscript — each P₄ molecule holds four atoms, so the atom's value quadruples"
 },
 {
  "value": 35.0,
  "misconception": "you added the subscript as if it were a mass — the 4 is a count and multiplies",
  "feedback": "This answer is wrong — you added the subscript as if it were a mass — the 4 is a count and multiplies"
 },
 {
  "value": 7.75,
  "misconception": "you divided by the subscript instead of multiplying — the 4 counts atoms per molecule",
  "feedback": "This answer is wrong — you divided by the subscript instead of multiplying — the 4 counts atoms per molecule"
 }
]
feedback
{
 "keyExplanation": "The subscript counts the atoms in one molecule: 4.\nmolar mass of P₄ = 4 × 31.0\nmolar mass of P₄ = 124.0, reported as 124 g/mol (3 significant figures)\nEach P₄ molecule weighs four phosphorus atoms' worth.",
 "anticipatedWrong": [
  {
   "value": 31.0,
   "misconception": "you ignored the subscript — each P₄ molecule holds four atoms, so the atom's value quadruples",
   "feedback": "This answer is wrong — you ignored the subscript — each P₄ molecule holds four atoms, so the atom's value quadruples"
  },
  {
   "value": 35.0,
   "misconception": "you added the subscript as if it were a mass — the 4 is a count and multiplies",
   "feedback": "This answer is wrong — you added the subscript as if it were a mass — the 4 is a count and multiplies"
  },
  {
   "value": 7.75,
   "misconception": "you divided by the subscript instead of multiplying — the 4 counts atoms per molecule",
   "feedback": "This answer is wrong — you divided by the subscript instead of multiplying — the 4 counts atoms per molecule"
  }
 ]
}
MOL-014 · medium · application · seq 12 · seed:U7-G1/pool[39] (= MOL-014 gateItems[0]) — FRESHENED per rows/_AUTHORING_PLAN.md row 12: base 3.0 -> 6.0 mol so no option value lands on the corpus-consumed 1.5; foil classes and wording carried · mcq4
A sample of potassium chloride, KCl, contains 6.0 mol. A second sample of potassium chloride has twice the mass of the first. How many moles does the second sample contain?
12.0 mol KEY
3.0 mol
6.0 mol
24.0 mol
MOL-012 · medium · application · seq 13 · authored (stage 2; RECORDED DEVIATION from plan row 13: K2SO4 -> BaSO4 to break the 12A/13A adjacent-potassium repeat the plan created; seed pool[33] CaCO3 was MF per its own placement note; seed working shell carried) · numeric
Barium sulfate has the formula BaSO₄. Use these molar masses: Ba 137.3, S 32.1, O 16.0 g/mol. Calculate BaSO₄'s molar mass, in g/mol, to one decimal place. Give your answer to 4 significant figures.
Answer: 233.4 g/mol KEY
also accepted: 233.4 g/mol
wrongValueDiagnostics
[
 {
  "value": 185.4,
  "misconception": "you ignored the subscript — the formula counts four oxygens, not one",
  "feedback": "This answer is wrong — you ignored the subscript — the formula counts four oxygens, not one"
 },
 {
  "value": 741.6,
  "misconception": "you multiplied every element by the subscript — the 4 belongs to oxygen alone",
  "feedback": "This answer is wrong — you multiplied every element by the subscript — the 4 belongs to oxygen alone"
 },
 {
  "value": 189.4,
  "misconception": "you added the subscript as if it were a mass — the 4 is a count and multiplies the oxygen value",
  "feedback": "This answer is wrong — you added the subscript as if it were a mass — the 4 is a count and multiplies the oxygen value"
 }
]
feedback
{
 "keyExplanation": "Count the atoms the formula unit shows: 1 barium, 1 sulfur, and 4 oxygens.\nmolar mass of BaSO₄ = 137.3 + 32.1 + 4 × 16.0\nmolar mass of BaSO₄ = 233.4 g/mol\nThe subscript 4 multiplies oxygen only; barium and sulfur appear once each.",
 "anticipatedWrong": [
  {
   "value": 185.4,
   "misconception": "you ignored the subscript — the formula counts four oxygens, not one",
   "feedback": "This answer is wrong — you ignored the subscript — the formula counts four oxygens, not one"
  },
  {
   "value": 741.6,
   "misconception": "you multiplied every element by the subscript — the 4 belongs to oxygen alone",
   "feedback": "This answer is wrong — you multiplied every element by the subscript — the 4 belongs to oxygen alone"
  },
  {
   "value": 189.4,
   "misconception": "you added the subscript as if it were a mass — the 4 is a count and multiplies the oxygen value",
   "feedback": "This answer is wrong — you added the subscript as if it were a mass — the 4 is a count and multiplies the oxygen value"
  }
 ]
}
MOL-013 · medium · application · seq 14 · seed:U7-G1/pool[36] (= MOL-013 gateItems[0]) — RE-CATIONED per rows/_AUTHORING_PLAN.md row 14 (magnesium = own-WE word); plan's 211.7 corrected to the canonical-constant sum 211.6 (RECORDED arithmetic correction); seed working and diagnostic classes carried · numeric
Calculate the molar mass of strontium nitrate, Sr(NO₃)₂. (molar masses: Sr 87.6, N 14.0, O 16.0 g/mol.) Give your answer in g/mol to one decimal place. Give your answer to 4 significant figures.
Answer: 211.6 g/mol KEY
also accepted: 211.6 g/mol
wrongValueDiagnostics
[
 {
  "value": 149.6,
  "misconception": "you ignored the subscript outside the parentheses — the 2 multiplies the whole NO₃ group",
  "feedback": "This answer is wrong — you ignored the subscript outside the parentheses — the 2 multiplies the whole NO₃ group"
 },
 {
  "value": 299.2,
  "misconception": "you multiplied the strontium as well — the outside subscript reaches only what sits inside the parentheses",
  "feedback": "This answer is wrong — you multiplied the strontium as well — the outside subscript reaches only what sits inside the parentheses"
 },
 {
  "value": 163.6,
  "misconception": "you applied the outside 2 only to the nitrogen — every atom inside the parentheses is doubled, all three oxygens included",
  "feedback": "This answer is wrong — you applied the outside 2 only to the nitrogen — every atom inside the parentheses is doubled, all three oxygens included"
 }
]
feedback
{
 "keyExplanation": "Inside the parentheses: 14.0 + 3 × 16.0 = 62.0\nMultiply by the outside subscript: 2 × 62.0 = 124.0\nAdd the atoms outside the parentheses: 87.6 + 124.0\nM = 211.6 g/mol",
 "anticipatedWrong": [
  {
   "value": 149.6,
   "misconception": "you ignored the subscript outside the parentheses — the 2 multiplies the whole NO₃ group",
   "feedback": "This answer is wrong — you ignored the subscript outside the parentheses — the 2 multiplies the whole NO₃ group"
  },
  {
   "value": 299.2,
   "misconception": "you multiplied the strontium as well — the outside subscript reaches only what sits inside the parentheses",
   "feedback": "This answer is wrong — you multiplied the strontium as well — the outside subscript reaches only what sits inside the parentheses"
  },
  {
   "value": 163.6,
   "misconception": "you applied the outside 2 only to the nitrogen — every atom inside the parentheses is doubled, all three oxygens included",
   "feedback": "This answer is wrong — you applied the outside 2 only to the nitrogen — every atom inside the parentheses is doubled, all three oxygens included"
  }
 ]
}
MOL-017 · medium · application · seq 15 · seed:U7-G1/pool[49] (= MOL-017 gateItems[0], N2O) — RE-INSTANCED per rows/_AUTHORING_PLAN.md row 15 (44.0 = corpus-consumed CO₂ constant; seed specNotes order re-instancing): propane 176.4/44.1; 5-line scaffold, sig-figs convention, and both seed diagnostics carried; decimal-slip diagnostic added · numeric
A storage tank holds 176.4 g of propane, C₃H₈ (M = 44.1 g/mol). How many moles of propane is that? Give your answer in moles to 3 significant figures.
Answer: 4.0 mol KEY
also accepted: 4.00 mol
wrongValueDiagnostics
[
 {
  "value": 0.25,
  "misconception": "you divided the molar mass by the mass instead of the mass by the molar mass",
  "feedback": "This answer is wrong — you divided the molar mass by the mass instead of the mass by the molar mass"
 },
 {
  "value": 7779.24,
  "misconception": "you multiplied the mass by the molar mass instead of dividing",
  "feedback": "This answer is wrong — you multiplied the mass by the molar mass instead of dividing"
 },
 {
  "value": 40.0,
  "misconception": "you slipped a decimal place in the division — check by multiplying back: this count times 44.1 g/mol comes to ten times the tank's mass",
  "feedback": "This answer is wrong — you slipped a decimal place in the division — check by multiplying back: this count times 44.1 g/mol comes to ten times the tank's mass"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nm = 176.4 g\nM = 44.1 g/mol\nWrite down the equation:\nn = m / M\nSubstitute in the values, and calculate:\nn = 176.4 / 44.1\nn = 4.00 mol\nSanity check: a lab-to-tank-scale mole count lands around 0.01 to 10 mol — 4.00 mol passes.",
 "anticipatedWrong": [
  {
   "value": 0.25,
   "misconception": "you divided the molar mass by the mass instead of the mass by the molar mass",
   "feedback": "This answer is wrong — you divided the molar mass by the mass instead of the mass by the molar mass"
  },
  {
   "value": 7779.24,
   "misconception": "you multiplied the mass by the molar mass instead of dividing",
   "feedback": "This answer is wrong — you multiplied the mass by the molar mass instead of dividing"
  },
  {
   "value": 40.0,
   "misconception": "you slipped a decimal place in the division — check by multiplying back: this count times 44.1 g/mol comes to ten times the tank's mass",
   "feedback": "This answer is wrong — you slipped a decimal place in the division — check by multiplying back: this count times 44.1 g/mol comes to ten times the tank's mass"
  }
 ]
}
MOL-018 · medium · application · seq 16 · seed:U7-G1/pool[52] (= MOL-018 gateItems[0], O2) — RE-INSTANCED per rows/_AUTHORING_PLAN.md row 16 (32.0 corpus-consumed); mole amount re-valued to the float-exact 2.50 (RECORDED); seed's supplied-M shell, 'Make m the subject' scaffold line, and both diagnostics carried; STAGE-3 re-value 2.50 -> 4.00 mol (2.50 sits in MOL-018 spec text; 4.00 scan-exempt, float-exact) · numeric
A plating bath needs 4.000 mol of copper(II) sulfate, CuSO₄ (M = 159.6 g/mol). What mass of copper(II) sulfate is that? Give your answer in grams to 1 decimal place.
Answer: 638.4 g KEY
also accepted: 638.4 g
wrongValueDiagnostics
[
 {
  "value": 0.02506,
  "misconception": "you divided the moles by the molar mass instead of multiplying them",
  "feedback": "This answer is wrong — you divided the moles by the molar mass instead of multiplying them"
 },
 {
  "value": 39.9,
  "misconception": "you divided the molar mass by the moles instead of multiplying them",
  "feedback": "This answer is wrong — you divided the molar mass by the moles instead of multiplying them"
 },
 {
  "value": 63.84,
  "misconception": "you slipped a decimal place in the product — 4.00 portions of 159.6 g must weigh over six hundred grams",
  "feedback": "This answer is wrong — you slipped a decimal place in the product — 4.00 portions of 159.6 g must weigh over six hundred grams"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nn = 4.000 mol\nM = 159.6 g/mol\nWrite down the equation:\nn = m / M\nMake m the subject:\nm = n × M\nSubstitute in the values, and calculate:\nm = 4.000 × 159.6\nm = 638.4 g\nSanity check: more than 1 mol must weigh more than the 159.6 g molar mass — 638.4 g passes.",
 "anticipatedWrong": [
  {
   "value": 0.02506,
   "misconception": "you divided the moles by the molar mass instead of multiplying them",
   "feedback": "This answer is wrong — you divided the moles by the molar mass instead of multiplying them"
  },
  {
   "value": 39.9,
   "misconception": "you divided the molar mass by the moles instead of multiplying them",
   "feedback": "This answer is wrong — you divided the molar mass by the moles instead of multiplying them"
  },
  {
   "value": 63.84,
   "misconception": "you slipped a decimal place in the product — 4.00 portions of 159.6 g must weigh over six hundred grams",
   "feedback": "This answer is wrong — you slipped a decimal place in the product — 4.00 portions of 159.6 g must weigh over six hundred grams"
  }
 ]
}
MOL-010 · hard · transfer · seq 17 · seed:U7-G1/pool[27] (= MOL-010 gateItems[0]) — carried near-verbatim; 6.022 × 10²³ = taught rule content (scoped adjudication class, stage-1 flag 7) · mcq4
A student calls the amu–g/mol match 'a lucky coincidence.' Which reply is right?
The match is built in: one gram is 6.022 × 10²³ amu — the same number as the particles in one mole. KEY
It really is a coincidence, noticed after the two units were fixed.
The match holds only for the lightest elements, where the numbers stay small.
The match comes from all atoms having nearly equal masses.
MOL-015 · hard · transfer · seq 18 · seed:U7-G1/pool[42] (= MOL-015 gateItems[0]) — packet mass FRESHENED 250.0 -> 300.0 g (250.0 = MOL-001 spec-text number, corpus scan 2026-09-08); the real compound M's 60.0/80.0 kept as the recorded COMPOUND-M CONSTANTS family (QB rule 8); wording and foil classes carried · mcq4
A farmer buys two 300.0 g packets of fertilizer compounds: one of urea, CO(NH₂)₂ (molar mass 60.0 g/mol), and one of ammonium nitrate, NH₄NO₃ (molar mass 80.0 g/mol). Which packet contains more moles of its compound?
The urea packet. KEY
Mass figures alone can never settle it — the packets' particle counts would have to be measured directly.
The two packets contain equal numbers of moles.
The ammonium nitrate packet, because its formula packs more atoms into each unit.
MOL-019 · hard · transfer · seq 19 · seed:U7-G1/pool[54] (= MOL-019 gateItems[0]) — RE-LEGGED per rows/_AUTHORING_PLAN.md row 19 + recorded deviations: LiCl and helium legs dropped (both words sit in MOL-019's OWN corpus, and the plan's LiCl values tied iron) -> SrCl₂ 396.5/158.6 and formula-only H₂ 2.0/2.0; N₂ and iron legs kept with fresh masses; seed's foil geometry (by-mass / upside-down / smallest-M) carried · mcq4
Four samples: 84.0 g of nitrogen gas, N₂ (M = 28.0 g/mol); 396.5 g of strontium chloride, SrCl₂ (M = 158.6 g/mol); 111.6 g of iron (M = 55.8 g/mol); 2.0 g of H₂ gas (M = 2.0 g/mol). Rank them from most moles to fewest.
nitrogen gas > strontium chloride > iron > H₂ KEY
strontium chloride > iron > nitrogen gas > H₂
H₂ > iron > strontium chloride > nitrogen gas
H₂ > nitrogen gas > iron > strontium chloride
MOL-020 · hard · transfer · seq 20 · seed:U7-G1/pool[57] (= MOL-020 gateItems[0], glucose) — RE-INSTANCED to an unnamed powdered compound at 54.0/108.0 (RECORDED deviation: glucose collides cross-form with 8C and 180.0 is corpus-consumed; unnamed-substance register = the bank's own pool[44] precedent); the gate's single nothing-wrong key kept at A per seed specNotes; STAGE-3 re-value 54.0/0.500 -> 324.0/3.00 (0.500 is a scanned non-integer shared with four lessons' spec text; 324.0 corpus-clean, division float-exact) · mcq4
A question asks how many moles are in 324 g of a powdered compound whose molar mass is 108.0 g/mol. A student writes: n = 324 / 108.0 = 3.00 mol. What, if anything, is the error?
Nothing is wrong — the equation, the operation, and the size all check out. KEY
The solution multiplied by the molar mass instead of dividing.
The division is upside down and should be 108.0 / 324.0.
The answer fails the size check, because a 324 g sample must hold less than 1 mol.
MOL-002 · medium · application · seq 21 · authored FRQ (plan row 21; MOL-002 prod_seeds empty; the spared mcq seed pool[5] is shape reference only; order count 260 chosen over the plan's 250-class to keep the 18A packet-mass value class free, RECORDED) · frq
A hardware supplier counts identical cotter pins by weighing.
A hardware supplier counts identical cotter pins by weighing. (a) The cotter pins in a bin have a total mass of 549.0 g on their own, and one pin has a mass of 4.5 g. How many pins does the bin hold? Show every stage of your working: the values you start from, the equation in words, the substitution, and the count. (b) A customer orders 260 pins. What total mass of pins should the packer weigh out? Show your working the same way. (c) Another packer answers part (a) by computing 4.5 / 549.0. Explain what is wrong with that division and state the corrected count. Plain numbers and words are expected throughout — no special symbols. Treat the given masses as exact.
Part a. The cotter pins in a bin have a total mass of 549.0 g on their own, and one pin has a mass of 4.5 g. How many pins does the bin hold? Show every stage of your working: the values you start from, the equation in words, the substitution, and the count.
Model: Values: total mass = 549.0 g; mass of one pin = 4.5 g. Equation in words: number of pins = total mass divided by mass of one pin. Substitution: 549.0 / 4.5. Count: 122 pins. Check: 122 pins of 4.5 g each weigh exactly 549.0 g.
Part b. A customer orders 260 pins. What total mass of pins should the packer weigh out? Show your working the same way.
Model: Turn the equation around: mass to weigh out = number of pins times mass of one pin = 260 times 4.5 = 1170.0 g.
Part c. Another packer answers part (a) by computing 4.5 / 549.0. Explain what is wrong with that division and state the corrected count. Plain numbers and words are expected throughout — no special symbols. Treat the given masses as exact.
Model: The division is upside down — the total mass goes on top. 4.5 / 549.0 is about 0.008, and a bin cannot hold a fraction of a pin. The corrected division is 549.0 / 4.5 = 122 pins. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
Full model answer. (a) Values: total mass = 549.0 g; mass of one pin = 4.5 g. Equation in words: number of pins = total mass divided by mass of one pin. Substitution: 549.0 / 4.5. Count: 122 pins. Check: 122 pins of 4.5 g each weigh exactly 549.0 g. (b) Turn the equation around: mass to weigh out = number of pins times mass of one pin = 260 times 4.5 = 1170.0 g. (c) The division is upside down — the total mass goes on top. 4.5 / 549.0 is about 0.008, and a bin cannot hold a fraction of a pin. The corrected division is 549.0 / 4.5 = 122 pins. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
MOL-013 · medium · application · seq 22 · authored FRQ (plan row 22; spared seed pool[38] Fe(OH)3 which-calculation mcq = case material for the (b) critique register; iron 19A/22A repeat RECORDED; wrong value 74.8 = the plan's 3-only-on-hydrogen class) · frq
(a) Calculate the molar mass of iron(III) hydroxide, Fe(OH)₃. (molar masses: Fe 55.8, O 16.0, H 1.0 g/mol.) Show every stage: the value inside the parentheses, the multiplication by the outside subscript, and the final sum. (b) A classmate reports M = 74.8 g/mol for the same compound. Identify exactly what the classmate did wrong and state the corrected molar mass. (c) Now calculate the molar mass of the compound (NH₄)₃PO₄. (molar masses: N 14.0, H 1.0, P 31.0, O 16.0 g/mol.) Show the same stages. Plain numbers and words are expected — no special symbols.
(a) Calculate the molar mass of iron(III) hydroxide, Fe(OH)₃. (molar masses: Fe 55.8, O 16.0, H 1.0 g/mol.) Show every stage: the value inside the parentheses, the multiplication by the outside subscript, and the final sum. (b) A classmate reports M = 74.8 g/mol for the same compound. Identify exactly what the classmate did wrong and state the corrected molar mass. (c) Now calculate the molar mass of the compound (NH₄)₃PO₄. (molar masses: N 14.0, H 1.0, P 31.0, O 16.0 g/mol.) Show the same stages. Plain numbers and words are expected — no special symbols.
Part a. Calculate the molar mass of iron(III) hydroxide, Fe(OH)₃. (molar masses: Fe 55.8, O 16.0, H 1.0 g/mol.) Show every stage: the value inside the parentheses, the multiplication by the outside subscript, and the final sum.
Model: Inside the parentheses: 16.0 + 1.0 = 17.0. Multiply by the outside subscript: 3 times 17.0 = 51.0. Add the iron, which sits outside the parentheses: 55.8 + 51.0 = 106.8. M of iron(III) hydroxide = 106.8 g/mol.
Part b. A classmate reports M = 74.8 g/mol for the same compound. Identify exactly what the classmate did wrong and state the corrected molar mass.
Model: The classmate applied the outside 3 only to the hydrogen: 55.8 + 16.0 + 3 times 1.0 = 74.8. The subscript outside parentheses multiplies EVERYTHING inside them — three oxygens and three hydrogens. Corrected: M = 106.8 g/mol.
Part c. Now calculate the molar mass of the compound (NH₄)₃PO₄. (molar masses: N 14.0, H 1.0, P 31.0, O 16.0 g/mol.) Show the same stages. Plain numbers and words are expected — no special symbols.
Model: Ammonium group: 14.0 + 4 times 1.0 = 18.0. The subscript after those parentheses triples the whole group: 3 times 18.0 = 54.0. Phosphate part: 31.0 + 4 times 16.0 = 95.0. Total: 54.0 + 95.0 = 149.0. M of ammonium phosphate = 149.0 g/mol. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
Full model answer. (a) Inside the parentheses: 16.0 + 1.0 = 17.0. Multiply by the outside subscript: 3 times 17.0 = 51.0. Add the iron, which sits outside the parentheses: 55.8 + 51.0 = 106.8. M of iron(III) hydroxide = 106.8 g/mol. (b) The classmate applied the outside 3 only to the hydrogen: 55.8 + 16.0 + 3 times 1.0 = 74.8. The subscript outside parentheses multiplies EVERYTHING inside them — three oxygens and three hydrogens. Corrected: M = 106.8 g/mol. (c) Ammonium group: 14.0 + 4 times 1.0 = 18.0. The subscript after those parentheses triples the whole group: 3 times 18.0 = 54.0. Phosphate part: 31.0 + 4 times 16.0 = 95.0. Total: 54.0 + 95.0 = 149.0. M of ammonium phosphate = 149.0 g/mol. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
MOL-021 · hard · application · seq 23 · seeds:U7-G1/pool[60]+pool[61] (James's demo_curated MOL-021 gate items, carried near-verbatim as parts (a)/(b)); part (c) = pool[62] restructured to free response with the pair freshened to the form's own two compounds (the seed's NaOH is MOL-021's own WE instance, MF RECORDED; plan's SnF2 candidate dropped — tin is form-B ledger property, RECORDED) · frq
(a) How many moles are in 202.2 g of potassium nitrate, KNO₃? (molar masses: K 39.1, N 14.0, O 16.0 g/mol.) Show both stages: first assemble the molar mass from the formula, then apply the mole equation. Give the answer in moles to 3 significant figures. (b) What is the mass, in grams, of 2.000 mol of aluminum sulfate, Al₂(SO₄)₃? (molar masses: Al 27.0, S 32.1, O 16.0 g/mol.) Show both stages, including how you rearrange the mole equation. Give the answer to 1 decimal place. (c) Bottle 1 holds 505.5 g of potassium nitrate; bottle 2 holds 1026.9 g of aluminum sulfate. Using the molar masses you found, work out which bottle holds more moles — show the working for both bottles, reporting each mole value to 3 significant figures. Plain numbers and words are expected — no special symbols.
(a) How many moles are in 202.2 g of potassium nitrate, KNO₃? (molar masses: K 39.1, N 14.0, O 16.0 g/mol.) Show both stages: first assemble the molar mass from the formula, then apply the mole equation. Give the answer in moles to 3 significant figures. (b) What is the mass, in grams, of 2.000 mol of aluminum sulfate, Al₂(SO₄)₃? (molar masses: Al 27.0, S 32.1, O 16.0 g/mol.) Show both stages, including how you rearrange the mole equation. Give the answer to 1 decimal place. (c) Bottle 1 holds 505.5 g of potassium nitrate; bottle 2 holds 1026.9 g of aluminum sulfate. Using the molar masses you found, work out which bottle holds more moles — show the working for both bottles, reporting each mole value to 3 significant figures. Plain numbers and words are expected — no special symbols.
Part a. How many moles are in 202.2 g of potassium nitrate, KNO₃? (molar masses: K 39.1, N 14.0, O 16.0 g/mol.) Show both stages: first assemble the molar mass from the formula, then apply the mole equation. Give the answer in moles to 3 significant figures.
Model: Stage 1 — molar mass of potassium nitrate: 39.1 + 14.0 + 3 times 16.0 = 101.1 g/mol. Stage 2 — moles: n = m/M = 202.2 / 101.1 = 2.00 mol.
Part b. What is the mass, in grams, of 2.000 mol of aluminum sulfate, Al₂(SO₄)₃? (molar masses: Al 27.0, S 32.1, O 16.0 g/mol.) Show both stages, including how you rearrange the mole equation. Give the answer to 1 decimal place.
Model: Stage 1 — molar mass of aluminum sulfate: inside the parentheses 32.1 + 4 times 16.0 = 96.1; tripled by the outside subscript to 288.3; plus two aluminum, 2 times 27.0 = 54.0; total 342.3 g/mol. Stage 2 — rearrange to make m the subject: m = n times M = 2.000 times 342.3 = 684.6 g.
Part c. Bottle 1 holds 505.5 g of potassium nitrate; bottle 2 holds 1026.9 g of aluminum sulfate. Using the molar masses you found, work out which bottle holds more moles — show the working for both bottles, reporting each mole value to 3 significant figures. Plain numbers and words are expected — no special symbols.
Model: Bottle 1: 505.5 / 101.1 = 5.00 mol. Bottle 2: 1026.9 / 342.3 = 3.00 mol. Bottle 1 holds more moles, even though bottle 2 is heavier — each mole of aluminum sulfate weighs far more. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
Full model answer. (a) Stage 1 — molar mass of potassium nitrate: 39.1 + 14.0 + 3 times 16.0 = 101.1 g/mol. Stage 2 — moles: n = m/M = 202.2 / 101.1 = 2.00 mol. (b) Stage 1 — molar mass of aluminum sulfate: inside the parentheses 32.1 + 4 times 16.0 = 96.1; tripled by the outside subscript to 288.3; plus two aluminum, 2 times 27.0 = 54.0; total 342.3 g/mol. Stage 2 — rearrange to make m the subject: m = n times M = 2.000 times 342.3 = 684.6 g. (c) Bottle 1: 505.5 / 101.1 = 5.00 mol. Bottle 2: 1026.9 / 342.3 = 3.00 mol. Bottle 1 holds more moles, even though bottle 2 is heavier — each mole of aluminum sulfate weighs far more. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.

Form B

MOL-004 · easy · recall · seq 1 · seed:U7-G1/pool[10] (= MOL-004 gateItems[1], spec text carried near-verbatim; element-free, no freshness flags) · mcq4
A student calls the mole 'the chemist's unit of mass.' Which correction is right?
The mole is a counting unit — one mole means one fixed number of particles. KEY
The student is right — mole amounts are read straight off a balance.
The mole is a unit of mass, but only for pure substances.
The mole is a counting unit, but its count changes from substance to substance.
MOL-007 · easy · recall · seq 2 · seed:U7-G1/pool[19] (= MOL-007 gateItems[1]) | STAGE-2 FRESHEN per must_freshen_instance + _AUTHORING_PLAN row 2: copper/63.5 -> cobalt/58.9 (copper + 63.5 reserved for the seq-10A pt-square figure seed; cobalt real M, corpus-clean in scope); weigh-out structure, option set, and misconception set carried; no copper/63.5 token remains in any field · mcq4
A student weighs out exactly 1 mol of cobalt, and the balance reads 58.9 g. Which value is cobalt's molar mass?
58.9 g/mol KEY
58.9 g
6.022 × 10²³ g/mol
1.0 g/mol
MOL-016 · easy · recall · seq 3 · seed:U7-G1/pool[46] (= MOL-016 gateItems[1], James) — verbatim (field mapping only) · mcq4
One quantity in n = m/M is measured in grams per mole. Which symbol is it?
M — the molar mass, in grams per mole KEY
m — the mass of the sample
n — the amount of substance in moles
No symbol — grams per mole only appears in the answer.
MOL-005 · easy · recall · seq 4 · seed:U7-G1/pool[13] (= MOL-005 gateItems[1], James) — must_freshen_instance applied: lead/aluminum → osmium/beryllium (see authoring_notes) · mcq4
A student writes: '1 mol of osmium contains fewer atoms than 1 mol of beryllium, because osmium atoms are heavier.' Which correction is right?
The counts are equal — one mole is 6.022 × 10²³ particles for any substance. KEY
The student is right — heavier atoms fill a mole with fewer of themselves.
The student has it backwards — 1 mol of osmium contains more atoms than 1 mol of beryllium.
The counts are equal, but only when the two samples also have equal masses.
MOL-001 · medium · application · seq 5 · seed:U7-G1.pool[1] (MOL-001 gateItems[1], particle-by-particle rule quote) — kept verbatim (element-free; stem opens with the taught rule line 'Chemical changes happen particle by particle' — A3 rule-line whitelist expected per stage-1 note) · mcq4
Chemical changes happen particle by particle. Why does that make a worked-out particle count necessary for a chemist who has weighed a sample?
Reactions happen particle by particle, and the balance's mass reading does not state the particle count. KEY
Equal masses of two substances hold equal numbers of particles, so the mass alone is enough.
Reactions depend on total mass, so the particle count never matters.
Weighing both reactants together makes the balance read out their particle counts.
MOL-002 · medium · application · seq 6 · seed:U7-G1/pool[4] (= MOL-002 gateItems[1]) — stem, worded 5-line scaffold, and inverted/product diagnostics carried near-verbatim; tolerance tightened 0.01 -> 0 with integer:true per _AUTHORING_PLAN row 6 (exact integer counts, RECORDED); off-by-decimal diagnostic added per the plan's fixed set; answer 80 display-only per the seed placement note (answer dicts not V-T7-scanned) · numeric
A crate holds identical flat washers. The washers alone have a total mass of 640.0 g, and one washer weighs 8.0 g on its own. How many washers are in the crate?
Answer: 80 washers KEY
also accepted: 80 washers
wrongValueDiagnostics
[
 {
  "value": 0.0125,
  "misconception": "you divided the mass of one washer by the total mass — the total goes on top",
  "feedback": "This answer is wrong — you divided the mass of one washer by the total mass — the total goes on top"
 },
 {
  "value": 5120,
  "misconception": "you multiplied the two masses instead of dividing",
  "feedback": "This answer is wrong — you multiplied the two masses instead of dividing"
 },
 {
  "value": 800,
  "misconception": "you slipped a decimal place in the division — check by multiplying back: this count times the mass of one washer comes to ten times the crate's total mass",
  "feedback": "This answer is wrong — you slipped a decimal place in the division — check by multiplying back: this count times the mass of one washer comes to ten times the crate's total mass"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\ntotal mass = 640.0 g\nmass of one washer = 8.0 g\nWrite down the equation:\nnumber of washers = total mass / mass of one washer\nSubstitute in the values, and calculate:\nnumber of washers = 640.0 / 8.0\nnumber of washers = 80",
 "anticipatedWrong": [
  {
   "value": 0.0125,
   "misconception": "you divided the mass of one washer by the total mass — the total goes on top",
   "feedback": "This answer is wrong — you divided the mass of one washer by the total mass — the total goes on top"
  },
  {
   "value": 5120,
   "misconception": "you multiplied the two masses instead of dividing",
   "feedback": "This answer is wrong — you multiplied the two masses instead of dividing"
  },
  {
   "value": 800,
   "misconception": "you slipped a decimal place in the division — check by multiplying back: this count times the mass of one washer comes to ten times the crate's total mass",
   "feedback": "This answer is wrong — you slipped a decimal place in the division — check by multiplying back: this count times the mass of one washer comes to ten times the crate's total mass"
  }
 ]
}
MOL-003 · medium · application · seq 7 · seed:U7-G1/pool[7] (= MOL-003 gateItems[1], acetone-vs-springs what-makes-the-difference) — carried verbatim; acetone = cross-scope WARN-class single use (MOL-001 fluency), plan-ledger record acetone(7B); number-free · mcq4
Weighing can count the molecules in a bottle of pure acetone but not the pieces in a bin of assorted springs. What makes the difference?
Acetone molecules share one steady average mass, while the springs' masses differ from piece to piece. KEY
The acetone sample is smaller, and small samples are countable.
Molecules, unlike springs, are small enough to be counted directly by laboratory instruments.
The springs are made of a much denser material than acetone, and counting by weighing fails for dense materials.
MOL-006 · medium · application · seq 8 · seed:U7-G1/pool[16] (= MOL-006 gateItems[1]) — verbatim (field mapping only; neon = cross-scope WARN-class single use (MOL-004/005 fluency), recorded neon(8B) in the _AUTHORING_PLAN ledger) · mcq4
Which particle does a mole of neon, Ne, count?
Atoms KEY
Molecules
Formula units
Ions
MOL-009 · medium · application · seq 9 · seed:U7-G1/pool[25] (= MOL-009 gateItems[1]) | STAGE-2 FRESHEN per must_freshen_instance + _AUTHORING_PLAN row 9: aluminum/27.0 -> chromium/52.0 (27.0 consumed by MOL-013 WE + MOL-015/019 fluency; chromium real M 52.0, corpus-clean in scope); read structure, option set, and misconception set carried; d-foil re-derived DOUBLED (104.0) instead of halved because 26.0 would collide with 15B's acetylene constant — misconception class unchanged; no aluminum/27.0/13.5 token remains in any field · mcq4
One chromium atom has an average mass of 52.0 amu. What is the mass of 1 mol of chromium atoms?
52.0 g KEY
52.0 amu
6.022 × 10²³ g
104.0 g
MOL-008 · medium · application · seq 10 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 10: fresh square, tin corpus-clean; ledger tin 10B/23B recorded) · numeric
The periodic-table square for tin is shown. State tin's molar mass, in g/mol. Give your answer to 4 significant figures.
FIGURE (pt-square-reference-table · stimulus) — title: "periodic-table square: tin" · rows: [["atomic number", "50"], ["symbol", "Sn"], ["average atomic mass", "118.7"]] · style: "single periodic-table element square, exactly as printed on the classroom periodic table"
raw spec
{
 "figureRequired": true,
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "pt-square-reference-table",
  "title": "periodic-table square: tin",
  "rows": [
   [
    "atomic number",
    "50"
   ],
   [
    "symbol",
    "Sn"
   ],
   [
    "average atomic mass",
    "118.7"
   ]
  ],
  "style": "single periodic-table element square, exactly as printed on the classroom periodic table"
 }
}
Answer: 118.7 g/mol KEY
also accepted: 118.7 g/mol
wrongValueDiagnostics
[
 {
  "value": 50,
  "misconception": "you read the atomic number — the smaller whole number counts protons and is not a mass",
  "feedback": "This answer is wrong — you read the atomic number — the smaller whole number counts protons and is not a mass"
 },
 {
  "value": 68.7,
  "misconception": "you subtracted the atomic number from the atomic mass — that neutron-count move belongs to isotope work, not molar mass",
  "feedback": "This answer is wrong — you subtracted the atomic number from the atomic mass — that neutron-count move belongs to isotope work, not molar mass"
 },
 {
  "value": 237.4,
  "misconception": "you doubled the printed value — the square's average atomic mass is already the whole molar mass for an element of single atoms",
  "feedback": "This answer is wrong — you doubled the printed value — the square's average atomic mass is already the whole molar mass for an element of single atoms"
 }
]
feedback
{
 "keyExplanation": "The square prints the atomic number 50 and the average atomic mass 118.7. The average atomic mass, read in grams per mole, is the molar mass: tin's is 118.7 g/mol. The 50 counts protons and is never a mass.",
 "anticipatedWrong": [
  {
   "value": 50,
   "misconception": "you read the atomic number — the smaller whole number counts protons and is not a mass",
   "feedback": "This answer is wrong — you read the atomic number — the smaller whole number counts protons and is not a mass"
  },
  {
   "value": 68.7,
   "misconception": "you subtracted the atomic number from the atomic mass — that neutron-count move belongs to isotope work, not molar mass",
   "feedback": "This answer is wrong — you subtracted the atomic number from the atomic mass — that neutron-count move belongs to isotope work, not molar mass"
  },
  {
   "value": 237.4,
   "misconception": "you doubled the printed value — the square's average atomic mass is already the whole molar mass for an element of single atoms",
   "feedback": "This answer is wrong — you doubled the printed value — the square's average atomic mass is already the whole molar mass for an element of single atoms"
  }
 ]
}
MOL-011 · medium · application · seq 11 · seed:U7-G1/pool[32] (= MOL-011 gateItems[2], mcq) — instance transferred to the numeric-ENTRY format per the quant-lesson rule (mcq option set becomes the anticipatedWrong diagnostics, near-verbatim); spared-seed transfer recorded in the plan · numeric
Iodine's periodic-table square prints 126.9. Iodine gas consists of I₂ molecules. Calculate the molar mass of I₂, in g/mol, to one decimal place. Give your answer to 4 significant figures.
Answer: 253.8 g/mol KEY
also accepted: 253.8 g/mol
wrongValueDiagnostics
[
 {
  "value": 126.9,
  "misconception": "you ignored the subscript — each I₂ molecule holds two atoms, so the atom's value doubles",
  "feedback": "This answer is wrong — you ignored the subscript — each I₂ molecule holds two atoms, so the atom's value doubles"
 },
 {
  "value": 128.9,
  "misconception": "you added the subscript as if it were a mass — the 2 is a count and multiplies",
  "feedback": "This answer is wrong — you added the subscript as if it were a mass — the 2 is a count and multiplies"
 },
 {
  "value": 63.45,
  "misconception": "you divided by the subscript instead of multiplying — the 2 counts atoms per molecule",
  "feedback": "This answer is wrong — you divided by the subscript instead of multiplying — the 2 counts atoms per molecule"
 }
]
feedback
{
 "keyExplanation": "The subscript counts the atoms in one molecule: 2.\nmolar mass of I₂ = 2 × 126.9\nmolar mass of I₂ = 253.8 g/mol\nEach iodine molecule weighs two iodine atoms' worth.",
 "anticipatedWrong": [
  {
   "value": 126.9,
   "misconception": "you ignored the subscript — each I₂ molecule holds two atoms, so the atom's value doubles",
   "feedback": "This answer is wrong — you ignored the subscript — each I₂ molecule holds two atoms, so the atom's value doubles"
  },
  {
   "value": 128.9,
   "misconception": "you added the subscript as if it were a mass — the 2 is a count and multiplies",
   "feedback": "This answer is wrong — you added the subscript as if it were a mass — the 2 is a count and multiplies"
  },
  {
   "value": 63.45,
   "misconception": "you divided by the subscript instead of multiplying — the 2 counts atoms per molecule",
   "feedback": "This answer is wrong — you divided by the subscript instead of multiplying — the 2 counts atoms per molecule"
  }
 ]
}
MOL-014 · medium · application · seq 12 · seed:U7-G1/pool[40] (= MOL-014 gateItems[1]) — RE-VALUED per rows/_AUTHORING_PLAN.md row 12: 75.0/225.0 -> 85.0/255.0 (75.0 sits in MOL-017 spec text, corpus scan 2026-09-08); the extra-grams foil re-derived to 170.0; structure and misconception classes kept · mcq4
Beaker 1 holds 85.0 g of ethanol. Beaker 2 holds 255.0 g of ethanol. How do their numbers of moles compare?
Beaker 2 holds three times as many moles as beaker 1. KEY
The two beakers hold equal numbers of moles, because the liquid is the same.
Beaker 2 holds 170.0 more moles than beaker 1.
Beaker 1 holds three times as many moles as beaker 2.
MOL-012 · medium · application · seq 13 · seed:U7-G1/pool[34] (= MOL-012 gateItems[1], glucose) — RE-COMPOUNDED per rows/_AUTHORING_PLAN.md row 13 (glucose/180.0 = MOL-018 fluency instance, MF); CS₂ written FORMULA-ONLY because 'carbon' sits in MOL-012's own corpus (probe 2026-09-08); seed working shell carried. NOTE: anticipatedWrong 44.1 coincides with 15A's supplied propane M — answer-side value, unscanned, ledger-recorded · numeric
The liquid solvent CS₂ is a compound of two elements. Use these molar masses: C 12.0, S 32.1 g/mol. Calculate CS₂'s molar mass, in g/mol, to one decimal place. Give your answer to 3 significant figures.
Answer: 76.2 g/mol KEY
also accepted: 76.2 g/mol
wrongValueDiagnostics
[
 {
  "value": 44.1,
  "misconception": "you ignored the subscript — the formula counts two S atoms, not one",
  "feedback": "This answer is wrong — you ignored the subscript — the formula counts two S atoms, not one"
 },
 {
  "value": 88.2,
  "misconception": "you multiplied every element by the subscript — the 2 belongs to S alone",
  "feedback": "This answer is wrong — you multiplied every element by the subscript — the 2 belongs to S alone"
 },
 {
  "value": 46.1,
  "misconception": "you added the subscript as if it were a mass — the 2 is a count and multiplies the S value",
  "feedback": "This answer is wrong — you added the subscript as if it were a mass — the 2 is a count and multiplies the S value"
 }
]
feedback
{
 "keyExplanation": "Count the atoms the formula shows: 1 carbon-family atom (C) and 2 sulfur-family atoms (S).\nmolar mass of CS₂ = 12.0 + 2 × 32.1\nmolar mass of CS₂ = 76.2 g/mol\nThe subscript 2 multiplies the S value only; C appears once.",
 "anticipatedWrong": [
  {
   "value": 44.1,
   "misconception": "you ignored the subscript — the formula counts two S atoms, not one",
   "feedback": "This answer is wrong — you ignored the subscript — the formula counts two S atoms, not one"
  },
  {
   "value": 88.2,
   "misconception": "you multiplied every element by the subscript — the 2 belongs to S alone",
   "feedback": "This answer is wrong — you multiplied every element by the subscript — the 2 belongs to S alone"
  },
  {
   "value": 46.1,
   "misconception": "you added the subscript as if it were a mass — the 2 is a count and multiplies the S value",
   "feedback": "This answer is wrong — you added the subscript as if it were a mass — the 2 is a count and multiplies the S value"
  }
 ]
}
MOL-013 · medium · application · seq 14 · seed:U7-G1/pool[37] (= MOL-013 gateItems[1], Al2(SO4)3) — RE-COMPOUNDED per rows/_AUTHORING_PLAN.md row 14 (the seed instance is reserved for the seq-23A demo FRQ part); leading-subscript + parentheses shape, working, and diagnostic classes carried; chromium 9B/14B repeat RECORDED · numeric
Calculate the molar mass of chromium(III) sulfate, Cr₂(SO₄)₃. (molar masses: Cr 52.0, S 32.1, O 16.0 g/mol.) Give your answer in g/mol to one decimal place. Give your answer to 4 significant figures.
Answer: 392.3 g/mol KEY
also accepted: 392.3 g/mol
wrongValueDiagnostics
[
 {
  "value": 200.1,
  "misconception": "you ignored the subscript outside the parentheses — the 3 multiplies the whole SO₄ group",
  "feedback": "This answer is wrong — you ignored the subscript outside the parentheses — the 3 multiplies the whole SO₄ group"
 },
 {
  "value": 340.3,
  "misconception": "you counted only one chromium — the subscript 2 after Cr doubles it",
  "feedback": "This answer is wrong — you counted only one chromium — the subscript 2 after Cr doubles it"
 },
 {
  "value": 600.3,
  "misconception": "you multiplied the chromium by 3 as well — the outside subscript reaches only what sits inside the parentheses",
  "feedback": "This answer is wrong — you multiplied the chromium by 3 as well — the outside subscript reaches only what sits inside the parentheses"
 }
]
feedback
{
 "keyExplanation": "Inside the parentheses: 32.1 + 4 × 16.0 = 96.1\nMultiply by the outside subscript: 3 × 96.1 = 288.3\nAdd the atoms outside the parentheses: 2 × 52.0 + 288.3\nM = 392.3 g/mol",
 "anticipatedWrong": [
  {
   "value": 200.1,
   "misconception": "you ignored the subscript outside the parentheses — the 3 multiplies the whole SO₄ group",
   "feedback": "This answer is wrong — you ignored the subscript outside the parentheses — the 3 multiplies the whole SO₄ group"
  },
  {
   "value": 340.3,
   "misconception": "you counted only one chromium — the subscript 2 after Cr doubles it",
   "feedback": "This answer is wrong — you counted only one chromium — the subscript 2 after Cr doubles it"
  },
  {
   "value": 600.3,
   "misconception": "you multiplied the chromium by 3 as well — the outside subscript reaches only what sits inside the parentheses",
   "feedback": "This answer is wrong — you multiplied the chromium by 3 as well — the outside subscript reaches only what sits inside the parentheses"
  }
 ]
}
MOL-017 · medium · application · seq 15 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 15: fresh triple; acetylene corpus-clean; 26.0 = canonical C₂H₂ molar mass, CONSTANTS-CLASS family entry for its coincidental MOL-008 spec-text hit) · numeric
A welding cylinder holds 91.0 g of acetylene, C₂H₂ (M = 26.0 g/mol). How many moles of acetylene is that? Give your answer in moles to 3 significant figures.
Answer: 3.5 mol KEY
also accepted: 3.50 mol
wrongValueDiagnostics
[
 {
  "value": 0.286,
  "misconception": "you divided the molar mass by the mass instead of the mass by the molar mass",
  "feedback": "This answer is wrong — you divided the molar mass by the mass instead of the mass by the molar mass"
 },
 {
  "value": 2366,
  "misconception": "you multiplied the mass by the molar mass instead of dividing",
  "feedback": "This answer is wrong — you multiplied the mass by the molar mass instead of dividing"
 },
 {
  "value": 35.0,
  "misconception": "you slipped a decimal place in the division — check by multiplying back: this count times 26.0 g/mol comes to ten times the cylinder's mass",
  "feedback": "This answer is wrong — you slipped a decimal place in the division — check by multiplying back: this count times 26.0 g/mol comes to ten times the cylinder's mass"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nm = 91.0 g\nM = 26.0 g/mol\nWrite down the equation:\nn = m / M\nSubstitute in the values, and calculate:\nn = 91.0 / 26.0\nn = 3.50 mol\nSanity check: a cylinder-scale mole count lands around 0.01 to 10 mol — 3.50 mol passes.",
 "anticipatedWrong": [
  {
   "value": 0.286,
   "misconception": "you divided the molar mass by the mass instead of the mass by the molar mass",
   "feedback": "This answer is wrong — you divided the molar mass by the mass instead of the mass by the molar mass"
  },
  {
   "value": 2366,
   "misconception": "you multiplied the mass by the molar mass instead of dividing",
   "feedback": "This answer is wrong — you multiplied the mass by the molar mass instead of dividing"
  },
  {
   "value": 35.0,
   "misconception": "you slipped a decimal place in the division — check by multiplying back: this count times 26.0 g/mol comes to ten times the cylinder's mass",
   "feedback": "This answer is wrong — you slipped a decimal place in the division — check by multiplying back: this count times 26.0 g/mol comes to ten times the cylinder's mass"
  }
 ]
}
MOL-018 · medium · application · seq 16 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 16; plan's 129.8 corrected to the canonical sum 129.9 and n re-valued to the float-exact 0.500 (RECORDED); cobalt 2B/16B same-form repeat RECORDED); STAGE-3 re-value 0.500 -> 2.00 mol (0.500 is a scanned non-integer shared with four lessons' spec text; 259.8 coincides with 23E's CsI molar mass in UNSCANNED model-answer text - RECORDED) · numeric
A recipe for a pigment calls for 2.000 mol of cobalt(II) chloride, CoCl₂ (M = 129.9 g/mol). What mass of cobalt(II) chloride is that? Give your answer in grams to 1 decimal place.
Answer: 259.8 g KEY
also accepted: 259.8 g
wrongValueDiagnostics
[
 {
  "value": 0.0154,
  "misconception": "you divided the moles by the molar mass instead of multiplying them",
  "feedback": "This answer is wrong — you divided the moles by the molar mass instead of multiplying them"
 },
 {
  "value": 64.95,
  "misconception": "you divided the molar mass by the moles instead of multiplying them",
  "feedback": "This answer is wrong — you divided the molar mass by the moles instead of multiplying them"
 },
 {
  "value": 25.98,
  "misconception": "you slipped a decimal place in the product — two portions of 129.9 g must weigh about 260 g",
  "feedback": "This answer is wrong — you slipped a decimal place in the product — two portions of 129.9 g must weigh about 260 g"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nn = 2.000 mol\nM = 129.9 g/mol\nWrite down the equation:\nn = m / M\nMake m the subject:\nm = n × M\nSubstitute in the values, and calculate:\nm = 2.000 × 129.9\nm = 259.8 g\nSanity check: more than 1 mol must weigh more than the 129.9 g molar mass — 259.8 g passes.",
 "anticipatedWrong": [
  {
   "value": 0.0154,
   "misconception": "you divided the moles by the molar mass instead of multiplying them",
   "feedback": "This answer is wrong — you divided the moles by the molar mass instead of multiplying them"
  },
  {
   "value": 64.95,
   "misconception": "you divided the molar mass by the moles instead of multiplying them",
   "feedback": "This answer is wrong — you divided the molar mass by the moles instead of multiplying them"
  },
  {
   "value": 25.98,
   "misconception": "you slipped a decimal place in the product — two portions of 129.9 g must weigh about 260 g",
   "feedback": "This answer is wrong — you slipped a decimal place in the product — two portions of 129.9 g must weigh about 260 g"
  }
 ]
}
MOL-010 · hard · transfer · seq 17 · seed:U7-G1/pool[28] (= MOL-010 gateItems[1], fluorine) — FRESHENED per rows/_AUTHORING_PLAN.md row 17 (19.0 = MOL-009/011 fluency corpus): vanadium 50.9; structure and foil classes carried; QB 3.9 sweep: fluorine/19.0 removed from all fields · mcq4
1 mol of vanadium atoms has a mass of 50.9 g. Where does the 50.9 come from?
Each vanadium atom averages 50.9 amu, and counting out 6.022 × 10²³ of them turns 50.9 amu into 50.9 g. KEY
From weighing a single vanadium atom in grams.
From a chemists' rule that one mole of every element must weigh a whole number of grams.
From the gram being defined as the mass of one vanadium atom.
MOL-015 · hard · transfer · seq 18 · seed:U7-G1/pool[43] (= MOL-015 gateItems[1], reverse inference) — carried near-verbatim (element-free; no freshness flags) · mcq4
Two samples have exactly equal masses, and sample A contains more moles than sample B. Which statement about the two substances' molar masses must be true?
Sample B's substance has the larger molar mass. KEY
Sample A's substance has the larger molar mass.
The two molar masses must be equal.
Nothing about the molar masses can be concluded.
MOL-019 · hard · transfer · seq 19 · seed:U7-G1/pool[55] (= MOL-019 gateItems[1]) — glucose leg REPLACED per plan by NaBr, RE-VALUED to 51.45 g so the upside-down division singles out NaBr (restores the seed's foil geometry the plan's 205.8 value broke by tying zinc); NaBr written FORMULA-ONLY ('sodium' is MOL-019-own corpus, RECORDED); ethane and zinc legs carried · mcq4
Three bottles: 90.0 g of ethane (M = 30.0 g/mol); 130.8 g of zinc (M = 65.4 g/mol); 51.45 g of NaBr (M = 102.9 g/mol). Which bottle holds the most moles?
The ethane bottle. KEY
The zinc bottle.
The NaBr bottle.
It cannot be decided from the masses and molar masses alone.
MOL-020 · hard · transfer · seq 20 · seed:U7-G1/pool[58] (= MOL-020 gateItems[1], CO) — RE-SUBSTANCED per rows/_AUTHORING_PLAN.md row 20 to hydrogen chloride at 6.00 mol (m = 36.5/6.00 = 6.08 g); error class, foil classes, and three-check explanation carried; QB 3.9 sweep: CO/28.0/3.00/9.33/84.0 removed from all fields · mcq4
A question asks for the mass of 6.00 mol of hydrogen chloride, HCl (M = 36.5 g/mol). A student writes: m = 36.5 / 6.00 = 6.08 g. What is the error?
The solution divided by the moles instead of multiplying them. KEY
Nothing is wrong with the solution.
The error is in the equation itself — the correct relation is n = M/m.
The solution copied the molar mass of the wrong gas.
MOL-002 · medium · application · seq 21 · authored FRQ (plan row 21; fresh thumbtack instance, values corpus-clean and float-exact) · frq
A stationery warehouse counts identical thumbtacks by weighing.
A stationery warehouse counts identical thumbtacks by weighing. (a) The thumbtacks in a box have a total mass of 391.0 g on their own, and one thumbtack has a mass of 2.3 g. How many thumbtacks are in the box? Show every stage of your working: the values you start from, the equation in words, the substitution, and the count. (b) A school orders 320 thumbtacks. What total mass of thumbtacks should be weighed out? Show your working the same way. (c) A coworker answers part (a) by computing 2.3 / 391.0. Explain what is wrong with that division and state the corrected count. Plain numbers and words are expected throughout — no special symbols. Treat the given masses as exact.
Part a. The thumbtacks in a box have a total mass of 391.0 g on their own, and one thumbtack has a mass of 2.3 g. How many thumbtacks are in the box? Show every stage of your working: the values you start from, the equation in words, the substitution, and the count.
Model: Values: total mass = 391.0 g; mass of one thumbtack = 2.3 g. Equation in words: number of thumbtacks = total mass divided by mass of one thumbtack. Substitution: 391.0 / 2.3. Count: 170 thumbtacks. Check: 170 thumbtacks of 2.3 g each weigh exactly 391.0 g.
Part b. A school orders 320 thumbtacks. What total mass of thumbtacks should be weighed out? Show your working the same way.
Model: Turn the equation around: mass to weigh out = number of thumbtacks times mass of one thumbtack = 320 times 2.3 = 736.0 g.
Part c. A coworker answers part (a) by computing 2.3 / 391.0. Explain what is wrong with that division and state the corrected count. Plain numbers and words are expected throughout — no special symbols. Treat the given masses as exact.
Model: The division is upside down — the total mass goes on top. 2.3 / 391.0 is about 0.006, and a box cannot hold a fraction of a thumbtack. The corrected division is 391.0 / 2.3 = 170 thumbtacks. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
Full model answer. (a) Values: total mass = 391.0 g; mass of one thumbtack = 2.3 g. Equation in words: number of thumbtacks = total mass divided by mass of one thumbtack. Substitution: 391.0 / 2.3. Count: 170 thumbtacks. Check: 170 thumbtacks of 2.3 g each weigh exactly 391.0 g. (b) Turn the equation around: mass to weigh out = number of thumbtacks times mass of one thumbtack = 320 times 2.3 = 736.0 g. (c) The division is upside down — the total mass goes on top. 2.3 / 391.0 is about 0.006, and a box cannot hold a fraction of a thumbtack. The corrected division is 391.0 / 2.3 = 170 thumbtacks. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
MOL-013 · medium · application · seq 22 · authored FRQ (plan row 22; zinc 19B/22B repeat RECORDED; plan's 386.1 corrected to the canonical-constant sum 386.2, RECORDED) · frq
(a) Calculate the molar mass of zinc nitrate, Zn(NO₃)₂. (molar masses: Zn 65.4, N 14.0, O 16.0 g/mol.) Show every stage: the value inside the parentheses, the multiplication by the outside subscript, and the final sum. (b) A labmate reports M = 127.4 g/mol for the same compound. Identify exactly what the labmate did wrong and state the corrected molar mass. (c) Now calculate the molar mass of zinc phosphate, Zn₃(PO₄)₂. (molar masses: Zn 65.4, P 31.0, O 16.0 g/mol.) Show the same stages. Plain numbers and words are expected — no special symbols.
(a) Calculate the molar mass of zinc nitrate, Zn(NO₃)₂. (molar masses: Zn 65.4, N 14.0, O 16.0 g/mol.) Show every stage: the value inside the parentheses, the multiplication by the outside subscript, and the final sum. (b) A labmate reports M = 127.4 g/mol for the same compound. Identify exactly what the labmate did wrong and state the corrected molar mass. (c) Now calculate the molar mass of zinc phosphate, Zn₃(PO₄)₂. (molar masses: Zn 65.4, P 31.0, O 16.0 g/mol.) Show the same stages. Plain numbers and words are expected — no special symbols.
Part a. Calculate the molar mass of zinc nitrate, Zn(NO₃)₂. (molar masses: Zn 65.4, N 14.0, O 16.0 g/mol.) Show every stage: the value inside the parentheses, the multiplication by the outside subscript, and the final sum.
Model: Inside the parentheses: 14.0 + 3 times 16.0 = 62.0. Multiply by the outside subscript: 2 times 62.0 = 124.0. Add the zinc, which sits outside the parentheses: 65.4 + 124.0 = 189.4. M of zinc nitrate = 189.4 g/mol.
Part b. A labmate reports M = 127.4 g/mol for the same compound. Identify exactly what the labmate did wrong and state the corrected molar mass.
Model: The labmate ignored the subscript outside the parentheses and counted the nitrate group once: 65.4 + 62.0 = 127.4. The outside 2 doubles the whole group. Corrected: M = 189.4 g/mol.
Part c. Now calculate the molar mass of zinc phosphate, Zn₃(PO₄)₂. (molar masses: Zn 65.4, P 31.0, O 16.0 g/mol.) Show the same stages. Plain numbers and words are expected — no special symbols.
Model: Three zincs: 3 times 65.4 = 196.2. Phosphate group: 31.0 + 4 times 16.0 = 95.0; the outside 2 doubles it: 190.0. Total: 196.2 + 190.0 = 386.2. M of zinc phosphate = 386.2 g/mol. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
Full model answer. (a) Inside the parentheses: 14.0 + 3 times 16.0 = 62.0. Multiply by the outside subscript: 2 times 62.0 = 124.0. Add the zinc, which sits outside the parentheses: 65.4 + 124.0 = 189.4. M of zinc nitrate = 189.4 g/mol. (b) The labmate ignored the subscript outside the parentheses and counted the nitrate group once: 65.4 + 62.0 = 127.4. The outside 2 doubles the whole group. Corrected: M = 189.4 g/mol. (c) Three zincs: 3 times 65.4 = 196.2. Phosphate group: 31.0 + 4 times 16.0 = 95.0; the outside 2 doubles it: 190.0. Total: 196.2 + 190.0 = 386.2. M of zinc phosphate = 386.2 g/mol. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
MOL-021 · hard · application · seq 23 · authored FRQ (plan row 23; SnF2 mass re-valued 470.1 -> 626.8 for float-exactness, RECORDED; tin 10B/23B + beryllium 4B/23B repeats RECORDED) · frq
(a) How many moles are in 626.8 g of tin(II) fluoride, SnF₂? (molar masses: Sn 118.7, F 19.0 g/mol.) Show both stages: first assemble the molar mass from the formula, then apply the mole equation. Give the answer in moles to 3 significant figures. (b) What is the mass, in grams, of 0.800 mol of beryllium hydroxide, Be(OH)₂? (molar masses: Be 9.0, O 16.0, H 1.0 g/mol.) Show both stages, including how you rearrange the mole equation. Give the answer to 3 significant figures. (c) Bottle 1 holds 313.4 g of tin(II) fluoride; bottle 2 holds 129.0 g of beryllium hydroxide. Using the molar masses you found, work out which bottle holds more moles — show the working for both bottles and report the mole values to 3 significant figures. Plain numbers and words are expected — no special symbols.
(a) How many moles are in 626.8 g of tin(II) fluoride, SnF₂? (molar masses: Sn 118.7, F 19.0 g/mol.) Show both stages: first assemble the molar mass from the formula, then apply the mole equation. Give the answer in moles to 3 significant figures. (b) What is the mass, in grams, of 0.800 mol of beryllium hydroxide, Be(OH)₂? (molar masses: Be 9.0, O 16.0, H 1.0 g/mol.) Show both stages, including how you rearrange the mole equation. Give the answer to 3 significant figures. (c) Bottle 1 holds 313.4 g of tin(II) fluoride; bottle 2 holds 129.0 g of beryllium hydroxide. Using the molar masses you found, work out which bottle holds more moles — show the working for both bottles and report the mole values to 3 significant figures. Plain numbers and words are expected — no special symbols.
Part a. How many moles are in 626.8 g of tin(II) fluoride, SnF₂? (molar masses: Sn 118.7, F 19.0 g/mol.) Show both stages: first assemble the molar mass from the formula, then apply the mole equation. Give the answer in moles to 3 significant figures.
Model: Stage 1 — molar mass of tin(II) fluoride: 118.7 + 2 times 19.0 = 156.7 g/mol. Stage 2 — moles: n = m/M = 626.8 / 156.7 = 4.00 mol.
Part b. What is the mass, in grams, of 0.800 mol of beryllium hydroxide, Be(OH)₂? (molar masses: Be 9.0, O 16.0, H 1.0 g/mol.) Show both stages, including how you rearrange the mole equation. Give the answer to 3 significant figures.
Model: Stage 1 — molar mass of beryllium hydroxide: inside the parentheses 16.0 + 1.0 = 17.0; doubled by the outside subscript to 34.0; plus the beryllium 9.0; total 43.0 g/mol. Stage 2 — rearrange to make m the subject: m = n times M = 0.800 times 43.0 = 34.4 g.
Part c. Bottle 1 holds 313.4 g of tin(II) fluoride; bottle 2 holds 129.0 g of beryllium hydroxide. Using the molar masses you found, work out which bottle holds more moles — show the working for both bottles and report the mole values to 3 significant figures. Plain numbers and words are expected — no special symbols.
Model: Bottle 1: 313.4 / 156.7 = 2.00 mol. Bottle 2: 129.0 / 43.0 = 3.00 mol. Bottle 2 holds more moles, even though bottle 1 is heavier — beryllium hydroxide's portions are much lighter. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
Full model answer. (a) Stage 1 — molar mass of tin(II) fluoride: 118.7 + 2 times 19.0 = 156.7 g/mol. Stage 2 — moles: n = m/M = 626.8 / 156.7 = 4.00 mol. (b) Stage 1 — molar mass of beryllium hydroxide: inside the parentheses 16.0 + 1.0 = 17.0; doubled by the outside subscript to 34.0; plus the beryllium 9.0; total 43.0 g/mol. Stage 2 — rearrange to make m the subject: m = n times M = 0.800 times 43.0 = 34.4 g. (c) Bottle 1: 313.4 / 156.7 = 2.00 mol. Bottle 2: 129.0 / 43.0 = 3.00 mol. Bottle 2 holds more moles, even though bottle 1 is heavier — beryllium hydroxide's portions are much lighter. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.

Form C

MOL-004 · easy · recall · seq 1 · seed:U7-G1/pool[11] (= MOL-004 gateItems[2]) — FRESHENED per rows/_AUTHORING_PLAN.md row 1: zinc -> gold (cross-form collision with 19B), table salt/NaCl -> the compound KBr (formula-only; potassium is a FORM-A-reserved word); structure, misconception classes, and trap set kept · mcq4
Sample A is 1 mol of gold. Sample B is 1 mol of the compound KBr. Compare the numbers of particles in the two samples.
Equal — the mole is the same fixed count for any substance. KEY
Sample A holds more particles, because gold is a metal.
Sample B holds more particles, because KBr is made of two elements.
The comparison needs both samples' masses.
MOL-007 · easy · recall · seq 2 · seed:U7-G1/pool[18] (= MOL-007 gateItems[0]) | STAGE-2 FRESHEN per must_freshen_instance + _AUTHORING_PLAN row 2: methane/16.0 -> nickel/58.7 (16.0 consumed by MOL-011/012/013/017 corpus, methane cross-scope; nickel real M, corpus-clean in scope); 'molecule' -> 'atom' in the single-particle foil because nickel is an atomic element (misconception unchanged); which-follows structure and option set carried; no methane/16.0 token remains in any field · mcq4
Nickel's molar mass is 58.7 g/mol. Which statement follows?
1 mol of nickel has a mass of 58.7 g. KEY
One nickel atom has a mass of 58.7 g.
58.7 mol of nickel has a mass of 1.0 g.
Any sample of nickel has a mass of 58.7 g.
MOL-016 · easy · recall · seq 3 · seed:U7-G1/pool[47] (= MOL-016 gateItems[2], James) — verbatim (field mapping only) · mcq4
A student writes: 'The number of moles is the molar mass divided by the mass.' What is wrong with the statement?
The division is flipped — n = m/M divides the mass by the molar mass. KEY
Nothing is wrong — the two divisions give the same number.
The operation is wrong — the two quantities should be multiplied.
The quantities are wrong — moles come from the mass and the particle count.
MOL-005 · easy · recall · seq 4 · authored (fresh facet per stage-2 plan: name-the-number recall; stem re-worded off the student-seen fluency ask, options fresh) · mcq4
Every mole contains the same fixed count of particles: 6.022 × 10²³. What is this count called?
Avogadro's number KEY
The mole
The molar mass
The atomic number
MOL-001 · medium · application · seq 5 · seed:U7-G1.pool[2] (MOL-001 gateItems[2], reports+route) — kept verbatim (element-free, number-free) · mcq4
What is the one quantity a balance reports, and what must be done to get a particle count from it?
It reports the sample's mass; the particle count must be worked out from that mass. KEY
It reports the particle count; the mass must be worked out from it.
It reports both, on different display settings.
It reports the mass, and no particle count can ever be obtained from it.
MOL-002 · medium · application · seq 6 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 6: fresh identical-object instance rivets 517.5/11.5 -> 45; 'rivet' zero digest hits, 517.5 corpus-clean and ledger-unique, 11.5 <= 12 scan-invisible; exact division float-verified; seed-parallel template, worded equation only) · numeric
A bin holds identical rivets. The rivets alone have a total mass of 517.5 g, and one rivet has a mass of 11.5 g. How many rivets are in the bin?
Answer: 45 rivets KEY
also accepted: 45 rivets
wrongValueDiagnostics
[
 {
  "value": 0.0222,
  "misconception": "you divided the mass of one rivet by the total mass — the total goes on top",
  "feedback": "This answer is wrong — you divided the mass of one rivet by the total mass — the total goes on top"
 },
 {
  "value": 5951.25,
  "misconception": "you multiplied the two masses instead of dividing",
  "feedback": "This answer is wrong — you multiplied the two masses instead of dividing"
 },
 {
  "value": 450,
  "misconception": "you slipped a decimal place in the division — check by multiplying back: this count times the mass of one rivet comes to ten times the bin's total mass",
  "feedback": "This answer is wrong — you slipped a decimal place in the division — check by multiplying back: this count times the mass of one rivet comes to ten times the bin's total mass"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\ntotal mass = 517.5 g\nmass of one rivet = 11.5 g\nWrite down the equation:\nnumber of rivets = total mass / mass of one rivet\nSubstitute in the values, and calculate:\nnumber of rivets = 517.5 / 11.5\nnumber of rivets = 45",
 "anticipatedWrong": [
  {
   "value": 0.0222,
   "misconception": "you divided the mass of one rivet by the total mass — the total goes on top",
   "feedback": "This answer is wrong — you divided the mass of one rivet by the total mass — the total goes on top"
  },
  {
   "value": 5951.25,
   "misconception": "you multiplied the two masses instead of dividing",
   "feedback": "This answer is wrong — you multiplied the two masses instead of dividing"
  },
  {
   "value": 450,
   "misconception": "you slipped a decimal place in the division — check by multiplying back: this count times the mass of one rivet comes to ten times the bin's total mass",
   "feedback": "This answer is wrong — you slipped a decimal place in the division — check by multiplying back: this count times the mass of one rivet comes to ten times the bin's total mass"
  }
 ]
}
MOL-003 · medium · application · seq 7 · seed:U7-G1/pool[8] (= MOL-003 gateItems[2], isotope 'average' clause on pure Cl2 — the lesson's own U02-retrieval facet per specNotes) — carried verbatim (HTML subscripts -> Unicode Cl₂ only); chlorine = cross-scope WARN-class single use (MOL-011 WE), plan-ledger record chlorine(7C) · mcq4
Chlorine atoms come in two isotopes with different masses. Why can weighing still count the molecules in a sample of pure gaseous chlorine, Cl₂?
The counting runs on the average particle mass, and the particles of a pure substance share one steady average mass. KEY
It cannot — the isotopes break counting by weighing.
The two isotopes always appear in exactly equal numbers.
The balance reads the mass of the lighter isotope only and ignores the heavier isotope completely.
MOL-006 · medium · application · seq 8 · seed:U7-G1/pool[17] (= MOL-006 gateItems[2]) — verbatim except seed sub-tag markup converted to Unicode subscripts (C₆H₁₂O₆); glucose = cross-scope WARN-class single use whose gate-wide home is THIS slot (recorded glucose(8C); the 19B glucose leg is REPLACED per the _AUTHORING_PLAN ledger, and the other glucose seeds 34/50/57 are spared or MF'd) · mcq4
Which particle does a mole of glucose, C₆H₁₂O₆, count?
Molecules KEY
Atoms
Formula units
Ions
MOL-009 · medium · application · seq 9 · seed:U7-G1/pool[24] (= MOL-009 gateItems[0]) | STAGE-2 FRESHEN per must_freshen_instance + _AUTHORING_PLAN row 9: iron/55.8 -> titanium/47.9 (55.8 consumed by MOL-015 fluency; titanium real M 47.9, corpus-clean in scope); wrong-scale correction structure, option set, and misconception set carried byte-level otherwise; no iron/55.8 token remains in any field · mcq4
A student writes: 'One titanium atom has an average mass of 47.9 g.' Titanium's molar mass is 47.9 g/mol. Which correction is right?
One titanium atom averages 47.9 amu — 47.9 g is the mass of a whole mole of titanium atoms. KEY
The student is right — atoms are weighed in grams.
One titanium atom has an average mass of 47.9 g/mol.
The number 47.9 belongs only to the mole scale, and the atom's amu value cannot be found from it.
MOL-008 · medium · application · seq 10 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 10: fresh square, manganese corpus-clean; ledger manganese 10C/13C (+22C) recorded) · numeric
The periodic-table square for manganese is shown. State manganese's molar mass, in g/mol. Give your answer to 3 significant figures.
FIGURE (pt-square-reference-table · stimulus) — title: "periodic-table square: manganese" · rows: [["atomic number", "25"], ["symbol", "Mn"], ["average atomic mass", "54.9"]] · style: "single periodic-table element square, exactly as printed on the classroom periodic table"
raw spec
{
 "figureRequired": true,
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "pt-square-reference-table",
  "title": "periodic-table square: manganese",
  "rows": [
   [
    "atomic number",
    "25"
   ],
   [
    "symbol",
    "Mn"
   ],
   [
    "average atomic mass",
    "54.9"
   ]
  ],
  "style": "single periodic-table element square, exactly as printed on the classroom periodic table"
 }
}
Answer: 54.9 g/mol KEY
also accepted: 54.9 g/mol
wrongValueDiagnostics
[
 {
  "value": 25,
  "misconception": "you read the atomic number — the smaller whole number counts protons and is not a mass",
  "feedback": "This answer is wrong — you read the atomic number — the smaller whole number counts protons and is not a mass"
 },
 {
  "value": 29.9,
  "misconception": "you subtracted the atomic number from the atomic mass — that neutron-count move belongs to isotope work, not molar mass",
  "feedback": "This answer is wrong — you subtracted the atomic number from the atomic mass — that neutron-count move belongs to isotope work, not molar mass"
 },
 {
  "value": 109.8,
  "misconception": "you doubled the printed value — the square's average atomic mass is already the whole molar mass for an element of single atoms",
  "feedback": "This answer is wrong — you doubled the printed value — the square's average atomic mass is already the whole molar mass for an element of single atoms"
 }
]
feedback
{
 "keyExplanation": "The square prints the atomic number 25 and the average atomic mass 54.9. The average atomic mass, read in grams per mole, is the molar mass: manganese's is 54.9 g/mol. The 25 counts protons and is never a mass.",
 "anticipatedWrong": [
  {
   "value": 25,
   "misconception": "you read the atomic number — the smaller whole number counts protons and is not a mass",
   "feedback": "This answer is wrong — you read the atomic number — the smaller whole number counts protons and is not a mass"
  },
  {
   "value": 29.9,
   "misconception": "you subtracted the atomic number from the atomic mass — that neutron-count move belongs to isotope work, not molar mass",
   "feedback": "This answer is wrong — you subtracted the atomic number from the atomic mass — that neutron-count move belongs to isotope work, not molar mass"
  },
  {
   "value": 109.8,
   "misconception": "you doubled the printed value — the square's average atomic mass is already the whole molar mass for an element of single atoms",
   "feedback": "This answer is wrong — you doubled the printed value — the square's average atomic mass is already the whole molar mass for an element of single atoms"
  }
 ]
}
MOL-011 · medium · application · seq 11 · seed:U7-G1/pool[30] (= MOL-011 gateItems[0], ozone) — re-instanced to S₈ per _AUTHORING_PLAN row 11 (the ozone instance moves to form D); printed value FIGURE-borne so the corpus-shared constant never sits in a scanned stem · numeric
Solid sulfur exists as S₈ molecules. Sulfur's periodic-table square is shown. Calculate the molar mass of S₈, in g/mol. Give your answer to 3 significant figures.
FIGURE (pt-square-reference-table · stimulus) — title: "periodic-table square: sulfur" · rows: [["atomic number", "16"], ["symbol", "S"], ["average atomic mass", "32.1"]] · style: "single periodic-table element square, exactly as printed on the classroom periodic table"
raw spec
{
 "figureRequired": true,
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "pt-square-reference-table",
  "title": "periodic-table square: sulfur",
  "rows": [
   [
    "atomic number",
    "16"
   ],
   [
    "symbol",
    "S"
   ],
   [
    "average atomic mass",
    "32.1"
   ]
  ],
  "style": "single periodic-table element square, exactly as printed on the classroom periodic table"
 }
}
Answer: 257 g/mol KEY
also accepted: 257 g/mol
wrongValueDiagnostics
[
 {
  "value": 32.1,
  "misconception": "you ignored the subscript — each S₈ molecule holds eight atoms, so the atom's value multiplies by 8",
  "feedback": "This answer is wrong — you ignored the subscript — each S₈ molecule holds eight atoms, so the atom's value multiplies by 8"
 },
 {
  "value": 40.1,
  "misconception": "you added the subscript as if it were a mass — the 8 is a count and multiplies",
  "feedback": "This answer is wrong — you added the subscript as if it were a mass — the 8 is a count and multiplies"
 },
 {
  "value": 4.0125,
  "misconception": "you divided by the subscript instead of multiplying — the 8 counts atoms per molecule",
  "feedback": "This answer is wrong — you divided by the subscript instead of multiplying — the 8 counts atoms per molecule"
 }
]
feedback
{
 "keyExplanation": "The square prints the average atomic mass 32.1. The subscript counts the atoms in one molecule: 8.\nmolar mass of S₈ = 8 × 32.1\nmolar mass of S₈ = 256.8, reported as 257 g/mol (3 significant figures)\nEach S₈ ring weighs eight sulfur atoms' worth.",
 "anticipatedWrong": [
  {
   "value": 32.1,
   "misconception": "you ignored the subscript — each S₈ molecule holds eight atoms, so the atom's value multiplies by 8",
   "feedback": "This answer is wrong — you ignored the subscript — each S₈ molecule holds eight atoms, so the atom's value multiplies by 8"
  },
  {
   "value": 40.1,
   "misconception": "you added the subscript as if it were a mass — the 8 is a count and multiplies",
   "feedback": "This answer is wrong — you added the subscript as if it were a mass — the 8 is a count and multiplies"
  },
  {
   "value": 4.0125,
   "misconception": "you divided by the subscript instead of multiplying — the 8 counts atoms per molecule",
   "feedback": "This answer is wrong — you divided by the subscript instead of multiplying — the 8 counts atoms per molecule"
  }
 ]
}
MOL-014 · medium · application · seq 12 · seed:U7-G1/pool[41] (= MOL-014 gateItems[2], boundary item) — carried near-verbatim (element-free, number-free; no freshness flags in seed_placement) · mcq4
A student doubles a sample's mass but also swaps the substance for a different one. Can the student still conclude that the number of moles doubled?
No — the doubling rule works only when the substance stays the same. KEY
Yes — doubling the mass always doubles the moles, whatever the substance.
Yes — mass and moles are the same quantity measured in different units.
No — the number of moles never changes when the mass changes.
MOL-012 · medium · application · seq 13 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 13: fresh two-element compound; manganese 10C/13C same-form repeat RECORDED; Cl 35.5 = canonical constant in MOL-011/012 spec text, CONSTANTS-CLASS family) · numeric
Manganese(II) chloride has the formula MnCl₂. Use these molar masses: Mn 54.9, Cl 35.5 g/mol. Calculate MnCl₂'s molar mass, in g/mol, to one decimal place. Give your answer to 4 significant figures.
Answer: 125.9 g/mol KEY
also accepted: 125.9 g/mol
wrongValueDiagnostics
[
 {
  "value": 90.4,
  "misconception": "you ignored the subscript — the formula counts two chlorines, not one",
  "feedback": "This answer is wrong — you ignored the subscript — the formula counts two chlorines, not one"
 },
 {
  "value": 180.8,
  "misconception": "you multiplied every element by the subscript — the 2 belongs to chlorine alone",
  "feedback": "This answer is wrong — you multiplied every element by the subscript — the 2 belongs to chlorine alone"
 },
 {
  "value": 92.4,
  "misconception": "you added the subscript as if it were a mass — the 2 is a count and multiplies the chlorine value",
  "feedback": "This answer is wrong — you added the subscript as if it were a mass — the 2 is a count and multiplies the chlorine value"
 }
]
feedback
{
 "keyExplanation": "Count the atoms the formula unit shows: 1 manganese and 2 chlorines.\nmolar mass of MnCl₂ = 54.9 + 2 × 35.5\nmolar mass of MnCl₂ = 125.9 g/mol\nThe subscript 2 multiplies the chlorine value only; manganese appears once.",
 "anticipatedWrong": [
  {
   "value": 90.4,
   "misconception": "you ignored the subscript — the formula counts two chlorines, not one",
   "feedback": "This answer is wrong — you ignored the subscript — the formula counts two chlorines, not one"
  },
  {
   "value": 180.8,
   "misconception": "you multiplied every element by the subscript — the 2 belongs to chlorine alone",
   "feedback": "This answer is wrong — you multiplied every element by the subscript — the 2 belongs to chlorine alone"
  },
  {
   "value": 92.4,
   "misconception": "you added the subscript as if it were a mass — the 2 is a count and multiplies the chlorine value",
   "feedback": "This answer is wrong — you added the subscript as if it were a mass — the 2 is a count and multiplies the chlorine value"
  }
 ]
}
MOL-013 · medium · application · seq 14 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 14: fresh hydroxide with parentheses; nickel 2C/14C same-form non-adjacent repeat RECORDED) · numeric
Calculate the molar mass of nickel(II) hydroxide, Ni(OH)₂. (molar masses: Ni 58.7, O 16.0, H 1.0 g/mol.) Give your answer in g/mol to one decimal place. Give your answer to 3 significant figures.
Answer: 92.7 g/mol KEY
also accepted: 92.7 g/mol
wrongValueDiagnostics
[
 {
  "value": 75.7,
  "misconception": "you ignored the subscript outside the parentheses — the 2 multiplies both the oxygen and the hydrogen",
  "feedback": "This answer is wrong — you ignored the subscript outside the parentheses — the 2 multiplies both the oxygen and the hydrogen"
 },
 {
  "value": 151.4,
  "misconception": "you multiplied the nickel as well — the outside subscript reaches only what sits inside the parentheses",
  "feedback": "This answer is wrong — you multiplied the nickel as well — the outside subscript reaches only what sits inside the parentheses"
 },
 {
  "value": 91.7,
  "misconception": "you applied the outside 2 only to the oxygen — every atom inside the parentheses is doubled, the hydrogen included",
  "feedback": "This answer is wrong — you applied the outside 2 only to the oxygen — every atom inside the parentheses is doubled, the hydrogen included"
 }
]
feedback
{
 "keyExplanation": "Inside the parentheses: 16.0 + 1.0 = 17.0\nMultiply by the outside subscript: 2 × 17.0 = 34.0\nAdd the atoms outside the parentheses: 58.7 + 34.0\nM = 92.7 g/mol",
 "anticipatedWrong": [
  {
   "value": 75.7,
   "misconception": "you ignored the subscript outside the parentheses — the 2 multiplies both the oxygen and the hydrogen",
   "feedback": "This answer is wrong — you ignored the subscript outside the parentheses — the 2 multiplies both the oxygen and the hydrogen"
  },
  {
   "value": 151.4,
   "misconception": "you multiplied the nickel as well — the outside subscript reaches only what sits inside the parentheses",
   "feedback": "This answer is wrong — you multiplied the nickel as well — the outside subscript reaches only what sits inside the parentheses"
  },
  {
   "value": 91.7,
   "misconception": "you applied the outside 2 only to the oxygen — every atom inside the parentheses is doubled, the hydrogen included",
   "feedback": "This answer is wrong — you applied the outside 2 only to the oxygen — every atom inside the parentheses is doubled, the hydrogen included"
  }
 ]
}
MOL-017 · medium · application · seq 15 · authored (stage 2; RECORDED DEVIATION from plan row 15: the planned SO₂ leg would put 'sulfur' cross-form against the plan's own 11E S₂ leg — re-authored on benzene 156.2/78.1, corpus-clean; the SO₂ instance moved to 19C formula-only) · numeric
A sealed drum holds 156.2 g of benzene, C₆H₆ (M = 78.1 g/mol). How many moles of benzene is that? Give your answer in moles to 3 significant figures.
Answer: 2.0 mol KEY
also accepted: 2.00 mol
wrongValueDiagnostics
[
 {
  "value": 0.5,
  "misconception": "you divided the molar mass by the mass instead of the mass by the molar mass",
  "feedback": "This answer is wrong — you divided the molar mass by the mass instead of the mass by the molar mass"
 },
 {
  "value": 12199.22,
  "misconception": "you multiplied the mass by the molar mass instead of dividing",
  "feedback": "This answer is wrong — you multiplied the mass by the molar mass instead of dividing"
 },
 {
  "value": 20.0,
  "misconception": "you slipped a decimal place in the division — check by multiplying back: this count times 78.1 g/mol comes to ten times the drum's mass",
  "feedback": "This answer is wrong — you slipped a decimal place in the division — check by multiplying back: this count times 78.1 g/mol comes to ten times the drum's mass"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nm = 156.2 g\nM = 78.1 g/mol\nWrite down the equation:\nn = m / M\nSubstitute in the values, and calculate:\nn = 156.2 / 78.1\nn = 2.00 mol\nSanity check: a drum-scale mole count lands around 0.01 to 10 mol — 2.00 mol passes.",
 "anticipatedWrong": [
  {
   "value": 0.5,
   "misconception": "you divided the molar mass by the mass instead of the mass by the molar mass",
   "feedback": "This answer is wrong — you divided the molar mass by the mass instead of the mass by the molar mass"
  },
  {
   "value": 12199.22,
   "misconception": "you multiplied the mass by the molar mass instead of dividing",
   "feedback": "This answer is wrong — you multiplied the mass by the molar mass instead of dividing"
  },
  {
   "value": 20.0,
   "misconception": "you slipped a decimal place in the division — check by multiplying back: this count times 78.1 g/mol comes to ten times the drum's mass",
   "feedback": "This answer is wrong — you slipped a decimal place in the division — check by multiplying back: this count times 78.1 g/mol comes to ten times the drum's mass"
  }
 ]
}
MOL-018 · medium · application · seq 16 · authored (stage 2; RECORDED DEVIATION from plan row 16: NiO would be form C's third nickel compound and both plan alternates collide cross-form — fresh germanium dioxide instance, gate-unique word) · numeric
A ceramics kiln run uses 2.000 mol of germanium dioxide, GeO₂ (M = 104.6 g/mol). What mass of germanium dioxide is that? Give your answer in grams to 1 decimal place.
Answer: 209.2 g KEY
also accepted: 209.2 g
wrongValueDiagnostics
[
 {
  "value": 0.01912,
  "misconception": "you divided the moles by the molar mass instead of multiplying them",
  "feedback": "This answer is wrong — you divided the moles by the molar mass instead of multiplying them"
 },
 {
  "value": 52.3,
  "misconception": "you divided the molar mass by the moles instead of multiplying them",
  "feedback": "This answer is wrong — you divided the molar mass by the moles instead of multiplying them"
 },
 {
  "value": 20.92,
  "misconception": "you slipped a decimal place in the product — two portions of 104.6 g must weigh about 200 g",
  "feedback": "This answer is wrong — you slipped a decimal place in the product — two portions of 104.6 g must weigh about 200 g"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nn = 2.000 mol\nM = 104.6 g/mol\nWrite down the equation:\nn = m / M\nMake m the subject:\nm = n × M\nSubstitute in the values, and calculate:\nm = 2.000 × 104.6\nm = 209.2 g\nSanity check: more than 1 mol must weigh more than the 104.6 g molar mass — 209.2 g passes.",
 "anticipatedWrong": [
  {
   "value": 0.01912,
   "misconception": "you divided the moles by the molar mass instead of multiplying them",
   "feedback": "This answer is wrong — you divided the moles by the molar mass instead of multiplying them"
  },
  {
   "value": 52.3,
   "misconception": "you divided the molar mass by the moles instead of multiplying them",
   "feedback": "This answer is wrong — you divided the molar mass by the moles instead of multiplying them"
  },
  {
   "value": 20.92,
   "misconception": "you slipped a decimal place in the product — two portions of 104.6 g must weigh about 200 g",
   "feedback": "This answer is wrong — you slipped a decimal place in the product — two portions of 104.6 g must weigh about 200 g"
  }
 ]
}
MOL-010 · hard · transfer · seq 17 · seed:U7-G1/pool[29] (= MOL-010 gateItems[2]) — carried near-verbatim (element-free; no freshness flags) · mcq4
Which fact makes an element's amu number reappear as its g/mol number?
One gram is 6.022 × 10²³ atomic mass units — the same number as the particles in one mole. KEY
Atoms of every element share one standard mass.
The amu and the gram are exactly equal units of mass, just written under different names.
A mole of any substance weighs exactly one gram.
MOL-015 · hard · transfer · seq 18 · seed:U7-G1/pool[44] (= MOL-015 gateItems[2]) — RE-VALUED per rows/_AUTHORING_PLAN.md row 18 (20.0/40.0/100.0 corpus-consumed): 27.5/55.0/110.0; unnamed substances keep every value free; subtracted-foil re-derived; structure carried · mcq4
Substance X has a molar mass of 27.5 g/mol. Substance Y has a molar mass of 55.0 g/mol. A chemist weighs out 110.0 g of each. How do the two mole counts compare?
The X sample holds twice as many moles as the Y sample. KEY
The Y sample holds twice as many moles as the X sample.
The two samples hold equal numbers of moles.
The X sample holds 27.5 more moles than the Y sample.
MOL-019 · hard · transfer · seq 19 · seed:U7-G1/pool[56] (= MOL-019 gateItems[2]) — RE-LEGGED per plan (copper leg = cross-form collision with 10A; 74.6/39.9-mass values corpus-consumed) + recorded deviation: the plan's CaS leg dropped ('calcium' is MOL-019-own corpus) for the formula-only SO₂ instance freed by row 15; argon leg re-massed; all divisions float-exact; four orderings pairwise distinct · mcq4
Three canisters: 128.2 g of SO₂ gas (M = 64.1 g/mol); 59.85 g of argon (M = 39.9 g/mol); 74.48 g of palladium (M = 106.4 g/mol). Rank them from most moles to fewest.
SO₂ > argon > palladium KEY
SO₂ > palladium > argon
palladium > argon > SO₂
argon > SO₂ > palladium
MOL-020 · hard · transfer · seq 20 · seed:U7-G1/pool[59] (= MOL-020 gateItems[2], argon) — RE-SUBSTANCED to krypton per plan (argon = MOL-015 WE; 20.0/39.9 corpus-consumed) and RE-MASSED 13.0 -> 9.5 g (RECORDED deviation: the plan's 13.0 sits in MOL-008 spec text); which-check format and foil classes carried · mcq4
A question asks how many moles are in 9.5 g of krypton (M = 83.8 g/mol). A student writes: n = 9.5 × 83.8 = 796.1 mol. Which check exposes the error most directly?
The size check — 9.5 g is less than one 83.8 g portion. KEY
No check is needed, because the answer is correct.
The rounding check — 796.1 mol should be written as 800 mol.
The values check — the molar mass of krypton was copied wrongly.
MOL-002 · medium · application · seq 21 · authored FRQ (plan row 21; order count re-valued 500 -> 480: 500 sits in MOL-002's own spec text, RECORDED; values corpus-clean and float-exact); STAGE-3 re-value: unit mass 0.75 -> 2.1 g (0.75 sits in MOL-017 spec text) so count = 80, order mass = 1008.0 g (float-exact, corpus-clean) · frq
An office-supply packer counts identical heavy-duty packing staples by weighing.
An office-supply packer counts identical heavy-duty packing staples by weighing. (a) The staples in a box have a total mass of 168.0 g on their own, and one staple has a mass of 2.1 g. How many staples are in the box? Show every stage of your working: the values you start from, the equation in words, the substitution, and the count. (b) An order calls for 480 staples. What total mass of staples should be weighed out? Show your working the same way. (c) A trainee answers part (a) by computing 2.1 / 168.0. Explain what is wrong with that division and state the corrected count. Plain numbers and words are expected throughout — no special symbols. Treat the given masses as exact.
Part a. The staples in a box have a total mass of 168.0 g on their own, and one staple has a mass of 2.1 g. How many staples are in the box? Show every stage of your working: the values you start from, the equation in words, the substitution, and the count.
Model: Values: total mass = 168.0 g; mass of one staple = 2.1 g. Equation in words: number of staples = total mass divided by mass of one staple. Substitution: 168.0 / 2.1. Count: 80 staples. Check: 80 staples of 2.1 g each weigh exactly 168.0 g.
Part b. An order calls for 480 staples. What total mass of staples should be weighed out? Show your working the same way.
Model: Turn the equation around: mass to weigh out = number of staples times mass of one staple = 480 times 2.1 = 1008.0 g.
Part c. A trainee answers part (a) by computing 2.1 / 168.0. Explain what is wrong with that division and state the corrected count. Plain numbers and words are expected throughout — no special symbols. Treat the given masses as exact.
Model: The division is upside down — the total mass goes on top. 2.1 / 168.0 is about 0.013, and a box cannot hold a fraction of a staple. The corrected division is 168.0 / 2.1 = 80 staples. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
Full model answer. (a) Values: total mass = 168.0 g; mass of one staple = 2.1 g. Equation in words: number of staples = total mass divided by mass of one staple. Substitution: 168.0 / 2.1. Count: 80 staples. Check: 80 staples of 2.1 g each weigh exactly 168.0 g. (b) Turn the equation around: mass to weigh out = number of staples times mass of one staple = 480 times 2.1 = 1008.0 g. (c) The division is upside down — the total mass goes on top. 2.1 / 168.0 is about 0.013, and a box cannot hold a fraction of a staple. The corrected division is 168.0 / 2.1 = 80 staples. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
MOL-013 · medium · application · seq 22 · authored FRQ (plan row 22; RECORDED DEVIATION: the plan's Ca3(PO4)2 dropped — 'calcium' is MOL-013's own corpus word — replaced by Mn2(SO4)3; gold 1C/22C + manganese 10C/13C/22C repeats RECORDED) · frq
(a) Calculate the molar mass of gold(III) hydroxide, Au(OH)₃. (molar masses: Au 197.0, O 16.0, H 1.0 g/mol.) Show every stage: the value inside the parentheses, the multiplication by the outside subscript, and the final sum. (b) A study partner reports M = 214.0 g/mol for the same compound. Identify exactly what the partner did wrong and state the corrected molar mass. (c) Now calculate the molar mass of manganese(III) sulfate, Mn₂(SO₄)₃. (molar masses: Mn 54.9, S 32.1, O 16.0 g/mol.) Show the same stages. Plain numbers and words are expected — no special symbols.
(a) Calculate the molar mass of gold(III) hydroxide, Au(OH)₃. (molar masses: Au 197.0, O 16.0, H 1.0 g/mol.) Show every stage: the value inside the parentheses, the multiplication by the outside subscript, and the final sum. (b) A study partner reports M = 214.0 g/mol for the same compound. Identify exactly what the partner did wrong and state the corrected molar mass. (c) Now calculate the molar mass of manganese(III) sulfate, Mn₂(SO₄)₃. (molar masses: Mn 54.9, S 32.1, O 16.0 g/mol.) Show the same stages. Plain numbers and words are expected — no special symbols.
Part a. Calculate the molar mass of gold(III) hydroxide, Au(OH)₃. (molar masses: Au 197.0, O 16.0, H 1.0 g/mol.) Show every stage: the value inside the parentheses, the multiplication by the outside subscript, and the final sum.
Model: Inside the parentheses: 16.0 + 1.0 = 17.0. Multiply by the outside subscript: 3 times 17.0 = 51.0. Add the gold, which sits outside the parentheses: 197.0 + 51.0 = 248.0. M of gold(III) hydroxide = 248.0 g/mol.
Part b. A study partner reports M = 214.0 g/mol for the same compound. Identify exactly what the partner did wrong and state the corrected molar mass.
Model: The partner ignored the parentheses and counted the hydroxide group once: 197.0 + 17.0 = 214.0. The outside 3 triples the whole group — three oxygens and three hydrogens. Corrected: M = 248.0 g/mol.
Part c. Now calculate the molar mass of manganese(III) sulfate, Mn₂(SO₄)₃. (molar masses: Mn 54.9, S 32.1, O 16.0 g/mol.) Show the same stages. Plain numbers and words are expected — no special symbols.
Model: Two manganese: 2 times 54.9 = 109.8. Sulfate group: 32.1 + 4 times 16.0 = 96.1; the outside 3 triples it: 288.3. Total: 109.8 + 288.3 = 398.1. M of manganese(III) sulfate = 398.1 g/mol. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
Full model answer. (a) Inside the parentheses: 16.0 + 1.0 = 17.0. Multiply by the outside subscript: 3 times 17.0 = 51.0. Add the gold, which sits outside the parentheses: 197.0 + 51.0 = 248.0. M of gold(III) hydroxide = 248.0 g/mol. (b) The partner ignored the parentheses and counted the hydroxide group once: 197.0 + 17.0 = 214.0. The outside 3 triples the whole group — three oxygens and three hydrogens. Corrected: M = 248.0 g/mol. (c) Two manganese: 2 times 54.9 = 109.8. Sulfate group: 32.1 + 4 times 16.0 = 96.1; the outside 3 triples it: 288.3. Total: 109.8 + 288.3 = 398.1. M of manganese(III) sulfate = 398.1 g/mol. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
MOL-021 · hard · application · seq 23 · authored FRQ (plan row 23, parenthesized slot filled: Ti(NO₃)₄; RECORDED: titanium 9C/23C repeat; selenium = gate-unique; bottle-2 mass equals its molar mass — the seed-62 one-portion register) · frq
(a) How many moles are in 555.0 g of selenium dioxide, SeO₂? (molar masses: Se 79.0, O 16.0 g/mol.) Show both stages: first assemble the molar mass from the formula, then apply the mole equation. Give the answer in moles to 3 significant figures. (b) What is the mass, in grams, of a 2.000 mol batch of titanium(IV) nitrate, Ti(NO₃)₄? (molar masses: Ti 47.9, N 14.0, O 16.0 g/mol.) Show both stages, including how you rearrange the mole equation. Give the answer to 1 decimal place. (c) Bottle 1 holds 444.0 g of selenium dioxide; bottle 2 holds 295.9 g of titanium(IV) nitrate. Using the molar masses you found, work out which bottle holds more moles — show the working for both bottles; state the two mole amounts to 3 significant figures. Plain numbers and words are expected — no special symbols.
(a) How many moles are in 555.0 g of selenium dioxide, SeO₂? (molar masses: Se 79.0, O 16.0 g/mol.) Show both stages: first assemble the molar mass from the formula, then apply the mole equation. Give the answer in moles to 3 significant figures. (b) What is the mass, in grams, of a 2.000 mol batch of titanium(IV) nitrate, Ti(NO₃)₄? (molar masses: Ti 47.9, N 14.0, O 16.0 g/mol.) Show both stages, including how you rearrange the mole equation. Give the answer to 1 decimal place. (c) Bottle 1 holds 444.0 g of selenium dioxide; bottle 2 holds 295.9 g of titanium(IV) nitrate. Using the molar masses you found, work out which bottle holds more moles — show the working for both bottles; state the two mole amounts to 3 significant figures. Plain numbers and words are expected — no special symbols.
Part a. How many moles are in 555.0 g of selenium dioxide, SeO₂? (molar masses: Se 79.0, O 16.0 g/mol.) Show both stages: first assemble the molar mass from the formula, then apply the mole equation. Give the answer in moles to 3 significant figures.
Model: Stage 1 — molar mass of selenium dioxide: 79.0 + 2 times 16.0 = 111.0 g/mol. Stage 2 — moles: n = m/M = 555.0 / 111.0 = 5.00 mol.
Part b. What is the mass, in grams, of a 2.000 mol batch of titanium(IV) nitrate, Ti(NO₃)₄? (molar masses: Ti 47.9, N 14.0, O 16.0 g/mol.) Show both stages, including how you rearrange the mole equation. Give the answer to 1 decimal place.
Model: Stage 1 — molar mass of titanium(IV) nitrate: inside the parentheses 14.0 + 3 times 16.0 = 62.0; multiplied by the outside subscript 4 to 248.0; plus the titanium 47.9; total 295.9 g/mol. Stage 2 — rearrange to make m the subject: m = n times M = 2.000 times 295.9 = 591.8 g.
Part c. Bottle 1 holds 444.0 g of selenium dioxide; bottle 2 holds 295.9 g of titanium(IV) nitrate. Using the molar masses you found, work out which bottle holds more moles — show the working for both bottles; state the two mole amounts to 3 significant figures. Plain numbers and words are expected — no special symbols.
Model: Bottle 1: 444.0 / 111.0 = 4.00 mol. Bottle 2: 295.9 / 295.9 = 1.00 mol — exactly one portion. Bottle 1 holds more moles. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
Full model answer. (a) Stage 1 — molar mass of selenium dioxide: 79.0 + 2 times 16.0 = 111.0 g/mol. Stage 2 — moles: n = m/M = 555.0 / 111.0 = 5.00 mol. (b) Stage 1 — molar mass of titanium(IV) nitrate: inside the parentheses 14.0 + 3 times 16.0 = 62.0; multiplied by the outside subscript 4 to 248.0; plus the titanium 47.9; total 295.9 g/mol. Stage 2 — rearrange to make m the subject: m = n times M = 2.000 times 295.9 = 591.8 g. (c) Bottle 1: 444.0 / 111.0 = 4.00 mol. Bottle 2: 295.9 / 295.9 = 1.00 mol — exactly one portion. Bottle 1 holds more moles. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.

Form D

MOL-004 · easy · recall · seq 1 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 1: fresh factor comparison in the blueprint's D/E register; octane corpus-clean; foil classes parallel A — inverted / elastic / reached-for-molar-mass) · mcq4
How does the number of molecules in 6 mol of octane, C₈H₁₈, compare with the number in 1 mol of octane?
Six times as many. KEY
One-sixth as many.
The same number.
It cannot be determined without octane's molar mass.
MOL-007 · easy · recall · seq 2 · authored (fresh definitional facet per blueprint case_target + _AUTHORING_PLAN row 2: gallium 69.7 unit-discrimination read; unit foils misconception-keyed to the taught g/mol rule line — plain-mass unit, amount unit, inverted ratio; gallium real M 69.7, corpus-clean in scope — 69.7 appears only in out-of-scope MOL-031/032) · mcq4
A data sheet lists gallium's molar mass as 69.7 but leaves the unit blank. Which unit completes the entry?
g/mol KEY
g
mol
mol/g
MOL-016 · easy · recall · seq 3 · authored (fresh facet: pick-the-equation among permutational corruptions; framing distinct from MOL-016's own fluency stem) · mcq4
A student's formula sheet needs one line linking the amount of substance n, the mass m, and the molar mass M. Which line is correct?
n = m/M — divide the mass by the molar mass KEY
n = M/m — divide the molar mass by the mass
n = m × M — multiply the mass by the molar mass
m = n/M — divide the moles by the molar mass
MOL-005 · easy · recall · seq 4 · authored (fresh instance per stage-2 plan: how-many-particles recall, options permute exponent/coefficient; mcq-not-numeric justified by the V-T6 e-notation gap that spared pool[12] — see authoring_notes) · mcq4
A sealed glass ampoule holds exactly 1 mol of rhodium. How many rhodium atoms are inside?
6.022 × 10²³ KEY
6.202 × 10²³
6.022 × 10²²
6.022 × 10²⁰
MOL-001 · medium · application · seq 5 · authored (stage 2 parallel claim-critique slot per _AUTHORING_PLAN row 5: museum-balance any-exhibit claim, substance-free and number-free; misconception classes mirror gateItems[0] — count-display / precision-cost / heavy-particles) · mcq4
A museum lab weighs every exhibit it receives. A staff member claims that a top-of-the-range balance would display the number of particles in any exhibit. What is wrong with the claim?
A balance of any grade measures mass only — a particle count must be worked out from the mass reading. KEY
Nothing is wrong — top-of-the-range balances offer a particle-count display.
The claim is wrong only for mixed-material exhibits — for a pure substance the balance would display the count.
The claim fails only for exhibits whose particles are unusually heavy.
MOL-002 · medium · application · seq 6 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 6 with RECORDED DEVIATION: plan's dowels 302.4/8.4 -> 36 is not float-exact (302.4/8.4 = 35.99999999999999), so the pair shifted to 306.0/8.5 -> 36 keeping the planned answer; 306.0 corpus-clean and ledger-unique, 8.5 <= 12; exact division float-verified; 'dowel' zero digest hits; U6-G1 row09 D deviation class) · numeric
A carton holds identical dowels. The dowels alone have a total mass of 306.0 g, and one dowel has a mass of 8.5 g. How many dowels are in the carton?
Answer: 36 dowels KEY
also accepted: 36 dowels
wrongValueDiagnostics
[
 {
  "value": 0.0278,
  "misconception": "you divided the mass of one dowel by the total mass — the total goes on top",
  "feedback": "This answer is wrong — you divided the mass of one dowel by the total mass — the total goes on top"
 },
 {
  "value": 2601,
  "misconception": "you multiplied the two masses instead of dividing",
  "feedback": "This answer is wrong — you multiplied the two masses instead of dividing"
 },
 {
  "value": 360,
  "misconception": "you slipped a decimal place in the division — check by multiplying back: this count times the mass of one dowel comes to ten times the carton's total mass",
  "feedback": "This answer is wrong — you slipped a decimal place in the division — check by multiplying back: this count times the mass of one dowel comes to ten times the carton's total mass"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\ntotal mass = 306.0 g\nmass of one dowel = 8.5 g\nWrite down the equation:\nnumber of dowels = total mass / mass of one dowel\nSubstitute in the values, and calculate:\nnumber of dowels = 306.0 / 8.5\nnumber of dowels = 36",
 "anticipatedWrong": [
  {
   "value": 0.0278,
   "misconception": "you divided the mass of one dowel by the total mass — the total goes on top",
   "feedback": "This answer is wrong — you divided the mass of one dowel by the total mass — the total goes on top"
  },
  {
   "value": 2601,
   "misconception": "you multiplied the two masses instead of dividing",
   "feedback": "This answer is wrong — you multiplied the two masses instead of dividing"
  },
  {
   "value": 360,
   "misconception": "you slipped a decimal place in the division — check by multiplying back: this count times the mass of one dowel comes to ten times the carton's total mass",
   "feedback": "This answer is wrong — you slipped a decimal place in the division — check by multiplying back: this count times the mass of one dowel comes to ten times the carton's total mass"
  }
 ]
}
MOL-003 · medium · application · seq 7 · authored (stage 2 per rows/_AUTHORING_PLAN.md row 7: pure tantalum wire vs mixed fasteners — fresh pure-vs-mixed contrast paralleling pool[7]'s structure and foil classes: sample-size / instruments-count / blamed-the-balance; tantalum corpus-clean, number-free; avoids MOL-003's own WE tokens spool/jar/nuts/bolts) · mcq4
Weighing can count the atoms in a reel of pure tantalum wire but not the pieces in a tub of mixed fasteners. What makes the difference?
Tantalum atoms share one steady average mass, while the fasteners' masses differ from piece to piece. KEY
The reel holds less material, and smaller samples are countable.
Atoms, unlike fasteners, are small enough to be counted directly by laboratory instruments.
The fasteners are too heavy for the balance to weigh accurately.
MOL-006 · medium · application · seq 8 · authored (stage 2 per _AUTHORING_PLAN row 8: fresh molecular case in the seeds' exact register — phosphine PH₃, word 'phosphine' only, NO 'phosphorus' (11A's reserved word); phosphine = zero hits in _scope_digest.txt, corpus-clean; foil classes mirror seed pool[17] — split-the-particle / ionic-branch / charged-particles) · mcq4
Which particle does a mole of phosphine, PH₃, count?
Molecules KEY
Atoms
Formula units
Ions
MOL-009 · medium · application · seq 9 · authored (fresh element on the Si-gate permutational pairing pattern per blueprint case_target + _AUTHORING_PLAN row 9: hafnium 178.5 pairing facet; the four amu/gram pairings form the closed permutation space — swap, both-amu, both-grams misconception-keyed to the taught scale rules; hafnium real M 178.5, word + value corpus-clean in scope) · mcq4
Hafnium's periodic-table square prints 178.5. Which pair of statements is right?
One hafnium atom: 178.5 amu. 1 mol of hafnium atoms: 178.5 g. KEY
One hafnium atom: 178.5 g. 1 mol of hafnium atoms: 178.5 amu.
One hafnium atom: 178.5 amu. 1 mol of hafnium atoms: 178.5 amu.
One hafnium atom: 178.5 g. 1 mol of hafnium atoms: 178.5 g.
MOL-008 · medium · application · seq 10 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 10: fresh square, cadmium corpus-clean; ledger cadmium 10D/14D recorded) · numeric
The periodic-table square for cadmium is shown. State cadmium's molar mass, in g/mol. Give your answer to 4 significant figures.
FIGURE (pt-square-reference-table · stimulus) — title: "periodic-table square: cadmium" · rows: [["atomic number", "48"], ["symbol", "Cd"], ["average atomic mass", "112.4"]] · style: "single periodic-table element square, exactly as printed on the classroom periodic table"
raw spec
{
 "figureRequired": true,
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "pt-square-reference-table",
  "title": "periodic-table square: cadmium",
  "rows": [
   [
    "atomic number",
    "48"
   ],
   [
    "symbol",
    "Cd"
   ],
   [
    "average atomic mass",
    "112.4"
   ]
  ],
  "style": "single periodic-table element square, exactly as printed on the classroom periodic table"
 }
}
Answer: 112.4 g/mol KEY
also accepted: 112.4 g/mol
wrongValueDiagnostics
[
 {
  "value": 48,
  "misconception": "you read the atomic number — the smaller whole number counts protons and is not a mass",
  "feedback": "This answer is wrong — you read the atomic number — the smaller whole number counts protons and is not a mass"
 },
 {
  "value": 64.4,
  "misconception": "you subtracted the atomic number from the atomic mass — that neutron-count move belongs to isotope work, not molar mass",
  "feedback": "This answer is wrong — you subtracted the atomic number from the atomic mass — that neutron-count move belongs to isotope work, not molar mass"
 },
 {
  "value": 224.8,
  "misconception": "you doubled the printed value — the square's average atomic mass is already the whole molar mass for an element of single atoms",
  "feedback": "This answer is wrong — you doubled the printed value — the square's average atomic mass is already the whole molar mass for an element of single atoms"
 }
]
feedback
{
 "keyExplanation": "The square prints the atomic number 48 and the average atomic mass 112.4. The average atomic mass, read in grams per mole, is the molar mass: cadmium's is 112.4 g/mol. The 48 counts protons and is never a mass.",
 "anticipatedWrong": [
  {
   "value": 48,
   "misconception": "you read the atomic number — the smaller whole number counts protons and is not a mass",
   "feedback": "This answer is wrong — you read the atomic number — the smaller whole number counts protons and is not a mass"
  },
  {
   "value": 64.4,
   "misconception": "you subtracted the atomic number from the atomic mass — that neutron-count move belongs to isotope work, not molar mass",
   "feedback": "This answer is wrong — you subtracted the atomic number from the atomic mass — that neutron-count move belongs to isotope work, not molar mass"
  },
  {
   "value": 224.8,
   "misconception": "you doubled the printed value — the square's average atomic mass is already the whole molar mass for an element of single atoms",
   "feedback": "This answer is wrong — you doubled the printed value — the square's average atomic mass is already the whole molar mass for an element of single atoms"
  }
 ]
}
MOL-011 · medium · application · seq 11 · seed:U7-G1/pool[30] (= MOL-011 gateItems[0]) — the seed's own ozone instance, with its must_freshen resolved structurally: the corpus-consumed 16.0 moves into the figure (unscanned) and the stem says only 'ozone', never the own-article word 'oxygen' (plan row 11 D) · numeric
Ozone exists as O₃ molecules. The periodic-table square for the element ozone is built from is shown. Calculate the molar mass of O₃, in g/mol, to one decimal place. Give your answer to 3 significant figures.
FIGURE (pt-square-reference-table · stimulus) — title: "periodic-table square" · rows: [["atomic number", "8"], ["symbol", "O"], ["average atomic mass", "16.0"]] · style: "single periodic-table element square, exactly as printed on the classroom periodic table"
raw spec
{
 "figureRequired": true,
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "pt-square-reference-table",
  "title": "periodic-table square",
  "rows": [
   [
    "atomic number",
    "8"
   ],
   [
    "symbol",
    "O"
   ],
   [
    "average atomic mass",
    "16.0"
   ]
  ],
  "style": "single periodic-table element square, exactly as printed on the classroom periodic table"
 }
}
Answer: 48.0 g/mol KEY
also accepted: 48.0 g/mol
wrongValueDiagnostics
[
 {
  "value": 16.0,
  "misconception": "you ignored the subscript — each O₃ molecule holds three atoms, so the atom's value triples",
  "feedback": "This answer is wrong — you ignored the subscript — each O₃ molecule holds three atoms, so the atom's value triples"
 },
 {
  "value": 19.0,
  "misconception": "you added the subscript as if it were a mass — the 3 is a count and multiplies",
  "feedback": "This answer is wrong — you added the subscript as if it were a mass — the 3 is a count and multiplies"
 },
 {
  "value": 5.33,
  "misconception": "you divided by the subscript instead of multiplying — the 3 counts atoms per molecule",
  "feedback": "This answer is wrong — you divided by the subscript instead of multiplying — the 3 counts atoms per molecule"
 }
]
feedback
{
 "keyExplanation": "The square prints the average atomic mass 16.0. The subscript counts the atoms in one molecule: 3.\nmolar mass of O₃ = 3 × 16.0\nmolar mass of O₃ = 48.0 g/mol\nEach ozone molecule weighs three atoms' worth of the element it is built from.",
 "anticipatedWrong": [
  {
   "value": 16.0,
   "misconception": "you ignored the subscript — each O₃ molecule holds three atoms, so the atom's value triples",
   "feedback": "This answer is wrong — you ignored the subscript — each O₃ molecule holds three atoms, so the atom's value triples"
  },
  {
   "value": 19.0,
   "misconception": "you added the subscript as if it were a mass — the 3 is a count and multiplies",
   "feedback": "This answer is wrong — you added the subscript as if it were a mass — the 3 is a count and multiplies"
  },
  {
   "value": 5.33,
   "misconception": "you divided by the subscript instead of multiplying — the 3 counts atoms per molecule",
   "feedback": "This answer is wrong — you divided by the subscript instead of multiplying — the 3 counts atoms per molecule"
  }
 ]
}
MOL-014 · medium · application · seq 12 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 12: fresh quarter-factor case on glycerin; foil classes = factor-twice / moles-as-property / reached-for-molar-mass) · mcq4
A bottle of glycerin contains 8.0 mol. A chemist pours out most of it, keeping exactly one-quarter of the original mass. How many moles does the kept portion contain?
2.0 mol KEY
4.0 mol
8.0 mol
It cannot be determined without glycerin's molar mass.
MOL-012 · medium · application · seq 13 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 13: fresh two-element compound; 13A/13D barium = same-seq A5 share, RECORDED; Ba 137.3 + F 19.0 = canonical constants, CONSTANTS-CLASS family) · numeric
Barium fluoride has the formula BaF₂. Use these molar masses: Ba 137.3, F 19.0 g/mol. Calculate BaF₂'s molar mass, in g/mol, to one decimal place. Give your answer to 4 significant figures.
Answer: 175.3 g/mol KEY
also accepted: 175.3 g/mol
wrongValueDiagnostics
[
 {
  "value": 156.3,
  "misconception": "you ignored the subscript — the formula counts two fluorines, not one",
  "feedback": "This answer is wrong — you ignored the subscript — the formula counts two fluorines, not one"
 },
 {
  "value": 312.6,
  "misconception": "you multiplied every element by the subscript — the 2 belongs to fluorine alone",
  "feedback": "This answer is wrong — you multiplied every element by the subscript — the 2 belongs to fluorine alone"
 },
 {
  "value": 158.3,
  "misconception": "you added the subscript as if it were a mass — the 2 is a count and multiplies the fluorine value",
  "feedback": "This answer is wrong — you added the subscript as if it were a mass — the 2 is a count and multiplies the fluorine value"
 }
]
feedback
{
 "keyExplanation": "Count the atoms the formula unit shows: 1 barium and 2 fluorines.\nmolar mass of BaF₂ = 137.3 + 2 × 19.0\nmolar mass of BaF₂ = 175.3 g/mol\nThe subscript 2 multiplies the fluorine value only; barium appears once.",
 "anticipatedWrong": [
  {
   "value": 156.3,
   "misconception": "you ignored the subscript — the formula counts two fluorines, not one",
   "feedback": "This answer is wrong — you ignored the subscript — the formula counts two fluorines, not one"
  },
  {
   "value": 312.6,
   "misconception": "you multiplied every element by the subscript — the 2 belongs to fluorine alone",
   "feedback": "This answer is wrong — you multiplied every element by the subscript — the 2 belongs to fluorine alone"
  },
  {
   "value": 158.3,
   "misconception": "you added the subscript as if it were a mass — the 2 is a count and multiplies the fluorine value",
   "feedback": "This answer is wrong — you added the subscript as if it were a mass — the 2 is a count and multiplies the fluorine value"
  }
 ]
}
MOL-013 · medium · application · seq 14 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 14: fresh nitrate with parentheses; cadmium 10D/14D same-form repeat RECORDED — the 10D square value is figure-borne, so 112.4 appears in one scanned stem only) · numeric
Calculate the molar mass of cadmium nitrate, Cd(NO₃)₂. (molar masses: Cd 112.4, N 14.0, O 16.0 g/mol.) Give your answer in g/mol to one decimal place. Give your answer to 4 significant figures.
Answer: 236.4 g/mol KEY
also accepted: 236.4 g/mol
wrongValueDiagnostics
[
 {
  "value": 174.4,
  "misconception": "you ignored the subscript outside the parentheses — the 2 multiplies the whole NO₃ group",
  "feedback": "This answer is wrong — you ignored the subscript outside the parentheses — the 2 multiplies the whole NO₃ group"
 },
 {
  "value": 348.8,
  "misconception": "you multiplied the cadmium as well — the outside subscript reaches only what sits inside the parentheses",
  "feedback": "This answer is wrong — you multiplied the cadmium as well — the outside subscript reaches only what sits inside the parentheses"
 },
 {
  "value": 188.4,
  "misconception": "you applied the outside 2 only to the nitrogen — every atom inside the parentheses is doubled, all three oxygens included",
  "feedback": "This answer is wrong — you applied the outside 2 only to the nitrogen — every atom inside the parentheses is doubled, all three oxygens included"
 }
]
feedback
{
 "keyExplanation": "Inside the parentheses: 14.0 + 3 × 16.0 = 62.0\nMultiply by the outside subscript: 2 × 62.0 = 124.0\nAdd the atoms outside the parentheses: 112.4 + 124.0\nM = 236.4 g/mol",
 "anticipatedWrong": [
  {
   "value": 174.4,
   "misconception": "you ignored the subscript outside the parentheses — the 2 multiplies the whole NO₃ group",
   "feedback": "This answer is wrong — you ignored the subscript outside the parentheses — the 2 multiplies the whole NO₃ group"
  },
  {
   "value": 348.8,
   "misconception": "you multiplied the cadmium as well — the outside subscript reaches only what sits inside the parentheses",
   "feedback": "This answer is wrong — you multiplied the cadmium as well — the outside subscript reaches only what sits inside the parentheses"
  },
  {
   "value": 188.4,
   "misconception": "you applied the outside 2 only to the nitrogen — every atom inside the parentheses is doubled, all three oxygens included",
   "feedback": "This answer is wrong — you applied the outside 2 only to the nitrogen — every atom inside the parentheses is doubled, all three oxygens included"
  }
 ]
}
MOL-017 · medium · application · seq 15 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 15: glycerol triple re-massed to the float-exact 184.2/92.1 (planned 276.3/92.1 fails float-exactness — U6-G1 row09-D precedent, RECORDED); glycerol is a distinct token from 12D's glycerin, noted) · numeric
A laboratory bottle holds 184.2 g of glycerol (M = 92.1 g/mol). How many moles of glycerol is that? Give your answer in moles to 3 significant figures.
Answer: 2.0 mol KEY
also accepted: 2.00 mol
wrongValueDiagnostics
[
 {
  "value": 0.5,
  "misconception": "you divided the molar mass by the mass instead of the mass by the molar mass",
  "feedback": "This answer is wrong — you divided the molar mass by the mass instead of the mass by the molar mass"
 },
 {
  "value": 16964.82,
  "misconception": "you multiplied the mass by the molar mass instead of dividing",
  "feedback": "This answer is wrong — you multiplied the mass by the molar mass instead of dividing"
 },
 {
  "value": 20.0,
  "misconception": "you slipped a decimal place in the division — check by multiplying back: this count times 92.1 g/mol comes to ten times the bottle's mass",
  "feedback": "This answer is wrong — you slipped a decimal place in the division — check by multiplying back: this count times 92.1 g/mol comes to ten times the bottle's mass"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nm = 184.2 g\nM = 92.1 g/mol\nWrite down the equation:\nn = m / M\nSubstitute in the values, and calculate:\nn = 184.2 / 92.1\nn = 2.00 mol\nSanity check: a bottle-scale mole count lands around 0.01 to 10 mol — 2.00 mol passes.",
 "anticipatedWrong": [
  {
   "value": 0.5,
   "misconception": "you divided the molar mass by the mass instead of the mass by the molar mass",
   "feedback": "This answer is wrong — you divided the molar mass by the mass instead of the mass by the molar mass"
  },
  {
   "value": 16964.82,
   "misconception": "you multiplied the mass by the molar mass instead of dividing",
   "feedback": "This answer is wrong — you multiplied the mass by the molar mass instead of dividing"
  },
  {
   "value": 20.0,
   "misconception": "you slipped a decimal place in the division — check by multiplying back: this count times 92.1 g/mol comes to ten times the bottle's mass",
   "feedback": "This answer is wrong — you slipped a decimal place in the division — check by multiplying back: this count times 92.1 g/mol comes to ten times the bottle's mass"
  }
 ]
}
MOL-018 · medium · application · seq 16 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 16's own preferred switch: CdS dropped for Ga₂O₃; gallium 2D/16D same-form repeat RECORDED (plan-anticipated); float-exact 0.500 × 187.4); STAGE-3 re-value 0.500 -> 2.00 mol (same scanned-non-integer class as 16B; float-exact) · numeric
A crystal-growth batch needs 2.000 mol of gallium oxide, Ga₂O₃ (M = 187.4 g/mol). What mass of gallium oxide is that? Give your answer in grams to 1 decimal place.
Answer: 374.8 g KEY
also accepted: 374.8 g
wrongValueDiagnostics
[
 {
  "value": 0.01067,
  "misconception": "you divided the moles by the molar mass instead of multiplying them",
  "feedback": "This answer is wrong — you divided the moles by the molar mass instead of multiplying them"
 },
 {
  "value": 93.7,
  "misconception": "you divided the molar mass by the moles instead of multiplying them",
  "feedback": "This answer is wrong — you divided the molar mass by the moles instead of multiplying them"
 },
 {
  "value": 37.48,
  "misconception": "you slipped a decimal place in the product — two portions of 187.4 g must weigh about 375 g",
  "feedback": "This answer is wrong — you slipped a decimal place in the product — two portions of 187.4 g must weigh about 375 g"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nn = 2.000 mol\nM = 187.4 g/mol\nWrite down the equation:\nn = m / M\nMake m the subject:\nm = n × M\nSubstitute in the values, and calculate:\nm = 2.000 × 187.4\nm = 374.8 g\nSanity check: more than 1 mol must weigh more than the 187.4 g molar mass — 374.8 g passes.",
 "anticipatedWrong": [
  {
   "value": 0.01067,
   "misconception": "you divided the moles by the molar mass instead of multiplying them",
   "feedback": "This answer is wrong — you divided the moles by the molar mass instead of multiplying them"
  },
  {
   "value": 93.7,
   "misconception": "you divided the molar mass by the moles instead of multiplying them",
   "feedback": "This answer is wrong — you divided the molar mass by the moles instead of multiplying them"
  },
  {
   "value": 37.48,
   "misconception": "you slipped a decimal place in the product — two portions of 187.4 g must weigh about 375 g",
   "feedback": "This answer is wrong — you slipped a decimal place in the product — two portions of 187.4 g must weigh about 375 g"
  }
 ]
}
MOL-010 · hard · transfer · seq 17 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 17: element-free heavy-atoms scope critique; foil classes = capped-regrouping / whole-number carve-out / shrunken mole) · mcq4
A classmate says the pattern 'an atom's amu value reappears as the element's g/mol value' must break down for very heavy atoms. Which reply is right?
The pattern holds for every element — one gram is 6.022 × 10²³ amu, whatever the atom's mass. KEY
The classmate is right — very heavy atoms are too massive for the amu-to-gram regrouping to work.
The pattern survives for heavy atoms only when their amu values round to whole numbers.
The pattern breaks down because a mole of a heavy element contains fewer particles.
MOL-015 · hard · transfer · seq 18 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 18: fresh equal-mass metal pair; shared mass RE-VALUED 150.0 -> 195.0 g (150.0 = the 12B seed-carried option value, plan ledger check); indium 18D/22D + bismuth 18D/22D repeats RECORDED) · mcq4
A metals dealer owns a 195.0 g bar of indium (molar mass 114.8 g/mol) and a 195.0 g bar of bismuth (molar mass 209.0 g/mol). Which bar contains more moles of atoms?
The indium bar. KEY
The bismuth bar.
The two bars contain equal numbers of moles, because their masses are equal.
Counting the atoms in each bar is the only way to decide.
MOL-019 · hard · transfer · seq 19 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 19: molybdenum/butane/PbS with fresh float-exact masses; molybdenum 19D/23D repeat RECORDED; lead = WARN-other single use; four orderings pairwise distinct) · mcq4
Three drums: 191.8 g of molybdenum (M = 95.9 g/mol); 87.15 g of butane, C₄H₁₀ (M = 58.1 g/mol); 119.65 g of lead(II) sulfide, PbS (M = 239.3 g/mol). Rank them from most moles to fewest.
molybdenum > butane > lead(II) sulfide KEY
molybdenum > lead(II) sulfide > butane
lead(II) sulfide > butane > molybdenum
butane > molybdenum > lead(II) sulfide
MOL-020 · hard · transfer · seq 20 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 20: fresh upside-down critique on citric acid 576.3/192.1 (corpus-clean; corrected division float-exact); foil classes = nothing-wrong / multiplied-misread / rounding-blame) · mcq4
A question asks how many moles are in 576.3 g of citric acid (M = 192.1 g/mol). A student writes: n = 192.1 / 576.3 = 0.333 mol. What is the error?
The division is upside down — n = m/M puts the mass on top: 576.3 / 192.1 = 3.000 mol. KEY
Nothing is wrong with the solution.
The solution multiplied by the molar mass instead of dividing.
The answer fails only because 0.333 should be rounded to 0.3.
MOL-002 · medium · application · seq 21 · authored FRQ (plan row 21; order count re-valued 150 -> 140: 150 collides with the 12B seed-carried option value 150.0, RECORDED; values corpus-clean and float-exact) · frq
A garment factory counts identical metal eyelets by weighing.
A garment factory counts identical metal eyelets by weighing. (a) The eyelets in a tray have a total mass of 217.0 g on their own, and one eyelet has a mass of 3.5 g. How many eyelets are in the tray? Show every stage of your working: the values you start from, the equation in words, the substitution, and the count. (b) A production run needs 140 eyelets. What total mass of eyelets should be weighed out? Show your working the same way. (c) A line worker answers part (a) by computing 3.5 / 217.0. Explain what is wrong with that division and state the corrected count. Plain numbers and words are expected throughout — no special symbols. Treat the given masses as exact.
Part a. The eyelets in a tray have a total mass of 217.0 g on their own, and one eyelet has a mass of 3.5 g. How many eyelets are in the tray? Show every stage of your working: the values you start from, the equation in words, the substitution, and the count.
Model: Values: total mass = 217.0 g; mass of one eyelet = 3.5 g. Equation in words: number of eyelets = total mass divided by mass of one eyelet. Substitution: 217.0 / 3.5. Count: 62 eyelets. Check: 62 eyelets of 3.5 g each weigh exactly 217.0 g.
Part b. A production run needs 140 eyelets. What total mass of eyelets should be weighed out? Show your working the same way.
Model: Turn the equation around: mass to weigh out = number of eyelets times mass of one eyelet = 140 times 3.5 = 490.0 g.
Part c. A line worker answers part (a) by computing 3.5 / 217.0. Explain what is wrong with that division and state the corrected count. Plain numbers and words are expected throughout — no special symbols. Treat the given masses as exact.
Model: The division is upside down — the total mass goes on top. 3.5 / 217.0 is about 0.016, and a tray cannot hold a fraction of an eyelet. The corrected division is 217.0 / 3.5 = 62 eyelets. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
Full model answer. (a) Values: total mass = 217.0 g; mass of one eyelet = 3.5 g. Equation in words: number of eyelets = total mass divided by mass of one eyelet. Substitution: 217.0 / 3.5. Count: 62 eyelets. Check: 62 eyelets of 3.5 g each weigh exactly 217.0 g. (b) Turn the equation around: mass to weigh out = number of eyelets times mass of one eyelet = 140 times 3.5 = 490.0 g. (c) The division is upside down — the total mass goes on top. 3.5 / 217.0 is about 0.016, and a tray cannot hold a fraction of an eyelet. The corrected division is 217.0 / 3.5 = 62 eyelets. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
MOL-013 · medium · application · seq 22 · authored FRQ (plan row 22; RECORDED DEVIATION: the plan's Ba(NO3)2 dropped — 'barium' is MOL-013's own corpus word — replaced by Bi(NO3)3 395.0; bismuth 18D/22D + indium 18D/22D repeats RECORDED) · frq
(a) Calculate the molar mass of bismuth(III) nitrate, Bi(NO₃)₃. (molar masses: Bi 209.0, N 14.0, O 16.0 g/mol.) Show every stage: the value inside the parentheses, the multiplication by the outside subscript, and the final sum. (b) A bench partner reports M = 271.0 g/mol for the same compound. Identify exactly what the partner did wrong and state the corrected molar mass. (c) Now calculate the molar mass of indium sulfate, In₂(SO₄)₃. (molar masses: In 114.8, S 32.1, O 16.0 g/mol.) Show the same stages. Plain numbers and words are expected — no special symbols.
(a) Calculate the molar mass of bismuth(III) nitrate, Bi(NO₃)₃. (molar masses: Bi 209.0, N 14.0, O 16.0 g/mol.) Show every stage: the value inside the parentheses, the multiplication by the outside subscript, and the final sum. (b) A bench partner reports M = 271.0 g/mol for the same compound. Identify exactly what the partner did wrong and state the corrected molar mass. (c) Now calculate the molar mass of indium sulfate, In₂(SO₄)₃. (molar masses: In 114.8, S 32.1, O 16.0 g/mol.) Show the same stages. Plain numbers and words are expected — no special symbols.
Part a. Calculate the molar mass of bismuth(III) nitrate, Bi(NO₃)₃. (molar masses: Bi 209.0, N 14.0, O 16.0 g/mol.) Show every stage: the value inside the parentheses, the multiplication by the outside subscript, and the final sum.
Model: Inside the parentheses: 14.0 + 3 times 16.0 = 62.0. Multiply by the outside subscript: 3 times 62.0 = 186.0. Add the bismuth, which sits outside the parentheses: 209.0 + 186.0 = 395.0. M of bismuth(III) nitrate = 395.0 g/mol.
Part b. A bench partner reports M = 271.0 g/mol for the same compound. Identify exactly what the partner did wrong and state the corrected molar mass.
Model: The partner ignored the subscript outside the parentheses and counted the nitrate group once: 209.0 + 62.0 = 271.0. The outside 3 triples the whole group. Corrected: M = 395.0 g/mol.
Part c. Now calculate the molar mass of indium sulfate, In₂(SO₄)₃. (molar masses: In 114.8, S 32.1, O 16.0 g/mol.) Show the same stages. Plain numbers and words are expected — no special symbols.
Model: Two indium: 2 times 114.8 = 229.6. Sulfate group: 32.1 + 4 times 16.0 = 96.1; the outside 3 triples it: 288.3. Total: 229.6 + 288.3 = 517.9. M of indium sulfate = 517.9 g/mol. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
Full model answer. (a) Inside the parentheses: 14.0 + 3 times 16.0 = 62.0. Multiply by the outside subscript: 3 times 62.0 = 186.0. Add the bismuth, which sits outside the parentheses: 209.0 + 186.0 = 395.0. M of bismuth(III) nitrate = 395.0 g/mol. (b) The partner ignored the subscript outside the parentheses and counted the nitrate group once: 209.0 + 62.0 = 271.0. The outside 3 triples the whole group. Corrected: M = 395.0 g/mol. (c) Two indium: 2 times 114.8 = 229.6. Sulfate group: 32.1 + 4 times 16.0 = 96.1; the outside 3 triples it: 288.3. Total: 229.6 + 288.3 = 517.9. M of indium sulfate = 517.9 g/mol. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
MOL-021 · hard · application · seq 23 · authored FRQ (plan row 23; MoO3 kept per plan (431.7 float-exact); parenthesized slot filled with Ce(SO₄)₂ (cerium gate-unique); molybdenum 19D/23D repeat RECORDED) · frq
(a) How many moles are in 431.7 g of molybdenum trioxide, MoO₃? (molar masses: Mo 95.9, O 16.0 g/mol.) Show both stages: first assemble the molar mass from the formula, then apply the mole equation. Give the answer in moles to 3 significant figures. (b) What is the mass, in grams, of exactly 2.000 mol of cerium(IV) sulfate, Ce(SO₄)₂? (molar masses: Ce 140.1, S 32.1, O 16.0 g/mol.) Show both stages, including how you rearrange the mole equation. Give the answer to 1 decimal place. (c) Drum 1 holds 719.5 g of molybdenum trioxide; drum 2 holds 996.9 g of cerium(IV) sulfate. Using the molar masses you found, work out which drum holds more moles — show the working for both drums, quoting each mole figure to 3 significant figures. Plain numbers and words are expected — no special symbols.
(a) How many moles are in 431.7 g of molybdenum trioxide, MoO₃? (molar masses: Mo 95.9, O 16.0 g/mol.) Show both stages: first assemble the molar mass from the formula, then apply the mole equation. Give the answer in moles to 3 significant figures. (b) What is the mass, in grams, of exactly 2.000 mol of cerium(IV) sulfate, Ce(SO₄)₂? (molar masses: Ce 140.1, S 32.1, O 16.0 g/mol.) Show both stages, including how you rearrange the mole equation. Give the answer to 1 decimal place. (c) Drum 1 holds 719.5 g of molybdenum trioxide; drum 2 holds 996.9 g of cerium(IV) sulfate. Using the molar masses you found, work out which drum holds more moles — show the working for both drums, quoting each mole figure to 3 significant figures. Plain numbers and words are expected — no special symbols.
Part a. How many moles are in 431.7 g of molybdenum trioxide, MoO₃? (molar masses: Mo 95.9, O 16.0 g/mol.) Show both stages: first assemble the molar mass from the formula, then apply the mole equation. Give the answer in moles to 3 significant figures.
Model: Stage 1 — molar mass of molybdenum trioxide: 95.9 + 3 times 16.0 = 143.9 g/mol. Stage 2 — moles: n = m/M = 431.7 / 143.9 = 3.00 mol.
Part b. What is the mass, in grams, of exactly 2.000 mol of cerium(IV) sulfate, Ce(SO₄)₂? (molar masses: Ce 140.1, S 32.1, O 16.0 g/mol.) Show both stages, including how you rearrange the mole equation. Give the answer to 1 decimal place.
Model: Stage 1 — molar mass of cerium(IV) sulfate: inside the parentheses 32.1 + 4 times 16.0 = 96.1; doubled by the outside subscript to 192.2; plus the cerium 140.1; total 332.3 g/mol. Stage 2 — rearrange to make m the subject: m = n times M = 2.000 times 332.3 = 664.6 g.
Part c. Drum 1 holds 719.5 g of molybdenum trioxide; drum 2 holds 996.9 g of cerium(IV) sulfate. Using the molar masses you found, work out which drum holds more moles — show the working for both drums, quoting each mole figure to 3 significant figures. Plain numbers and words are expected — no special symbols.
Model: Drum 1: 719.5 / 143.9 = 5.00 mol. Drum 2: 996.9 / 332.3 = 3.00 mol. Drum 1 holds more moles, even though drum 2 is heavier — each mole of cerium(IV) sulfate weighs far more. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
Full model answer. (a) Stage 1 — molar mass of molybdenum trioxide: 95.9 + 3 times 16.0 = 143.9 g/mol. Stage 2 — moles: n = m/M = 431.7 / 143.9 = 3.00 mol. (b) Stage 1 — molar mass of cerium(IV) sulfate: inside the parentheses 32.1 + 4 times 16.0 = 96.1; doubled by the outside subscript to 192.2; plus the cerium 140.1; total 332.3 g/mol. Stage 2 — rearrange to make m the subject: m = n times M = 2.000 times 332.3 = 664.6 g. (c) Drum 1: 719.5 / 143.9 = 5.00 mol. Drum 2: 996.9 / 332.3 = 3.00 mol. Drum 1 holds more moles, even though drum 2 is heavier — each mole of cerium(IV) sulfate weighs far more. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.

Form E

MOL-004 · easy · recall · seq 1 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 1: fresh equal-count compare on platinum vs methanol; foil classes parallel pool[11] with the counted-elements trap recast as counted-atoms-inside-the-molecule) · mcq4
Sample A is 1 mol of platinum. Sample B is 1 mol of methanol, CH₃OH. Compare the numbers of particles in the two samples.
Equal — the mole is the same fixed count for any substance. KEY
Sample A holds more particles, because platinum is a metal.
Sample B holds more particles, because each methanol molecule contains several atoms.
The comparison needs both samples' masses.
MOL-007 · easy · recall · seq 2 · authored (fresh read-back facet per blueprint case_target + _AUTHORING_PLAN row 2: mercury 200.6 read-back in the article WE register (CO2 44.0 -> 44.0 g), pinned to exactly 1 mol per specNotes — no m = n × M; foils misconception-keyed: per-mole-unit slip, Avogadro-count-as-mass (rule-content value, seq-4 adjudication class), mole-count-as-mass; mercury real M 200.6, corpus-clean in scope) · mcq4
Mercury's molar mass is 200.6 g/mol. What is the mass of exactly 1 mol of mercury?
200.6 g KEY
200.6 g/mol
6.022 × 10²³ g
1.0 g
MOL-016 · easy · recall · seq 3 · authored (fresh facet: unit of the result, g ÷ g/mol → mol; framing and distractor set distinct from MOL-016's own fluency item) · mcq4
A student divides a sample's mass, in grams, by the substance's molar mass, in grams per mole, and gets a number. Which unit belongs on that number?
mol KEY
g
g/mol
No unit — the units cancel completely.
MOL-005 · easy · recall · seq 4 · authored (fresh facet per stage-2 plan: 'changes with substance' correction, substance-free) · mcq4
A student says that Avogadro's number must be looked up separately for each substance, the way a molar mass is. Which correction is right?
There is nothing to look up — one fixed count, 6.022 × 10²³ particles per mole, works for every substance. KEY
The student is right — every substance has its own Avogadro's number that must be looked up.
The number is the same only for substances whose particles are single atoms, not molecules.
The number stays fixed for solids and liquids, but for gases it changes with the container.
MOL-001 · medium · application · seq 5 · authored (stage 2 parallel claim-critique slot per _AUTHORING_PLAN row 5: 'particle count mode' ad critique, substance-free and number-free; count-display / sensitivity foils mirror gateItems[0], cut-the-route foil mirrors gateItems[2]) · mcq4
An advertisement for a new laboratory balance promises a 'particle count mode' that detects and displays how many particles are in a weighed sample. What is wrong with the advertised mode?
A balance of any sensitivity senses mass only — it cannot detect particles; the count must be worked out from the mass reading. KEY
Nothing is wrong — a balance just needs the right software to count the particles in a weighed sample.
The mode only needs the display re-labeled — the mass number and the particle count are the same number in different units.
The mode is impossible because a particle count can never be obtained from a measured mass.
MOL-002 · medium · application · seq 6 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 6: fresh identical-object instance grommets 175.5/6.5 -> 27; 'grommet' zero digest hits, 175.5 corpus-clean and ledger-unique, 6.5 <= 12 scan-invisible; exact division float-verified; answer-side 27 vs corpus 27.0 recorded for the stage-2 ledger; seed-parallel template, worded equation only) · numeric
A canister holds identical rubber grommets. The grommets alone have a total mass of 175.5 g, and one grommet has a mass of 6.5 g. How many grommets are in the canister?
Answer: 27 grommets KEY
also accepted: 27 grommets
wrongValueDiagnostics
[
 {
  "value": 0.037,
  "misconception": "you divided the mass of one grommet by the total mass — the total goes on top",
  "feedback": "This answer is wrong — you divided the mass of one grommet by the total mass — the total goes on top"
 },
 {
  "value": 1140.75,
  "misconception": "you multiplied the two masses instead of dividing",
  "feedback": "This answer is wrong — you multiplied the two masses instead of dividing"
 },
 {
  "value": 270,
  "misconception": "you slipped a decimal place in the division — check by multiplying back: this count times the mass of one grommet comes to ten times the canister's total mass",
  "feedback": "This answer is wrong — you slipped a decimal place in the division — check by multiplying back: this count times the mass of one grommet comes to ten times the canister's total mass"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\ntotal mass = 175.5 g\nmass of one grommet = 6.5 g\nWrite down the equation:\nnumber of grommets = total mass / mass of one grommet\nSubstitute in the values, and calculate:\nnumber of grommets = 175.5 / 6.5\nnumber of grommets = 27",
 "anticipatedWrong": [
  {
   "value": 0.037,
   "misconception": "you divided the mass of one grommet by the total mass — the total goes on top",
   "feedback": "This answer is wrong — you divided the mass of one grommet by the total mass — the total goes on top"
  },
  {
   "value": 1140.75,
   "misconception": "you multiplied the two masses instead of dividing",
   "feedback": "This answer is wrong — you multiplied the two masses instead of dividing"
  },
  {
   "value": 270,
   "misconception": "you slipped a decimal place in the division — check by multiplying back: this count times the mass of one grommet comes to ten times the canister's total mass",
   "feedback": "This answer is wrong — you slipped a decimal place in the division — check by multiplying back: this count times the mass of one grommet comes to ten times the canister's total mass"
  }
 ]
}
MOL-003 · medium · application · seq 7 · authored (stage 2 per rows/_AUTHORING_PLAN.md row 7: rubidium two-isotopes average facet — mirrors pool[8]'s taught 'average' clause on a fresh element with two real natural isotopes (85Rb/87Rb); key byte-identical to C per the same-seq parallel-key precedent (row01 C/E); foil classes parallel pool[8]: isotopes-break-it / even-split / selective-balance; rubidium corpus-clean, number-free) · mcq4
Rubidium atoms come in two isotopes with different masses. Why can weighing still count the atoms in a sample of pure rubidium?
The counting runs on the average particle mass, and the particles of a pure substance share one steady average mass. KEY
It cannot — the two isotopes break counting by weighing.
The two isotopes always appear in exactly equal numbers.
The balance reads the mass of the lighter isotope only and ignores the heavier isotope completely.
MOL-006 · medium · application · seq 8 · authored (stage 2 per _AUTHORING_PLAN row 8: ionic-class transfer case on ammonium chloride, NH₄Cl — the stem states the ion composition because the metal-with-nonmetal cue is absent and polyatomic-ion retrieval must not gate a MOL-006 item (blind-packet self-containment kept); the assessed boundary is MOL-006's own: ion-BUILT compound -> counted in neutral formula units, never in the ions inside. LEDGER: ammonium = note-class cross-scope WARN ((NH₄)₂S in MOL-013 fluency), form-E NH₄ repeat 8E/14E ((NH₄)₂SO₄) plan-RECORDED, non-adjacent; NH₄Cl formula fresh; 'chloride' not a scanned lexicon token) · mcq4
Ammonium chloride, NH₄Cl, is built from ammonium ions, NH₄⁺, and chloride ions, Cl⁻. Which particle does a mole of ammonium chloride count?
Formula units KEY
Ions
Molecules
Atoms
MOL-009 · medium · application · seq 9 · authored (reverse-read facet rotation per blueprint case_target + _AUTHORING_PLAN row 9: scandium 45.0, g/mol -> single-atom amu — the taught pairing read in the opposite direction from B; 45.0 freed by 20A's re-valuation (seed 57's 45.0 dropped) and verified corpus-clean; foils misconception-keyed: gram-scale-on-an-atom, Avogadro-count-as-mass (rule-content value, DISTRACTOR-only per MOL-009 specNotes), halved shared number (seed-25 d-foil class, 22.5 corpus-clean); option set homogeneous — all mass values; scandium real M 45.0) · mcq4
Scandium's molar mass is 45.0 g/mol. What is the average mass of one scandium atom?
45.0 amu KEY
45.0 g
6.022 × 10²³ amu
22.5 amu
MOL-008 · medium · application · seq 10 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 10: fresh square, tungsten corpus-clean; gate-unique word) · numeric
The periodic-table square for tungsten is shown. State tungsten's molar mass, in g/mol. Give your answer to 4 significant figures.
FIGURE (pt-square-reference-table · stimulus) — title: "periodic-table square: tungsten" · rows: [["atomic number", "74"], ["symbol", "W"], ["average atomic mass", "183.8"]] · style: "single periodic-table element square, exactly as printed on the classroom periodic table"
raw spec
{
 "figureRequired": true,
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "pt-square-reference-table",
  "title": "periodic-table square: tungsten",
  "rows": [
   [
    "atomic number",
    "74"
   ],
   [
    "symbol",
    "W"
   ],
   [
    "average atomic mass",
    "183.8"
   ]
  ],
  "style": "single periodic-table element square, exactly as printed on the classroom periodic table"
 }
}
Answer: 183.8 g/mol KEY
also accepted: 183.8 g/mol
wrongValueDiagnostics
[
 {
  "value": 74,
  "misconception": "you read the atomic number — the smaller whole number counts protons and is not a mass",
  "feedback": "This answer is wrong — you read the atomic number — the smaller whole number counts protons and is not a mass"
 },
 {
  "value": 109.8,
  "misconception": "you subtracted the atomic number from the atomic mass — that neutron-count move belongs to isotope work, not molar mass",
  "feedback": "This answer is wrong — you subtracted the atomic number from the atomic mass — that neutron-count move belongs to isotope work, not molar mass"
 },
 {
  "value": 367.6,
  "misconception": "you doubled the printed value — the square's average atomic mass is already the whole molar mass for an element of single atoms",
  "feedback": "This answer is wrong — you doubled the printed value — the square's average atomic mass is already the whole molar mass for an element of single atoms"
 }
]
feedback
{
 "keyExplanation": "The square prints the atomic number 74 and the average atomic mass 183.8. The average atomic mass, read in grams per mole, is the molar mass: tungsten's is 183.8 g/mol. The 74 counts protons and is never a mass.",
 "anticipatedWrong": [
  {
   "value": 74,
   "misconception": "you read the atomic number — the smaller whole number counts protons and is not a mass",
   "feedback": "This answer is wrong — you read the atomic number — the smaller whole number counts protons and is not a mass"
  },
  {
   "value": 109.8,
   "misconception": "you subtracted the atomic number from the atomic mass — that neutron-count move belongs to isotope work, not molar mass",
   "feedback": "This answer is wrong — you subtracted the atomic number from the atomic mass — that neutron-count move belongs to isotope work, not molar mass"
  },
  {
   "value": 367.6,
   "misconception": "you doubled the printed value — the square's average atomic mass is already the whole molar mass for an element of single atoms",
   "feedback": "This answer is wrong — you doubled the printed value — the square's average atomic mass is already the whole molar mass for an element of single atoms"
  }
 ]
}
MOL-011 · medium · application · seq 11 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 11: S₂ hot-vapor case in the nature-fixed molecular-element set; same-seq sulfur share with form C = A5 WARN, recorded; printed value figure-borne) · numeric
At high temperature, sulfur vapor consists of S₂ molecules. The periodic-table square for sulfur is shown. Calculate the molar mass of S₂, in g/mol, to one decimal place. Give your answer to 3 significant figures.
FIGURE (pt-square-reference-table · stimulus) — title: "periodic-table square: sulfur" · rows: [["atomic number", "16"], ["symbol", "S"], ["average atomic mass", "32.1"]] · style: "single periodic-table element square, exactly as printed on the classroom periodic table"
raw spec
{
 "figureRequired": true,
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "pt-square-reference-table",
  "title": "periodic-table square: sulfur",
  "rows": [
   [
    "atomic number",
    "16"
   ],
   [
    "symbol",
    "S"
   ],
   [
    "average atomic mass",
    "32.1"
   ]
  ],
  "style": "single periodic-table element square, exactly as printed on the classroom periodic table"
 }
}
Answer: 64.2 g/mol KEY
also accepted: 64.2 g/mol
wrongValueDiagnostics
[
 {
  "value": 32.1,
  "misconception": "you ignored the subscript — each S₂ molecule holds two atoms, so the atom's value doubles",
  "feedback": "This answer is wrong — you ignored the subscript — each S₂ molecule holds two atoms, so the atom's value doubles"
 },
 {
  "value": 34.1,
  "misconception": "you added the subscript as if it were a mass — the 2 is a count and multiplies",
  "feedback": "This answer is wrong — you added the subscript as if it were a mass — the 2 is a count and multiplies"
 },
 {
  "value": 16.05,
  "misconception": "you divided by the subscript instead of multiplying — the 2 counts atoms per molecule",
  "feedback": "This answer is wrong — you divided by the subscript instead of multiplying — the 2 counts atoms per molecule"
 }
]
feedback
{
 "keyExplanation": "The square prints the average atomic mass 32.1. The subscript counts the atoms in one molecule: 2.\nmolar mass of S₂ = 2 × 32.1\nmolar mass of S₂ = 64.2 g/mol\nEach hot-vapor S₂ molecule weighs two sulfur atoms' worth.",
 "anticipatedWrong": [
  {
   "value": 32.1,
   "misconception": "you ignored the subscript — each S₂ molecule holds two atoms, so the atom's value doubles",
   "feedback": "This answer is wrong — you ignored the subscript — each S₂ molecule holds two atoms, so the atom's value doubles"
  },
  {
   "value": 34.1,
   "misconception": "you added the subscript as if it were a mass — the 2 is a count and multiplies",
   "feedback": "This answer is wrong — you added the subscript as if it were a mass — the 2 is a count and multiplies"
  },
  {
   "value": 16.05,
   "misconception": "you divided by the subscript instead of multiplying — the 2 counts atoms per molecule",
   "feedback": "This answer is wrong — you divided by the subscript instead of multiplying — the 2 counts atoms per molecule"
  }
 ]
}
MOL-014 · medium · application · seq 12 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 12: fresh trimmed-block quarter case on naphthalene 88.8 -> 22.2 g, masses corpus-clean; foil classes parallel D) · mcq4
A block of naphthalene has a mass of 88.8 g. A technician trims the block down to 22.2 g. How does the number of moles in the trimmed block compare with the original?
One-quarter as many moles. KEY
The same number of moles — the substance has not changed.
Four times as many moles.
It cannot be determined without naphthalene's molar mass.
MOL-012 · medium · application · seq 13 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 13: fresh two-element compound; zirconium 13E/19E same-form repeat RECORDED; O 16.0 = canonical constant, CONSTANTS-CLASS family) · numeric
Zirconium dioxide has the formula ZrO₂. Use these molar masses: Zr 91.2, O 16.0 g/mol. Calculate ZrO₂'s molar mass, in g/mol, to one decimal place. Give your answer to 4 significant figures.
Answer: 123.2 g/mol KEY
also accepted: 123.2 g/mol
wrongValueDiagnostics
[
 {
  "value": 107.2,
  "misconception": "you ignored the subscript — the formula counts two oxygens, not one",
  "feedback": "This answer is wrong — you ignored the subscript — the formula counts two oxygens, not one"
 },
 {
  "value": 214.4,
  "misconception": "you multiplied every element by the subscript — the 2 belongs to oxygen alone",
  "feedback": "This answer is wrong — you multiplied every element by the subscript — the 2 belongs to oxygen alone"
 },
 {
  "value": 109.2,
  "misconception": "you added the subscript as if it were a mass — the 2 is a count and multiplies the oxygen value",
  "feedback": "This answer is wrong — you added the subscript as if it were a mass — the 2 is a count and multiplies the oxygen value"
 }
]
feedback
{
 "keyExplanation": "Count the atoms the formula unit shows: 1 zirconium and 2 oxygens.\nmolar mass of ZrO₂ = 91.2 + 2 × 16.0\nmolar mass of ZrO₂ = 123.2 g/mol\nThe subscript 2 multiplies the oxygen value only; zirconium appears once.",
 "anticipatedWrong": [
  {
   "value": 107.2,
   "misconception": "you ignored the subscript — the formula counts two oxygens, not one",
   "feedback": "This answer is wrong — you ignored the subscript — the formula counts two oxygens, not one"
  },
  {
   "value": 214.4,
   "misconception": "you multiplied every element by the subscript — the 2 belongs to oxygen alone",
   "feedback": "This answer is wrong — you multiplied every element by the subscript — the 2 belongs to oxygen alone"
  },
  {
   "value": 109.2,
   "misconception": "you added the subscript as if it were a mass — the 2 is a count and multiplies the oxygen value",
   "feedback": "This answer is wrong — you added the subscript as if it were a mass — the 2 is a count and multiplies the oxygen value"
  }
 ]
}
MOL-013 · medium · application · seq 14 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 14: leading-parentheses variant; ammonium 8E/14E same-form non-adjacent repeat RECORDED (note-class word); answer 132.1 corpus-clean) · numeric
Calculate the molar mass of the fertilizer salt (NH₄)₂SO₄. (molar masses: N 14.0, H 1.0, S 32.1, O 16.0 g/mol.) Give your answer in g/mol to one decimal place. Give your answer to 4 significant figures.
Answer: 132.1 g/mol KEY
also accepted: 132.1 g/mol
wrongValueDiagnostics
[
 {
  "value": 114.1,
  "misconception": "you counted only one NH₄ group — the subscript 2 after the parentheses doubles the whole group",
  "feedback": "This answer is wrong — you counted only one NH₄ group — the subscript 2 after the parentheses doubles the whole group"
 },
 {
  "value": 228.2,
  "misconception": "you applied the 2 to the sulfate part as well — the subscript reaches only the group inside its own parentheses",
  "feedback": "This answer is wrong — you applied the 2 to the sulfate part as well — the subscript reaches only the group inside its own parentheses"
 },
 {
  "value": 116.1,
  "misconception": "you added the subscript as if it were a mass — the 2 is a count and multiplies the NH₄ group's value",
  "feedback": "This answer is wrong — you added the subscript as if it were a mass — the 2 is a count and multiplies the NH₄ group's value"
 }
]
feedback
{
 "keyExplanation": "Inside the parentheses: 14.0 + 4 × 1.0 = 18.0\nMultiply by the subscript after the parentheses: 2 × 18.0 = 36.0\nAdd the atoms outside the parentheses: 36.0 + 32.1 + 4 × 16.0\nM = 132.1 g/mol",
 "anticipatedWrong": [
  {
   "value": 114.1,
   "misconception": "you counted only one NH₄ group — the subscript 2 after the parentheses doubles the whole group",
   "feedback": "This answer is wrong — you counted only one NH₄ group — the subscript 2 after the parentheses doubles the whole group"
  },
  {
   "value": 228.2,
   "misconception": "you applied the 2 to the sulfate part as well — the subscript reaches only the group inside its own parentheses",
   "feedback": "This answer is wrong — you applied the 2 to the sulfate part as well — the subscript reaches only the group inside its own parentheses"
  },
  {
   "value": 116.1,
   "misconception": "you added the subscript as if it were a mass — the 2 is a count and multiplies the NH₄ group's value",
   "feedback": "This answer is wrong — you added the subscript as if it were a mass — the 2 is a count and multiplies the NH₄ group's value"
  }
 ]
}
MOL-017 · medium · application · seq 15 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 15: fresh triple; ethylene glycol corpus-clean; float-exact 310.5/62.1) · numeric
A coolant reservoir holds 310.5 g of ethylene glycol (M = 62.1 g/mol). How many moles of ethylene glycol is that? Give your answer in moles to 3 significant figures.
Answer: 5.0 mol KEY
also accepted: 5.00 mol
wrongValueDiagnostics
[
 {
  "value": 0.2,
  "misconception": "you divided the molar mass by the mass instead of the mass by the molar mass",
  "feedback": "This answer is wrong — you divided the molar mass by the mass instead of the mass by the molar mass"
 },
 {
  "value": 19282.05,
  "misconception": "you multiplied the mass by the molar mass instead of dividing",
  "feedback": "This answer is wrong — you multiplied the mass by the molar mass instead of dividing"
 },
 {
  "value": 50.0,
  "misconception": "you slipped a decimal place in the division — check by multiplying back: this count times 62.1 g/mol comes to ten times the reservoir's mass",
  "feedback": "This answer is wrong — you slipped a decimal place in the division — check by multiplying back: this count times 62.1 g/mol comes to ten times the reservoir's mass"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nm = 310.5 g\nM = 62.1 g/mol\nWrite down the equation:\nn = m / M\nSubstitute in the values, and calculate:\nn = 310.5 / 62.1\nn = 5.00 mol\nSanity check: a reservoir-scale mole count lands around 0.01 to 10 mol — 5.00 mol passes.",
 "anticipatedWrong": [
  {
   "value": 0.2,
   "misconception": "you divided the molar mass by the mass instead of the mass by the molar mass",
   "feedback": "This answer is wrong — you divided the molar mass by the mass instead of the mass by the molar mass"
  },
  {
   "value": 19282.05,
   "misconception": "you multiplied the mass by the molar mass instead of dividing",
   "feedback": "This answer is wrong — you multiplied the mass by the molar mass instead of dividing"
  },
  {
   "value": 50.0,
   "misconception": "you slipped a decimal place in the division — check by multiplying back: this count times 62.1 g/mol comes to ten times the reservoir's mass",
   "feedback": "This answer is wrong — you slipped a decimal place in the division — check by multiplying back: this count times 62.1 g/mol comes to ten times the reservoir's mass"
  }
 ]
}
MOL-018 · medium · application · seq 16 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 16: CsCl 1.25 × 168.4 = 210.5 (float-exact); cesium 16E/23E same-form repeat RECORDED) · numeric
An optics lab orders 1.250 mol of cesium chloride, CsCl (M = 168.4 g/mol). What mass of cesium chloride is that? Give your answer in grams to 1 decimal place.
Answer: 210.5 g KEY
also accepted: 210.5 g
wrongValueDiagnostics
[
 {
  "value": 0.007423,
  "misconception": "you divided the moles by the molar mass instead of multiplying them",
  "feedback": "This answer is wrong — you divided the moles by the molar mass instead of multiplying them"
 },
 {
  "value": 134.72,
  "misconception": "you divided the molar mass by the moles instead of multiplying them",
  "feedback": "This answer is wrong — you divided the molar mass by the moles instead of multiplying them"
 },
 {
  "value": 21.05,
  "misconception": "you slipped a decimal place in the product — 1.25 portions of 168.4 g must weigh more than one portion, not far less",
  "feedback": "This answer is wrong — you slipped a decimal place in the product — 1.25 portions of 168.4 g must weigh more than one portion, not far less"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nn = 1.250 mol\nM = 168.4 g/mol\nWrite down the equation:\nn = m / M\nMake m the subject:\nm = n × M\nSubstitute in the values, and calculate:\nm = 1.250 × 168.4\nm = 210.5 g\nSanity check: more than 1 mol must weigh more than the 168.4 g molar mass — 210.5 g passes.",
 "anticipatedWrong": [
  {
   "value": 0.007423,
   "misconception": "you divided the moles by the molar mass instead of multiplying them",
   "feedback": "This answer is wrong — you divided the moles by the molar mass instead of multiplying them"
  },
  {
   "value": 134.72,
   "misconception": "you divided the molar mass by the moles instead of multiplying them",
   "feedback": "This answer is wrong — you divided the molar mass by the moles instead of multiplying them"
  },
  {
   "value": 21.05,
   "misconception": "you slipped a decimal place in the product — 1.25 portions of 168.4 g must weigh more than one portion, not far less",
   "feedback": "This answer is wrong — you slipped a decimal place in the product — 1.25 portions of 168.4 g must weigh more than one portion, not far less"
  }
 ]
}
MOL-010 · hard · transfer · seq 17 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 17: boron 10.8 derivation walk (10.8 scan-invisible); boron 17E/20E same-form repeat RECORDED; foil classes parallel B) · mcq4
Boron's atoms average 10.8 amu. Which line of reasoning correctly gets from that fact to boron's molar mass?
One gram is 6.022 × 10²³ amu, so one mole of boron atoms — that same count — weighs 10.8 g. KEY
A single boron atom weighs 10.8 g, so a mole of them weighs 10.8 g/mol times more.
Chemists round 10.8 up to the nearest whole number of grams to set the molar mass.
The gram is defined so that one boron atom weighs exactly 10.8 of them.
MOL-015 · hard · transfer · seq 18 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 18: fresh close-call pair niobium/yttrium at 275.0 g; yttrium 18E/22E repeat RECORDED; close-call foil probes the rule's edge) · mcq4
A supplier stocks two 275.0 g ingots: one of niobium (molar mass 92.9 g/mol) and one of yttrium (molar mass 88.9 g/mol). Which ingot contains more moles of atoms?
The yttrium ingot. KEY
The niobium ingot, because its molar mass is larger.
The two ingots contain equal numbers of moles, because their masses are equal.
The difference is too small for the molar masses to decide it.
MOL-019 · hard · transfer · seq 19 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 19: xenon/mercury/zirconium with fresh float-exact masses (zirconium's mass equals its molar mass — the one-portion case); mercury 2E/19E and zirconium 13E/19E repeats RECORDED; four orderings pairwise distinct) · mcq4
Three flasks: 262.6 g of xenon (M = 131.3 g/mol); 300.9 g of mercury (M = 200.6 g/mol); 91.2 g of zirconium (M = 91.2 g/mol). Rank them from most moles to fewest.
xenon > mercury > zirconium KEY
mercury > xenon > zirconium
zirconium > mercury > xenon
zirconium > xenon > mercury
MOL-020 · hard · transfer · seq 20 · authored (stage 2, per rows/_AUTHORING_PLAN.md row 20: fresh m = n/M slip on boron (values 7.00/10.8/0.648 scan-exempt; corrected 75.6 in the key text, RECORDED); boron 17E/20E same-form repeat RECORDED; foil classes = nothing-wrong / values-blame / wrong-direction) · mcq4
A question asks for the mass of 7.00 mol of boron (M = 10.8 g/mol). A student writes: m = 7.00 / 10.8 = 0.648 g. What is the error?
The moles were divided by the molar mass instead of multiplied. KEY
Nothing is wrong with the solution.
The molar mass of boron was copied wrongly.
The solution should have used Avogadro's number instead of the molar mass to reach grams.
MOL-002 · medium · application · seq 21 · authored FRQ (plan row 21; fresh drywall-anchor instance, values corpus-clean and float-exact) · frq
A building-supply store counts identical drywall anchors by weighing.
A building-supply store counts identical drywall anchors by weighing. (a) The anchors in a bucket have a total mass of 333.5 g on their own, and one anchor has a mass of 5.75 g. How many anchors are in the bucket? Show every stage of your working: the values you start from, the equation in words, the substitution, and the count. (b) A contractor orders 120 anchors. What total mass of anchors should be weighed out? Show your working the same way. (c) A clerk answers part (a) by computing 5.75 / 333.5. Explain what is wrong with that division and state the corrected count. Plain numbers and words are expected throughout — no special symbols. Treat the given masses as exact.
Part a. The anchors in a bucket have a total mass of 333.5 g on their own, and one anchor has a mass of 5.75 g. How many anchors are in the bucket? Show every stage of your working: the values you start from, the equation in words, the substitution, and the count.
Model: Values: total mass = 333.5 g; mass of one anchor = 5.75 g. Equation in words: number of anchors = total mass divided by mass of one anchor. Substitution: 333.5 / 5.75. Count: 58 anchors. Check: 58 anchors of 5.75 g each weigh exactly 333.5 g.
Part b. A contractor orders 120 anchors. What total mass of anchors should be weighed out? Show your working the same way.
Model: Turn the equation around: mass to weigh out = number of anchors times mass of one anchor = 120 times 5.75 = 690.0 g.
Part c. A clerk answers part (a) by computing 5.75 / 333.5. Explain what is wrong with that division and state the corrected count. Plain numbers and words are expected throughout — no special symbols. Treat the given masses as exact.
Model: The division is upside down — the total mass goes on top. 5.75 / 333.5 is about 0.017, and a bucket cannot hold a fraction of an anchor. The corrected division is 333.5 / 5.75 = 58 anchors. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
Full model answer. (a) Values: total mass = 333.5 g; mass of one anchor = 5.75 g. Equation in words: number of anchors = total mass divided by mass of one anchor. Substitution: 333.5 / 5.75. Count: 58 anchors. Check: 58 anchors of 5.75 g each weigh exactly 333.5 g. (b) Turn the equation around: mass to weigh out = number of anchors times mass of one anchor = 120 times 5.75 = 690.0 g. (c) The division is upside down — the total mass goes on top. 5.75 / 333.5 is about 0.017, and a bucket cannot hold a fraction of an anchor. The corrected division is 333.5 / 5.75 = 58 anchors. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
MOL-013 · medium · application · seq 22 · authored FRQ (plan row 22; yttrium 18E/22E + scandium 9E/22E repeats RECORDED; Sc 45.0 = CONSTANTS-CLASS family entry, spec-text hits MOL-002/MOL-012) · frq
(a) Calculate the molar mass of yttrium hydroxide, Y(OH)₃. (molar masses: Y 88.9, O 16.0, H 1.0 g/mol.) Show every stage: the value inside the parentheses, the multiplication by the outside subscript, and the final sum. (b) A classmate reports M = 105.9 g/mol for the same compound. Identify exactly what the classmate did wrong and state the corrected molar mass. (c) Now calculate the molar mass of scandium sulfate, Sc₂(SO₄)₃. (molar masses: Sc 45.0, S 32.1, O 16.0 g/mol.) Show the same stages. Plain numbers and words are expected — no special symbols.
(a) Calculate the molar mass of yttrium hydroxide, Y(OH)₃. (molar masses: Y 88.9, O 16.0, H 1.0 g/mol.) Show every stage: the value inside the parentheses, the multiplication by the outside subscript, and the final sum. (b) A classmate reports M = 105.9 g/mol for the same compound. Identify exactly what the classmate did wrong and state the corrected molar mass. (c) Now calculate the molar mass of scandium sulfate, Sc₂(SO₄)₃. (molar masses: Sc 45.0, S 32.1, O 16.0 g/mol.) Show the same stages. Plain numbers and words are expected — no special symbols.
Part a. Calculate the molar mass of yttrium hydroxide, Y(OH)₃. (molar masses: Y 88.9, O 16.0, H 1.0 g/mol.) Show every stage: the value inside the parentheses, the multiplication by the outside subscript, and the final sum.
Model: Inside the parentheses: 16.0 + 1.0 = 17.0. Multiply by the outside subscript: 3 times 17.0 = 51.0. Add the yttrium, which sits outside the parentheses: 88.9 + 51.0 = 139.9. M of yttrium hydroxide = 139.9 g/mol.
Part b. A classmate reports M = 105.9 g/mol for the same compound. Identify exactly what the classmate did wrong and state the corrected molar mass.
Model: The classmate ignored the parentheses and counted the hydroxide group once: 88.9 + 17.0 = 105.9. The outside 3 triples the whole group — three oxygens and three hydrogens. Corrected: M = 139.9 g/mol.
Part c. Now calculate the molar mass of scandium sulfate, Sc₂(SO₄)₃. (molar masses: Sc 45.0, S 32.1, O 16.0 g/mol.) Show the same stages. Plain numbers and words are expected — no special symbols.
Model: Two scandium: 2 times 45.0 = 90.0. Sulfate group: 32.1 + 4 times 16.0 = 96.1; the outside 3 triples it: 288.3. Total: 90.0 + 288.3 = 378.3. M of scandium sulfate = 378.3 g/mol. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
Full model answer. (a) Inside the parentheses: 16.0 + 1.0 = 17.0. Multiply by the outside subscript: 3 times 17.0 = 51.0. Add the yttrium, which sits outside the parentheses: 88.9 + 51.0 = 139.9. M of yttrium hydroxide = 139.9 g/mol. (b) The classmate ignored the parentheses and counted the hydroxide group once: 88.9 + 17.0 = 105.9. The outside 3 triples the whole group — three oxygens and three hydrogens. Corrected: M = 139.9 g/mol. (c) Two scandium: 2 times 45.0 = 90.0. Sulfate group: 32.1 + 4 times 16.0 = 96.1; the outside 3 triples it: 288.3. Total: 90.0 + 288.3 = 378.3. M of scandium sulfate = 378.3 g/mol. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
MOL-021 · hard · application · seq 23 · authored FRQ (plan row 23; the E-form parenthesized slot filled with La(OH)₃ per the plan's rebalance instruction (lanthanum gate-unique); cesium 16E/23E repeat RECORDED; 'iodide' distinct token from 11B iodine, noted; I 126.9 = CONSTANTS-CLASS cross-form entry) · frq
(a) How many moles are in 519.6 g of cesium iodide, CsI? (molar masses: Cs 132.9, I 126.9 g/mol.) Show both stages: first assemble the molar mass from the formula, then apply the mole equation. Give the answer in moles to 3 significant figures. (b) What is the mass, in grams, of 3.000 mol of lanthanum hydroxide, La(OH)₃? (molar masses: La 138.9, O 16.0, H 1.0 g/mol.) Show both stages, including how you rearrange the mole equation. Give the answer to 1 decimal place. (c) Jar 1 holds 649.5 g of cesium iodide; jar 2 holds 189.9 g of lanthanum hydroxide. Using the molar masses you found, work out which jar holds more moles — show the working for both jars, giving the mole results to 3 significant figures. Plain numbers and words are expected — no special symbols.
(a) How many moles are in 519.6 g of cesium iodide, CsI? (molar masses: Cs 132.9, I 126.9 g/mol.) Show both stages: first assemble the molar mass from the formula, then apply the mole equation. Give the answer in moles to 3 significant figures. (b) What is the mass, in grams, of 3.000 mol of lanthanum hydroxide, La(OH)₃? (molar masses: La 138.9, O 16.0, H 1.0 g/mol.) Show both stages, including how you rearrange the mole equation. Give the answer to 1 decimal place. (c) Jar 1 holds 649.5 g of cesium iodide; jar 2 holds 189.9 g of lanthanum hydroxide. Using the molar masses you found, work out which jar holds more moles — show the working for both jars, giving the mole results to 3 significant figures. Plain numbers and words are expected — no special symbols.
Part a. How many moles are in 519.6 g of cesium iodide, CsI? (molar masses: Cs 132.9, I 126.9 g/mol.) Show both stages: first assemble the molar mass from the formula, then apply the mole equation. Give the answer in moles to 3 significant figures.
Model: Stage 1 — molar mass of cesium iodide: 132.9 + 126.9 = 259.8 g/mol. Stage 2 — moles: n = m/M = 519.6 / 259.8 = 2.00 mol.
Part b. What is the mass, in grams, of 3.000 mol of lanthanum hydroxide, La(OH)₃? (molar masses: La 138.9, O 16.0, H 1.0 g/mol.) Show both stages, including how you rearrange the mole equation. Give the answer to 1 decimal place.
Model: Stage 1 — molar mass of lanthanum hydroxide: inside the parentheses 16.0 + 1.0 = 17.0; tripled by the outside subscript to 51.0; plus the lanthanum 138.9; total 189.9 g/mol. Stage 2 — rearrange to make m the subject: m = n times M = 3.000 times 189.9 = 569.7 g.
Part c. Jar 1 holds 649.5 g of cesium iodide; jar 2 holds 189.9 g of lanthanum hydroxide. Using the molar masses you found, work out which jar holds more moles — show the working for both jars, giving the mole results to 3 significant figures. Plain numbers and words are expected — no special symbols.
Model: Jar 1: 649.5 / 259.8 = 2.50 mol. Jar 2: 189.9 / 189.9 = 1.00 mol — exactly one portion. Jar 1 holds more moles. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.
Full model answer. (a) Stage 1 — molar mass of cesium iodide: 132.9 + 126.9 = 259.8 g/mol. Stage 2 — moles: n = m/M = 519.6 / 259.8 = 2.00 mol. (b) Stage 1 — molar mass of lanthanum hydroxide: inside the parentheses 16.0 + 1.0 = 17.0; tripled by the outside subscript to 51.0; plus the lanthanum 138.9; total 189.9 g/mol. Stage 2 — rearrange to make m the subject: m = n times M = 3.000 times 189.9 = 569.7 g. (c) Jar 1: 649.5 / 259.8 = 2.50 mol. Jar 2: 189.9 / 189.9 = 1.00 mol — exactly one portion. Jar 1 holds more moles. Grading note: plain numbers and words are expected; nothing the student writes needs subscripts or superscripts.

End of Topic Test: Particles and moles & Chained conversions (u07-t2 · 5 forms × 22 items · 110 total)

David's notes for James:
full notes file
# u07-t2 — notes/queries from David's side (U7-G2)

1. CALC-HEAVY FRQ SPLIT - frq_per_form = 2, OVERRIDING your gate data's frq_per_form=1 baseline (confirm the composition): mole conversions are NAMED in QB-v2's calc-heavy list; the count is 2 because the scope has exactly two free-response-only constructs - FRQ-1 (seq 21, MOL-027) = the forward two-stage chain m -> N with the written mole CHECKPOINT, FRQ-2 (seq 22, MOL-030, terminal capstone) = the route DECISION whose execution stage is the BACKWARD chain N -> m, so both directions are covered with zero redundancy. A third FRQ would re-measure single-step arithmetic already production-measured by the 7 numeric-ENTRY rows, or violate MOL-029's recognition-only fence. D5: min(20, 6 x 9 lessons) = 20 -> the standard 20 objective + FRQs, no scaling. Full adjudication text in blueprint.composition.frq_split_adjudication.
2. INTERACTION FORMATS (confirm): per form 13 mcq / 7 numeric ENTRY (seq 4 n->N and seq 11 m->N in scientific notation, seq 6 N->n in mol, seq 8/9 moles-of-atoms in mol incl. one PARENTHESES formula, seq 12 N->m in g, seq 15 the single-step route-choice case in mol - HS numeric-entry rule, never numeric-answer mcq) / 2 FRQ. The e-notation entry convention ('Enter your answer in scientific notation, such as 4.8e21; do not type units.') is your own MOL-023/027/030 fluency sentence, carried verbatim.
3. V-T6 SCIENTIFIC-NOTATION EXTENSION - the gate's structural ask (your + David's sign-off wanted): particle-count answers (~1e22-1e25) cannot pass the shipped V-T6 numeric contract (plain-numeral display regex; repr() e-notates floats >= 1e16), and _shared is read-only for gate agents. The proposed ADDITIVE branch (gated on answer.notation == 'scientific') is fully specified in gate_config.adjudication_notes, and the gate-local validate_with_adjudications.py ENFORCES it locally on every seq-4/11 slot (display regex with nonzero lead digit, mantissa sig figs == sigFigs, parsed display == keyed value, never a bare power of ten, value an exact n x 6.022e23 product; unit + non-empty acceptedAlternates) - enforcement, not relief. A V-T6 fail line is excused ONLY for a slot that passes every local check.
4. SIG-FIG / PRECISION CONVENTIONS (confirm - REVISED by the authoritative QC's loop-1 ruling, recorded in gate_config): displays and keyed values are rounded to the sig figs the least-precise given supports (seq 4 = 3 sf; seq 11 = 3 or 4 sf per form's givens; seq 12 = 4 sf); every stem's precision ask matches its key; tolerance = half a unit in the last assessed digit and still accepts the unrounded exact product (verified numerically per slot). Mole-entry slots seq 6/8/9 carry NO precision ask and keep minimal exact displays (e.g. 1.4 / 6 / 12; integer:true where applicable; padded spellings as acceptedAlternates) because value-compare grading cannot assess trailing zeros. FRQ finals ANNOUNCE precision + the sanity check in the stem. Every numeric item exports precisionAssessed + gradingNote + wrongValueDiagnostics.
5. SUPPLIED-CONSTANT ADJUDICATION (recorded): every calculation stem supplies 'Nₐ = 6.022 × 10²³ mol⁻¹' (course rule - the constant is always given, never recalled), so the token family {6.022, 10^22..10^25 normalization artifacts, 4.8, bare exponents 21-25} recurs on-screen for every student on every form and carries zero instance signal. Recorded as gate_config.adjudicated_instance_families 'supplied-constant-exponent' (78 slots) and enforced by the gate-local wrapper over the UNTOUCHED shared validator (U5-G1/U6-G1 precedent); substance-bearing shared windows stayed armed and the three genuine seq-19 collisions were fixed in CONTENT, never excused.
6. CHEMISTRY-FIXED MOLAR-MASS AT A#18 (confirm): form A seq 18 keeps YOUR CH4/N2O equal-moles seed pair; D8 mandates supplying M in-stem and 16.0/44.0 are chemistry facts that also sit in other scope lessons' text - scoped adjudication for exactly those two tokens at that one slot; every other M in the gate is corpus-fresh and gate-unique (asserted mechanically at build).
7. NOTA-SHAPED SEED DISTRACTOR REPLACED (opportunistic rule, confirm): pool 0's 'None of them - every quantity carries a unit' -> 'Both N and n stand for plain counts - the equation needs no units.' (misconception: stripping n of its mole unit). Only none-shaped option in the gate. Same seed's bare 'N'/'n' options were disambiguated to 'Capital N'/'Small n' (they collapse under the validator's case-folded duplicate check), and pool 13's 'm = n / M' foil was reworded in words (punctuation-stripped collapse against the key) - format normalizations only, your misconception structure kept.
8. SEED CENSUS (SEED RULING - confirm): 21 of your 27 pool items placed (form A 17 / B 3 / C 1, incl. the demo-curated route-table item kept near-verbatim at A#20's row family), 6 spares with recorded reasons; prod_seeds empty -> both FRQ families authored fresh x5 to your two-stage scaffold conventions. Freshness deviations recorded at build: propane->butane (10A) and N2O4->strontium chromate (19A) because their name+formula grams sit verbatim in MOL-025's fluency corpus; fructose dropped (formula identical to glucose's); sodium ended up UNUSED gate-wide (the pencilled sodium carbonate row became five single-use organic molecules).
9. FRQ TYPED-WORK CONVENTIONS (confirm): model answers/rubrics show the full two-stage scaffold but avoid literal scaffold-marker strings (V-T9 boundary); counts written in e-notation and Avogadro's number spelled out in words (V-T12 - typed answers never require sub/superscripts); the mole CHECKPOINT is written on its own line in every model answer (authoritative-QC rubric law, loop 2); seq-21 rubrics require the asked mole-window sanity check specifically (a count-range check is NOT an accepted alternative).
10. MOL-029 FENCE HONORED (confirm): factor-label is RECOGNITION-only per its own spec fence - one mcq row (seq 3), no numeric or FRQ ever builds a factor-label chain.
11. ACCEPTED ADVISORY CLASSES (recorded, no fix - David's call to reopen): (a) cross-item route cueing (route-recognition mcqs at seq 2/10/13 vs numeric executions 11/12 and FRQs 21/22) adjudicated blueprint-inherent - recognition and execution of the SAME taught route are distinct blueprint rows and the FRQ stems themselves mandate the two-stage scaffold; (b) house longest-option / taught-wording-echo style (V-T4 quantified balance passes; 176/176 blind key picks, 0 defensible-distractor fails); (c) A#11 and A#21 share the same final answer within form A (3.01e23) - instance-design wart, left as-is for blast-radius reasons; (d) authoritative-QC loop-3 objection reading MOL-025/027/030 specNotes as a CLOSED substance inventory was overruled by its own loop-4 re-read (fresh instances are MANDATED by the freshness law; identical fresh-instance siblings passed).
12. PACKAGING INSTANCE SWEEP (recorded, 0 defects): the U5-G1-canon formula/name-register sweep (the register the shared word-scan cannot see) found no memory-answerability defects - NH3 at A#7 (your seed), O2 at B#6 and CH4 at A#18 are word-register-visible substances the shared validator already scanned (ammonia/oxygen/methane are ledger-owned to exactly those rows) with fresh values throughout, and the shared 'dinitrogen' prefix word is naming-system vocabulary (scope teaches dinitrogen tetroxide N2O4; items use different N compounds). Banned-instance rule (course-wide: no fissile/weapons/dose framings): 0 hits across all 110 items - the palette is classroom-safe throughout.
13. CRASH-RESUME PROVENANCE (disclosure, no action needed): the authoring fleet was killed twice mid-run by laptop restarts - once after the stage-2 informal validator run (resumed at stage 3) and once after the stage-7 finalize wrote forms_final.json + validator_run_final.txt (resumed at packaging). Each resume re-audited the full disk state (every JSON artifact json.loads-verified, 5x22 form census, forms_final payload re-asserted byte-identical to forms_v1) before running only the missing stages; no truncated or inconsistent artifact was found.

Form A

MOL-022 · easy · recall · seq 1 · seed:U7-G2/pool[0] (= MOL-022 gateItems[0]) — none-shaped distractor REPLACED with the specified units-misconception distractor (opportunistic rule; seed_placement flag); bare N/n option texts disambiguated to 'Capital N'/'Small n' (V-T3's case-folding normalization collapses bare N and n to the same option — recorded format normalization) · mcq4
In the equation N = n × Nₐ, which symbol stands for a plain count of particles, with no unit?
Capital N KEY
Small n
The constant Nₐ
Both N and n stand for plain counts — the equation needs no units.
MOL-026 · easy · recall · seq 2 · seed:U7-G2/pool[14] (= MOL-026 gateItems[2]) — verbatim; helium owner seq 2 · mcq4
A question asks for the number of atoms in 2.0 g of helium. Which two steps, in order, answer it?
First n = m/M to find the moles, then N = n × Nₐ to find the number of atoms. KEY
First n = N/Nₐ to find the moles, then m = n × M to find the mass.
Multiply the mass by Nₐ in one step.
Only N = n × Nₐ is needed, because helium's atoms are its particles.
MOL-029 · easy · recall · seq 3 · seed:U7-G2/pool[22] (= MOL-029 gateItems[1]) — FRESHENED per seed_placement (CO₂/44.0 are MOL-029's own fluency corpus -> potassium bromide / 119.0; structure + misconceptions kept) · mcq4
A recipe needs the mass of 2.0 mol of potassium bromide, KBr (M = 119.0 g/mol). Which factor-label line carries out the step m = n × M?
2.0 mol × (119.0 g / 1 mol) KEY
2.0 mol × (1 mol / 119.0 g)
2.0 mol × (6.022 × 10²³ formula units / 1 mol)
119.0 g × (2.0 mol / 1 g)
MOL-023 · easy · application · seq 4 · seed:U7-G2/pool[4] (= MOL-023 gateItems[1], numeric) — instance kept; CONTRACT-NORMALIZED: seed graded exact 2.4088e24 but displayed rounded 2.409e24; our contract grades the displayed value (R22) -> 5 sf ask, display = exact mantissa (seed_placement contract_normalization) · numeric
How many formula units are in 4.00 mol of potassium carbonate, K₂CO₃? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer to 3 significant figures. Enter your answer in scientific notation, such as 4.8e21; do not type units.
Answer: 2.41e+24 formula units KEY
also accepted: 2.41E24, 2.41e+24, 2.41 × 10²⁴
wrongValueDiagnostics
[
 {
  "value": 6.642311524410495e-24,
  "misconception": "you divided by Avogadro's number instead of multiplying — each mole contributes 6.022 × 10²³ particles",
  "feedback": "This answer is wrong — you divided by Avogadro's number instead of multiplying — each mole contributes 6.022 × 10²³ particles"
 },
 {
  "value": 6.022e+23,
  "misconception": "you ignored the mole count and answered Nₐ itself",
  "feedback": "This answer is wrong — you ignored the mole count and answered Nₐ itself"
 },
 {
  "value": 2.4088e+23,
  "misconception": "you slipped one power of ten in the exponent",
  "feedback": "This answer is wrong — you slipped one power of ten in the exponent"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nn = 4.00 mol\nNₐ = 6.022 × 10²³ mol⁻¹\nWrite down the equation:\nN = n × Nₐ\nSubstitute in the values, and calculate:\nN = 4.00 × 6.022 × 10²³\nN = 2.4088 × 10²⁴ = 2.41 × 10²⁴ formula units (3 significant figures, set by the 3-significant-figure mole value)\nSanity check: a lab-scale count lands around 10²² to 10²⁵ — 2.41 × 10²⁴ passes.",
 "anticipatedWrong": [
  {
   "value": 6.642311524410495e-24,
   "misconception": "you divided by Avogadro's number instead of multiplying — each mole contributes 6.022 × 10²³ particles",
   "feedback": "This answer is wrong — you divided by Avogadro's number instead of multiplying — each mole contributes 6.022 × 10²³ particles"
  },
  {
   "value": 6.022e+23,
   "misconception": "you ignored the mole count and answered Nₐ itself",
   "feedback": "This answer is wrong — you ignored the mole count and answered Nₐ itself"
  },
  {
   "value": 2.4088e+23,
   "misconception": "you slipped one power of ten in the exponent",
   "feedback": "This answer is wrong — you slipped one power of ten in the exponent"
  }
 ]
}
MOL-023 · medium · application · seq 5 · authored (row 5: too-small class; rule line 5 is the tested content) · mcq4
A student calculates the number of molecules in 2.0 mol of a bottled gas, and the calculator shows 2.8 × 10⁻²⁴. Which taught check flags the answer as wrong?
A lab-scale count lands around 10²² to 10²⁵ — an answer smaller than 1 means the division went the wrong way. KEY
Nothing flags it — a calculator result can be trusted once the values are typed in correctly.
The answer is wrong only because it should have come out negative.
Every particle count must equal exactly 6.022 × 10²³.
MOL-024 · medium · application · seq 6 · seed:U7-G2/pool[7] (= MOL-024 gateItems[1], numeric) — FRESHENED per seed_placement: 'carbon monoxide' -> krypton (carbon is MOL-024's own corpus and unownable gate-wide); count/answer kept · numeric
A sample contains 4.5165 × 10²³ atoms of krypton. How many moles is that? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer in moles.
Answer: 0.75 mol KEY
also accepted: 0.750, .75
wrongValueDiagnostics
[
 {
  "value": 2.7198363e+47,
  "misconception": "you multiplied by Avogadro's number instead of dividing — this direction travels toward moles",
  "feedback": "This answer is wrong — you multiplied by Avogadro's number instead of dividing — this direction travels toward moles"
 },
 {
  "value": 1.3333333333333333,
  "misconception": "you divided Avogadro's number by the particle count instead of the other way around",
  "feedback": "This answer is wrong — you divided Avogadro's number by the particle count instead of the other way around"
 },
 {
  "value": 7.5,
  "misconception": "you slipped one power of ten in the division",
  "feedback": "This answer is wrong — you slipped one power of ten in the division"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nN = 4.5165 × 10²³ atoms\nNₐ = 6.022 × 10²³ mol⁻¹\nWrite down the equation:\nN = n × Nₐ\nMake n the subject:\nn = N/Nₐ\nSubstitute in the values, and calculate:\nn = 4.5165 × 10²³ / (6.022 × 10²³)\nn = 0.750 mol\nSanity check: a count below 6.022 × 10²³ means less than 1 mol, and above it more than 1 mol.",
 "anticipatedWrong": [
  {
   "value": 2.7198363e+47,
   "misconception": "you multiplied by Avogadro's number instead of dividing — this direction travels toward moles",
   "feedback": "This answer is wrong — you multiplied by Avogadro's number instead of dividing — this direction travels toward moles"
  },
  {
   "value": 1.3333333333333333,
   "misconception": "you divided Avogadro's number by the particle count instead of the other way around",
   "feedback": "This answer is wrong — you divided Avogadro's number by the particle count instead of the other way around"
  },
  {
   "value": 7.5,
   "misconception": "you slipped one power of ten in the division",
   "feedback": "This answer is wrong — you slipped one power of ten in the division"
  }
 ]
}
MOL-024 · medium · application · seq 7 · seed:U7-G2/pool[8] (= MOL-024 gateItems[2]) — counts FRESHENED per seed_placement (9.033e23 is fluency corpus and 4.8176e24 sits in MOL-027's fluency checks); ammonia + sulfur dioxide kept (owners seq 7); misconception classes kept · mcq4
Beaker A holds 1.11407 × 10²⁴ ammonia molecules (NH₃). Beaker B holds 2.22814 × 10²⁴ sulfur dioxide molecules (SO₂). Which beaker holds more moles? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Beaker B — 3.7 mol compared with 1.85 mol. KEY
Beaker A — 1.85 mol compared with 0.37 mol.
The beakers hold equal amounts, because both counts sit in the same power of ten.
Beaker B — but only because sulfur dioxide's molecules are heavier.
MOL-025 · medium · application · seq 8 · seed:U7-G2/pool[9] (= MOL-025 gateItems[0]) — RESTRUCTURED per seed_placement (triple collision: own-article 0.50, own-corpus hydrogen ask, unownable ethanol); construct + misconception space kept on phosphorus tribromide · numeric
A sample of 2.00 mol of phosphorus tribromide, PBr₃, contains how many moles of bromine atoms? Give your answer in mol.
Answer: 6.0 mol KEY
also accepted: 6.00, 6.0
wrongValueDiagnostics
[
 {
  "value": 2.0,
  "misconception": "you ignored the subscript and gave the moles of the compound itself",
  "feedback": "This answer is wrong — you ignored the subscript and gave the moles of the compound itself"
 },
 {
  "value": 0.6666666666666666,
  "misconception": "you divided the moles by the subscript instead of multiplying",
  "feedback": "This answer is wrong — you divided the moles by the subscript instead of multiplying"
 },
 {
  "value": 8.0,
  "misconception": "you multiplied by the molecule's total atom count (4) instead of bromine's subscript (3)",
  "feedback": "This answer is wrong — you multiplied by the molecule's total atom count (4) instead of bromine's subscript (3)"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmoles of PBr₃ = 2.00 mol\nsubscript of bromine in PBr₃ = 3\nWrite down the equation:\nmoles of bromine atoms = subscript × moles of compound\nSubstitute in the values, and calculate:\nmoles of bromine atoms = 3 × 2.00\nmoles of bromine atoms = 6.00 mol",
 "anticipatedWrong": [
  {
   "value": 2.0,
   "misconception": "you ignored the subscript and gave the moles of the compound itself",
   "feedback": "This answer is wrong — you ignored the subscript and gave the moles of the compound itself"
  },
  {
   "value": 0.6666666666666666,
   "misconception": "you divided the moles by the subscript instead of multiplying",
   "feedback": "This answer is wrong — you divided the moles by the subscript instead of multiplying"
  },
  {
   "value": 8.0,
   "misconception": "you multiplied by the molecule's total atom count (4) instead of bromine's subscript (3)",
   "feedback": "This answer is wrong — you multiplied by the molecule's total atom count (4) instead of bromine's subscript (3)"
  }
 ]
}
MOL-025 · medium · application · seq 9 · seed:U7-G2/pool[11] (= MOL-025 gateItems[2]) — RESTRUCTURED per seed_placement (own-article 0.50 mole value; own-corpus oxygen ask) -> magnesium chlorate, chlorine ask; parentheses construct + ignored-outer-subscript foil space kept as anticipatedWrong · numeric
How many moles of chlorine atoms are in 3.00 mol of magnesium chlorate, Mg(ClO₃)₂? Give your answer in mol.
Answer: 6.0 mol KEY
also accepted: 6.00, 6.0
wrongValueDiagnostics
[
 {
  "value": 3.0,
  "misconception": "you ignored the subscript outside the parentheses — it multiplies everything inside",
  "feedback": "This answer is wrong — you ignored the subscript outside the parentheses — it multiplies everything inside"
 },
 {
  "value": 12.0,
  "misconception": "you applied the outer subscript twice",
  "feedback": "This answer is wrong — you applied the outer subscript twice"
 }
]
feedback
{
 "keyExplanation": "The subscript outside the parentheses multiplies everything inside, so each formula unit of Mg(ClO₃)₂ carries 2 × 1 = 2 chlorine atoms.\nWrite down the equation:\nmoles of Cl atoms = subscript × moles of compound\nSubstitute in the values, and calculate:\nmoles of Cl atoms = 2 × 3.00\nmoles of Cl atoms = 6.00 mol",
 "anticipatedWrong": [
  {
   "value": 3.0,
   "misconception": "you ignored the subscript outside the parentheses — it multiplies everything inside",
   "feedback": "This answer is wrong — you ignored the subscript outside the parentheses — it multiplies everything inside"
  },
  {
   "value": 12.0,
   "misconception": "you applied the outer subscript twice",
   "feedback": "This answer is wrong — you applied the outer subscript twice"
  }
 ]
}
MOL-026 · medium · application · seq 10 · seed:U7-G2/pool[13] (= MOL-026 gateItems[1]) — count FRESHENED per seed_placement (3.011e23 is WE corpus in four scope lessons); the seed's propane SWAPPED to butane — propane-with-formula grams sit verbatim in MOL-025's fluency corpus (butane owner seq 10); the m = n ÷ M foil is reworded in words (the ÷/× symbols vanish under option normalization, V-T3 duplicate class) · mcq4
A question asks for the mass of 8.7319 × 10²³ molecules of butane, C₄H₁₀. Which two steps, in order, answer it?
First n = N/Nₐ to find the moles, then m = n × M to find the mass. KEY
First N = n × Nₐ to find the particle count, then n = m/M to find the moles.
Multiply the particle count by the molar mass in one step.
First n = N/Nₐ to find the moles, then divide the moles by the molar mass to find the mass.
MOL-027 · medium · application · seq 11 · seed:U7-G2/pool[15] (= MOL-027 gateItems[0]) — CONVERTED mcq->numeric ENTRY (HS rule); instance kept (glucose owner seq 11); computed distractors carried as anticipatedWrong · numeric
How many molecules are in 90.0 g of glucose, C₆H₁₂O₆? (M of C₆H₁₂O₆ = 180.0 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer to 3 significant figures. Enter your answer in scientific notation, such as 4.8e21; do not type units.
Answer: 3.01e+23 molecules KEY
also accepted: 3.01E23, 3.01e+23, 3.01 × 10²³
wrongValueDiagnostics
[
 {
  "value": 5.4198e+25,
  "misconception": "you skipped the mass-to-moles step and multiplied the mass by Avogadro's number",
  "feedback": "This answer is wrong — you skipped the mass-to-moles step and multiplied the mass by Avogadro's number"
 },
 {
  "value": 1.2044e+24,
  "misconception": "you divided the molar mass by the mass in step 1, inverting the mole value",
  "feedback": "This answer is wrong — you divided the molar mass by the mass in step 1, inverting the mole value"
 },
 {
  "value": 8.302889405513118e-25,
  "misconception": "you divided by Avogadro's number instead of multiplying — a count below one fails any sense check",
  "feedback": "This answer is wrong — you divided by Avogadro's number instead of multiplying — a count below one fails any sense check"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nm = 90.0 g\nM = 180.0 g/mol\nNₐ = 6.022 × 10²³ mol⁻¹\nStep 1 — convert mass to moles:\nn = m/M\nn = 90.0 / 180.0\nn = 0.500 mol   (the mole checkpoint — lab-scale values land between about 0.01 and 10 mol)\nStep 2 — convert moles to molecules:\nN = n × Nₐ\nN = 0.500 × 6.022 × 10²³\nN = 3.011 × 10²³ = 3.01 × 10²³ molecules (3 significant figures, set by the least-precise given)",
 "anticipatedWrong": [
  {
   "value": 5.4198e+25,
   "misconception": "you skipped the mass-to-moles step and multiplied the mass by Avogadro's number",
   "feedback": "This answer is wrong — you skipped the mass-to-moles step and multiplied the mass by Avogadro's number"
  },
  {
   "value": 1.2044e+24,
   "misconception": "you divided the molar mass by the mass in step 1, inverting the mole value",
   "feedback": "This answer is wrong — you divided the molar mass by the mass in step 1, inverting the mole value"
  },
  {
   "value": 8.302889405513118e-25,
   "misconception": "you divided by Avogadro's number instead of multiplying — a count below one fails any sense check",
   "feedback": "This answer is wrong — you divided by Avogadro's number instead of multiplying — a count below one fails any sense check"
  }
 ]
}
MOL-028 · medium · application · seq 12 · seed:U7-G2/pool[18] (= MOL-028 gateItems[0]) — RESTRUCTURED per seed_placement (ethanol unownable; 1.506e23 is MOL-030 fluency corpus); the seed's misconception space (full-mole mass / M÷n / count×M) is what is kept, on an allocator-verified fresh instance · numeric
A sample of barium chloride, BaCl₂ (M = 208.2 g/mol), contains 7.2264 × 10²³ formula units. What is the sample's mass, in grams? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer to 4 significant figures.
Answer: 249.8 g KEY
also accepted: 249.8, 249.8 g
wrongValueDiagnostics
[
 {
  "value": 208.2,
  "misconception": "you gave one full mole's mass — the sample is not 1 mol; divide the count by Nₐ first",
  "feedback": "This answer is wrong — you gave one full mole's mass — the sample is not 1 mol; divide the count by Nₐ first"
 },
 {
  "value": 173.5,
  "misconception": "you divided the molar mass by the moles instead of multiplying them",
  "feedback": "This answer is wrong — you divided the molar mass by the moles instead of multiplying them"
 },
 {
  "value": 1.50453648e+26,
  "misconception": "you skipped the particles-to-moles step and multiplied the particle count by the molar mass",
  "feedback": "This answer is wrong — you skipped the particles-to-moles step and multiplied the particle count by the molar mass"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nN = 7.2264 × 10²³ formula units\nM = 208.2 g/mol\nNₐ = 6.022 × 10²³ mol⁻¹\nStep 1 — convert formula units to moles:\nN = n × Nₐ\nMake n the subject:\nn = N/Nₐ\nn = 7.2264 × 10²³ / (6.022 × 10²³)\nn = 1.200 mol   (the mole checkpoint)\nStep 2 — convert moles to mass:\nn = m/M\nMake m the subject:\nm = n × M\nm = 1.200 × 208.2\nm = 249.84 g\nReported to 4 significant figures (set by M and Nₐ): m = 249.8 g",
 "anticipatedWrong": [
  {
   "value": 208.2,
   "misconception": "you gave one full mole's mass — the sample is not 1 mol; divide the count by Nₐ first",
   "feedback": "This answer is wrong — you gave one full mole's mass — the sample is not 1 mol; divide the count by Nₐ first"
  },
  {
   "value": 173.5,
   "misconception": "you divided the molar mass by the moles instead of multiplying them",
   "feedback": "This answer is wrong — you divided the molar mass by the moles instead of multiplying them"
  },
  {
   "value": 1.50453648e+26,
   "misconception": "you skipped the particles-to-moles step and multiplied the particle count by the molar mass",
   "feedback": "This answer is wrong — you skipped the particles-to-moles step and multiplied the particle count by the molar mass"
  }
 ]
}
MOL-026 · medium · application · seq 13 · seed:U7-G2/pool[12] (= MOL-026 gateItems[0]) — FRESHENED per seed_placement (sodium chloride is MOL-026's own fluency corpus; 5.85 is MOL-028 fluency corpus) -> calcium fluoride 15.6 g; critique structure, key register and misconception set kept · mcq4
To find the number of formula units in 15.6 g of calcium fluoride, CaF₂ (M = 78.1 g/mol), a student writes one line: N = m × Nₐ. What is wrong with the student's route?
The mass must first become moles with n = m/M — Nₐ counts particles per mole, not per gram. KEY
Nothing is wrong — multiplying the mass by Nₐ gives the particle count directly.
The student should have divided the mass by Nₐ instead of multiplying.
The student should have multiplied Nₐ by the molar mass first.
MOL-028 · medium · application · seq 14 · authored (row 14; diagnosis class 'fullmole'; allocator-verified fresh instance) · mcq4
A stockroom bottle of barium oxide, BaO (M = 153.3 g/mol), holds 2.16792 × 10²⁴ formula units. Asked for its mass, a student reports 153.3 g. (Nₐ = 6.022 × 10²³ mol⁻¹.) Which mistake did the student make?
The student gave one full mole's mass instead of using the computed mole value. KEY
The student multiplied the raw particle count by the molar mass, skipping the moles stage.
The student divided the molar mass by the mole value instead of multiplying them.
The student slipped one power of ten when dividing the count by Avogadro's number.
MOL-030 · medium · application · seq 15 · authored (row 15; capstone single-step case — M supplied but unused IS the route-choice test; allocator-verified) · numeric
A bottle holds 1.17429 × 10²⁴ molecules of citric acid, C₆H₈O₇ (M = 192.1 g/mol). How many moles is that? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer in moles to 3 significant figures.
Answer: 1.95 mol KEY
also accepted: 1.95
wrongValueDiagnostics
[
 {
  "value": 374.595,
  "misconception": "you ran a second stage the question never asked for — multiplying by the molar mass gives the mass, not the moles",
  "feedback": "This answer is wrong — you ran a second stage the question never asked for — multiplying by the molar mass gives the mass, not the moles"
 },
 {
  "value": 7.07157438e+47,
  "misconception": "you multiplied by Avogadro's number — moles and particle count are route neighbors, and this direction divides",
  "feedback": "This answer is wrong — you multiplied by Avogadro's number — moles and particle count are route neighbors, and this direction divides"
 },
 {
  "value": 0.5128205128205128,
  "misconception": "you divided Avogadro's number by the particle count instead of the other way around",
  "feedback": "This answer is wrong — you divided Avogadro's number by the particle count instead of the other way around"
 }
]
feedback
{
 "keyExplanation": "Ask the two route questions first: the question GIVES a particle count and ASKS for moles.\nCount and moles sit next to each other on the route, so ONE equation finishes the job — the supplied molar mass is never needed.\nN = n × Nₐ\nMake n the subject:\nn = N/Nₐ\nn = 1.17429 × 10²⁴ / (6.022 × 10²³)\nn = 1.95 mol",
 "anticipatedWrong": [
  {
   "value": 374.595,
   "misconception": "you ran a second stage the question never asked for — multiplying by the molar mass gives the mass, not the moles",
   "feedback": "This answer is wrong — you ran a second stage the question never asked for — multiplying by the molar mass gives the mass, not the moles"
  },
  {
   "value": 7.07157438e+47,
   "misconception": "you multiplied by Avogadro's number — moles and particle count are route neighbors, and this direction divides",
   "feedback": "This answer is wrong — you multiplied by Avogadro's number — moles and particle count are route neighbors, and this direction divides"
  },
  {
   "value": 0.5128205128205128,
   "misconception": "you divided Avogadro's number by the particle count instead of the other way around",
   "feedback": "This answer is wrong — you divided Avogadro's number by the particle count instead of the other way around"
  }
 ]
}
MOL-023 · medium · application · seq 16 · seed:U7-G2/pool[5] (= MOL-023 gateItems[2]) — FRESHENED per seed_placement (helium owner seq 2, nitrogen owner seq 20, corpus claim values) -> argon vs xenon with exempt-integer mole amounts and a value-free key; misconception classes kept (heavier-particles / equal-pure / size-reasoning replaces the power-slip foil, recorded) · mcq4
Flask 1 holds 3.00 mol of argon. Flask 2 holds 2.00 mol of xenon. Which flask contains more particles of gas? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Flask 1 — 3.00 mol means more particles than 2.00 mol, whatever each particle weighs. KEY
Flask 2 — xenon's particles are heavier, so there must be more of them in its flask.
The flasks hold equal counts, because both hold pure gases and every pure gas carries one mole's worth of particles.
Flask 1 — but only because argon's particles are smaller, so more of them fit inside.
MOL-025 · hard · transfer · seq 17 · seed:U7-G2/pool[10] (= MOL-025 gateItems[1]) — RESTRUCTURED per seed_placement (both seed substances carry MOL-025's own-corpus nitrogen/carbon/oxygen words; element ask moved to a metal per the element-ask fence); the seed's foil set and subscript-beats-moles trap are kept · mcq4
Sample 1 is 2.0 mol of lithium oxide, Li₂O. Sample 2 is 3.0 mol of lithium chlorate, LiClO₃. Which sample contains more moles of lithium atoms?
Sample 1 — 4.0 mol of lithium atoms compared with 3.0 mol. KEY
Sample 2, because 3.0 mol of compound is more than 2.0 mol of compound.
The two samples hold equal moles of lithium atoms.
It cannot be decided without the masses of the two samples.
MOL-027 · hard · transfer · seq 18 · seed:U7-G2/pool[16] (= MOL-027 gateItems[1]) — pair KEPT (methane owner seq 18); masses/values FRESHENED per seed_placement (0.25/1.51 are own-fluency corpus) to 48.0/132.0 -> 3.0 mol; M tokens 16.0/44.0 = recorded chemistry-fixed EXPECTED-FIRE (stage-3 scoped adjudication) · mcq4
Sample 1 is 48.0 g of methane, CH₄ (M = 16.0 g/mol). Sample 2 is 132.0 g of dinitrogen monoxide, N₂O (M = 44.0 g/mol). Which sample contains more molecules? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Neither — both samples contain 3.0 mol, so their molecule counts are equal. KEY
Sample 2 — at 132.0 g it is simply the heavier sample.
Sample 1 — its particles are lighter, so the sample must pack in more of them.
Only the two sample volumes could settle it.
MOL-028 · hard · transfer · seq 19 · seed:U7-G2/pool[20] (= MOL-028 gateItems[2]) — RESTRUCTURED per seed_placement (carbon monoxide side unownable; 3.011e23 corpus; N₂O₄ dropped too — its name+formula grams sit in MOL-025's fluency corpus) -> diborane vs strontium chromate; the seed's count-vs-molar-mass trade-off, foil classes and 2:1 count ratio are kept; key made value-light · mcq4
Sample 1 is 2.04748 × 10²⁴ particles of diborane, B₂H₆ (M = 27.7 g/mol). Sample 2 is 1.02374 × 10²⁴ particles of strontium chromate, SrCrO₄ (M = 203.6 g/mol). Which sample has the greater mass? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Sample 2 — each of its particles is more than twice as heavy, so half the count still gives the greater mass. KEY
Sample 1, because it contains twice as many particles.
The two samples have equal masses, because the count and the molar mass trade off exactly.
Settling it would need volume information that is not given.
MOL-030 · hard · transfer · seq 20 · seed:U7-G2/pool[26] = demo_curated[2] (MOL-030 gateItems[2]) — DEMO-CURATED near-verbatim keep (nitrogen+hydrogen owners seq 20); ONE value-format normalization per seed_placement: key/explanation counts restated at integer-mantissa rounding ('3.01'/'6.02' -> 'about 3.0/6.0 × 10²³'; '3.01' is MOL-027 fluency corpus); logic, foils and misconceptions untouched · mcq4
Sample 1 is 14.0 g of nitrogen gas, N₂ (M = 28.0 g/mol). Sample 2 is 2.0 g of hydrogen gas, H₂ (M = 2.0 g/mol). Which sample contains more molecules? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Sample 2 — about 6.0 × 10²³ molecules compared with about 3.0 × 10²³. KEY
Sample 1, because 14.0 g is the greater mass.
Sample 1, because each of its molecules is heavier than each hydrogen molecule.
The two samples contain equal numbers of molecules.
MOL-027 · medium · application · seq 21 · authored FRQ (row 21; prod_seeds empty -> fresh instance; calc-heavy FRQ-1) · frq
How many formula units are in 61.3 g of potassium chlorate, KClO₃? (M = 122.6 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Show every stage of your working: the values you start from, each stage's equation, the substitution, and the mole value written on its own line between the stages. Give the final count in scientific notation (for example 4.8e21) to 3 significant figures with its particle word, then add a one-line sanity check of the mole value against the lab-scale mole window you were taught.
How many formula units are in 61.3 g of potassium chlorate, KClO₃? (M = 122.6 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Show every stage of your working: the values you start from, each stage's equation, the substitution, and the mole value written on its own line between the stages. Give the final count in scientific notation (for example 4.8e21) to 3 significant figures with its particle word, then add a one-line sanity check of the mole value against the lab-scale mole window you were taught.
Part a. How many formula units are in 61.3 g of potassium chlorate, KClO₃? (M = 122.6 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Show every stage of your working: the values you start from, each stage's equation, the substitution, and the mole value written on its own line between the stages. Give the final count in scientific notation (for example 4.8e21) to 3 significant figures with its particle word, then add a one-line sanity check of the mole value against the lab-scale mole window you were taught.
Model: Values: m = 61.3 g; M = 122.6 g/mol; Avogadro's number = 6.022e23 per mole. Stage 1 (mass to moles): n = m/M = 61.3 / 122.6 = 0.500 mol. Mole checkpoint: 0.500 mol. Stage 2 (moles to formula units): N = n x Avogadro's number = 0.500 x 6.022e23 = 3.011e23, reported as 3.01e23 formula units (3 significant figures, set by the least-precise given). Sanity check: the mole value 0.500 mol sits inside the taught lab-scale mole window of about 0.01 to 10 mol. Grading note: plain e-notation is expected (the taught entry convention); no subscripts or superscripts are required in anything the student writes.
Full model answer. Values: m = 61.3 g; M = 122.6 g/mol; Avogadro's number = 6.022e23 per mole. Stage 1 (mass to moles): n = m/M = 61.3 / 122.6 = 0.500 mol. Mole checkpoint: 0.500 mol. Stage 2 (moles to formula units): N = n x Avogadro's number = 0.500 x 6.022e23 = 3.011e23, reported as 3.01e23 formula units (3 significant figures, set by the least-precise given). Sanity check: the mole value 0.500 mol sits inside the taught lab-scale mole window of about 0.01 to 10 mol. Grading note: plain e-notation is expected (the taught entry convention); no subscripts or superscripts are required in anything the student writes.
MOL-030 · hard · transfer · seq 22 · authored FRQ (row 22; capstone route-decision + backward execution; allocator-verified fresh instance; prod_seeds empty) · frq
A stockroom question reads: "A bottle of barium chromate, BaCrO₄, holds 1.62594 × 10²⁴ formula units. What is its mass?" (M = 253.3 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.)
A stockroom question reads: "A bottle of barium chromate, BaCrO₄, holds 1.62594 × 10²⁴ formula units. What is its mass?" (M = 253.3 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) (a) State which quantity the question gives you and which quantity it asks for, and name the two equations you will use, in order, each rearranged for its unknown. (b) Carry out the calculation, showing the working for both stages and writing the mole value on its own line between them. Give the mass in grams to 4 significant figures, and close by checking that the mole checkpoint lands within the everyday lab-scale amounts you were taught.
Part a. State which quantity the question gives you and which quantity it asks for, and name the two equations you will use, in order, each rearranged for its unknown.
Model: The question gives a particle count and asks for a mass — opposite ends of the route, so two stages through moles. Stage 1: the particle equation rearranged for n, n = N divided by Avogadro's number. Stage 2: the mole equation rearranged for m, m = n x M.
Part b. Carry out the calculation, showing the working for both stages and writing the mole value on its own line between them. Give the mass in grams to 4 significant figures, and close by checking that the mole checkpoint lands within the everyday lab-scale amounts you were taught.
Model: Stage 1: n = 1.62594e24 / 6.022e23 = 2.700 mol. Mole checkpoint: 2.700 mol. Stage 2: m = 2.700 x 253.3 = 683.91 g, reported as 683.9 g (4 significant figures). The mass of the sample is 683.9 g. The checkpoint 2.700 mol sits inside the taught range for lab-scale samples. Grading note: plain e-notation is expected; no subscripts or superscripts are required in anything the student writes.
Full model answer. (a) The question gives a particle count and asks for a mass — opposite ends of the route, so two stages through moles. Stage 1: the particle equation rearranged for n, n = N divided by Avogadro's number. Stage 2: the mole equation rearranged for m, m = n x M. (b) Stage 1: n = 1.62594e24 / 6.022e23 = 2.700 mol. Mole checkpoint: 2.700 mol. Stage 2: m = 2.700 x 253.3 = 683.91 g, reported as 683.9 g (4 significant figures). The mass of the sample is 683.9 g. The checkpoint 2.700 mol sits inside the taught range for lab-scale samples. Grading note: plain e-notation is expected; no subscripts or superscripts are required in anything the student writes.

Form B

MOL-022 · easy · recall · seq 1 · seed:U7-G2/pool[1] (= MOL-022 gateItems[1]) — verbatim (symbolic, instance-free) · mcq4
A student reads N = n × Nₐ and says that small n is the particle count. Which correction is right?
Capital N is the particle count; small n is the amount in moles. KEY
No correction is needed — capital and small n mean the same thing in this equation.
Small n is the particle count, and capital N is the amount in moles.
Neither symbol is the particle count — the particle count is Nₐ.
MOL-026 · easy · recall · seq 2 · authored (row 2; fresh instance acetylene 6.4 g) · mcq4
A question asks for the number of molecules in 6.4 g of acetylene, C₂H₂. Which two steps, in order, answer it?
First n = m/M to find the moles, then N = n × Nₐ to find the number of molecules. KEY
First n = N/Nₐ to find the moles, then m = n × M to find the mass.
Multiply the mass by Nₐ in one step.
First N = n × Nₐ to find the number of molecules, then n = m/M to find the moles.
MOL-029 · easy · recall · seq 3 · seed:U7-G2/pool[23] (= MOL-029 gateItems[2]) — verbatim (instance-free layout judgment) · mcq4
One student solves a conversion with two equation stages and a written mole value. Another solves the same conversion with a single factor-label line. Both work correctly, and neither rounds any value before the final answer. How do their answers compare?
The answers are identical, because the factor-label line runs the same steps through moles in a different layout. KEY
The factor-label answer is more accurate, because it avoids rounding at the mole step.
The answers differ, because the factor-label line skips the mole step entirely.
The equation answer is more accurate, because a written mole line locks in extra precision.
MOL-023 · easy · application · seq 4 · authored (row 4; fresh instance sucrose 8.00 mol; exact 5 sf mantissa — display string matches a corpus float but sits answer-side, outside V-T7's scanned surface; n=3.00 avoided: 'are in 3.00 mol' would collide with row 9A's gram window) · numeric
How many molecules are in 8.00 mol of sucrose, C₁₂H₂₂O₁₁? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer to 3 significant figures. Enter your answer in scientific notation, such as 4.8e21; do not type units.
Answer: 4.82e+24 molecules KEY
also accepted: 4.82E24, 4.82e+24, 4.82 × 10²⁴
wrongValueDiagnostics
[
 {
  "value": 1.328462304882099e-23,
  "misconception": "you divided by Avogadro's number instead of multiplying — each mole contributes 6.022 × 10²³ particles",
  "feedback": "This answer is wrong — you divided by Avogadro's number instead of multiplying — each mole contributes 6.022 × 10²³ particles"
 },
 {
  "value": 6.022e+23,
  "misconception": "you ignored the mole count and answered Nₐ itself",
  "feedback": "This answer is wrong — you ignored the mole count and answered Nₐ itself"
 },
 {
  "value": 4.8176e+23,
  "misconception": "you slipped one power of ten in the exponent",
  "feedback": "This answer is wrong — you slipped one power of ten in the exponent"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nn = 8.00 mol\nNₐ = 6.022 × 10²³ mol⁻¹\nWrite down the equation:\nN = n × Nₐ\nSubstitute in the values, and calculate:\nN = 8.00 × 6.022 × 10²³\nN = 4.8176 × 10²⁴ = 4.82 × 10²⁴ molecules (3 significant figures, set by the 3-significant-figure mole value)\nSanity check: a lab-scale count lands around 10²² to 10²⁵ — 4.82 × 10²⁴ passes.",
 "anticipatedWrong": [
  {
   "value": 1.328462304882099e-23,
   "misconception": "you divided by Avogadro's number instead of multiplying — each mole contributes 6.022 × 10²³ particles",
   "feedback": "This answer is wrong — you divided by Avogadro's number instead of multiplying — each mole contributes 6.022 × 10²³ particles"
  },
  {
   "value": 6.022e+23,
   "misconception": "you ignored the mole count and answered Nₐ itself",
   "feedback": "This answer is wrong — you ignored the mole count and answered Nₐ itself"
  },
  {
   "value": 4.8176e+23,
   "misconception": "you slipped one power of ten in the exponent",
   "feedback": "This answer is wrong — you slipped one power of ten in the exponent"
  }
 ]
}
MOL-023 · medium · application · seq 5 · authored (row 5: absurdly-large class — double multiplication) · mcq4
After a conversion, a student's screen shows 5.1 × 10⁴⁷ molecules for a flask-sized sample. Which taught check flags the answer as wrong?
A lab-scale count lands around 10²² to 10²⁵ — a count in the 10⁴⁷ region is far beyond any lab-scale sample. KEY
No check applies — very large counts are expected when Avogadro's number is involved.
The answer is wrong because it is bigger than Nₐ itself — a single flask tops out at one mole of particles.
The answer is wrong because it is not a whole number of moles.
MOL-024 · medium · application · seq 6 · seed:U7-G2/pool[6] (= MOL-024 gateItems[0]) — CONVERTED mcq->numeric ENTRY (HS rule) with the 1.8066e24 corpus count freshened to 2.7099e24; computed distractors carried as anticipatedWrong · numeric
A cylinder holds 2.7099 × 10²⁴ molecules of oxygen gas, O₂. How many moles is that? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer in moles.
Answer: 4.5 mol KEY
also accepted: 4.50
wrongValueDiagnostics
[
 {
  "value": 1.63190178e+48,
  "misconception": "you multiplied by Avogadro's number instead of dividing — this direction travels toward moles",
  "feedback": "This answer is wrong — you multiplied by Avogadro's number instead of dividing — this direction travels toward moles"
 },
 {
  "value": 0.2222222222222222,
  "misconception": "you divided Avogadro's number by the particle count instead of the other way around",
  "feedback": "This answer is wrong — you divided Avogadro's number by the particle count instead of the other way around"
 },
 {
  "value": 45.0,
  "misconception": "you slipped one power of ten in the division",
  "feedback": "This answer is wrong — you slipped one power of ten in the division"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nN = 2.7099 × 10²⁴ molecules\nNₐ = 6.022 × 10²³ mol⁻¹\nWrite down the equation:\nN = n × Nₐ\nMake n the subject:\nn = N/Nₐ\nSubstitute in the values, and calculate:\nn = 2.7099 × 10²⁴ / (6.022 × 10²³)\nn = 4.50 mol\nSanity check: a count below 6.022 × 10²³ means less than 1 mol, and above it more than 1 mol.",
 "anticipatedWrong": [
  {
   "value": 1.63190178e+48,
   "misconception": "you multiplied by Avogadro's number instead of dividing — this direction travels toward moles",
   "feedback": "This answer is wrong — you multiplied by Avogadro's number instead of dividing — this direction travels toward moles"
  },
  {
   "value": 0.2222222222222222,
   "misconception": "you divided Avogadro's number by the particle count instead of the other way around",
   "feedback": "This answer is wrong — you divided Avogadro's number by the particle count instead of the other way around"
  },
  {
   "value": 45.0,
   "misconception": "you slipped one power of ten in the division",
   "feedback": "This answer is wrong — you slipped one power of ten in the division"
  }
 ]
}
MOL-024 · medium · application · seq 7 · authored (row 7; fresh pair potassium sulfate / ammonium chloride) · mcq4
Jar 1 holds 3.9143 × 10²³ formula units of potassium sulfate. Jar 2 holds 2.1077 × 10²³ formula units of ammonium chloride. Which jar holds more moles? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Jar 1 — 0.65 mol compared with 0.35 mol. KEY
Jar 2 — 3.5 mol compared with 0.65 mol.
The jars hold equal amounts, because both are white ionic solids.
Answering would need the two molar masses as well.
MOL-025 · medium · application · seq 8 · authored (row 8; fresh instance aluminum iodide; aluminum+iodine owners seq 8) · numeric
A sample of 4.00 mol of aluminum iodide, AlI₃, holds how many moles of iodine atoms? Give your answer in mol.
Answer: 12.0 mol KEY
also accepted: 12.0
wrongValueDiagnostics
[
 {
  "value": 4.0,
  "misconception": "you ignored the subscript and gave the moles of the compound itself",
  "feedback": "This answer is wrong — you ignored the subscript and gave the moles of the compound itself"
 },
 {
  "value": 1.3333333333333333,
  "misconception": "you divided the moles by the subscript instead of multiplying",
  "feedback": "This answer is wrong — you divided the moles by the subscript instead of multiplying"
 },
 {
  "value": 16.0,
  "misconception": "you multiplied by the formula unit's total atom count (4) instead of iodine's subscript (3)",
  "feedback": "This answer is wrong — you multiplied by the formula unit's total atom count (4) instead of iodine's subscript (3)"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmoles of AlI₃ = 4.00 mol\nsubscript of iodine in AlI₃ = 3\nWrite down the equation:\nmoles of iodine atoms = subscript × moles of compound\nSubstitute in the values, and calculate:\nmoles of iodine atoms = 3 × 4.00\nmoles of iodine atoms = 12.0 mol",
 "anticipatedWrong": [
  {
   "value": 4.0,
   "misconception": "you ignored the subscript and gave the moles of the compound itself",
   "feedback": "This answer is wrong — you ignored the subscript and gave the moles of the compound itself"
  },
  {
   "value": 1.3333333333333333,
   "misconception": "you divided the moles by the subscript instead of multiplying",
   "feedback": "This answer is wrong — you divided the moles by the subscript instead of multiplying"
  },
  {
   "value": 16.0,
   "misconception": "you multiplied by the formula unit's total atom count (4) instead of iodine's subscript (3)",
   "feedback": "This answer is wrong — you multiplied by the formula unit's total atom count (4) instead of iodine's subscript (3)"
  }
 ]
}
MOL-025 · medium · application · seq 9 · authored (row 9; fresh parentheses instance barium chlorate) · numeric
A jar holds 4.00 mol of barium chlorate, Ba(ClO₃)₂. How many moles of atoms of chlorine does the jar hold? Give your answer in mol.
Answer: 8.0 mol KEY
also accepted: 8.00, 8.0
wrongValueDiagnostics
[
 {
  "value": 4.0,
  "misconception": "you ignored the subscript outside the parentheses — it multiplies everything inside",
  "feedback": "This answer is wrong — you ignored the subscript outside the parentheses — it multiplies everything inside"
 },
 {
  "value": 16.0,
  "misconception": "you applied the outer subscript twice",
  "feedback": "This answer is wrong — you applied the outer subscript twice"
 }
]
feedback
{
 "keyExplanation": "The subscript outside the parentheses multiplies everything inside, so each formula unit of Ba(ClO₃)₂ carries 2 × 1 = 2 chlorine atoms.\nWrite down the equation:\nmoles of Cl atoms = subscript × moles of compound\nSubstitute in the values, and calculate:\nmoles of Cl atoms = 2 × 4.00\nmoles of Cl atoms = 8.00 mol",
 "anticipatedWrong": [
  {
   "value": 4.0,
   "misconception": "you ignored the subscript outside the parentheses — it multiplies everything inside",
   "feedback": "This answer is wrong — you ignored the subscript outside the parentheses — it multiplies everything inside"
  },
  {
   "value": 16.0,
   "misconception": "you applied the outer subscript twice",
   "feedback": "This answer is wrong — you applied the outer subscript twice"
  }
 ]
}
MOL-026 · medium · application · seq 10 · authored (row 10; fresh instance ammonium sulfide — lithium fluoride moved off: LiF seeds row 3C and the substance registry keeps one home per substance) · mcq4
A question asks for the mass of 6.9253 × 10²³ formula units of ammonium sulfide, (NH₄)₂S. Which two steps, in order, answer it?
First n = N/Nₐ to find the moles, then m = n × M to find the mass. KEY
First N = n × Nₐ to find the particle count, then n = m/M to find the moles.
Multiply the particle count by the molar mass in one step.
First m = n × M to find the mass, then n = N/Nₐ to find the moles.
MOL-027 · medium · application · seq 11 · authored (row 11; fresh instance naphthalene; naphthalene owner seq 11) · numeric
How many molecules are in 192.3 g of naphthalene, C₁₀H₈? (M of C₁₀H₈ = 128.2 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer to 4 significant figures. Enter your answer in scientific notation, such as 4.8e21; do not type units.
Answer: 9.033e+23 molecules KEY
also accepted: 9.033E23, 9.033e+23, 9.033 × 10²³
wrongValueDiagnostics
[
 {
  "value": 1.1580306e+26,
  "misconception": "you skipped the mass-to-moles step and multiplied the mass by Avogadro's number",
  "feedback": "This answer is wrong — you skipped the mass-to-moles step and multiplied the mass by Avogadro's number"
 },
 {
  "value": 4.014666666666667e+23,
  "misconception": "you divided the molar mass by the mass in step 1, inverting the mole value",
  "feedback": "This answer is wrong — you divided the molar mass by the mass in step 1, inverting the mole value"
 },
 {
  "value": 2.4908668216539354e-24,
  "misconception": "you divided by Avogadro's number instead of multiplying — a count below one fails any sense check",
  "feedback": "This answer is wrong — you divided by Avogadro's number instead of multiplying — a count below one fails any sense check"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nm = 192.3 g\nM = 128.2 g/mol\nNₐ = 6.022 × 10²³ mol⁻¹\nStep 1 — convert mass to moles:\nn = m/M\nn = 192.3 / 128.2\nn = 1.500 mol   (the mole checkpoint — lab-scale values land between about 0.01 and 10 mol)\nStep 2 — convert moles to molecules:\nN = n × Nₐ\nN = 1.500 × 6.022 × 10²³\nN = 9.033 × 10²³ molecules",
 "anticipatedWrong": [
  {
   "value": 1.1580306e+26,
   "misconception": "you skipped the mass-to-moles step and multiplied the mass by Avogadro's number",
   "feedback": "This answer is wrong — you skipped the mass-to-moles step and multiplied the mass by Avogadro's number"
  },
  {
   "value": 4.014666666666667e+23,
   "misconception": "you divided the molar mass by the mass in step 1, inverting the mole value",
   "feedback": "This answer is wrong — you divided the molar mass by the mass in step 1, inverting the mole value"
  },
  {
   "value": 2.4908668216539354e-24,
   "misconception": "you divided by Avogadro's number instead of multiplying — a count below one fails any sense check",
   "feedback": "This answer is wrong — you divided by Avogadro's number instead of multiplying — a count below one fails any sense check"
  }
 ]
}
MOL-028 · medium · application · seq 12 · authored (row 12; allocator-verified fresh instance) · numeric
A sample of strontium sulfate, SrSO₄ (M = 183.7 g/mol), contains 1.44528 × 10²⁴ formula units. What is the sample's mass, in grams? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer to 4 significant figures.
Answer: 440.9 g KEY
also accepted: 440.9, 440.9 g
wrongValueDiagnostics
[
 {
  "value": 183.7,
  "misconception": "you gave one full mole's mass — the sample is not 1 mol; divide the count by Nₐ first",
  "feedback": "This answer is wrong — you gave one full mole's mass — the sample is not 1 mol; divide the count by Nₐ first"
 },
 {
  "value": 76.54166666666667,
  "misconception": "you divided the molar mass by the moles instead of multiplying them",
  "feedback": "This answer is wrong — you divided the molar mass by the moles instead of multiplying them"
 },
 {
  "value": 2.65497936e+26,
  "misconception": "you skipped the particles-to-moles step and multiplied the particle count by the molar mass",
  "feedback": "This answer is wrong — you skipped the particles-to-moles step and multiplied the particle count by the molar mass"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nN = 1.44528 × 10²⁴ formula units\nM = 183.7 g/mol\nNₐ = 6.022 × 10²³ mol⁻¹\nStep 1 — convert formula units to moles:\nN = n × Nₐ\nMake n the subject:\nn = N/Nₐ\nn = 1.44528 × 10²⁴ / (6.022 × 10²³)\nn = 2.400 mol   (the mole checkpoint)\nStep 2 — convert moles to mass:\nn = m/M\nMake m the subject:\nm = n × M\nm = 2.400 × 183.7\nm = 440.88 g\nReported to 4 significant figures (set by M and Nₐ): m = 440.9 g",
 "anticipatedWrong": [
  {
   "value": 183.7,
   "misconception": "you gave one full mole's mass — the sample is not 1 mol; divide the count by Nₐ first",
   "feedback": "This answer is wrong — you gave one full mole's mass — the sample is not 1 mol; divide the count by Nₐ first"
  },
  {
   "value": 76.54166666666667,
   "misconception": "you divided the molar mass by the moles instead of multiplying them",
   "feedback": "This answer is wrong — you divided the molar mass by the moles instead of multiplying them"
  },
  {
   "value": 2.65497936e+26,
   "misconception": "you skipped the particles-to-moles step and multiplied the particle count by the molar mass",
   "feedback": "This answer is wrong — you skipped the particles-to-moles step and multiplied the particle count by the molar mass"
  }
 ]
}
MOL-026 · medium · application · seq 13 · authored (row 13; broken-route class: constant misapplied to the mass step) · mcq4
To find the number of formula units in 21.3 g of potassium hydroxide, KOH (M = 56.1 g/mol), a student writes one line: n = Nₐ / m. What is wrong with the student's route?
The route is upside down — the moles come from n = m/M; Nₐ never divides by a mass. KEY
Nothing is wrong — any arrangement of the three quantities gives the moles.
The student wrote the right quantities and just needs to multiply the result by 10²³ at the end.
The student should write n = Nₐ × m instead.
MOL-028 · medium · application · seq 14 · authored (row 14; diagnosis class 'skip'; allocator-verified fresh instance) · mcq4
A stockroom bottle of lithium nitride, Li₃N (M = 34.8 g/mol), holds 9.9363 × 10²³ formula units. Asked for its mass, a student reports 3.4578324 × 10²⁵ g. (Nₐ = 6.022 × 10²³ mol⁻¹.) Which mistake did the student make?
The student multiplied the raw particle count by the molar mass, skipping the moles stage. KEY
The student gave one full mole's mass instead of using the computed mole value.
The student divided the molar mass by the mole value instead of multiplying them.
The student slipped one power of ten when dividing the count by Avogadro's number.
MOL-030 · medium · application · seq 15 · authored (row 15; capstone single-step case — M supplied but unused IS the route-choice test; allocator-verified) · numeric
A bottle holds 1.23451 × 10²⁴ molecules of ascorbic acid, C₆H₈O₆ (M = 176.1 g/mol). How many moles is that? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer in moles to 3 significant figures.
Answer: 2.05 mol KEY
also accepted: 2.05
wrongValueDiagnostics
[
 {
  "value": 361.005,
  "misconception": "you ran a second stage the question never asked for — multiplying by the molar mass gives the mass, not the moles",
  "feedback": "This answer is wrong — you ran a second stage the question never asked for — multiplying by the molar mass gives the mass, not the moles"
 },
 {
  "value": 7.43421922e+47,
  "misconception": "you multiplied by Avogadro's number — moles and particle count are route neighbors, and this direction divides",
  "feedback": "This answer is wrong — you multiplied by Avogadro's number — moles and particle count are route neighbors, and this direction divides"
 },
 {
  "value": 0.4878048780487805,
  "misconception": "you divided Avogadro's number by the particle count instead of the other way around",
  "feedback": "This answer is wrong — you divided Avogadro's number by the particle count instead of the other way around"
 }
]
feedback
{
 "keyExplanation": "Ask the two route questions first: the question GIVES a particle count and ASKS for moles.\nCount and moles sit next to each other on the route, so ONE equation finishes the job — the supplied molar mass is never needed.\nN = n × Nₐ\nMake n the subject:\nn = N/Nₐ\nn = 1.23451 × 10²⁴ / (6.022 × 10²³)\nn = 2.05 mol",
 "anticipatedWrong": [
  {
   "value": 361.005,
   "misconception": "you ran a second stage the question never asked for — multiplying by the molar mass gives the mass, not the moles",
   "feedback": "This answer is wrong — you ran a second stage the question never asked for — multiplying by the molar mass gives the mass, not the moles"
  },
  {
   "value": 7.43421922e+47,
   "misconception": "you multiplied by Avogadro's number — moles and particle count are route neighbors, and this direction divides",
   "feedback": "This answer is wrong — you multiplied by Avogadro's number — moles and particle count are route neighbors, and this direction divides"
  },
  {
   "value": 0.4878048780487805,
   "misconception": "you divided Avogadro's number by the particle count instead of the other way around",
   "feedback": "This answer is wrong — you divided Avogadro's number by the particle count instead of the other way around"
  }
 ]
}
MOL-023 · medium · application · seq 16 · authored (row 16; fresh gas pair ozone / silane — free-token names) · mcq4
Flask 1 holds 5.00 mol of ozone, O₃. Flask 2 holds 4.00 mol of silicon tetrahydride, SiH₄. Which flask contains more particles of gas? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Flask 1 — 5.00 mol means more particles than 4.00 mol, whatever each particle weighs. KEY
Flask 2 — silicon tetrahydride, SiH₄'s particles are heavier, so there must be more of them in its flask.
The flasks hold equal counts, because both hold pure gases and every pure gas carries one mole's worth of particles.
Flask 1 — but only because ozone, O₃'s particles are smaller, so more of them fit inside.
MOL-025 · hard · transfer · seq 17 · authored (row 17; fresh pair, subscript-3 vs unwritten-1) · mcq4
Sample 1 is 1.0 mol of strontium phosphate, Sr₃(PO₄)₂. Sample 2 is 2.0 mol of strontium carbonate, SrCO₃. Which sample contains more moles of strontium atoms?
Sample 1 — 3.0 mol of strontium atoms compared with 2.0 mol. KEY
Sample 2, because 2.0 mol of compound is more than 1.0 mol of compound.
The two samples hold equal moles of strontium atoms.
It cannot be decided without the masses of the two samples.
MOL-027 · hard · transfer · seq 18 · authored (row 18; fresh equal-moles pair sorbitol / boron trichloride — fructose dropped: its formula C₆H₁₂O₆ duplicates glucose's at seq 11, a cross-seq formula-gram collision) · mcq4
Sample 1 is 364.4 g of sorbitol, C₆H₁₄O₆ (M = 182.2 g/mol). Sample 2 is 234.4 g of boron trichloride, BCl₃ (M = 117.2 g/mol). Which sample contains more molecules? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Neither — both samples contain 2.0 mol, so their molecule counts are equal. KEY
Sample 1 — at 364.4 g it is simply the heavier sample.
Sample 2 — its particles are lighter, so the sample must pack in more of them.
Only the two sample volumes could settle it.
MOL-028 · hard · transfer · seq 19 · authored (row 19; fresh trade-off pair; acetone owner seq 19) · mcq4
Sample 1 is 2.64968 × 10²⁴ particles of acetone, C₃H₆O (M = 58.1 g/mol). Sample 2 is 1.32484 × 10²⁴ particles of potassium iodate, KIO₃ (M = 214.0 g/mol). Which sample has the greater mass? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Sample 2 — each of its particles is more than twice as heavy, so half the count still gives the greater mass. KEY
Sample 1, because it contains twice as many particles.
The two samples have equal masses, because the count and the molar mass trade off exactly.
Settling it would need volume information that is not given.
MOL-030 · hard · transfer · seq 20 · authored (row 20; fresh mixed pair potassium chromate / lithium nitrate — mass = M signals one full mole) · mcq4
Sample 1 is 97.1 g of potassium chromate, K₂CrO₄ (M = 194.2 g/mol). Sample 2 is 68.9 g of lithium nitrate, LiNO₃ (M = 68.9 g/mol). Which sample contains more formula units? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Sample 2 — a full mole of particles against half a mole, despite the smaller mass. KEY
Sample 1 — the bigger mass, 97.1 g, must mean more particles.
The counts come out equal — the bigger molar mass exactly offsets the smaller mass.
Sample 1 — heavier particles pack more matter into every single particle.
MOL-027 · medium · application · seq 21 · authored FRQ (row 21; fresh instance) · frq
How many formula units are in 316.6 g of potassium sulfite, K₂SO₃? (M = 158.3 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Show every stage of your working: the values you start from, each stage's equation, the substitution, and the mole value written on its own line between the stages. Give the final count in scientific notation (for example 4.8e21) to 4 significant figures with its particle word, then add a one-line sanity check of the mole value against the lab-scale mole window you were taught.
How many formula units are in 316.6 g of potassium sulfite, K₂SO₃? (M = 158.3 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Show every stage of your working: the values you start from, each stage's equation, the substitution, and the mole value written on its own line between the stages. Give the final count in scientific notation (for example 4.8e21) to 4 significant figures with its particle word, then add a one-line sanity check of the mole value against the lab-scale mole window you were taught.
Part a. How many formula units are in 316.6 g of potassium sulfite, K₂SO₃? (M = 158.3 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Show every stage of your working: the values you start from, each stage's equation, the substitution, and the mole value written on its own line between the stages. Give the final count in scientific notation (for example 4.8e21) to 4 significant figures with its particle word, then add a one-line sanity check of the mole value against the lab-scale mole window you were taught.
Model: Values: m = 316.6 g; M = 158.3 g/mol; Avogadro's number = 6.022e23 per mole. Stage 1 (mass to moles): n = m/M = 316.6 / 158.3 = 2.000 mol. Mole checkpoint: 2.000 mol. Stage 2 (moles to formula units): N = n x Avogadro's number = 2.000 x 6.022e23 = 1.2044e24, reported as 1.204e24 formula units (4 significant figures, set by the least-precise given). Sanity check: the mole value 2.000 mol sits inside the taught lab-scale mole window of about 0.01 to 10 mol. Grading note: plain e-notation is expected (the taught entry convention); no subscripts or superscripts are required in anything the student writes.
Full model answer. Values: m = 316.6 g; M = 158.3 g/mol; Avogadro's number = 6.022e23 per mole. Stage 1 (mass to moles): n = m/M = 316.6 / 158.3 = 2.000 mol. Mole checkpoint: 2.000 mol. Stage 2 (moles to formula units): N = n x Avogadro's number = 2.000 x 6.022e23 = 1.2044e24, reported as 1.204e24 formula units (4 significant figures, set by the least-precise given). Sanity check: the mole value 2.000 mol sits inside the taught lab-scale mole window of about 0.01 to 10 mol. Grading note: plain e-notation is expected (the taught entry convention); no subscripts or superscripts are required in anything the student writes.
MOL-030 · hard · transfer · seq 22 · authored FRQ (row 22; capstone route-decision + backward execution; allocator-verified fresh instance; prod_seeds empty) · frq
A stockroom question reads: "A bottle of ammonium phosphate, (NH₄)₃PO₄, holds 1.74638 × 10²⁴ formula units. What is its mass?" (M = 149.1 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.)
A stockroom question reads: "A bottle of ammonium phosphate, (NH₄)₃PO₄, holds 1.74638 × 10²⁴ formula units. What is its mass?" (M = 149.1 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) (a) State which quantity the question gives you and which quantity it asks for, and name the two equations you will use, in order, each rearranged for its unknown. (b) Carry out the calculation, showing the working for both stages and writing the mole value on its own line between them. Give the mass in grams to 4 significant figures, and close by checking that the mole checkpoint lands within the everyday lab-scale amounts you were taught.
Part a. State which quantity the question gives you and which quantity it asks for, and name the two equations you will use, in order, each rearranged for its unknown.
Model: The question gives a particle count and asks for a mass — opposite ends of the route, so two stages through moles. Stage 1: the particle equation rearranged for n, n = N divided by Avogadro's number. Stage 2: the mole equation rearranged for m, m = n x M.
Part b. Carry out the calculation, showing the working for both stages and writing the mole value on its own line between them. Give the mass in grams to 4 significant figures, and close by checking that the mole checkpoint lands within the everyday lab-scale amounts you were taught.
Model: Stage 1: n = 1.74638e24 / 6.022e23 = 2.900 mol. Mole checkpoint: 2.900 mol. Stage 2: m = 2.900 x 149.1 = 432.39 g, reported as 432.4 g (4 significant figures). The mass of the sample is 432.4 g. The checkpoint 2.900 mol sits inside the taught range for lab-scale samples. Grading note: plain e-notation is expected; no subscripts or superscripts are required in anything the student writes.
Full model answer. (a) The question gives a particle count and asks for a mass — opposite ends of the route, so two stages through moles. Stage 1: the particle equation rearranged for n, n = N divided by Avogadro's number. Stage 2: the mole equation rearranged for m, m = n x M. (b) Stage 1: n = 1.74638e24 / 6.022e23 = 2.900 mol. Mole checkpoint: 2.900 mol. Stage 2: m = 2.900 x 149.1 = 432.39 g, reported as 432.4 g (4 significant figures). The mass of the sample is 432.4 g. The checkpoint 2.900 mol sits inside the taught range for lab-scale samples. Grading note: plain e-notation is expected; no subscripts or superscripts are required in anything the student writes.

Form C

MOL-022 · easy · recall · seq 1 · seed:U7-G2/pool[2] (= MOL-022 gateItems[2]) — verbatim (symbolic, instance-free) · mcq4
In N = n × Nₐ, what unit does the quantity n carry?
mol — moles. KEY
No unit — n is a plain count.
g — grams.
particles per mole.
MOL-026 · easy · recall · seq 2 · authored (row 2; fresh instance potassium chloride 14.9 g) · mcq4
A question asks for the number of formula units in 14.9 g of potassium chloride, KCl. Which two steps, in order, answer it?
First n = m/M to find the moles, then N = n × Nₐ to find the number of formula units. KEY
First n = N/Nₐ to find the moles, then m = n × M to find the mass.
Multiply the mass by Nₐ in one step.
First m = n × M to find the mass, then N = n × Nₐ to find the number of formula units.
MOL-029 · easy · recall · seq 3 · authored (row 3 case rotation: single-factor n = m/M match on lithium fluoride) · mcq4
For lithium fluoride, LiF, one step of a solution is written as 7.4 g × (1 mol / 25.9 g). Which equation step does the line carry out?
n = m/M — dividing the mass by the molar mass to get moles. KEY
m = n × M — multiplying moles by the molar mass to get a mass.
N = n × Nₐ — multiplying moles by Avogadro's number to get a particle count.
n = N/Nₐ — dividing a particle count by Avogadro's number to get moles.
MOL-023 · easy · application · seq 4 · authored (row 4; fresh instance dinitrogen pentoxide 6.00 mol) · numeric
How many molecules are in 6.00 mol of dinitrogen pentoxide, N₂O₅? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer to 3 significant figures. Enter your answer in scientific notation, such as 4.8e21; do not type units.
Answer: 3.61e+24 molecules KEY
also accepted: 3.61E24, 3.61e+24, 3.61 × 10²⁴
wrongValueDiagnostics
[
 {
  "value": 9.963467286615742e-24,
  "misconception": "you divided by Avogadro's number instead of multiplying — each mole contributes 6.022 × 10²³ particles",
  "feedback": "This answer is wrong — you divided by Avogadro's number instead of multiplying — each mole contributes 6.022 × 10²³ particles"
 },
 {
  "value": 6.022e+23,
  "misconception": "you ignored the mole count and answered Nₐ itself",
  "feedback": "This answer is wrong — you ignored the mole count and answered Nₐ itself"
 },
 {
  "value": 3.6132e+23,
  "misconception": "you slipped one power of ten in the exponent",
  "feedback": "This answer is wrong — you slipped one power of ten in the exponent"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nn = 6.00 mol\nNₐ = 6.022 × 10²³ mol⁻¹\nWrite down the equation:\nN = n × Nₐ\nSubstitute in the values, and calculate:\nN = 6.00 × 6.022 × 10²³\nN = 3.6132 × 10²⁴ = 3.61 × 10²⁴ molecules (3 significant figures, set by the 3-significant-figure mole value)\nSanity check: a lab-scale count lands around 10²² to 10²⁵ — 3.61 × 10²⁴ passes.",
 "anticipatedWrong": [
  {
   "value": 9.963467286615742e-24,
   "misconception": "you divided by Avogadro's number instead of multiplying — each mole contributes 6.022 × 10²³ particles",
   "feedback": "This answer is wrong — you divided by Avogadro's number instead of multiplying — each mole contributes 6.022 × 10²³ particles"
  },
  {
   "value": 6.022e+23,
   "misconception": "you ignored the mole count and answered Nₐ itself",
   "feedback": "This answer is wrong — you ignored the mole count and answered Nₐ itself"
  },
  {
   "value": 3.6132e+23,
   "misconception": "you slipped one power of ten in the exponent",
   "feedback": "This answer is wrong — you slipped one power of ten in the exponent"
  }
 ]
}
MOL-023 · medium · application · seq 5 · authored (row 5: two-reports comparison against the range) · mcq4
Two students compute the number of molecules in 3.0 mol of a refrigerant gas. One reports 1.8 × 10²⁴ and the other reports 1.8 × 10³⁵. Without recomputing, which report passes the taught range check, and why?
1.8 × 10²⁴ — lab-scale counts land around 10²² to 10²⁵, and three moles sits near 10²⁴. KEY
1.8 × 10³⁵ — more moles should give a bigger power of ten, so the bigger answer is safer.
Both — the two answers share the same digits, so both pass the check.
Neither — a correct answer must come out as a whole multiple of 6.022 × 10²³.
MOL-024 · medium · application · seq 6 · authored (row 6; fresh instance strontium chloride 8.4308e23) · numeric
A crystal contains 8.4308 × 10²³ formula units of strontium chloride, SrCl₂. How many moles is that? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer in moles.
Answer: 1.4 mol KEY
also accepted: 1.40
wrongValueDiagnostics
[
 {
  "value": 5.07702776e+47,
  "misconception": "you multiplied by Avogadro's number instead of dividing — this direction travels toward moles",
  "feedback": "This answer is wrong — you multiplied by Avogadro's number instead of dividing — this direction travels toward moles"
 },
 {
  "value": 0.7142857142857143,
  "misconception": "you divided Avogadro's number by the particle count instead of the other way around",
  "feedback": "This answer is wrong — you divided Avogadro's number by the particle count instead of the other way around"
 },
 {
  "value": 14.0,
  "misconception": "you slipped one power of ten in the division",
  "feedback": "This answer is wrong — you slipped one power of ten in the division"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nN = 8.4308 × 10²³ formula units\nNₐ = 6.022 × 10²³ mol⁻¹\nWrite down the equation:\nN = n × Nₐ\nMake n the subject:\nn = N/Nₐ\nSubstitute in the values, and calculate:\nn = 8.4308 × 10²³ / (6.022 × 10²³)\nn = 1.40 mol\nSanity check: a count below 6.022 × 10²³ means less than 1 mol, and above it more than 1 mol.",
 "anticipatedWrong": [
  {
   "value": 5.07702776e+47,
   "misconception": "you multiplied by Avogadro's number instead of dividing — this direction travels toward moles",
   "feedback": "This answer is wrong — you multiplied by Avogadro's number instead of dividing — this direction travels toward moles"
  },
  {
   "value": 0.7142857142857143,
   "misconception": "you divided Avogadro's number by the particle count instead of the other way around",
   "feedback": "This answer is wrong — you divided Avogadro's number by the particle count instead of the other way around"
  },
  {
   "value": 14.0,
   "misconception": "you slipped one power of ten in the division",
   "feedback": "This answer is wrong — you slipped one power of ten in the division"
  }
 ]
}
MOL-024 · medium · application · seq 7 · authored (row 7; fresh pair neon / fluorine; neon+fluorine owners seq 7) · mcq4
Bulb 1 holds 5.7209 × 10²³ atoms of neon. Bulb 2 holds 4.2154 × 10²³ molecules of fluorine gas, F₂. Which bulb holds more moles of gas? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Bulb 1 — 0.95 mol compared with 0.7 mol. KEY
Bulb 2 — each fluorine molecule contains two atoms, so its count really doubles.
The bulbs hold equal amounts, because both counts start with digits near 5.
Bulb 2 — the heavier gas must be the one with more moles.
MOL-025 · medium · application · seq 8 · authored (row 8; UNWRITTEN-SUBSCRIPT-IS-1 case per the article's rule line 6) · numeric
A sample of 6.00 mol of iodine monobromide, IBr, carries how many moles of atoms of bromine? Give your answer in mol.
Answer: 6.0 mol KEY
also accepted: 6.00, 6.0
wrongValueDiagnostics
[
 {
  "value": 12.0,
  "misconception": "you doubled the moles — the unwritten subscript on Br is 1, so the atom moles equal the compound moles",
  "feedback": "This answer is wrong — you doubled the moles — the unwritten subscript on Br is 1, so the atom moles equal the compound moles"
 },
 {
  "value": 3.0,
  "misconception": "you halved the moles — nothing in IBr divides the amount; the subscript on Br is an unwritten 1",
  "feedback": "This answer is wrong — you halved the moles — nothing in IBr divides the amount; the subscript on Br is an unwritten 1"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmoles of IBr = 6.00 mol\nsubscript of bromine in IBr = 1 (a symbol with no written subscript counts as 1)\nWrite down the equation:\nmoles of bromine atoms = subscript × moles of compound\nSubstitute in the values, and calculate:\nmoles of bromine atoms = 1 × 6.00\nmoles of bromine atoms = 6.00 mol",
 "anticipatedWrong": [
  {
   "value": 12.0,
   "misconception": "you doubled the moles — the unwritten subscript on Br is 1, so the atom moles equal the compound moles",
   "feedback": "This answer is wrong — you doubled the moles — the unwritten subscript on Br is 1, so the atom moles equal the compound moles"
  },
  {
   "value": 3.0,
   "misconception": "you halved the moles — nothing in IBr divides the amount; the subscript on Br is an unwritten 1",
   "feedback": "This answer is wrong — you halved the moles — nothing in IBr divides the amount; the subscript on Br is an unwritten 1"
  }
 ]
}
MOL-025 · medium · application · seq 9 · authored (row 9; fresh parentheses instance strontium perchlorate; n freshened off the bare-power answer 10.0 flagged by the informal V-T6 run) · numeric
A bottle holds 7.00 mol of strontium perchlorate, Sr(ClO₄)₂. Counting atoms, what amount of chlorine, in moles, does it hold? Give your answer in mol.
Answer: 14.0 mol KEY
also accepted: 14.0
wrongValueDiagnostics
[
 {
  "value": 7.0,
  "misconception": "you ignored the subscript outside the parentheses — it multiplies everything inside",
  "feedback": "This answer is wrong — you ignored the subscript outside the parentheses — it multiplies everything inside"
 },
 {
  "value": 28.0,
  "misconception": "you applied the outer subscript twice",
  "feedback": "This answer is wrong — you applied the outer subscript twice"
 }
]
feedback
{
 "keyExplanation": "The subscript outside the parentheses multiplies everything inside, so each formula unit of Sr(ClO₄)₂ carries 2 × 1 = 2 chlorine atoms.\nWrite down the equation:\nmoles of Cl atoms = subscript × moles of compound\nSubstitute in the values, and calculate:\nmoles of Cl atoms = 2 × 7.00\nmoles of Cl atoms = 14.0 mol",
 "anticipatedWrong": [
  {
   "value": 7.0,
   "misconception": "you ignored the subscript outside the parentheses — it multiplies everything inside",
   "feedback": "This answer is wrong — you ignored the subscript outside the parentheses — it multiplies everything inside"
  },
  {
   "value": 28.0,
   "misconception": "you applied the outer subscript twice",
   "feedback": "This answer is wrong — you applied the outer subscript twice"
  }
 ]
}
MOL-026 · medium · application · seq 10 · authored (row 10; fresh instance lithium chloride) · mcq4
A question asks for the mass of 5.3596 × 10²³ formula units of lithium chloride, LiCl. Which two steps, in order, answer it?
First n = N/Nₐ to find the moles, then m = n × M to find the mass. KEY
First N = n × Nₐ to find the particle count, then n = m/M to find the moles.
Multiply the particle count by the molar mass in one step.
Divide the particle count by the molar mass in one step.
MOL-027 · medium · application · seq 11 · authored (row 11; fresh instance potassium permanganate) · numeric
How many formula units are in 39.5 g of potassium permanganate, KMnO₄? (M of KMnO₄ = 158.0 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer to 3 significant figures. Enter your answer in scientific notation, such as 4.8e21; do not type units.
Answer: 1.51e+23 formula units KEY
also accepted: 1.51E23, 1.51e+23, 1.51 × 10²³
wrongValueDiagnostics
[
 {
  "value": 2.37869e+25,
  "misconception": "you skipped the mass-to-moles step and multiplied the mass by Avogadro's number",
  "feedback": "This answer is wrong — you skipped the mass-to-moles step and multiplied the mass by Avogadro's number"
 },
 {
  "value": 2.4088e+24,
  "misconception": "you divided the molar mass by the mass in step 1, inverting the mole value",
  "feedback": "This answer is wrong — you divided the molar mass by the mass in step 1, inverting the mole value"
 },
 {
  "value": 4.151444702756559e-25,
  "misconception": "you divided by Avogadro's number instead of multiplying — a count below one fails any sense check",
  "feedback": "This answer is wrong — you divided by Avogadro's number instead of multiplying — a count below one fails any sense check"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nm = 39.5 g\nM = 158.0 g/mol\nNₐ = 6.022 × 10²³ mol⁻¹\nStep 1 — convert mass to moles:\nn = m/M\nn = 39.5 / 158.0\nn = 0.250 mol   (the mole checkpoint — lab-scale values land between about 0.01 and 10 mol)\nStep 2 — convert moles to formula units:\nN = n × Nₐ\nN = 0.250 × 6.022 × 10²³\nN = 1.5055 × 10²³ = 1.51 × 10²³ formula units (3 significant figures, set by the least-precise given)",
 "anticipatedWrong": [
  {
   "value": 2.37869e+25,
   "misconception": "you skipped the mass-to-moles step and multiplied the mass by Avogadro's number",
   "feedback": "This answer is wrong — you skipped the mass-to-moles step and multiplied the mass by Avogadro's number"
  },
  {
   "value": 2.4088e+24,
   "misconception": "you divided the molar mass by the mass in step 1, inverting the mole value",
   "feedback": "This answer is wrong — you divided the molar mass by the mass in step 1, inverting the mole value"
  },
  {
   "value": 4.151444702756559e-25,
   "misconception": "you divided by Avogadro's number instead of multiplying — a count below one fails any sense check",
   "feedback": "This answer is wrong — you divided by Avogadro's number instead of multiplying — a count below one fails any sense check"
  }
 ]
}
MOL-028 · medium · application · seq 12 · authored (row 12; allocator-verified fresh instance) · numeric
A sample of barium bromide, BaBr₂ (M = 297.1 g/mol), contains 1.56572 × 10²⁴ formula units. What is the sample's mass, in grams? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer to 4 significant figures.
Answer: 772.5 g KEY
also accepted: 772.5, 772.5 g
wrongValueDiagnostics
[
 {
  "value": 297.1,
  "misconception": "you gave one full mole's mass — the sample is not 1 mol; divide the count by Nₐ first",
  "feedback": "This answer is wrong — you gave one full mole's mass — the sample is not 1 mol; divide the count by Nₐ first"
 },
 {
  "value": 114.26923076923077,
  "misconception": "you divided the molar mass by the moles instead of multiplying them",
  "feedback": "This answer is wrong — you divided the molar mass by the moles instead of multiplying them"
 },
 {
  "value": 4.65175412e+26,
  "misconception": "you skipped the particles-to-moles step and multiplied the particle count by the molar mass",
  "feedback": "This answer is wrong — you skipped the particles-to-moles step and multiplied the particle count by the molar mass"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nN = 1.56572 × 10²⁴ formula units\nM = 297.1 g/mol\nNₐ = 6.022 × 10²³ mol⁻¹\nStep 1 — convert formula units to moles:\nN = n × Nₐ\nMake n the subject:\nn = N/Nₐ\nn = 1.56572 × 10²⁴ / (6.022 × 10²³)\nn = 2.600 mol   (the mole checkpoint)\nStep 2 — convert moles to mass:\nn = m/M\nMake m the subject:\nm = n × M\nm = 2.600 × 297.1\nm = 772.46 g\nReported to 4 significant figures (set by M and Nₐ): m = 772.5 g",
 "anticipatedWrong": [
  {
   "value": 297.1,
   "misconception": "you gave one full mole's mass — the sample is not 1 mol; divide the count by Nₐ first",
   "feedback": "This answer is wrong — you gave one full mole's mass — the sample is not 1 mol; divide the count by Nₐ first"
  },
  {
   "value": 114.26923076923077,
   "misconception": "you divided the molar mass by the moles instead of multiplying them",
   "feedback": "This answer is wrong — you divided the molar mass by the moles instead of multiplying them"
  },
  {
   "value": 4.65175412e+26,
   "misconception": "you skipped the particles-to-moles step and multiplied the particle count by the molar mass",
   "feedback": "This answer is wrong — you skipped the particles-to-moles step and multiplied the particle count by the molar mass"
  }
 ]
}
MOL-026 · medium · application · seq 13 · authored (row 13; broken-route class: constant-first scramble) · mcq4
To find the number of molecules in 33.9 g of boron trifluoride, BF₃ (M = 67.8 g/mol), a student writes one line: m ÷ Nₐ, then × M. What is wrong with the student's route?
Nₐ cannot act on a mass — the mass becomes moles with n = m/M, and only then does any other step apply. KEY
Nothing is wrong — dividing by the big constant scales the mass down correctly.
The route is wrong only because the two operations are in the wrong order.
The student should divide by M as well as by Nₐ.
MOL-028 · medium · application · seq 14 · authored (row 14; diagnosis class 'divM'; allocator-verified fresh instance) · mcq4
A stockroom bottle of potassium nitrate, KNO₃ (M = 101.1 g/mol), holds 2.52924 × 10²⁴ formula units. Asked for its mass, a student reports 24.1 g. (Nₐ = 6.022 × 10²³ mol⁻¹.) Which mistake did the student make?
The student divided the molar mass by the mole value instead of multiplying them. KEY
The student gave one full mole's mass instead of using the computed mole value.
The student multiplied the raw particle count by the molar mass, skipping the moles stage.
The student slipped one power of ten when dividing the count by Avogadro's number.
MOL-030 · medium · application · seq 15 · authored (row 15; capstone single-step case — M supplied but unused IS the route-choice test; allocator-verified) · numeric
A bottle holds 6.14244 × 10²³ molecules of camphor, C₁₀H₁₆O (M = 152.2 g/mol). How many moles is that? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer in moles to 3 significant figures.
Answer: 1.02 mol KEY
also accepted: 1.02
wrongValueDiagnostics
[
 {
  "value": 155.244,
  "misconception": "you ran a second stage the question never asked for — multiplying by the molar mass gives the mass, not the moles",
  "feedback": "This answer is wrong — you ran a second stage the question never asked for — multiplying by the molar mass gives the mass, not the moles"
 },
 {
  "value": 3.698977368e+47,
  "misconception": "you multiplied by Avogadro's number — moles and particle count are route neighbors, and this direction divides",
  "feedback": "This answer is wrong — you multiplied by Avogadro's number — moles and particle count are route neighbors, and this direction divides"
 },
 {
  "value": 0.9803921568627451,
  "misconception": "you divided Avogadro's number by the particle count instead of the other way around",
  "feedback": "This answer is wrong — you divided Avogadro's number by the particle count instead of the other way around"
 }
]
feedback
{
 "keyExplanation": "Ask the two route questions first: the question GIVES a particle count and ASKS for moles.\nCount and moles sit next to each other on the route, so ONE equation finishes the job — the supplied molar mass is never needed.\nN = n × Nₐ\nMake n the subject:\nn = N/Nₐ\nn = 6.14244 × 10²³ / (6.022 × 10²³)\nn = 1.02 mol",
 "anticipatedWrong": [
  {
   "value": 155.244,
   "misconception": "you ran a second stage the question never asked for — multiplying by the molar mass gives the mass, not the moles",
   "feedback": "This answer is wrong — you ran a second stage the question never asked for — multiplying by the molar mass gives the mass, not the moles"
  },
  {
   "value": 3.698977368e+47,
   "misconception": "you multiplied by Avogadro's number — moles and particle count are route neighbors, and this direction divides",
   "feedback": "This answer is wrong — you multiplied by Avogadro's number — moles and particle count are route neighbors, and this direction divides"
  },
  {
   "value": 0.9803921568627451,
   "misconception": "you divided Avogadro's number by the particle count instead of the other way around",
   "feedback": "This answer is wrong — you divided Avogadro's number by the particle count instead of the other way around"
  }
 ]
}
MOL-023 · medium · application · seq 16 · authored (row 16; fresh gas pair tetrafluoromethane / hexafluoroethane — free-token names) · mcq4
Flask 1 holds 7.00 mol of tetrafluoromethane, CF₄. Flask 2 holds 6.00 mol of hexafluoroethane, C₂F₆. Which flask contains more particles of gas? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Flask 1 — 7.00 mol means more particles than 6.00 mol, whatever each particle weighs. KEY
Flask 2 — hexafluoroethane, C₂F₆'s particles are heavier, so there must be more of them in its flask.
The flasks hold equal counts, because both hold pure gases and every pure gas carries one mole's worth of particles.
Flask 2 — each C₂F₆ molecule contains more atoms, so its flask holds more particles of gas.
MOL-025 · hard · transfer · seq 17 · authored (row 17; fresh pair boron oxide / boron nitride) · mcq4
Sample 1 is 2.0 mol of boron oxide, B₂O₃. Sample 2 is 3.0 mol of boron nitride, BN. Which sample contains more moles of boron atoms?
Sample 1 — 4.0 mol of boron atoms compared with 3.0 mol. KEY
Sample 2, because 3.0 mol of compound is more than 2.0 mol of compound.
The two samples hold equal moles of boron atoms.
It cannot be decided without the masses of the two samples.
MOL-027 · hard · transfer · seq 18 · authored (row 18; fresh equal-moles pair, molecular vs ionic — mixed particle words per the MOL-023 comparison precedent) · mcq4
Sample 1 is 416.4 g of silicon tetrafluoride, SiF₄ (M = 104.1 g/mol). Sample 2 is 535.2 g of lithium iodide, LiI (M = 133.8 g/mol). Which sample contains more particles (molecules or formula units)? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Neither — both samples contain 4.0 mol, so their particle counts are equal. KEY
Sample 2 — at 535.2 g it is simply the heavier sample.
Sample 1 — its particles are lighter, so the sample must pack in more of them.
Only the two sample volumes could settle it.
MOL-028 · hard · transfer · seq 19 · authored (row 19; fresh trade-off pair) | stage3 patch: name-first reorder kills the material-bearing '1024 particles of <cation>' shared window (V-T7; substances/values unchanged) · mcq4
Sample 1 is lithium hydroxide, LiOH (M = 23.9 g/mol); the sample holds 2.28836 × 10²⁴ particles. Sample 2 is potassium periodate, KIO₄ (M = 230.0 g/mol); that sample holds 1.14418 × 10²⁴ particles. Which sample has the greater mass? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Sample 2 — each of its particles is more than twice as heavy, so half the count still gives the greater mass. KEY
Sample 1, because it contains twice as many particles.
The two samples have equal masses, because the count and the molar mass trade off exactly.
Settling it would need volume information that is not given.
MOL-030 · hard · transfer · seq 20 · authored (row 20; fresh mixed pair strontium nitrate / lithium bromide) · mcq4
Sample 1 is 105.8 g of strontium nitrate, Sr(NO₃)₂ (M = 211.6 g/mol). Sample 2 is 86.8 g of lithium bromide, LiBr (M = 86.8 g/mol). Which sample contains more formula units? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Sample 2 — a full mole of particles against half a mole, despite the smaller mass. KEY
Sample 1 — the bigger mass, 105.8 g, must mean more particles.
The counts come out equal — the bigger molar mass exactly offsets the smaller mass.
Sample 1 — heavier particles pack more matter into every single particle.
MOL-027 · medium · application · seq 21 · authored FRQ (row 21; fresh instance) · frq
How many formula units are in 396.3 g of ammonium sulfate, (NH₄)₂SO₄? (M = 132.1 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Show every stage of your working: the values you start from, each stage's equation, the substitution, and the mole value written on its own line between the stages. Give the final count in scientific notation (for example 4.8e21) to 4 significant figures with its particle word, then add a one-line sanity check of the mole value against the lab-scale mole window you were taught.
How many formula units are in 396.3 g of ammonium sulfate, (NH₄)₂SO₄? (M = 132.1 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Show every stage of your working: the values you start from, each stage's equation, the substitution, and the mole value written on its own line between the stages. Give the final count in scientific notation (for example 4.8e21) to 4 significant figures with its particle word, then add a one-line sanity check of the mole value against the lab-scale mole window you were taught.
Part a. How many formula units are in 396.3 g of ammonium sulfate, (NH₄)₂SO₄? (M = 132.1 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Show every stage of your working: the values you start from, each stage's equation, the substitution, and the mole value written on its own line between the stages. Give the final count in scientific notation (for example 4.8e21) to 4 significant figures with its particle word, then add a one-line sanity check of the mole value against the lab-scale mole window you were taught.
Model: Values: m = 396.3 g; M = 132.1 g/mol; Avogadro's number = 6.022e23 per mole. Stage 1 (mass to moles): n = m/M = 396.3 / 132.1 = 3.000 mol. Mole checkpoint: 3.000 mol. Stage 2 (moles to formula units): N = n x Avogadro's number = 3.000 x 6.022e23 = 1.8066e24, reported as 1.807e24 formula units (4 significant figures, set by the least-precise given). Sanity check: the mole value 3.000 mol sits inside the taught lab-scale mole window of about 0.01 to 10 mol. Grading note: plain e-notation is expected (the taught entry convention); no subscripts or superscripts are required in anything the student writes.
Full model answer. Values: m = 396.3 g; M = 132.1 g/mol; Avogadro's number = 6.022e23 per mole. Stage 1 (mass to moles): n = m/M = 396.3 / 132.1 = 3.000 mol. Mole checkpoint: 3.000 mol. Stage 2 (moles to formula units): N = n x Avogadro's number = 3.000 x 6.022e23 = 1.8066e24, reported as 1.807e24 formula units (4 significant figures, set by the least-precise given). Sanity check: the mole value 3.000 mol sits inside the taught lab-scale mole window of about 0.01 to 10 mol. Grading note: plain e-notation is expected (the taught entry convention); no subscripts or superscripts are required in anything the student writes.
MOL-030 · hard · transfer · seq 22 · authored FRQ (row 22; capstone route-decision + backward execution; allocator-verified fresh instance; prod_seeds empty) · frq
A stockroom question reads: "A bottle of lithium bromate, LiBrO₃, holds 1.29473 × 10²⁴ formula units. What is its mass?" (M = 134.8 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.)
A stockroom question reads: "A bottle of lithium bromate, LiBrO₃, holds 1.29473 × 10²⁴ formula units. What is its mass?" (M = 134.8 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) (a) State which quantity the question gives you and which quantity it asks for, and name the two equations you will use, in order, each rearranged for its unknown. (b) Carry out the calculation, showing the working for both stages and writing the mole value on its own line between them. Give the mass in grams to 4 significant figures, and close by checking that the mole checkpoint lands within the everyday lab-scale amounts you were taught.
Part a. State which quantity the question gives you and which quantity it asks for, and name the two equations you will use, in order, each rearranged for its unknown.
Model: The question gives a particle count and asks for a mass — opposite ends of the route, so two stages through moles. Stage 1: the particle equation rearranged for n, n = N divided by Avogadro's number. Stage 2: the mole equation rearranged for m, m = n x M.
Part b. Carry out the calculation, showing the working for both stages and writing the mole value on its own line between them. Give the mass in grams to 4 significant figures, and close by checking that the mole checkpoint lands within the everyday lab-scale amounts you were taught.
Model: Stage 1: n = 1.29473e24 / 6.022e23 = 2.150 mol. Mole checkpoint: 2.150 mol. Stage 2: m = 2.150 x 134.8 = 289.82 g, reported as 289.8 g (4 significant figures). The mass of the sample is 289.8 g. The checkpoint 2.150 mol sits inside the taught range for lab-scale samples. Grading note: plain e-notation is expected; no subscripts or superscripts are required in anything the student writes.
Full model answer. (a) The question gives a particle count and asks for a mass — opposite ends of the route, so two stages through moles. Stage 1: the particle equation rearranged for n, n = N divided by Avogadro's number. Stage 2: the mole equation rearranged for m, m = n x M. (b) Stage 1: n = 1.29473e24 / 6.022e23 = 2.150 mol. Mole checkpoint: 2.150 mol. Stage 2: m = 2.150 x 134.8 = 289.82 g, reported as 289.8 g (4 significant figures). The mass of the sample is 289.8 g. The checkpoint 2.150 mol sits inside the taught range for lab-scale samples. Grading note: plain e-notation is expected; no subscripts or superscripts are required in anything the student writes.

Form D

MOL-022 · easy · recall · seq 1 · authored (row 1 case rotation: fixed-constant read; symbolic, instance-free) · mcq4
A student says every symbol in N = n × Nₐ takes a different value for every different sample. Which symbol is actually fixed?
Nₐ — Avogadro's number is a fixed constant, the same for every sample. KEY
N — every sample contains the same number of particles.
n — one mole is the same amount, so n is the same for every sample.
N and n are both fixed — only Nₐ changes between samples.
MOL-026 · easy · recall · seq 2 · authored (row 2; fresh instance dinitrogen trioxide 15.2 g) · mcq4
A question asks for the number of molecules in 15.2 g of dinitrogen trioxide, N₂O₃. Which two steps, in order, answer it?
First n = m/M to find the moles, then N = n × Nₐ to find the number of molecules. KEY
First n = N/Nₐ to find the moles, then m = n × M to find the mass.
Multiply the mass by Nₐ in one step.
Divide the mass by Nₐ in one step.
MOL-029 · easy · recall · seq 3 · authored (row 3 case rotation: which-unit-cancels on the particle factor) · mcq4
In the factor-label line 6.0 mol × (6.022 × 10²³ formula units / 1 mol) for barium sulfate, which unit cancels, and which unit remains?
Moles cancel, and formula units remain. KEY
Formula units cancel, and moles remain.
Nothing cancels, because the two quantities have different units.
Grams cancel, and moles remain.
MOL-023 · easy · application · seq 4 · authored (row 4; fresh instance lithium carbonate 7.00 mol) · numeric
How many formula units are in 7.00 mol of lithium carbonate, Li₂CO₃? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer to 3 significant figures. Enter your answer in scientific notation, such as 4.8e21; do not type units.
Answer: 4.22e+24 formula units KEY
also accepted: 4.22E24, 4.22e+24, 4.22 × 10²⁴
wrongValueDiagnostics
[
 {
  "value": 1.1624045167718365e-23,
  "misconception": "you divided by Avogadro's number instead of multiplying — each mole contributes 6.022 × 10²³ particles",
  "feedback": "This answer is wrong — you divided by Avogadro's number instead of multiplying — each mole contributes 6.022 × 10²³ particles"
 },
 {
  "value": 6.022e+23,
  "misconception": "you ignored the mole count and answered Nₐ itself",
  "feedback": "This answer is wrong — you ignored the mole count and answered Nₐ itself"
 },
 {
  "value": 4.2154e+23,
  "misconception": "you slipped one power of ten in the exponent",
  "feedback": "This answer is wrong — you slipped one power of ten in the exponent"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nn = 7.00 mol\nNₐ = 6.022 × 10²³ mol⁻¹\nWrite down the equation:\nN = n × Nₐ\nSubstitute in the values, and calculate:\nN = 7.00 × 6.022 × 10²³\nN = 4.2154 × 10²⁴ = 4.22 × 10²⁴ formula units (3 significant figures, set by the 3-significant-figure mole value)\nSanity check: a lab-scale count lands around 10²² to 10²⁵ — 4.22 × 10²⁴ passes.",
 "anticipatedWrong": [
  {
   "value": 1.1624045167718365e-23,
   "misconception": "you divided by Avogadro's number instead of multiplying — each mole contributes 6.022 × 10²³ particles",
   "feedback": "This answer is wrong — you divided by Avogadro's number instead of multiplying — each mole contributes 6.022 × 10²³ particles"
  },
  {
   "value": 6.022e+23,
   "misconception": "you ignored the mole count and answered Nₐ itself",
   "feedback": "This answer is wrong — you ignored the mole count and answered Nₐ itself"
  },
  {
   "value": 4.2154e+23,
   "misconception": "you slipped one power of ten in the exponent",
   "feedback": "This answer is wrong — you slipped one power of ten in the exponent"
  }
 ]
}
MOL-023 · medium · application · seq 5 · authored (row 5: pick-the-plausible-magnitude; options are magnitude CLAIMS with reasons) · mcq4
A question asks how many molecules a beaker-sized sample of a molecular solid holds. Which reported count could be right for such a sample?
7.7 × 10²³ molecules — it sits inside the taught lab-scale range of about 10²² to 10²⁵. KEY
7.7 × 10¹¹ molecules — reporting a smaller count is the safer call when unsure.
7.7 × 10⁻²³ molecules — counts smaller than 1 are allowed for very light gases.
7.7 × 10⁴⁴ molecules — big samples reach counts like this.
MOL-024 · medium · application · seq 6 · authored (row 6; fresh instance cesium iodide 6.6242e23 — cesium family window owner seq 6) · numeric
A crystal contains 6.6242 × 10²³ formula units of cesium iodide, CsI. How many moles is that? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer in moles.
Answer: 1.1 mol KEY
also accepted: 1.10
wrongValueDiagnostics
[
 {
  "value": 3.98909324e+47,
  "misconception": "you multiplied by Avogadro's number instead of dividing — this direction travels toward moles",
  "feedback": "This answer is wrong — you multiplied by Avogadro's number instead of dividing — this direction travels toward moles"
 },
 {
  "value": 0.9090909090909091,
  "misconception": "you divided Avogadro's number by the particle count instead of the other way around",
  "feedback": "This answer is wrong — you divided Avogadro's number by the particle count instead of the other way around"
 },
 {
  "value": 11.0,
  "misconception": "you slipped one power of ten in the division",
  "feedback": "This answer is wrong — you slipped one power of ten in the division"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nN = 6.6242 × 10²³ formula units\nNₐ = 6.022 × 10²³ mol⁻¹\nWrite down the equation:\nN = n × Nₐ\nMake n the subject:\nn = N/Nₐ\nSubstitute in the values, and calculate:\nn = 6.6242 × 10²³ / (6.022 × 10²³)\nn = 1.10 mol\nSanity check: a count below 6.022 × 10²³ means less than 1 mol, and above it more than 1 mol.",
 "anticipatedWrong": [
  {
   "value": 3.98909324e+47,
   "misconception": "you multiplied by Avogadro's number instead of dividing — this direction travels toward moles",
   "feedback": "This answer is wrong — you multiplied by Avogadro's number instead of dividing — this direction travels toward moles"
  },
  {
   "value": 0.9090909090909091,
   "misconception": "you divided Avogadro's number by the particle count instead of the other way around",
   "feedback": "This answer is wrong — you divided Avogadro's number by the particle count instead of the other way around"
  },
  {
   "value": 11.0,
   "misconception": "you slipped one power of ten in the division",
   "feedback": "This answer is wrong — you slipped one power of ten in the division"
  }
 ]
}
MOL-024 · medium · application · seq 7 · authored (row 7; fresh pair potassium bromate / potassium dichromate — same-family jars keep the family gram window at seq 7 only) · mcq4
Vial 1 holds 9.3341 × 10²³ formula units of potassium bromate. Vial 2 holds 8.1297 × 10²³ formula units of potassium dichromate. Which vial holds more moles? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Vial 1 — 1.55 mol compared with 1.35 mol. KEY
Vial 2 — 1.35 mol beats 0.155 mol.
The vials hold equal amounts, because both counts sit above Avogadro's number.
Deciding also needs the two molar masses, not just the counts.
MOL-025 · medium · application · seq 8 · authored (row 8; fresh instance tetraphosphorus decoxide, subscript-4 read) · numeric
A sample of 3.00 mol of tetraphosphorus decoxide, P₄O₁₀, contains how many moles of atoms of phosphorus? Give your answer in mol.
Answer: 12.0 mol KEY
also accepted: 12.0
wrongValueDiagnostics
[
 {
  "value": 3.0,
  "misconception": "you ignored the subscript and gave the moles of the compound itself",
  "feedback": "This answer is wrong — you ignored the subscript and gave the moles of the compound itself"
 },
 {
  "value": 0.75,
  "misconception": "you divided the moles by the subscript instead of multiplying",
  "feedback": "This answer is wrong — you divided the moles by the subscript instead of multiplying"
 },
 {
  "value": 30.0,
  "misconception": "you used oxygen's subscript instead of phosphorus's",
  "feedback": "This answer is wrong — you used oxygen's subscript instead of phosphorus's"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmoles of P₄O₁₀ = 3.00 mol\nsubscript of phosphorus in P₄O₁₀ = 4\nWrite down the equation:\nmoles of phosphorus atoms = subscript × moles of compound\nSubstitute in the values, and calculate:\nmoles of phosphorus atoms = 4 × 3.00\nmoles of phosphorus atoms = 12.0 mol",
 "anticipatedWrong": [
  {
   "value": 3.0,
   "misconception": "you ignored the subscript and gave the moles of the compound itself",
   "feedback": "This answer is wrong — you ignored the subscript and gave the moles of the compound itself"
  },
  {
   "value": 0.75,
   "misconception": "you divided the moles by the subscript instead of multiplying",
   "feedback": "This answer is wrong — you divided the moles by the subscript instead of multiplying"
  },
  {
   "value": 30.0,
   "misconception": "you used oxygen's subscript instead of phosphorus's",
   "feedback": "This answer is wrong — you used oxygen's subscript instead of phosphorus's"
  }
 ]
}
MOL-025 · medium · application · seq 9 · authored (row 9; fresh parentheses instance zinc chlorate; zinc owner seq 9) · numeric
The chlorine atoms in a 2.00 mol batch of zinc chlorate, Zn(ClO₃)₂, amount to how many moles? Give your answer in mol.
Answer: 4.0 mol KEY
also accepted: 4.00, 4.0
wrongValueDiagnostics
[
 {
  "value": 2.0,
  "misconception": "you ignored the subscript outside the parentheses — it multiplies everything inside",
  "feedback": "This answer is wrong — you ignored the subscript outside the parentheses — it multiplies everything inside"
 },
 {
  "value": 8.0,
  "misconception": "you applied the outer subscript twice",
  "feedback": "This answer is wrong — you applied the outer subscript twice"
 }
]
feedback
{
 "keyExplanation": "The subscript outside the parentheses multiplies everything inside, so each formula unit of Zn(ClO₃)₂ carries 2 × 1 = 2 chlorine atoms.\nWrite down the equation:\nmoles of Cl atoms = subscript × moles of compound\nSubstitute in the values, and calculate:\nmoles of Cl atoms = 2 × 2.00\nmoles of Cl atoms = 4.00 mol",
 "anticipatedWrong": [
  {
   "value": 2.0,
   "misconception": "you ignored the subscript outside the parentheses — it multiplies everything inside",
   "feedback": "This answer is wrong — you ignored the subscript outside the parentheses — it multiplies everything inside"
  },
  {
   "value": 8.0,
   "misconception": "you applied the outer subscript twice",
   "feedback": "This answer is wrong — you applied the outer subscript twice"
  }
 ]
}
MOL-026 · medium · application · seq 10 · authored (row 10; fresh instance barium sulfite — barium SULFIDE is row 20E's anchor instance) · mcq4
A question asks for the mass of 7.8286 × 10²³ formula units of barium sulfite, BaSO₃. Which two steps, in order, answer it?
First n = N/Nₐ to find the moles, then m = n × M to find the mass. KEY
First N = n × Nₐ to find the particle count, then n = m/M to find the moles.
Multiply the particle count by the molar mass in one step.
First n = m/M to find the moles, then N = n × Nₐ to find the particle count.
MOL-027 · medium · application · seq 11 · authored (row 11; fresh instance lithium sulfate) · numeric
How many formula units are in 219.8 g of lithium sulfate, Li₂SO₄? (M of Li₂SO₄ = 109.9 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer to 4 significant figures. Enter your answer in scientific notation, such as 4.8e21; do not type units.
Answer: 1.204e+24 formula units KEY
also accepted: 1.204E24, 1.204e+24, 1.204 × 10²⁴
wrongValueDiagnostics
[
 {
  "value": 1.3236356e+26,
  "misconception": "you skipped the mass-to-moles step and multiplied the mass by Avogadro's number",
  "feedback": "This answer is wrong — you skipped the mass-to-moles step and multiplied the mass by Avogadro's number"
 },
 {
  "value": 3.011e+23,
  "misconception": "you divided the molar mass by the mass in step 1, inverting the mole value",
  "feedback": "This answer is wrong — you divided the molar mass by the mass in step 1, inverting the mole value"
 },
 {
  "value": 3.321155762205247e-24,
  "misconception": "you divided by Avogadro's number instead of multiplying — a count below one fails any sense check",
  "feedback": "This answer is wrong — you divided by Avogadro's number instead of multiplying — a count below one fails any sense check"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nm = 219.8 g\nM = 109.9 g/mol\nNₐ = 6.022 × 10²³ mol⁻¹\nStep 1 — convert mass to moles:\nn = m/M\nn = 219.8 / 109.9\nn = 2.000 mol   (the mole checkpoint — lab-scale values land between about 0.01 and 10 mol)\nStep 2 — convert moles to formula units:\nN = n × Nₐ\nN = 2.000 × 6.022 × 10²³\nN = 1.2044 × 10²⁴ = 1.204 × 10²⁴ formula units (4 significant figures, set by the least-precise given)",
 "anticipatedWrong": [
  {
   "value": 1.3236356e+26,
   "misconception": "you skipped the mass-to-moles step and multiplied the mass by Avogadro's number",
   "feedback": "This answer is wrong — you skipped the mass-to-moles step and multiplied the mass by Avogadro's number"
  },
  {
   "value": 3.011e+23,
   "misconception": "you divided the molar mass by the mass in step 1, inverting the mole value",
   "feedback": "This answer is wrong — you divided the molar mass by the mass in step 1, inverting the mole value"
  },
  {
   "value": 3.321155762205247e-24,
   "misconception": "you divided by Avogadro's number instead of multiplying — a count below one fails any sense check",
   "feedback": "This answer is wrong — you divided by Avogadro's number instead of multiplying — a count below one fails any sense check"
  }
 ]
}
MOL-028 · medium · application · seq 12 · authored (row 12; allocator-verified fresh instance) · numeric
A sample of lithium phosphate, Li₃PO₄ (M = 115.8 g/mol), contains 7.5275 × 10²³ formula units. What is the sample's mass, in grams? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer to 4 significant figures.
Answer: 144.8 g KEY
also accepted: 144.8, 144.8 g
wrongValueDiagnostics
[
 {
  "value": 115.8,
  "misconception": "you gave one full mole's mass — the sample is not 1 mol; divide the count by Nₐ first",
  "feedback": "This answer is wrong — you gave one full mole's mass — the sample is not 1 mol; divide the count by Nₐ first"
 },
 {
  "value": 92.64,
  "misconception": "you divided the molar mass by the moles instead of multiplying them",
  "feedback": "This answer is wrong — you divided the molar mass by the moles instead of multiplying them"
 },
 {
  "value": 8.716845e+25,
  "misconception": "you skipped the particles-to-moles step and multiplied the particle count by the molar mass",
  "feedback": "This answer is wrong — you skipped the particles-to-moles step and multiplied the particle count by the molar mass"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nN = 7.5275 × 10²³ formula units\nM = 115.8 g/mol\nNₐ = 6.022 × 10²³ mol⁻¹\nStep 1 — convert formula units to moles:\nN = n × Nₐ\nMake n the subject:\nn = N/Nₐ\nn = 7.5275 × 10²³ / (6.022 × 10²³)\nn = 1.250 mol   (the mole checkpoint)\nStep 2 — convert moles to mass:\nn = m/M\nMake m the subject:\nm = n × M\nm = 1.250 × 115.8\nm = 144.75 g\nReported to 4 significant figures (set by M and Nₐ): m = 144.8 g",
 "anticipatedWrong": [
  {
   "value": 115.8,
   "misconception": "you gave one full mole's mass — the sample is not 1 mol; divide the count by Nₐ first",
   "feedback": "This answer is wrong — you gave one full mole's mass — the sample is not 1 mol; divide the count by Nₐ first"
  },
  {
   "value": 92.64,
   "misconception": "you divided the molar mass by the moles instead of multiplying them",
   "feedback": "This answer is wrong — you divided the molar mass by the moles instead of multiplying them"
  },
  {
   "value": 8.716845e+25,
   "misconception": "you skipped the particles-to-moles step and multiplied the particle count by the molar mass",
   "feedback": "This answer is wrong — you skipped the particles-to-moles step and multiplied the particle count by the molar mass"
  }
 ]
}
MOL-026 · medium · application · seq 13 · authored (row 13; broken-route class: both constants, no mole checkpoint) · mcq4
To find the number of formula units in 24.9 g of strontium fluoride, SrF₂ (M = 125.6 g/mol), a student writes one line: N = (m / Nₐ) × M. What is wrong with the student's route?
The mole value is missing — the route is n = m/M first, then N = n × Nₐ; Nₐ never divides a mass. KEY
Nothing is wrong — the M at the end converts the result into a particle count.
The student's only error is dividing by Nₐ instead of multiplying by it.
The route works, but only for elements rather than compounds.
MOL-028 · medium · application · seq 14 · authored (row 14; diagnosis class 'slip'; allocator-verified fresh instance) · mcq4
A stockroom bottle of barium carbonate, BaCO₃ (M = 197.3 g/mol), holds 1.26462 × 10²⁴ formula units. Asked for its mass, a student computes n = 1.26462 × 10²⁴ / (6.022 × 10²⁴) and reports 41.4 g. (Nₐ = 6.022 × 10²³ mol⁻¹.) Which mistake did the student make?
The student slipped one power of ten when dividing the count by Avogadro's number. KEY
The student gave one full mole's mass instead of using the computed mole value.
The student multiplied the raw particle count by the molar mass, skipping the moles stage.
The student divided the molar mass by the mole value instead of multiplying them.
MOL-030 · medium · application · seq 15 · authored (row 15; capstone single-step case — M supplied but unused IS the route-choice test; allocator-verified) · numeric
A bottle holds 6.26288 × 10²³ molecules of benzoic acid, C₇H₆O₂ (M = 122.1 g/mol). How many moles is that? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer in moles to 3 significant figures.
Answer: 1.04 mol KEY
also accepted: 1.04
wrongValueDiagnostics
[
 {
  "value": 126.984,
  "misconception": "you ran a second stage the question never asked for — multiplying by the molar mass gives the mass, not the moles",
  "feedback": "This answer is wrong — you ran a second stage the question never asked for — multiplying by the molar mass gives the mass, not the moles"
 },
 {
  "value": 3.771506336e+47,
  "misconception": "you multiplied by Avogadro's number — moles and particle count are route neighbors, and this direction divides",
  "feedback": "This answer is wrong — you multiplied by Avogadro's number — moles and particle count are route neighbors, and this direction divides"
 },
 {
  "value": 0.9615384615384616,
  "misconception": "you divided Avogadro's number by the particle count instead of the other way around",
  "feedback": "This answer is wrong — you divided Avogadro's number by the particle count instead of the other way around"
 }
]
feedback
{
 "keyExplanation": "Ask the two route questions first: the question GIVES a particle count and ASKS for moles.\nCount and moles sit next to each other on the route, so ONE equation finishes the job — the supplied molar mass is never needed.\nN = n × Nₐ\nMake n the subject:\nn = N/Nₐ\nn = 6.26288 × 10²³ / (6.022 × 10²³)\nn = 1.04 mol",
 "anticipatedWrong": [
  {
   "value": 126.984,
   "misconception": "you ran a second stage the question never asked for — multiplying by the molar mass gives the mass, not the moles",
   "feedback": "This answer is wrong — you ran a second stage the question never asked for — multiplying by the molar mass gives the mass, not the moles"
  },
  {
   "value": 3.771506336e+47,
   "misconception": "you multiplied by Avogadro's number — moles and particle count are route neighbors, and this direction divides",
   "feedback": "This answer is wrong — you multiplied by Avogadro's number — moles and particle count are route neighbors, and this direction divides"
  },
  {
   "value": 0.9615384615384616,
   "misconception": "you divided Avogadro's number by the particle count instead of the other way around",
   "feedback": "This answer is wrong — you divided Avogadro's number by the particle count instead of the other way around"
  }
 ]
}
MOL-023 · medium · application · seq 16 · authored (row 16; fresh gas pair butadiene / cyclopropane — free-token names) · mcq4
Flask 1 holds 9.00 mol of butadiene, C₄H₆. Flask 2 holds 8.00 mol of cyclopropane, C₃H₆. Which flask contains more particles of gas? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Flask 1 — 9.00 mol means more particles than 8.00 mol, whatever each particle weighs. KEY
Flask 2 — cyclopropane's lighter particles mean more of them are packed into its flask.
The flasks hold equal counts, because both hold pure gases and every pure gas carries one mole's worth of particles.
Flask 2 — each cyclopropane molecule contains fewer atoms, so its flask must hold more separate molecules.
MOL-025 · hard · transfer · seq 17 · authored (row 17; fresh pair; lead owner seq 17) · mcq4
Sample 1 is 1.0 mol of lead(II) phosphate, Pb₃(PO₄)₂. Sample 2 is 2.0 mol of lead(II) chloride, PbCl₂. Which sample contains more moles of lead atoms?
Sample 1 — 3.0 mol of lead atoms compared with 2.0 mol. KEY
Sample 2, because 2.0 mol of compound is more than 1.0 mol of compound.
The two samples hold equal moles of lead atoms.
It cannot be decided without the masses of the two samples.
MOL-027 · hard · transfer · seq 18 · authored (row 18; fresh equal-moles pair strontium oxide / potassium oxide) · mcq4
Sample 1 is 310.8 g of strontium oxide, SrO (M = 103.6 g/mol). Sample 2 is 282.6 g of potassium oxide, K₂O (M = 94.2 g/mol). Which sample contains more formula units? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Neither — both samples contain 3.0 mol, so their particle counts are equal. KEY
Sample 1 — at 310.8 g it is simply the heavier sample.
Sample 2 — its particles are lighter, so the sample must pack in more of them.
Only the two sample volumes could settle it.
MOL-028 · hard · transfer · seq 19 · authored (row 19; fresh trade-off pair) · mcq4
Sample 1 is 2.77012 × 10²⁴ particles of ammonium fluoride, NH₄F (M = 37.0 g/mol). Sample 2 is 1.38506 × 10²⁴ particles of barium nitrate, Ba(NO₃)₂ (M = 261.3 g/mol). Which sample has the greater mass? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Sample 2 — each of its particles is more than twice as heavy, so half the count still gives the greater mass. KEY
Sample 1, because it contains twice as many particles.
The two samples have equal masses, because the count and the molar mass trade off exactly.
Settling it would need volume information that is not given.
MOL-030 · hard · transfer · seq 20 · authored (row 20; fresh mixed pair strontium bromide / formaldehyde) · mcq4
Sample 1 is 123.7 g of strontium bromide, SrBr₂ (M = 247.4 g/mol). Sample 2 is 30.0 g of formaldehyde, CH₂O (M = 30.0 g/mol). Which sample contains more particles (molecules or formula units)? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Sample 2 — a full mole of particles against half a mole, despite the smaller mass. KEY
Sample 1 — the bigger mass, 123.7 g, must mean more particles.
The counts come out equal — the bigger molar mass exactly offsets the smaller mass.
Sample 1 — heavier particles pack more matter into every single particle.
MOL-027 · medium · application · seq 21 · authored FRQ (row 21; fresh instance — lithium phosphate dropped: its formula fragment ₃PO₄ collides with ammonium phosphate's at row 14) · frq
How many formula units are in 454.5 g of lithium iodate, LiIO₃? (M = 181.8 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Show every stage of your working: the values you start from, each stage's equation, the substitution, and the mole value written on its own line between the stages. Give the final count in scientific notation (for example 4.8e21) to 4 significant figures with its particle word, then add a one-line sanity check of the mole value against the lab-scale mole window you were taught.
How many formula units are in 454.5 g of lithium iodate, LiIO₃? (M = 181.8 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Show every stage of your working: the values you start from, each stage's equation, the substitution, and the mole value written on its own line between the stages. Give the final count in scientific notation (for example 4.8e21) to 4 significant figures with its particle word, then add a one-line sanity check of the mole value against the lab-scale mole window you were taught.
Part a. How many formula units are in 454.5 g of lithium iodate, LiIO₃? (M = 181.8 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Show every stage of your working: the values you start from, each stage's equation, the substitution, and the mole value written on its own line between the stages. Give the final count in scientific notation (for example 4.8e21) to 4 significant figures with its particle word, then add a one-line sanity check of the mole value against the lab-scale mole window you were taught.
Model: Values: m = 454.5 g; M = 181.8 g/mol; Avogadro's number = 6.022e23 per mole. Stage 1 (mass to moles): n = m/M = 454.5 / 181.8 = 2.500 mol. Mole checkpoint: 2.500 mol. Stage 2 (moles to formula units): N = n x Avogadro's number = 2.500 x 6.022e23 = 1.5055e24, reported as 1.506e24 formula units (4 significant figures, set by the least-precise given). Sanity check: the mole value 2.500 mol sits inside the taught lab-scale mole window of about 0.01 to 10 mol. Grading note: plain e-notation is expected (the taught entry convention); no subscripts or superscripts are required in anything the student writes.
Full model answer. Values: m = 454.5 g; M = 181.8 g/mol; Avogadro's number = 6.022e23 per mole. Stage 1 (mass to moles): n = m/M = 454.5 / 181.8 = 2.500 mol. Mole checkpoint: 2.500 mol. Stage 2 (moles to formula units): N = n x Avogadro's number = 2.500 x 6.022e23 = 1.5055e24, reported as 1.506e24 formula units (4 significant figures, set by the least-precise given). Sanity check: the mole value 2.500 mol sits inside the taught lab-scale mole window of about 0.01 to 10 mol. Grading note: plain e-notation is expected (the taught entry convention); no subscripts or superscripts are required in anything the student writes.
MOL-030 · hard · transfer · seq 22 · authored FRQ (row 22; capstone route-decision + backward execution; allocator-verified fresh instance; prod_seeds empty) · frq
A stockroom question reads: "A bottle of strontium sulfide, SrS, holds 1.86682 × 10²⁴ formula units. What is its mass?" (M = 119.7 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.)
A stockroom question reads: "A bottle of strontium sulfide, SrS, holds 1.86682 × 10²⁴ formula units. What is its mass?" (M = 119.7 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) (a) State which quantity the question gives you and which quantity it asks for, and name the two equations you will use, in order, each rearranged for its unknown. (b) Carry out the calculation, showing the working for both stages and writing the mole value on its own line between them. Give the mass in grams to 4 significant figures, and close by checking that the mole checkpoint lands within the everyday lab-scale amounts you were taught.
Part a. State which quantity the question gives you and which quantity it asks for, and name the two equations you will use, in order, each rearranged for its unknown.
Model: The question gives a particle count and asks for a mass — opposite ends of the route, so two stages through moles. Stage 1: the particle equation rearranged for n, n = N divided by Avogadro's number. Stage 2: the mole equation rearranged for m, m = n x M.
Part b. Carry out the calculation, showing the working for both stages and writing the mole value on its own line between them. Give the mass in grams to 4 significant figures, and close by checking that the mole checkpoint lands within the everyday lab-scale amounts you were taught.
Model: Stage 1: n = 1.86682e24 / 6.022e23 = 3.100 mol. Mole checkpoint: 3.100 mol. Stage 2: m = 3.100 x 119.7 = 371.07 g, reported as 371.1 g (4 significant figures). The mass of the sample is 371.1 g. The checkpoint 3.100 mol sits inside the taught range for lab-scale samples. Grading note: plain e-notation is expected; no subscripts or superscripts are required in anything the student writes.
Full model answer. (a) The question gives a particle count and asks for a mass — opposite ends of the route, so two stages through moles. Stage 1: the particle equation rearranged for n, n = N divided by Avogadro's number. Stage 2: the mole equation rearranged for m, m = n x M. (b) Stage 1: n = 1.86682e24 / 6.022e23 = 3.100 mol. Mole checkpoint: 3.100 mol. Stage 2: m = 3.100 x 119.7 = 371.07 g, reported as 371.1 g (4 significant figures). The mass of the sample is 371.1 g. The checkpoint 3.100 mol sits inside the taught range for lab-scale samples. Grading note: plain e-notation is expected; no subscripts or superscripts are required in anything the student writes.

Form E

MOL-022 · easy · recall · seq 1 · authored (row 1 case rotation: unit-misuse correction; symbolic) · mcq4
As the final line of a calculation a student writes: N = 4.0 mol. What is wrong with the line?
The unit is wrong — N is a plain count with no unit; the amount in moles is small n. KEY
Nothing is wrong — any quantity in the equation may carry mol.
The mol is fine — the mistake is writing N where the constant Nₐ belongs.
The line is wrong only because a mole amount must be a whole number.
MOL-026 · easy · recall · seq 2 · authored (row 2; fresh instance propene 8.6 g) · mcq4
A question asks for the number of molecules in 8.6 g of propene, C₃H₆. Which two steps, in order, answer it?
First n = m/M to find the moles, then N = n × Nₐ to find the number of molecules. KEY
First n = N/Nₐ to find the moles, then m = n × M to find the mass.
Multiply the mass by Nₐ in one step.
First N = n × Nₐ to find the number of molecules, then n = m/M to find the moles.
MOL-029 · easy · recall · seq 3 · authored (row 3 case rotation: single-factor m = n × M match on urea) · mcq4
One line of a worked solution reads 4.0 mol × (60.1 g / 1 mol), for urea, CO(NH₂)₂. Which equation step does the line carry out?
m = n × M — multiplying the moles by the molar mass to get a mass. KEY
n = m/M — dividing a mass by the molar mass to get moles.
N = n × Nₐ — multiplying moles by Avogadro's number to get a particle count.
n = N/Nₐ — dividing a particle count by Avogadro's number to get moles.
MOL-023 · easy · application · seq 4 · authored (row 4; fresh instance silver nitrate 9.00 mol; silver owner seq 4) · numeric
How many formula units are in 9.00 mol of silver nitrate, AgNO₃? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer to 3 significant figures. Enter your answer in scientific notation, such as 4.8e21; do not type units.
Answer: 5.42e+24 formula units KEY
also accepted: 5.42E24, 5.42e+24, 5.42 × 10²⁴
wrongValueDiagnostics
[
 {
  "value": 1.4945200929923614e-23,
  "misconception": "you divided by Avogadro's number instead of multiplying — each mole contributes 6.022 × 10²³ particles",
  "feedback": "This answer is wrong — you divided by Avogadro's number instead of multiplying — each mole contributes 6.022 × 10²³ particles"
 },
 {
  "value": 6.022e+23,
  "misconception": "you ignored the mole count and answered Nₐ itself",
  "feedback": "This answer is wrong — you ignored the mole count and answered Nₐ itself"
 },
 {
  "value": 5.4198e+23,
  "misconception": "you slipped one power of ten in the exponent",
  "feedback": "This answer is wrong — you slipped one power of ten in the exponent"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nn = 9.00 mol\nNₐ = 6.022 × 10²³ mol⁻¹\nWrite down the equation:\nN = n × Nₐ\nSubstitute in the values, and calculate:\nN = 9.00 × 6.022 × 10²³\nN = 5.4198 × 10²⁴ = 5.42 × 10²⁴ formula units (3 significant figures, set by the 3-significant-figure mole value)\nSanity check: a lab-scale count lands around 10²² to 10²⁵ — 5.42 × 10²⁴ passes.",
 "anticipatedWrong": [
  {
   "value": 1.4945200929923614e-23,
   "misconception": "you divided by Avogadro's number instead of multiplying — each mole contributes 6.022 × 10²³ particles",
   "feedback": "This answer is wrong — you divided by Avogadro's number instead of multiplying — each mole contributes 6.022 × 10²³ particles"
  },
  {
   "value": 6.022e+23,
   "misconception": "you ignored the mole count and answered Nₐ itself",
   "feedback": "This answer is wrong — you ignored the mole count and answered Nₐ itself"
  },
  {
   "value": 5.4198e+23,
   "misconception": "you slipped one power of ten in the exponent",
   "feedback": "This answer is wrong — you slipped one power of ten in the exponent"
  }
 ]
}
MOL-023 · medium · application · seq 5 · authored (row 5: too-large-but-plausible-looking class) · mcq4
A student reports 4.7 × 10²⁶ molecules for a sealed test-tube sample and argues the answer must be right because the calculator showed it. Which taught check says otherwise?
A lab-scale count lands around 10²² to 10²⁵ — 10²⁶ is beyond that range. KEY
None — the taught check is there to catch counts that come out below 1, and this count is well above 1.
The count fails because it is not an exact multiple of 6.022 × 10²³.
The count fails because a sealed tube must hold a whole number of moles.
MOL-024 · medium · application · seq 6 · authored (row 6; fresh instance titanium strip 3.3121e23 atoms; titanium owner seq 6) · numeric
A strip of metal contains 3.3121 × 10²³ atoms of pure titanium. How many moles is that? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer in moles.
Answer: 0.55 mol KEY
also accepted: 0.550, .55
wrongValueDiagnostics
[
 {
  "value": 1.99454662e+47,
  "misconception": "you multiplied by Avogadro's number instead of dividing — this direction travels toward moles",
  "feedback": "This answer is wrong — you multiplied by Avogadro's number instead of dividing — this direction travels toward moles"
 },
 {
  "value": 1.8181818181818181,
  "misconception": "you divided Avogadro's number by the particle count instead of the other way around",
  "feedback": "This answer is wrong — you divided Avogadro's number by the particle count instead of the other way around"
 },
 {
  "value": 5.5,
  "misconception": "you slipped one power of ten in the division",
  "feedback": "This answer is wrong — you slipped one power of ten in the division"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nN = 3.3121 × 10²³ atoms\nNₐ = 6.022 × 10²³ mol⁻¹\nWrite down the equation:\nN = n × Nₐ\nMake n the subject:\nn = N/Nₐ\nSubstitute in the values, and calculate:\nn = 3.3121 × 10²³ / (6.022 × 10²³)\nn = 0.550 mol\nSanity check: a count below 6.022 × 10²³ means less than 1 mol, and above it more than 1 mol.",
 "anticipatedWrong": [
  {
   "value": 1.99454662e+47,
   "misconception": "you multiplied by Avogadro's number instead of dividing — this direction travels toward moles",
   "feedback": "This answer is wrong — you multiplied by Avogadro's number instead of dividing — this direction travels toward moles"
  },
  {
   "value": 1.8181818181818181,
   "misconception": "you divided Avogadro's number by the particle count instead of the other way around",
   "feedback": "This answer is wrong — you divided Avogadro's number by the particle count instead of the other way around"
  },
  {
   "value": 5.5,
   "misconception": "you slipped one power of ten in the division",
   "feedback": "This answer is wrong — you slipped one power of ten in the division"
  }
 ]
}
MOL-024 · medium · application · seq 7 · authored (row 7; fresh pair dinitrogen difluoride / sulfur hexafluoride — sulfur owner seq 7) · mcq4
Flask 1 holds 6.3231 × 10²³ molecules of dinitrogen difluoride, N₂F₂. Flask 2 holds 5.1187 × 10²³ molecules of sulfur hexafluoride, SF₆. Which flask holds more moles? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Flask 1 — 1.05 mol compared with 0.85 mol. KEY
Flask 2 — 8.5 mol beats 1.05 mol.
Both flasks hold the same amount, because each molecule of both gases contains fluorine.
Flask 1 — its molecules carry more atoms each, and that is what tips the moles.
MOL-025 · medium · application · seq 8 · authored (row 8; fresh instance potassium sulfide, metal-element ask) · numeric
A sample of 6.00 mol of potassium sulfide, K₂S, holds how many moles of potassium atoms? Give your answer in mol.
Answer: 12.0 mol KEY
also accepted: 12.0
wrongValueDiagnostics
[
 {
  "value": 6.0,
  "misconception": "you ignored the subscript and gave the moles of the compound itself",
  "feedback": "This answer is wrong — you ignored the subscript and gave the moles of the compound itself"
 },
 {
  "value": 3.0,
  "misconception": "you divided the moles by the subscript instead of multiplying",
  "feedback": "This answer is wrong — you divided the moles by the subscript instead of multiplying"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmoles of K₂S = 6.00 mol\nsubscript of potassium in K₂S = 2\nWrite down the equation:\nmoles of potassium atoms = subscript × moles of compound\nSubstitute in the values, and calculate:\nmoles of potassium atoms = 2 × 6.00\nmoles of potassium atoms = 12.0 mol",
 "anticipatedWrong": [
  {
   "value": 6.0,
   "misconception": "you ignored the subscript and gave the moles of the compound itself",
   "feedback": "This answer is wrong — you ignored the subscript and gave the moles of the compound itself"
  },
  {
   "value": 3.0,
   "misconception": "you divided the moles by the subscript instead of multiplying",
   "feedback": "This answer is wrong — you divided the moles by the subscript instead of multiplying"
  }
 ]
}
MOL-025 · medium · application · seq 9 · authored (row 9; fresh parentheses instance iron(III) chlorate, outer-3 case; iron owner seq 9) · numeric
Exactly 3.00 mol of iron(III) chlorate, Fe(ClO₃)₃, is weighed out. How many moles does its chlorine-atom count come to? Give your answer in mol.
Answer: 9.0 mol KEY
also accepted: 9.00, 9.0
wrongValueDiagnostics
[
 {
  "value": 3.0,
  "misconception": "you ignored the subscript outside the parentheses — it multiplies everything inside",
  "feedback": "This answer is wrong — you ignored the subscript outside the parentheses — it multiplies everything inside"
 },
 {
  "value": 27.0,
  "misconception": "you applied the outer subscript twice",
  "feedback": "This answer is wrong — you applied the outer subscript twice"
 }
]
feedback
{
 "keyExplanation": "The subscript outside the parentheses multiplies everything inside, so each formula unit of Fe(ClO₃)₃ carries 3 × 1 = 3 chlorine atoms.\nWrite down the equation:\nmoles of Cl atoms = subscript × moles of compound\nSubstitute in the values, and calculate:\nmoles of Cl atoms = 3 × 3.00\nmoles of Cl atoms = 9.00 mol",
 "anticipatedWrong": [
  {
   "value": 3.0,
   "misconception": "you ignored the subscript outside the parentheses — it multiplies everything inside",
   "feedback": "This answer is wrong — you ignored the subscript outside the parentheses — it multiplies everything inside"
  },
  {
   "value": 27.0,
   "misconception": "you applied the outer subscript twice",
   "feedback": "This answer is wrong — you applied the outer subscript twice"
  }
 ]
}
MOL-026 · medium · application · seq 10 · authored (row 10; fresh instance ammonium bromide) · mcq4
A question asks for the mass of 2.8905 × 10²³ formula units of ammonium bromide, NH₄Br. Which two steps, in order, answer it?
First n = N/Nₐ to find the moles, then m = n × M to find the mass. KEY
First N = n × Nₐ to find the particle count, then n = m/M to find the moles.
Multiply the particle count by the molar mass in one step.
First N = n × Nₐ to find the particle count, then m = n ÷ M to find the mass.
MOL-027 · medium · application · seq 11 · authored (row 11; fresh instance barium fluoride stock jar) · numeric
How many formula units are in 525.9 g of barium fluoride, BaF₂? (M of BaF₂ = 175.3 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer to 4 significant figures. Enter your answer in scientific notation, such as 4.8e21; do not type units.
Answer: 1.807e+24 formula units KEY
also accepted: 1.807E24, 1.807e+24, 1.807 × 10²⁴
wrongValueDiagnostics
[
 {
  "value": 3.1669698e+26,
  "misconception": "you skipped the mass-to-moles step and multiplied the mass by Avogadro's number",
  "feedback": "This answer is wrong — you skipped the mass-to-moles step and multiplied the mass by Avogadro's number"
 },
 {
  "value": 2.0073333333333334e+23,
  "misconception": "you divided the molar mass by the mass in step 1, inverting the mole value",
  "feedback": "This answer is wrong — you divided the molar mass by the mass in step 1, inverting the mole value"
 },
 {
  "value": 4.981733643307871e-24,
  "misconception": "you divided by Avogadro's number instead of multiplying — a count below one fails any sense check",
  "feedback": "This answer is wrong — you divided by Avogadro's number instead of multiplying — a count below one fails any sense check"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nm = 525.9 g\nM = 175.3 g/mol\nNₐ = 6.022 × 10²³ mol⁻¹\nStep 1 — convert mass to moles:\nn = m/M\nn = 525.9 / 175.3\nn = 3.000 mol   (the mole checkpoint — lab-scale values land between about 0.01 and 10 mol)\nStep 2 — convert moles to formula units:\nN = n × Nₐ\nN = 3.000 × 6.022 × 10²³\nN = 1.8066 × 10²⁴ = 1.807 × 10²⁴ formula units (4 significant figures, set by the least-precise given)",
 "anticipatedWrong": [
  {
   "value": 3.1669698e+26,
   "misconception": "you skipped the mass-to-moles step and multiplied the mass by Avogadro's number",
   "feedback": "This answer is wrong — you skipped the mass-to-moles step and multiplied the mass by Avogadro's number"
  },
  {
   "value": 2.0073333333333334e+23,
   "misconception": "you divided the molar mass by the mass in step 1, inverting the mole value",
   "feedback": "This answer is wrong — you divided the molar mass by the mass in step 1, inverting the mole value"
  },
  {
   "value": 4.981733643307871e-24,
   "misconception": "you divided by Avogadro's number instead of multiplying — a count below one fails any sense check",
   "feedback": "This answer is wrong — you divided by Avogadro's number instead of multiplying — a count below one fails any sense check"
  }
 ]
}
MOL-028 · medium · application · seq 12 · authored (row 12; allocator-verified fresh instance) · numeric
A sample of ammonium iodide, NH₄I (M = 144.9 g/mol), contains 1.92704 × 10²⁴ formula units. What is the sample's mass, in grams? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer to 4 significant figures.
Answer: 463.7 g KEY
also accepted: 463.7, 463.7 g
wrongValueDiagnostics
[
 {
  "value": 144.9,
  "misconception": "you gave one full mole's mass — the sample is not 1 mol; divide the count by Nₐ first",
  "feedback": "This answer is wrong — you gave one full mole's mass — the sample is not 1 mol; divide the count by Nₐ first"
 },
 {
  "value": 45.28125,
  "misconception": "you divided the molar mass by the moles instead of multiplying them",
  "feedback": "This answer is wrong — you divided the molar mass by the moles instead of multiplying them"
 },
 {
  "value": 2.79228096e+26,
  "misconception": "you skipped the particles-to-moles step and multiplied the particle count by the molar mass",
  "feedback": "This answer is wrong — you skipped the particles-to-moles step and multiplied the particle count by the molar mass"
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nN = 1.92704 × 10²⁴ formula units\nM = 144.9 g/mol\nNₐ = 6.022 × 10²³ mol⁻¹\nStep 1 — convert formula units to moles:\nN = n × Nₐ\nMake n the subject:\nn = N/Nₐ\nn = 1.92704 × 10²⁴ / (6.022 × 10²³)\nn = 3.200 mol   (the mole checkpoint)\nStep 2 — convert moles to mass:\nn = m/M\nMake m the subject:\nm = n × M\nm = 3.200 × 144.9\nm = 463.68 g\nReported to 4 significant figures (set by M and Nₐ): m = 463.7 g",
 "anticipatedWrong": [
  {
   "value": 144.9,
   "misconception": "you gave one full mole's mass — the sample is not 1 mol; divide the count by Nₐ first",
   "feedback": "This answer is wrong — you gave one full mole's mass — the sample is not 1 mol; divide the count by Nₐ first"
  },
  {
   "value": 45.28125,
   "misconception": "you divided the molar mass by the moles instead of multiplying them",
   "feedback": "This answer is wrong — you divided the molar mass by the moles instead of multiplying them"
  },
  {
   "value": 2.79228096e+26,
   "misconception": "you skipped the particles-to-moles step and multiplied the particle count by the molar mass",
   "feedback": "This answer is wrong — you skipped the particles-to-moles step and multiplied the particle count by the molar mass"
  }
 ]
}
MOL-026 · medium · application · seq 13 · authored (row 13; broken-route class: stops at the checkpoint) · mcq4
To find the number of formula units in 43.4 g of barium iodide, BaI₂ (M = 391.1 g/mol), a student writes one line: N = m / M. What is wrong with the student's route?
The route stops at moles — m/M gives the mole value n, and a second step, N = n × Nₐ, is still needed. KEY
Nothing is wrong — dividing by the molar mass gives the particle count directly.
The student should have multiplied m by M instead.
The line is wrong because a particle count has to be measured with an instrument, not calculated from a mass.
MOL-028 · medium · application · seq 14 · authored (row 14; diagnosis class 'fullmole'; allocator-verified fresh instance) · mcq4
A stockroom bottle of strontium hydroxide, Sr(OH)₂ (M = 121.6 g/mol), holds 1.05385 × 10²⁴ formula units. Asked for its mass, a student reports 121.6 g. (Nₐ = 6.022 × 10²³ mol⁻¹.) Which mistake did the student make?
The student gave one full mole's mass instead of using the computed mole value. KEY
The student multiplied the raw particle count by the molar mass, skipping the moles stage.
The student divided the molar mass by the mole value instead of multiplying them.
The student slipped one power of ten when dividing the count by Avogadro's number.
MOL-030 · medium · application · seq 15 · authored (row 15; capstone single-step case — M supplied but unused IS the route-choice test; allocator-verified) · numeric
A bottle holds 6.50376 × 10²³ molecules of vanillin, C₈H₈O₃ (M = 152.1 g/mol). How many moles is that? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer in moles to 3 significant figures.
Answer: 1.08 mol KEY
also accepted: 1.08
wrongValueDiagnostics
[
 {
  "value": 164.268,
  "misconception": "you ran a second stage the question never asked for — multiplying by the molar mass gives the mass, not the moles",
  "feedback": "This answer is wrong — you ran a second stage the question never asked for — multiplying by the molar mass gives the mass, not the moles"
 },
 {
  "value": 3.916564272e+47,
  "misconception": "you multiplied by Avogadro's number — moles and particle count are route neighbors, and this direction divides",
  "feedback": "This answer is wrong — you multiplied by Avogadro's number — moles and particle count are route neighbors, and this direction divides"
 },
 {
  "value": 0.9259259259259259,
  "misconception": "you divided Avogadro's number by the particle count instead of the other way around",
  "feedback": "This answer is wrong — you divided Avogadro's number by the particle count instead of the other way around"
 }
]
feedback
{
 "keyExplanation": "Ask the two route questions first: the question GIVES a particle count and ASKS for moles.\nCount and moles sit next to each other on the route, so ONE equation finishes the job — the supplied molar mass is never needed.\nN = n × Nₐ\nMake n the subject:\nn = N/Nₐ\nn = 6.50376 × 10²³ / (6.022 × 10²³)\nn = 1.08 mol",
 "anticipatedWrong": [
  {
   "value": 164.268,
   "misconception": "you ran a second stage the question never asked for — multiplying by the molar mass gives the mass, not the moles",
   "feedback": "This answer is wrong — you ran a second stage the question never asked for — multiplying by the molar mass gives the mass, not the moles"
  },
  {
   "value": 3.916564272e+47,
   "misconception": "you multiplied by Avogadro's number — moles and particle count are route neighbors, and this direction divides",
   "feedback": "This answer is wrong — you multiplied by Avogadro's number — moles and particle count are route neighbors, and this direction divides"
  },
  {
   "value": 0.9259259259259259,
   "misconception": "you divided Avogadro's number by the particle count instead of the other way around",
   "feedback": "This answer is wrong — you divided Avogadro's number by the particle count instead of the other way around"
  }
 ]
}
MOL-023 · medium · application · seq 16 · authored (row 16; fresh gas pair isobutane / difluoromethane (refrigerant register) — free-token names) · mcq4
Flask 1 holds 11.00 mol of isobutane, C₄H₁₀. Flask 2 holds 10.00 mol of difluoromethane, CH₂F₂. Which flask contains more particles of gas? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Flask 1 — 11.00 mol means more particles than 10.00 mol, whatever each particle weighs. KEY
Flask 2 — difluoromethane, CH₂F₂'s particles are heavier, so there must be more of them in its flask.
The flasks hold equal counts, because both hold pure gases and every pure gas carries one mole's worth of particles.
Flask 1 — but only because isobutane, C₄H₁₀'s particles are smaller, so more of them fit inside.
MOL-025 · hard · transfer · seq 17 · authored (row 17; fresh pair; copper owner seq 17) · mcq4
Sample 1 is 2.0 mol of copper(I) oxide, Cu₂O. Sample 2 is 3.0 mol of copper(II) sulfate, CuSO₄. Which sample contains more moles of copper atoms?
Sample 1 — 4.0 mol of copper atoms compared with 3.0 mol. KEY
Sample 2, because 3.0 mol of compound is more than 2.0 mol of compound.
The two samples hold equal moles of copper atoms.
It cannot be decided without the masses of the two samples.
MOL-027 · hard · transfer · seq 18 · authored (row 18; fresh equal-moles pair ammonium carbonate / silicon tetrachloride) · mcq4
Sample 1 is 480.5 g of ammonium carbonate, (NH₄)₂CO₃ (M = 96.1 g/mol). Sample 2 is 849.5 g of silicon tetrachloride, SiCl₄ (M = 169.9 g/mol). Which sample contains more particles (molecules or formula units)? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Neither — both samples contain 5.0 mol, so their particle counts are equal. KEY
Sample 2 — at 849.5 g it is simply the heavier sample.
Sample 1 — its particles are lighter, so the sample must pack in more of them.
Only the two sample volumes could settle it.
MOL-028 · hard · transfer · seq 19 · authored (row 19; fresh trade-off pair) | stage3 patch: name-first reorder kills the material-bearing '1024 particles of <cation>' shared window (V-T7; substances/values unchanged) · mcq4
Sample 1 is 3.37232 × 10²⁴ particles of lithium sulfide, Li₂S (M = 45.9 g/mol). Sample 2 is strontium iodide, SrI₂ (M = 341.4 g/mol), counted at 1.68616 × 10²⁴ particles. Which sample has the greater mass? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Sample 2 — each of its particles is more than twice as heavy, so half the count still gives the greater mass. KEY
Sample 1, because it contains twice as many particles.
The two samples have equal masses, because the count and the molar mass trade off exactly.
Settling it would need volume information that is not given.
MOL-030 · hard · transfer · seq 20 · authored (row 20; fresh mixed pair barium sulfide / propyne, mass-equals-M anchor) · mcq4
Sample 1 is 169.4 g of barium sulfide, BaS (M = 169.4 g/mol). Sample 2 is 120.3 g of propyne, C₃H₄ (M = 40.1 g/mol). Which sample contains more particles? (Nₐ = 6.022 × 10²³ mol⁻¹.)
Sample 2 — 3.0 mol of molecules against 1.0 mol of formula units, despite the smaller mass. KEY
Sample 1 — the far bigger mass, 169.4 g, must mean more particles.
The counts match exactly — the larger molar mass cancels out the smaller mass.
Sample 1 — a heavy formula unit simply counts for more than a light molecule.
MOL-027 · medium · application · seq 21 · authored FRQ (row 21; fresh instance) · frq
How many formula units are in 685.2 g of barium hydroxide, Ba(OH)₂? (M = 171.3 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Show every stage of your working: the values you start from, each stage's equation, the substitution, and the mole value written on its own line between the stages. Give the final count in scientific notation (for example 4.8e21) to 4 significant figures with its particle word, then add a one-line sanity check of the mole value against the lab-scale mole window you were taught.
How many formula units are in 685.2 g of barium hydroxide, Ba(OH)₂? (M = 171.3 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Show every stage of your working: the values you start from, each stage's equation, the substitution, and the mole value written on its own line between the stages. Give the final count in scientific notation (for example 4.8e21) to 4 significant figures with its particle word, then add a one-line sanity check of the mole value against the lab-scale mole window you were taught.
Part a. How many formula units are in 685.2 g of barium hydroxide, Ba(OH)₂? (M = 171.3 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Show every stage of your working: the values you start from, each stage's equation, the substitution, and the mole value written on its own line between the stages. Give the final count in scientific notation (for example 4.8e21) to 4 significant figures with its particle word, then add a one-line sanity check of the mole value against the lab-scale mole window you were taught.
Model: Values: m = 685.2 g; M = 171.3 g/mol; Avogadro's number = 6.022e23 per mole. Stage 1 (mass to moles): n = m/M = 685.2 / 171.3 = 4.000 mol. Mole checkpoint: 4.000 mol. Stage 2 (moles to formula units): N = n x Avogadro's number = 4.000 x 6.022e23 = 2.4088e24, reported as 2.409e24 formula units (4 significant figures, set by the least-precise given). Sanity check: the mole value 4.000 mol sits inside the taught lab-scale mole window of about 0.01 to 10 mol. Grading note: plain e-notation is expected (the taught entry convention); no subscripts or superscripts are required in anything the student writes.
Full model answer. Values: m = 685.2 g; M = 171.3 g/mol; Avogadro's number = 6.022e23 per mole. Stage 1 (mass to moles): n = m/M = 685.2 / 171.3 = 4.000 mol. Mole checkpoint: 4.000 mol. Stage 2 (moles to formula units): N = n x Avogadro's number = 4.000 x 6.022e23 = 2.4088e24, reported as 2.409e24 formula units (4 significant figures, set by the least-precise given). Sanity check: the mole value 4.000 mol sits inside the taught lab-scale mole window of about 0.01 to 10 mol. Grading note: plain e-notation is expected (the taught entry convention); no subscripts or superscripts are required in anything the student writes.
MOL-030 · hard · transfer · seq 22 · authored FRQ (row 22; capstone route-decision + backward execution; allocator-verified fresh instance; prod_seeds empty) · frq
A stockroom question reads: "A bottle of potassium thiosulfate, K₂S₂O₃, holds 1.98726 × 10²⁴ formula units. What is its mass?" (M = 190.3 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.)
A stockroom question reads: "A bottle of potassium thiosulfate, K₂S₂O₃, holds 1.98726 × 10²⁴ formula units. What is its mass?" (M = 190.3 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) (a) State which quantity the question gives you and which quantity it asks for, and name the two equations you will use, in order, each rearranged for its unknown. (b) Carry out the calculation, showing the working for both stages and writing the mole value on its own line between them. Give the mass in grams to 4 significant figures, and close by checking that the mole checkpoint lands within the everyday lab-scale amounts you were taught.
Part a. State which quantity the question gives you and which quantity it asks for, and name the two equations you will use, in order, each rearranged for its unknown.
Model: The question gives a particle count and asks for a mass — opposite ends of the route, so two stages through moles. Stage 1: the particle equation rearranged for n, n = N divided by Avogadro's number. Stage 2: the mole equation rearranged for m, m = n x M.
Part b. Carry out the calculation, showing the working for both stages and writing the mole value on its own line between them. Give the mass in grams to 4 significant figures, and close by checking that the mole checkpoint lands within the everyday lab-scale amounts you were taught.
Model: Stage 1: n = 1.98726e24 / 6.022e23 = 3.300 mol. Mole checkpoint: 3.300 mol. Stage 2: m = 3.300 x 190.3 = 627.99 g, reported as 628.0 g (4 significant figures). The mass of the sample is 628.0 g. The checkpoint 3.300 mol sits inside the taught range for lab-scale samples. Grading note: plain e-notation is expected; no subscripts or superscripts are required in anything the student writes.
Full model answer. (a) The question gives a particle count and asks for a mass — opposite ends of the route, so two stages through moles. Stage 1: the particle equation rearranged for n, n = N divided by Avogadro's number. Stage 2: the mole equation rearranged for m, m = n x M. (b) Stage 1: n = 1.98726e24 / 6.022e23 = 3.300 mol. Mole checkpoint: 3.300 mol. Stage 2: m = 3.300 x 190.3 = 627.99 g, reported as 628.0 g (4 significant figures). The mass of the sample is 628.0 g. The checkpoint 3.300 mol sits inside the taught range for lab-scale samples. Grading note: plain e-notation is expected; no subscripts or superscripts are required in anything the student writes.

End of Topic Test: Percent composition (u07-t3 · 5 forms × 22 items · 110 total)

David's notes for James:
full notes file
# u07-t3 — notes/queries from David's side (U7-G3)

1. CALC-HEAVY FRQ SPLIT — frq_per_form = 2, overriding the seed file's frq_per_form=1 baseline (confirm): percent composition is calc-heavy per QB-v2, and the scope carries exactly two free-response-only constructs — FRQ-1 (seq 21, MOL-035) = the formula-route FULL composition with the total-100 check (the check itself is FRQ-assessed here, per MOL-035 extraNotes: gate items assess the full set and the check), FRQ-2 (seq 22, MOL-033) = the measured-sample route incl. the rest-is-the-other-element move. Full adjudication in blueprint.composition.
2. SEQ-10 MODALITY (confirm): the numeric-ENTRY row frames the full percent breakdown and collects ONE element's percent, exactly per MOL-035 extraNotes ('Fluency numeric items frame the full-composition task and collect one element's percent'); the total-100 deliverable is graded by FRQ-1 part (c). Both blind rounds and sol loop 1 flagged the framing before the extraNotes citation was added to gradingNote — adjudicated to the lesson note; the stem's total-100 line is worded as a self-check.
3. SEQ-18 DIAGNOSTIC KEY (confirm the rule-line intent): the item keys 'the ratio was inverted' from a reported >100% result, per MOL-032 rule line 6 verbatim ('Every percent must land between 0 and 100 — a result over 100 means the ratio was inverted'). A blind reviewer flagged this as logically overstrong (other errors could exceed 100); course-local definitions held. If the rule line is meant softer, say so and we re-key the wording.
4. PRECISION CONVENTIONS (confirm): percents at 3 significant figures per your fluency convention; scaled masses in seq 7 and seq 17 are exact whole-number multiples of the stated 1-dp per-100 share, and the stems now demand 'masses to 1 decimal place' (sol read 539.1-g-style values as overprecise vs a 3-sf percent; the stated 1-dp demand resolves it). The idiom is a recorded gate_config task-idiom adjudication.
5. SEQ-17 SHARE COUNTS: every form scales the per-100 reading by a WHOLE number of 100 g shares (4/11/14/10/13), extending your MOL-031 gate seed's 'half that share' meaning-level precedent — sol ruled a 4.5x count a calculation routine outside MOL-031's fence.
6. A7 TAUGHT CONSTANTS: atomic masses, molar masses and the rule numbers 100/99.9/100.1 recur across forms by necessity (MOL-032/035 extraNotes bind every calc stem to supply them); every other number is engineered globally unique across all 110 items. The recorded, context-verified family lives in gate_config.json.
7. SYMBOL CONVENTION carried from U7-G2: target elements are referenced by symbol in calc stems/options; spelled element names are scope material tokens on the quant lessons.

Form A

MOL-031 · easy · recall · seq 1 · seedless authored row (per-100 reading case family; fluency-style read-correctly item) · mcq4
Barium chloride is 65.9% barium by mass. Which option below reads it correctly?
Every 100 g of barium chloride contains 65.9 g of barium. KEY
65.9% of the atoms in barium chloride are barium atoms.
Any sample of barium chloride contains 65.9 g of barium.
65.9% of the volume of any barium chloride sample is barium.
MOL-032 · easy · recall · seq 2 · seed:U7-G3.pool[5] (MOL-032 gateItems[2], H2O2 setup) — instance freshened (H2O2 reserved for seq 16; own-lesson element words dropped for symbols) · mcq4
A student must find the percent by mass of Li in lithium oxide, Li₂O. (atomic masses: Li 6.9; O 16.0; M of Li₂O = 29.8 g/mol.) Which setup is correct?
(2 × 6.9 ÷ 29.8) × 100 KEY
(6.9 ÷ 29.8) × 100
(29.8 ÷ (2 × 6.9)) × 100
(2 × 6.9 ÷ 16.0) × 100
MOL-034 · easy · recall · seq 3 · seed:U7-G3.pool[11] (MOL-034 gateItems[2]) — kept verbatim (instance-free; mechanism key is the taught rule line) · mcq4
Why does a pure compound show the same percent composition in every sample?
Because every molecule or formula unit of a compound contains exactly the same atoms, so the mass ratio is the same in any sample. KEY
Because chemists manufacture compounds so carefully that every batch leaving a factory matches every other batch exactly.
Because every sample of a compound is measured on the same kind of balance, and matching instruments give matching ratios.
Because the atoms in a larger sample adjust their individual masses until the overall ratio settles at the usual value.
MOL-031 · easy · application · seq 4 · seed:U7-G3.pool[0] (MOL-031 gateItems[0], fertilizer 35.0%/100 kg) — instance freshened (35.0 = MOL-032 fluency corpus; farmer/fertilizer scenario = MOL-031 hook corpus) · mcq4
A supplier sells a 100 kg drum of magnesium oxide, which is 60.3% Mg by mass. How much Mg does the drum contain?
60.3 kg of Mg — the percent is the share in every 100 units of the total mass. KEY
60.3 g of Mg — percent composition is always stated in grams.
100 kg of Mg — the compound is bought for its Mg content.
39.7 kg of Mg — the stated percent is the share of everything else.
MOL-032 · medium · application · seq 5 · seed:U7-G3.pool[3] (MOL-032 gateItems[0], NH4NO3 %O) — converted to numeric ENTRY per convert_to_numeric_entry + instance freshened (NH4NO3/60.0/80.0/48.0 = MOL-033 fluency corpus) · numeric
What is the percent by mass of Al in aluminum oxide, Al₂O₃? (atomic masses: Al 27.0; O 16.0; M of Al₂O₃ = 102.0 g/mol.) Then give your answer as a percent to 3 significant figures.
Answer: 52.9 % KEY
also accepted: 52.9%
wrongValueDiagnostics
[
 {
  "value": 26.5,
  "misconception": "you forgot to multiply by the subscript — the formula carries 2 Al atoms",
  "feedback": "This answer is wrong — you forgot to multiply by the subscript — the formula carries 2 Al atoms."
 },
 {
  "value": 189.0,
  "misconception": "you inverted the ratio — a percent over 100 is impossible",
  "feedback": "This answer is wrong — you inverted the ratio — a percent over 100 is impossible."
 },
 {
  "value": 0.5294,
  "misconception": "you forgot to multiply by 100",
  "feedback": "This answer is wrong — you forgot to multiply by 100."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of Al in one mole = 2 × 27.0 = 54.0 g\nM = 102.0 g/mol\nWrite down the equation:\n% Al = (mass of Al in one mole ÷ M) × 100\nSubstitute in the values, and calculate:\n% Al = (54.0 ÷ 102.0) × 100\n% Al = 52.9%"
}
MOL-033 · medium · application · seq 6 · seed:U7-G3.pool[7] (MOL-033 gateItems[1], ammonia 14/17) — converted to numeric ENTRY + instance freshened (82.4 = MOL-031 fluency corpus; ammonia = scope material) · numeric
A 40.7 g sample of zinc oxide is analyzed and found to contain 32.7 g of Zn. What percent of the sample's mass is Zn? Give your answer as a percent to 3 significant figures.
Answer: 80.3 % KEY
also accepted: 80.3%
wrongValueDiagnostics
[
 {
  "value": 124.0,
  "misconception": "you inverted the ratio — a percent over 100 is impossible",
  "feedback": "This answer is wrong — you inverted the ratio — a percent over 100 is impossible."
 },
 {
  "value": 0.8034,
  "misconception": "you forgot to multiply by 100",
  "feedback": "This answer is wrong — you forgot to multiply by 100."
 },
 {
  "value": 19.7,
  "misconception": "you found the other element's share instead of Zn's",
  "feedback": "This answer is wrong — you found the other element's share instead of Zn's."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of Zn in the sample = 32.7 g\nmass of the sample = 40.7 g\nWrite down the equation:\n% Zn = (mass of Zn in the sample ÷ mass of the sample) × 100\nSubstitute in the values, and calculate:\n% Zn = (32.7 ÷ 40.7) × 100\n% Zn = 80.3%"
}
MOL-034 · medium · application · seq 7 · seed:U7-G3.pool[10] (MOL-034 gateItems[1], ammonia 82.4% scaling) — instance freshened (82.4 = MOL-031 fluency corpus; ammonia = scope material) · mcq4
Methanol is 37.5% C by mass, so 100.0 g of methanol contains 37.5 g of C. Reporting masses to 1 decimal place, what does a 300.0 g sample of methanol contain?
112.5 g of C, and the sample is still 37.5% C. KEY
112.5 g of C, and the sample is now 112.5% C.
37.5 g of C, because a compound always holds the same mass of each element.
37.5 g of C, and the sample is now 12.5% C.
MOL-031 · medium · application · seq 8 · seed:U7-G3.pool[1] (MOL-031 gateItems[1], ethylene 85.7/14.3) — instance freshened (stem spelled carbon/hydrogen = own-lesson corpus materials) · mcq4
The compound BaBr₂ contains only two elements, Ba and Br. It is 46.2% Ba by mass. What percent of its mass is Br?
53.8% KEY
46.2%
7.60%
It cannot be found without the atomic masses.
MOL-033 · medium · application · seq 9 · seed:U7-G3.pool[6] (MOL-033 gateItems[0], ethane 16/4) — converted to numeric ENTRY + instance freshened (16.0/20.0 = WE corpus; ethane/carbon/hydrogen words) · numeric
A 39.9 g sample of iron(III) oxide is analyzed: it contains 12.0 g of O and 27.9 g of Fe. What percent of the sample's mass is Fe? Give your answer as a percent to 3 significant figures.
Answer: 69.9 % KEY
also accepted: 69.9%
wrongValueDiagnostics
[
 {
  "value": 232.5,
  "misconception": "you divided the Fe mass by the other element's mass instead of by the whole sample's mass",
  "feedback": "This answer is wrong — you divided the Fe mass by the other element's mass instead of by the whole sample's mass."
 },
 {
  "value": 30.1,
  "misconception": "you computed O's share instead of Fe's",
  "feedback": "This answer is wrong — you computed O's share instead of Fe's."
 },
 {
  "value": 0.6992,
  "misconception": "you forgot to multiply by 100",
  "feedback": "This answer is wrong — you forgot to multiply by 100."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of Fe in the sample = 27.9 g\nmass of the sample = 39.9 g\nWrite down the equation:\n% Fe = (mass of Fe in the sample ÷ mass of the sample) × 100\nSubstitute in the values, and calculate:\n% Fe = (27.9 ÷ 39.9) × 100\n% Fe = 69.9%\nThe divisor is always the whole sample's 39.9 g — dividing by the O mass gives an element-to-element ratio, not a percent of the compound."
}
MOL-035 · medium · application · seq 10 · authored (numeric ENTRY capstone row; fluency collect-one-element modality) · numeric
Work out the full percent breakdown of calcium nitrate, Ca(NO₃)₂ — as a self-check, your percents should total 100. (atomic masses: Ca 40.1; N 14.0; O 16.0; M of Ca(NO₃)₂ = 164.1 g/mol.) Enter your percent for O, to 3 significant figures.
Answer: 58.5 % KEY
also accepted: 58.5%
wrongValueDiagnostics
[
 {
  "value": 9.75,
  "misconception": "you forgot to multiply by the subscript — the formula carries 6 O atoms",
  "feedback": "This answer is wrong — you forgot to multiply by the subscript — the formula carries 6 O atoms."
 },
 {
  "value": 171.0,
  "misconception": "you inverted the ratio — a percent over 100 is impossible",
  "feedback": "This answer is wrong — you inverted the ratio — a percent over 100 is impossible."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of Ca in one mole = 1 × 40.1 = 40.1 g\nmass of N in one mole = 2 × 14.0 = 28.0 g\nmass of O in one mole = 6 × 16.0 = 96.0 g\nM = 164.1 g/mol\nWrite down the equation, and run it once per element:\n% Ca = (40.1 ÷ 164.1) × 100 = 24.4%\n% N = (28.0 ÷ 164.1) × 100 = 17.1%\n% O = (96.0 ÷ 164.1) × 100 = 58.5%\nCheck the total: 100.0 ✓"
}
MOL-032 · medium · application · seq 11 · seed:U7-G3.pool[4] (MOL-032 gateItems[1], acetic acid %O) — converted to numeric ENTRY + instance freshened (53.3 and M 60.0 = MOL-035 fluency corpus) · numeric
What is the percent by mass of S in sulfur dioxide, SO₂? (atomic masses: S 32.1; O 16.0; M of SO₂ = 64.1 g/mol.) Report your answer as a percent, to 3 significant figures.
Answer: 50.1 % KEY
also accepted: 50.1%
wrongValueDiagnostics
[
 {
  "value": 100.0,
  "misconception": "you divided by the other elements' combined mass instead of the whole compound's molar mass",
  "feedback": "This answer is wrong — you divided by the other elements' combined mass instead of the whole compound's molar mass."
 },
 {
  "value": 200.0,
  "misconception": "you inverted the ratio — a percent over 100 is impossible",
  "feedback": "This answer is wrong — you inverted the ratio — a percent over 100 is impossible."
 },
 {
  "value": 0.5008,
  "misconception": "you forgot to multiply by 100",
  "feedback": "This answer is wrong — you forgot to multiply by 100."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of S in one mole = 1 × 32.1 = 32.1 g\nM = 64.1 g/mol\nWrite down the equation:\n% S = (mass of S in one mole ÷ M) × 100\nSubstitute in the values, and calculate:\n% S = (32.1 ÷ 64.1) × 100\n% S = 50.1%"
}
MOL-034 · medium · application · seq 12 · authored (two-sources fixedness application; fresh instances — fluency CO2/water family is student-seen) · mcq4
Two shipments of pure strontium chloride come from two different suppliers. How do their percents of Sr by mass compare?
They are identical, because a pure compound's formula fixes the mass share of every element, no matter where or how the compound was made. KEY
They differ slightly, because where and how a compound was made leaves a lasting trace in its makeup that analysis can detect.
The larger shipment has the higher percent of Sr, because it simply contains more Sr than the smaller one does.
Neither percent is knowable until each has been through its own laboratory analysis, since only a measurement settles a composition.
MOL-033 · medium · application · seq 13 · seed:U7-G3.pool[8] (MOL-033 gateItems[2], glucose rest-is-oxygen) — converted to numeric ENTRY + instance freshened (glucose = own-lesson fluency material; 53.3 corpus) · numeric
A 177.8 g sample of manganese(II) hydroxide is broken down: it holds 109.8 g of Mn plus 64.0 g of O; everything else is H. What percent of the sample's mass is H? Give your answer as a percent to 3 significant figures.
Answer: 2.25 % KEY
also accepted: 2.25%
wrongValueDiagnostics
[
 {
  "value": 97.8,
  "misconception": "you computed the combined share of Mn and O — the question asks for the H that makes up the rest",
  "feedback": "This answer is wrong — you computed the combined share of Mn and O — the question asks for the H that makes up the rest."
 },
 {
  "value": 61.8,
  "misconception": "you computed Mn's share instead of H's",
  "feedback": "This answer is wrong — you computed Mn's share instead of H's."
 },
 {
  "value": 4.0,
  "misconception": "you gave the H mass in grams as a percent without dividing by the sample's mass",
  "feedback": "This answer is wrong — you gave the H mass in grams as a percent without dividing by the sample's mass."
 }
]
feedback
{
 "keyExplanation": "The element masses add up to the sample's mass, so the H is 177.8 − 109.8 − 64.0 = 4.0 g.\nWrite down the equation:\n% H = (mass of H in the sample ÷ mass of the sample) × 100\nSubstitute in the values, and calculate:\n% H = (4.0 ÷ 177.8) × 100\n% H = 2.25%"
}
MOL-035 · medium · application · seq 14 · seed:U7-G3.pool[12]+demo_curated[0] (MOL-035, CO2 full composition) — instance freshened (27.3 = MOL-034 fluency corpus; carbon/oxygen words = own-lesson materials) · mcq4
What is the full percent composition of CrCl₃? (atomic masses: Cr 52.0; Cl 35.5; M of CrCl₃ = 158.5 g/mol)
32.8% Cr and 67.2% Cl KEY
67.2% Cr and 32.8% Cl
42.3% Cr and 57.7% Cl
59.4% Cr and 40.6% Cl
MOL-033 · medium · application · seq 15 · authored (which-division discrimination; the lesson's hammered misconception) · mcq4
A 286.8 g sample of AgCl is taken apart: 215.8 g is Ag and 71.0 g is Cl. Which calculation gives the percent of the sample's mass that comes from Cl?
(71.0 ÷ 286.8) × 100 KEY
(71.0 ÷ 215.8) × 100
(215.8 ÷ 286.8) × 100
(286.8 ÷ 71.0) × 100
MOL-035 · medium · application · seq 16 · seed:U7-G3.pool[14]+demo_curated[2] (MOL-035, H2O2 full composition, equal-count trap) — instance freshened (hydrogen/oxygen words = own-lesson materials) · mcq4
Give the full percent composition of MnO. (atomic masses: Mn 54.9; O 16.0; M of MnO = 70.9 g/mol)
77.4% Mn and 22.6% O KEY
22.6% Mn and 77.4% O
Half the mass from Mn and half from O, because the formula holds one atom of each.
54.9% Mn and 16.0% O
MOL-031 · hard · transfer · seq 17 · seed:U7-G3.pool[2] (MOL-031 gateItems[2], ethane 50 g claim) — instance freshened (20.0 = MOL-035 WE corpus; 50 = MOL-031 hook corpus; ethane/hydrogen words) · mcq4
A student says: '400.0 g of pentane contains 16.7 g of H, because pentane is 16.7% H by mass.' What mistake did the student make? (Report masses to 1 decimal place.)
Percent by mass is grams per 100 g: 400.0 g is four such shares, so the sample contains 4 × 16.7 = 66.8 g of H. KEY
Nothing is wrong — a stated percent by mass gives the grams of the element in any sample, whatever the sample's size.
The claim fails only because the H percent of pentane changes with the sample's size, rising as the sample grows.
The claim fails because a percent can never be turned into a mass — the two kinds of quantity cannot be linked.
MOL-032 · hard · transfer · seq 18 · authored (bounds-rule diagnosis; rule line 6 of MOL-032) · mcq4
A student computes the mass percent of Pb in Pb₃O₄ (atomic masses: Pb 207.2; O 16.0; M of Pb₃O₄ = 685.6 g/mol) and reports 110.3%. Without seeing any of the work, which statement must be true?
The ratio was inverted — every percent must land between 0 and 100, and only dividing M by the element's contribution can push the result past 100. KEY
The student forgot to multiply by the subscript when finding the element's contribution to one mole of the compound, before dividing by M.
The student forgot to multiply by 100 at the end, leaving the answer as a bare ratio instead of a percent of the whole.
Nothing is certainly wrong — Pb₃O₄ could simply be very rich in Pb, and a compound's richest element can pass 100.
MOL-034 · hard · transfer · seq 19 · seed:U7-G3.pool[9] (MOL-034 gateItems[0], doubling claim) — kept (instance-free); case rotates across forms · mcq4
A student claims: 'If you double a sample of a pure compound, you also double the percent of each element in it.' Which correction fixes the claim?
The action doubles each element's grams but leaves each percent unchanged — the grams all scale together, so no share of the total can shift. KEY
The claim is right for the heaviest element only — its percent shifts as the sample changes while the lighter elements' percents stay fixed.
The claim is right for solids but wrong for liquids and gases, whose particles rearrange too freely to hold one fixed ratio.
The percents stay the same only when the changed sample comes from the same batch as the original, made in the same production run.
MOL-035 · hard · transfer · seq 20 · seed:U7-G3.pool[13]+demo_curated[1] (MOL-035, 110-total error-spotting) — numbers freshened (40.0/6.7 = own-lesson fluency corpus; element words dropped); verdict case rotates · mcq4
For a compound of three elements, a student calculates the percents as 52.6%, 13.9%, and 43.5%. Without redoing the arithmetic, what can be concluded?
The percents total 110.0, and the percents of all the elements in a compound must total 100 — one of the calculations carries an error. KEY
Nothing shows an error — each of the three percents lies between 0 and 100, so the set is acceptable as it stands.
The largest percent must belong to the heaviest element, and without the formula nothing confirms that it does here.
The smallest percent, 13.9%, is too small to be genuine — no element contributes that little.
MOL-035 · hard · application · seq 21 · authored (FRQ#1, formula-route capstone; 5-line scaffold per element; fresh instance per form) · frq
Consider H₃PO₄, known as phosphoric acid. (atomic masses: H 1.0; P 31.0; O 16.0; M of H₃PO₄ = 98.0 g/mol.)
Consider H₃PO₄, known as phosphoric acid. (atomic masses: H 1.0; P 31.0; O 16.0; M of H₃PO₄ = 98.0 g/mol.) (a) For each element in the formula, state the mass that element contributes to one mole of the compound. (b) Work through the percent share of each element, showing your working. Give each percent to 3 significant figures. (c) Check your answer by totaling the percents, state the total, and say what the total tells you about your calculations.
Part a. For each element in the formula, state the mass that element contributes to one mole of the compound.
Model: mass of H in one mole = 3 × 1.0 = 3.0 g mass of P in one mole = 1 × 31.0 = 31.0 g mass of O in one mole = 4 × 16.0 = 64.0 g M = 98.0 g/mol
Part b. Work through the percent share of each element, showing your working. Give each percent to 3 significant figures.
Model: Run the equation once per element: % H = (mass of H in one mole ÷ M) × 100 % H = (3.0 ÷ 98.0) × 100 % H = 3.06% % P = (mass of P in one mole ÷ M) × 100 % P = (31.0 ÷ 98.0) × 100 % P = 31.6% % O = (mass of O in one mole ÷ M) × 100 % O = (64.0 ÷ 98.0) × 100 % O = 65.3%
Part c. Check your answer by totaling the percents, state the total, and say what the total tells you about your calculations.
Model: Check the total: 3.06 + 31.6 + 65.3 = 99.96 The percents of all the elements in a compound must total 100, and small rounding differences (totals from 99.9 to 100.1) are acceptable — a total of 99.96 passes the check.
Full model answer. (a) mass of H in one mole = 3 × 1.0 = 3.0 g mass of P in one mole = 1 × 31.0 = 31.0 g mass of O in one mole = 4 × 16.0 = 64.0 g M = 98.0 g/mol (b) Run the equation once per element: % H = (mass of H in one mole ÷ M) × 100 % H = (3.0 ÷ 98.0) × 100 % H = 3.06% % P = (mass of P in one mole ÷ M) × 100 % P = (31.0 ÷ 98.0) × 100 % P = 31.6% % O = (mass of O in one mole ÷ M) × 100 % O = (64.0 ÷ 98.0) × 100 % O = 65.3% (c) Check the total: 3.06 + 31.6 + 65.3 = 99.96 The percents of all the elements in a compound must total 100, and small rounding differences (totals from 99.9 to 100.1) are acceptable — a total of 99.96 passes the check.
MOL-033 · hard · application · seq 22 · authored (FRQ#2, measured-sample route; real composition data, identity unnamed; fresh instance per form) · frq
A 207.3 g sample of an unidentified pure compound is broken apart in an analysis. The analysis finds 117.3 g of K and 18.0 g of C; the rest of the sample's mass is the third element, O.
A 207.3 g sample of an unidentified pure compound is broken apart in an analysis. The analysis finds 117.3 g of K and 18.0 g of C; the rest of the sample's mass is the third element, O. (a) Find the mass of O in the sample, and state the fact about the element masses that lets you find it. (b) Calculate what percent of the sample's mass the O supplies. Give the percent to 3 significant figures, and show every step of your working. (c) Calculate the other two percents, total all three, and state what the total shows about your work.
Part a. Find the mass of O in the sample, and state the fact about the element masses that lets you find it.
Model: The element masses in a sample add up to the whole sample's mass. mass of O = 207.3 − 117.3 − 18.0 = 72.0 g
Part b. Calculate what percent of the sample's mass the O supplies. Give the percent to 3 significant figures, and show every step of your working.
Model: mass of O in the sample = 72.0 g mass of the sample = 207.3 g % O = (mass of O in the sample ÷ mass of the sample) × 100 % O = (72.0 ÷ 207.3) × 100 % O = 34.7%
Part c. Calculate the other two percents, total all three, and state what the total shows about your work.
Model: % K = (117.3 ÷ 207.3) × 100 = 56.6% % C = (18.0 ÷ 207.3) × 100 = 8.68% Check the total: 34.7 + 56.6 + 8.68 = 100.0 The percents of all the elements in a compound must total 100 (small rounding differences like 99.9 or 100.1 are acceptable), so the set passes the check — every share was taken of the whole sample.
Full model answer. (a) The element masses in a sample add up to the whole sample's mass. mass of O = 207.3 − 117.3 − 18.0 = 72.0 g (b) mass of O in the sample = 72.0 g mass of the sample = 207.3 g % O = (mass of O in the sample ÷ mass of the sample) × 100 % O = (72.0 ÷ 207.3) × 100 % O = 34.7% (c) % K = (117.3 ÷ 207.3) × 100 = 56.6% % C = (18.0 ÷ 207.3) × 100 = 8.68% Check the total: 34.7 + 56.6 + 8.68 = 100.0 The percents of all the elements in a compound must total 100 (small rounding differences like 99.9 or 100.1 are acceptable), so the set passes the check — every share was taken of the whole sample.

Form B

MOL-031 · easy · recall · seq 1 · authored (parallel reading slot; formula-and-symbol instance per the naming convention) · mcq4
The compound BaF₂ is 78.3% Ba by mass. Which option below reads it correctly?
Every 100 g of BaF₂ contains 78.3 g of Ba. KEY
78.3% of the atoms in BaF₂ are Ba atoms.
Any sample of BaF₂ contains 78.3 g of Ba.
78.3% of the volume of any BaF₂ sample is Ba.
MOL-032 · easy · recall · seq 2 · authored (parallel to seq-2 seed structure) · mcq4
A student must find the percent by mass of Li in Li₂S. (atomic masses: Li 6.9; S 32.1; M of Li₂S = 45.9 g/mol.) Which setup is correct?
(2 × 6.9 ÷ 45.9) × 100 KEY
(6.9 ÷ 45.9) × 100
(45.9 ÷ (2 × 6.9)) × 100
(2 × 6.9 ÷ 32.1) × 100
MOL-034 · easy · recall · seq 3 · authored (fresh framing; same verbatim mechanism key) · mcq4
Two samples of one pure compound differ greatly in mass, yet their percent compositions match. Which statement explains why?
Because every molecule or formula unit of a compound contains exactly the same atoms, so the mass ratio is the same in any sample. KEY
Because both samples were drawn from one well-mixed container, and thorough mixing spreads the elements evenly through both.
Because the mass differences cancel out when the percents are calculated, hiding whatever gap the samples started with.
Because large and small samples are weighed with the same units, and shared units force the two ratios into agreement.
MOL-031 · easy · application · seq 4 · authored (parallel to seq-4 seed structure) · mcq4
A supplier sells a 100 g jar of the compound MgCl₂, which is 25.5% Mg by mass. How much Mg does the jar contain?
25.5 g of Mg — the percent is the share in every 100 units of the total mass. KEY
25.5 kg of Mg — percent shares are always quoted per 100 kg.
100 g of Mg — the compound is bought for its Mg content.
74.5 g of Mg — the stated percent is the share of everything else.
MOL-032 · medium · application · seq 5 · authored (parallel numeric slot) · numeric
What is the percent by mass of F in AlF₃? (atomic masses: Al 27.0; F 19.0; M of AlF₃ = 84.0 g/mol.) Then give your answer as a percent to 3 significant figures.
Answer: 67.9 % KEY
also accepted: 67.9%
wrongValueDiagnostics
[
 {
  "value": 22.6,
  "misconception": "you forgot to multiply by the subscript — the formula carries 3 F atoms",
  "feedback": "This answer is wrong — you forgot to multiply by the subscript — the formula carries 3 F atoms."
 },
 {
  "value": 147.0,
  "misconception": "you inverted the ratio — a percent over 100 is impossible",
  "feedback": "This answer is wrong — you inverted the ratio — a percent over 100 is impossible."
 },
 {
  "value": 0.6786,
  "misconception": "you forgot to multiply by 100",
  "feedback": "This answer is wrong — you forgot to multiply by 100."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of F in one mole = 3 × 19.0 = 57.0 g\nM = 84.0 g/mol\nWrite down the equation:\n% F = (mass of F in one mole ÷ M) × 100\nSubstitute in the values, and calculate:\n% F = (57.0 ÷ 84.0) × 100\n% F = 67.9%"
}
MOL-033 · medium · application · seq 6 · authored (parallel numeric slot) · numeric
A 195.0 g sample of the compound ZnS is analyzed and found to contain 130.8 g of Zn. What percent of the sample's mass is Zn? Give your answer as a percent to 3 significant figures.
Answer: 67.1 % KEY
also accepted: 67.1%
wrongValueDiagnostics
[
 {
  "value": 149.0,
  "misconception": "you inverted the ratio — a percent over 100 is impossible",
  "feedback": "This answer is wrong — you inverted the ratio — a percent over 100 is impossible."
 },
 {
  "value": 0.6708,
  "misconception": "you forgot to multiply by 100",
  "feedback": "This answer is wrong — you forgot to multiply by 100."
 },
 {
  "value": 32.9,
  "misconception": "you found the other element's share instead of Zn's",
  "feedback": "This answer is wrong — you found the other element's share instead of Zn's."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of Zn in the sample = 130.8 g\nmass of the sample = 195.0 g\nWrite down the equation:\n% Zn = (mass of Zn in the sample ÷ mass of the sample) × 100\nSubstitute in the values, and calculate:\n% Zn = (130.8 ÷ 195.0) × 100\n% Zn = 67.1%"
}
MOL-034 · medium · application · seq 7 · authored (parallel scaling slot) · mcq4
Titanium dioxide is 59.9% Ti by mass, so 100.0 g of titanium dioxide contains 59.9 g of Ti. Reporting masses to 1 decimal place, what does a 900.0 g sample of titanium dioxide contain?
539.1 g of Ti, and the sample is still 59.9% Ti. KEY
539.1 g of Ti, and the sample is now 539.1% Ti.
59.9 g of Ti, because a compound always holds the same mass of each element.
59.9 g of Ti, and the sample is now 6.66% Ti.
MOL-031 · medium · application · seq 8 · authored (parallel leftover-share slot) · mcq4
The compound SrBr₂ contains only two elements, Sr and Br. It is 35.4% Sr by mass. What percent of its mass is Br?
64.6% KEY
35.4%
29.2%
It cannot be found without the atomic masses.
MOL-033 · medium · application · seq 9 · authored (parallel numeric slot) · numeric
A 175.8 g sample of FeS is analyzed: it contains 64.2 g of S and 111.6 g of Fe. What percent of the sample's mass is Fe? Give your answer as a percent to 3 significant figures.
Answer: 63.5 % KEY
also accepted: 63.5%
wrongValueDiagnostics
[
 {
  "value": 174.0,
  "misconception": "you divided the Fe mass by the other element's mass instead of by the whole sample's mass",
  "feedback": "This answer is wrong — you divided the Fe mass by the other element's mass instead of by the whole sample's mass."
 },
 {
  "value": 36.5,
  "misconception": "you computed S's share instead of Fe's",
  "feedback": "This answer is wrong — you computed S's share instead of Fe's."
 },
 {
  "value": 0.6348,
  "misconception": "you forgot to multiply by 100",
  "feedback": "This answer is wrong — you forgot to multiply by 100."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of Fe in the sample = 111.6 g\nmass of the sample = 175.8 g\nWrite down the equation:\n% Fe = (mass of Fe in the sample ÷ mass of the sample) × 100\nSubstitute in the values, and calculate:\n% Fe = (111.6 ÷ 175.8) × 100\n% Fe = 63.5%\nThe divisor is always the whole sample's 175.8 g — dividing by the S mass gives an element-to-element ratio, not a percent of the compound."
}
MOL-035 · medium · application · seq 10 · authored (numeric ENTRY capstone row; fluency collect-one-element modality) · numeric
Work out the full percent breakdown of CaSO₄ — as a self-check, your percents should total 100. (atomic masses: Ca 40.1; S 32.1; O 16.0; M of CaSO₄ = 136.2 g/mol.) Enter your percent for O, to 3 significant figures.
Answer: 47.0 % KEY
also accepted: 47.0%
wrongValueDiagnostics
[
 {
  "value": 11.7,
  "misconception": "you forgot to multiply by the subscript — the formula carries 4 O atoms",
  "feedback": "This answer is wrong — you forgot to multiply by the subscript — the formula carries 4 O atoms."
 },
 {
  "value": 213.0,
  "misconception": "you inverted the ratio — a percent over 100 is impossible",
  "feedback": "This answer is wrong — you inverted the ratio — a percent over 100 is impossible."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of Ca in one mole = 1 × 40.1 = 40.1 g\nmass of S in one mole = 1 × 32.1 = 32.1 g\nmass of O in one mole = 4 × 16.0 = 64.0 g\nM = 136.2 g/mol\nWrite down the equation, and run it once per element:\n% Ca = (40.1 ÷ 136.2) × 100 = 29.4%\n% S = (32.1 ÷ 136.2) × 100 = 23.6%\n% O = (64.0 ÷ 136.2) × 100 = 47.0%\nCheck the total: 100.0 ✓"
}
MOL-032 · medium · application · seq 11 · authored (parallel numeric slot) · numeric
What is the percent by mass of O in sulfur trioxide, SO₃? (atomic masses: S 32.1; O 16.0; M of SO₃ = 80.1 g/mol.) Report your answer as a percent, to 3 significant figures.
Answer: 59.9 % KEY
also accepted: 59.9%
wrongValueDiagnostics
[
 {
  "value": 20.0,
  "misconception": "you forgot to multiply by the subscript — the formula carries 3 O atoms",
  "feedback": "This answer is wrong — you forgot to multiply by the subscript — the formula carries 3 O atoms."
 },
 {
  "value": 167.0,
  "misconception": "you inverted the ratio — a percent over 100 is impossible",
  "feedback": "This answer is wrong — you inverted the ratio — a percent over 100 is impossible."
 },
 {
  "value": 0.5993,
  "misconception": "you forgot to multiply by 100",
  "feedback": "This answer is wrong — you forgot to multiply by 100."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of O in one mole = 3 × 16.0 = 48.0 g\nM = 80.1 g/mol\nWrite down the equation:\n% O = (mass of O in one mole ÷ M) × 100\nSubstitute in the values, and calculate:\n% O = (48.0 ÷ 80.1) × 100\n% O = 59.9%"
}
MOL-034 · medium · application · seq 12 · authored (two-sources fixedness application; fresh instances — fluency CO2/water family is student-seen) · mcq4
Two batches of pure gold(III) chloride come from two different refineries. How do their percents of Au by mass compare?
They are identical, because a pure compound's formula fixes the mass share of every element, no matter where or how the compound was made. KEY
They differ slightly, because where and how a compound was made leaves a lasting trace in its makeup that analysis can detect.
The larger batch has the higher percent of Au, because it simply contains more Au than the smaller one does.
Neither percent is knowable until each has been through its own laboratory analysis, since only a measurement settles a composition.
MOL-033 · medium · application · seq 13 · authored (parallel numeric slot) · numeric
A 453.0 g sample of MnSO₄ is broken down: it holds 164.7 g of Mn plus 96.3 g of S; everything else is O. What percent of the sample's mass is O? Give your answer as a percent to 3 significant figures.
Answer: 42.4 % KEY
also accepted: 42.4%
wrongValueDiagnostics
[
 {
  "value": 57.6,
  "misconception": "you computed the combined share of Mn and S — the question asks for the O that makes up the rest",
  "feedback": "This answer is wrong — you computed the combined share of Mn and S — the question asks for the O that makes up the rest."
 },
 {
  "value": 36.4,
  "misconception": "you computed Mn's share instead of O's",
  "feedback": "This answer is wrong — you computed Mn's share instead of O's."
 },
 {
  "value": 192.0,
  "misconception": "you gave the O mass in grams as a percent without dividing by the sample's mass",
  "feedback": "This answer is wrong — you gave the O mass in grams as a percent without dividing by the sample's mass."
 }
]
feedback
{
 "keyExplanation": "The element masses add up to the sample's mass, so the O is 453.0 − 164.7 − 96.3 = 192.0 g.\nWrite down the equation:\n% O = (mass of O in the sample ÷ mass of the sample) × 100\nSubstitute in the values, and calculate:\n% O = (192.0 ÷ 453.0) × 100\n% O = 42.4%"
}
MOL-035 · medium · application · seq 14 · authored (parallel full-set slot) · mcq4
What is the full percent composition of SiBr₄? (atomic masses: Si 28.1; Br 79.9; M of SiBr₄ = 347.7 g/mol)
8.08% Si and 91.9% Br KEY
91.9% Si and 8.08% Br
15.0% Si and 85.0% Br
26.0% Si and 74.0% Br
MOL-033 · medium · application · seq 15 · authored (which-division discrimination; the lesson's hammered misconception) · mcq4
A 563.4 g sample of AgBr is taken apart: 323.7 g is Ag and 239.7 g is Br. Which calculation gives the percent of the sample's mass that comes from Br?
(239.7 ÷ 563.4) × 100 KEY
(239.7 ÷ 323.7) × 100
(323.7 ÷ 563.4) × 100
(563.4 ÷ 239.7) × 100
MOL-035 · medium · application · seq 16 · authored (parallel full-set slot; atom-count-trap ratio rotates) · mcq4
Give the full percent composition of SiO₂. (atomic masses: Si 28.1; O 16.0; M of SiO₂ = 60.1 g/mol)
46.8% Si and 53.2% O KEY
53.2% Si and 46.8% O
A third of the mass from Si and two thirds from O, because the formula holds one Si atom for every two O atoms.
23.4% Si and 76.6% O
MOL-031 · hard · transfer · seq 17 · authored (parallel claim-repair slot) · mcq4
A student says: '1100.0 g of hexane contains 16.3 g of H, because hexane is 16.3% H by mass.' What mistake did the student make? (Report masses to 1 decimal place.)
Percent by mass is grams per 100 g: 1100.0 g is eleven such shares, so the sample contains 11 × 16.3 = 179.3 g of H. KEY
Nothing is wrong — a stated percent by mass gives the grams of the element in any sample, whatever the sample's size.
The claim fails only because the H percent of hexane changes with the sample's size, rising as the sample grows.
The claim fails because a percent can never be turned into a mass — the two kinds of quantity cannot be linked.
MOL-032 · hard · transfer · seq 18 · authored (bounds-rule diagnosis; rule line 6 of MOL-032) · mcq4
A student computes the mass percent of I in MgI₂ (atomic masses: Mg 24.3; I 126.9; M of MgI₂ = 278.1 g/mol) and reports 109.6%. Without seeing any of the work, which statement must be true?
The ratio was inverted — every percent must land between 0 and 100, and only dividing M by the element's contribution can push the result past 100. KEY
The student forgot to multiply by the subscript when finding the element's contribution to one mole of the compound, before dividing by M.
The student forgot to multiply by 100 at the end, leaving the answer as a bare ratio instead of a percent of the whole.
Nothing is certainly wrong — MgI₂ could simply be very rich in I, and a compound's richest element can pass 100.
MOL-034 · hard · transfer · seq 19 · authored (scaling-claim rotation per blueprint case_target) · mcq4
A student claims: 'If you triple a sample of a pure compound, you also triple the percent of each element in it.' Which correction fixes the claim?
The action triples each element's grams but leaves each percent unchanged — the grams all scale together, so no share of the total can shift. KEY
The claim is right for the heaviest element only — its percent shifts as the sample changes while the lighter elements' percents stay fixed.
The claim is right for solids but wrong for liquids and gases, whose particles rearrange too freely to hold one fixed ratio.
The percents stay the same only when the changed sample comes from the same batch as the original, made in the same production run.
MOL-035 · hard · transfer · seq 20 · authored (parallel error-spotting slot; case rotates) · mcq4
For a compound of three elements, a student calculates the percents as 38.4%, 9.2%, and 43.8%. Without redoing the arithmetic, what can be concluded?
The percents total 91.4, and the percents of all the elements in a compound must total 100 — one of the calculations carries an error. KEY
Nothing shows an error — each of the three percents lies between 0 and 100, so the set is acceptable as it stands.
The largest percent must belong to the heaviest element, and without the formula nothing confirms that it does here.
The smallest percent, 9.2%, is too small to be genuine — no element contributes that little.
MOL-035 · hard · application · seq 21 · authored (FRQ#1, formula-route capstone; 5-line scaffold per element; fresh instance per form) · frq
Consider the compound SrCO₃. (atomic masses: Sr 87.6; C 12.0; O 16.0; M of SrCO₃ = 147.6 g/mol.)
Consider the compound SrCO₃. (atomic masses: Sr 87.6; C 12.0; O 16.0; M of SrCO₃ = 147.6 g/mol.) (a) For each element in the formula, state the mass that element contributes to one mole of the compound. (b) Work through the percent share of each element, showing your working. Give each percent to 3 significant figures. (c) Check your answer by totaling the percents, state the total, and say what the total tells you about your calculations.
Part a. For each element in the formula, state the mass that element contributes to one mole of the compound.
Model: mass of Sr in one mole = 1 × 87.6 = 87.6 g mass of C in one mole = 1 × 12.0 = 12.0 g mass of O in one mole = 3 × 16.0 = 48.0 g M = 147.6 g/mol
Part b. Work through the percent share of each element, showing your working. Give each percent to 3 significant figures.
Model: Run the equation once per element: % Sr = (mass of Sr in one mole ÷ M) × 100 % Sr = (87.6 ÷ 147.6) × 100 % Sr = 59.3% % C = (mass of C in one mole ÷ M) × 100 % C = (12.0 ÷ 147.6) × 100 % C = 8.13% % O = (mass of O in one mole ÷ M) × 100 % O = (48.0 ÷ 147.6) × 100 % O = 32.5%
Part c. Check your answer by totaling the percents, state the total, and say what the total tells you about your calculations.
Model: Check the total: 59.3 + 8.13 + 32.5 = 99.93 The percents of all the elements in a compound must total 100, and small rounding differences (totals from 99.9 to 100.1) are acceptable — a total of 99.93 passes the check.
Full model answer. (a) mass of Sr in one mole = 1 × 87.6 = 87.6 g mass of C in one mole = 1 × 12.0 = 12.0 g mass of O in one mole = 3 × 16.0 = 48.0 g M = 147.6 g/mol (b) Run the equation once per element: % Sr = (mass of Sr in one mole ÷ M) × 100 % Sr = (87.6 ÷ 147.6) × 100 % Sr = 59.3% % C = (mass of C in one mole ÷ M) × 100 % C = (12.0 ÷ 147.6) × 100 % C = 8.13% % O = (mass of O in one mole ÷ M) × 100 % O = (48.0 ÷ 147.6) × 100 % O = 32.5% (c) Check the total: 59.3 + 8.13 + 32.5 = 99.93 The percents of all the elements in a compound must total 100, and small rounding differences (totals from 99.9 to 100.1) are acceptable — a total of 99.93 passes the check.
MOL-033 · hard · application · seq 22 · authored (FRQ#2, measured-sample route; real composition data, identity unnamed; fresh instance per form) · frq
A 1167.0 g sample of an unidentified pure compound is broken apart in an analysis. The analysis finds 686.5 g of Ba and 160.5 g of S; the rest of the sample's mass is the third element, O.
A 1167.0 g sample of an unidentified pure compound is broken apart in an analysis. The analysis finds 686.5 g of Ba and 160.5 g of S; the rest of the sample's mass is the third element, O. (a) Find the mass of O in the sample, and state the fact about the element masses that lets you find it. (b) Calculate what percent of the sample's mass the O supplies. Give the percent to 3 significant figures, and show every step of your working. (c) Calculate the other two percents, total all three, and state what the total shows about your work.
Part a. Find the mass of O in the sample, and state the fact about the element masses that lets you find it.
Model: The element masses in a sample add up to the whole sample's mass. mass of O = 1167.0 − 686.5 − 160.5 = 320.0 g
Part b. Calculate what percent of the sample's mass the O supplies. Give the percent to 3 significant figures, and show every step of your working.
Model: mass of O in the sample = 320.0 g mass of the sample = 1167.0 g % O = (mass of O in the sample ÷ mass of the sample) × 100 % O = (320.0 ÷ 1167.0) × 100 % O = 27.4%
Part c. Calculate the other two percents, total all three, and state what the total shows about your work.
Model: % Ba = (686.5 ÷ 1167.0) × 100 = 58.8% % S = (160.5 ÷ 1167.0) × 100 = 13.8% Check the total: 27.4 + 58.8 + 13.8 = 100.0 The percents of all the elements in a compound must total 100 (small rounding differences like 99.9 or 100.1 are acceptable), so the set passes the check — every share was taken of the whole sample.
Full model answer. (a) The element masses in a sample add up to the whole sample's mass. mass of O = 1167.0 − 686.5 − 160.5 = 320.0 g (b) mass of O in the sample = 320.0 g mass of the sample = 1167.0 g % O = (mass of O in the sample ÷ mass of the sample) × 100 % O = (320.0 ÷ 1167.0) × 100 % O = 27.4% (c) % Ba = (686.5 ÷ 1167.0) × 100 = 58.8% % S = (160.5 ÷ 1167.0) × 100 = 13.8% Check the total: 27.4 + 58.8 + 13.8 = 100.0 The percents of all the elements in a compound must total 100 (small rounding differences like 99.9 or 100.1 are acceptable), so the set passes the check — every share was taken of the whole sample.

Form C

MOL-031 · easy · recall · seq 1 · authored (parallel reading slot; formula-and-symbol instance per the naming convention) · mcq4
The compound BaI₂ is 35.1% Ba by mass. Which option below reads it correctly?
Every 100 g of BaI₂ contains 35.1 g of Ba. KEY
35.1% of the atoms in BaI₂ are Ba atoms.
Any sample of BaI₂ contains 35.1 g of Ba.
35.1% of the volume of any BaI₂ sample is Ba.
MOL-032 · easy · recall · seq 2 · authored (parallel to seq-2 seed structure) · mcq4
A student must find the percent by mass of Li in Li₂CO₃. (atomic masses: Li 6.9; C 12.0; O 16.0; M of Li₂CO₃ = 73.8 g/mol.) Which setup is correct?
(2 × 6.9 ÷ 73.8) × 100 KEY
(6.9 ÷ 73.8) × 100
(73.8 ÷ (2 × 6.9)) × 100
(2 × 6.9 ÷ 12.0) × 100
MOL-034 · easy · recall · seq 3 · authored (fresh framing; same verbatim mechanism key) · mcq4
What makes percent composition a property of a compound itself, rather than of any one sample of it?
Because every molecule or formula unit of a compound contains exactly the same atoms, so the mass ratio is the same in any sample. KEY
Every laboratory agrees by convention to quote percent compositions for one standard sample size, so the values always match.
The percents are averaged over many separate samples before being reported, which smooths away sample-to-sample variation.
Compounds are only ever sold in fixed package sizes, so the samples chemists analyze all start out with matching masses.
MOL-031 · easy · application · seq 4 · authored (parallel to seq-4 seed structure) · mcq4
A supplier sells a 100 kg sack of the compound MgF₂, which is 39.0% Mg by mass. How much Mg does the sack contain?
39.0 kg of Mg — the percent is the share in every 100 units of the total mass. KEY
39.0 g of Mg — percent composition is always stated in grams.
100 kg of Mg — the compound is bought for its Mg content.
61.0 kg of Mg — the stated percent is the share of everything else.
MOL-032 · medium · application · seq 5 · authored (parallel numeric slot) · numeric
What is the percent by mass of Al in Al₂S₃? (atomic masses: Al 27.0; S 32.1; M of Al₂S₃ = 150.3 g/mol.) Then give your answer as a percent to 3 significant figures.
Answer: 35.9 % KEY
also accepted: 35.9%
wrongValueDiagnostics
[
 {
  "value": 18.0,
  "misconception": "you forgot to multiply by the subscript — the formula carries 2 Al atoms",
  "feedback": "This answer is wrong — you forgot to multiply by the subscript — the formula carries 2 Al atoms."
 },
 {
  "value": 278.0,
  "misconception": "you inverted the ratio — a percent over 100 is impossible",
  "feedback": "This answer is wrong — you inverted the ratio — a percent over 100 is impossible."
 },
 {
  "value": 0.3593,
  "misconception": "you forgot to multiply by 100",
  "feedback": "This answer is wrong — you forgot to multiply by 100."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of Al in one mole = 2 × 27.0 = 54.0 g\nM = 150.3 g/mol\nWrite down the equation:\n% Al = (mass of Al in one mole ÷ M) × 100\nSubstitute in the values, and calculate:\n% Al = (54.0 ÷ 150.3) × 100\n% Al = 35.9%"
}
MOL-033 · medium · application · seq 6 · authored (parallel numeric slot) · numeric
A 258.5 g sample of the compound ZnF₂ is analyzed and found to contain 163.5 g of Zn. What percent of the sample's mass is Zn? Give your answer as a percent to 3 significant figures.
Answer: 63.2 % KEY
also accepted: 63.2%
wrongValueDiagnostics
[
 {
  "value": 158.0,
  "misconception": "you inverted the ratio — a percent over 100 is impossible",
  "feedback": "This answer is wrong — you inverted the ratio — a percent over 100 is impossible."
 },
 {
  "value": 0.6325,
  "misconception": "you forgot to multiply by 100",
  "feedback": "This answer is wrong — you forgot to multiply by 100."
 },
 {
  "value": 36.8,
  "misconception": "you found the other element's share instead of Zn's",
  "feedback": "This answer is wrong — you found the other element's share instead of Zn's."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of Zn in the sample = 163.5 g\nmass of the sample = 258.5 g\nWrite down the equation:\n% Zn = (mass of Zn in the sample ÷ mass of the sample) × 100\nSubstitute in the values, and calculate:\n% Zn = (163.5 ÷ 258.5) × 100\n% Zn = 63.2%"
}
MOL-034 · medium · application · seq 7 · authored (parallel scaling slot) · mcq4
Chromium(III) oxide is 68.4% Cr by mass, so 100.0 g of chromium(III) oxide contains 68.4 g of Cr. Reporting masses to 1 decimal place, what does a 700.0 g sample of chromium(III) oxide contain?
478.8 g of Cr, and the sample is still 68.4% Cr. KEY
478.8 g of Cr, and the sample is now 478.8% Cr.
68.4 g of Cr, because a compound always holds the same mass of each element.
68.4 g of Cr, and the sample is now 9.77% Cr.
MOL-031 · medium · application · seq 8 · authored (parallel leftover-share slot) · mcq4
The compound SrS contains only two elements, Sr and S. It is 73.2% Sr by mass. What percent of its mass is S?
26.8% KEY
73.2%
46.4%
It cannot be found without the atomic masses.
MOL-033 · medium · application · seq 9 · authored (parallel numeric slot) · numeric
A 169.2 g sample of FeF₃ is analyzed: it contains 85.5 g of F and 83.7 g of Fe. What percent of the sample's mass is Fe? Give your answer as a percent to 3 significant figures.
Answer: 49.5 % KEY
also accepted: 49.5%
wrongValueDiagnostics
[
 {
  "value": 97.9,
  "misconception": "you divided the Fe mass by the other element's mass instead of by the whole sample's mass",
  "feedback": "This answer is wrong — you divided the Fe mass by the other element's mass instead of by the whole sample's mass."
 },
 {
  "value": 50.5,
  "misconception": "you computed F's share instead of Fe's",
  "feedback": "This answer is wrong — you computed F's share instead of Fe's."
 },
 {
  "value": 0.4947,
  "misconception": "you forgot to multiply by 100",
  "feedback": "This answer is wrong — you forgot to multiply by 100."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of Fe in the sample = 83.7 g\nmass of the sample = 169.2 g\nWrite down the equation:\n% Fe = (mass of Fe in the sample ÷ mass of the sample) × 100\nSubstitute in the values, and calculate:\n% Fe = (83.7 ÷ 169.2) × 100\n% Fe = 49.5%\nThe divisor is always the whole sample's 169.2 g — dividing by the F mass gives an element-to-element ratio, not a percent of the compound."
}
MOL-035 · medium · application · seq 10 · authored (numeric ENTRY capstone row; fluency collect-one-element modality) · numeric
Work out the full percent breakdown of Ca₃N₂ — as a self-check, your percents should total 100. (atomic masses: Ca 40.1; N 14.0; M of Ca₃N₂ = 148.3 g/mol.) Enter your percent for N, to 3 significant figures.
Answer: 18.9 % KEY
also accepted: 18.9%
wrongValueDiagnostics
[
 {
  "value": 9.44,
  "misconception": "you forgot to multiply by the subscript — the formula carries 2 N atoms",
  "feedback": "This answer is wrong — you forgot to multiply by the subscript — the formula carries 2 N atoms."
 },
 {
  "value": 530.0,
  "misconception": "you inverted the ratio — a percent over 100 is impossible",
  "feedback": "This answer is wrong — you inverted the ratio — a percent over 100 is impossible."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of Ca in one mole = 3 × 40.1 = 120.3 g\nmass of N in one mole = 2 × 14.0 = 28.0 g\nM = 148.3 g/mol\nWrite down the equation, and run it once per element:\n% Ca = (120.3 ÷ 148.3) × 100 = 81.1%\n% N = (28.0 ÷ 148.3) × 100 = 18.9%\nCheck the total: 100.0 ✓"
}
MOL-032 · medium · application · seq 11 · authored (parallel numeric slot) · numeric
What is the percent by mass of S in sulfuric acid, H₂SO₄? (atomic masses: H 1.0; S 32.1; O 16.0; M of H₂SO₄ = 98.1 g/mol.) Report your answer as a percent, to 3 significant figures.
Answer: 32.7 % KEY
also accepted: 32.7%
wrongValueDiagnostics
[
 {
  "value": 48.6,
  "misconception": "you divided by the other elements' combined mass instead of the whole compound's molar mass",
  "feedback": "This answer is wrong — you divided by the other elements' combined mass instead of the whole compound's molar mass."
 },
 {
  "value": 306.0,
  "misconception": "you inverted the ratio — a percent over 100 is impossible",
  "feedback": "This answer is wrong — you inverted the ratio — a percent over 100 is impossible."
 },
 {
  "value": 0.3272,
  "misconception": "you forgot to multiply by 100",
  "feedback": "This answer is wrong — you forgot to multiply by 100."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of S in one mole = 1 × 32.1 = 32.1 g\nM = 98.1 g/mol\nWrite down the equation:\n% S = (mass of S in one mole ÷ M) × 100\nSubstitute in the values, and calculate:\n% S = (32.1 ÷ 98.1) × 100\n% S = 32.7%"
}
MOL-034 · medium · application · seq 12 · authored (two-sources fixedness application; fresh instances — fluency CO2/water family is student-seen) · mcq4
Two lots of pure urea come from a plant in one country and a plant in another. How do their percents of N by mass compare?
They are identical, because a pure compound's formula fixes the mass share of every element, no matter where or how the compound was made. KEY
They differ slightly, because where and how a compound was made leaves a lasting trace in its makeup that analysis can detect.
The larger lot has the higher percent of N, because it simply contains more N than the smaller one does.
Neither percent is knowable until each has been through its own laboratory analysis, since only a measurement settles a composition.
MOL-033 · medium · application · seq 13 · authored (parallel numeric slot) · numeric
A 715.6 g sample of Mn(NO₃)₂ undergoes a full analysis: it holds 219.6 g of Mn plus 112.0 g of N; everything else is O. What percent of the sample's mass is O? Give your answer as a percent to 3 significant figures.
Answer: 53.7 % KEY
also accepted: 53.7%
wrongValueDiagnostics
[
 {
  "value": 46.3,
  "misconception": "you computed the combined share of Mn and N — the question asks for the O that makes up the rest",
  "feedback": "This answer is wrong — you computed the combined share of Mn and N — the question asks for the O that makes up the rest."
 },
 {
  "value": 30.7,
  "misconception": "you computed Mn's share instead of O's",
  "feedback": "This answer is wrong — you computed Mn's share instead of O's."
 },
 {
  "value": 384.0,
  "misconception": "you gave the O mass in grams as a percent without dividing by the sample's mass",
  "feedback": "This answer is wrong — you gave the O mass in grams as a percent without dividing by the sample's mass."
 }
]
feedback
{
 "keyExplanation": "The element masses add up to the sample's mass, so the O is 715.6 − 219.6 − 112.0 = 384.0 g.\nWrite down the equation:\n% O = (mass of O in the sample ÷ mass of the sample) × 100\nSubstitute in the values, and calculate:\n% O = (384.0 ÷ 715.6) × 100\n% O = 53.7%"
}
MOL-035 · medium · application · seq 14 · authored (parallel full-set slot) · mcq4
What is the full percent composition of TiCl₄? (atomic masses: Ti 47.9; Cl 35.5; M of TiCl₄ = 189.9 g/mol)
25.2% Ti and 74.8% Cl KEY
74.8% Ti and 25.2% Cl
12.6% Ti and 87.4% Cl
57.4% Ti and 42.6% Cl
MOL-033 · medium · application · seq 15 · authored (which-division discrimination; the lesson's hammered misconception) · mcq4
A 939.2 g sample of AgI is taken apart: 431.6 g is Ag and 507.6 g is I. Which calculation gives the percent of the sample's mass that comes from I?
(507.6 ÷ 939.2) × 100 KEY
(507.6 ÷ 431.6) × 100
(431.6 ÷ 939.2) × 100
(939.2 ÷ 507.6) × 100
MOL-035 · medium · application · seq 16 · authored (parallel full-set slot; atom-count-trap ratio rotates) · mcq4
Give the full percent composition of SF₆. (atomic masses: S 32.1; F 19.0; M of SF₆ = 146.1 g/mol)
22.0% S and 78.0% F KEY
78.0% S and 22.0% F
14.3% S and 85.7% F
11.0% S and 89.0% F
MOL-031 · hard · transfer · seq 17 · authored (parallel claim-repair slot) · mcq4
A student says: '1400.0 g of octane contains 15.8 g of H, because octane is 15.8% H by mass.' What mistake did the student make? (Report masses to 1 decimal place.)
Percent by mass is grams per 100 g: 1400.0 g is fourteen such shares, so the sample contains 14 × 15.8 = 221.2 g of H. KEY
Nothing is wrong — a stated percent by mass gives the grams of the element in any sample, whatever the sample's size.
The claim fails only because the H percent of octane changes with the sample's size, rising as the sample grows.
The claim fails because a percent can never be turned into a mass — the two kinds of quantity cannot be linked.
MOL-032 · hard · transfer · seq 18 · authored (bounds-rule diagnosis; rule line 6 of MOL-032) · mcq4
A student computes the mass percent of Br in CaBr₂ (atomic masses: Ca 40.1; Br 79.9; M of CaBr₂ = 199.9 g/mol) and reports 125.1%. Without seeing any of the work, which statement must be true?
The ratio was inverted — every percent must land between 0 and 100, and only dividing M by the element's contribution can push the result past 100. KEY
The student forgot to multiply by the subscript when finding the element's contribution to one mole of the compound, before dividing by M.
The student forgot to multiply by 100 at the end, leaving the answer as a bare ratio instead of a percent of the whole.
Nothing is certainly wrong — CaBr₂ could simply be very rich in Br, and a compound's richest element can pass 100.
MOL-034 · hard · transfer · seq 19 · authored (scaling-claim rotation per blueprint case_target) · mcq4
A student claims: 'If you halve a sample of a pure compound, you also halve the percent of each element in it.' Which correction fixes the claim?
The action halves each element's grams but leaves each percent unchanged — the grams all scale together, so no share of the total can shift. KEY
The claim is right for the heaviest element only — its percent shifts as the sample changes while the lighter elements' percents stay fixed.
The claim is right for solids but wrong for liquids and gases, whose particles rearrange too freely to hold one fixed ratio.
The percents stay the same only when the changed sample comes from the same batch as the original, made in the same production run.
MOL-035 · hard · transfer · seq 20 · authored (parallel error-spotting slot; case rotates) · mcq4
For a compound of three elements, a student calculates the percents as 44.6%, 8.3%, and 47.2%. Without redoing the arithmetic, what can be concluded?
No arithmetic slip is certain — the percents total 100.1, and small rounding differences, like 99.9 or 100.1, are acceptable. KEY
The total 100.1 proves a subscript error — the percents of a compound must land exactly on 100 with no gap at all.
The percents must all be recalculated from scratch before any conclusion about an error is possible.
The smallest percent, 8.3%, is too small to be genuine — no element contributes that little.
MOL-035 · hard · application · seq 21 · authored (FRQ#1, formula-route capstone; 5-line scaffold per element; fresh instance per form) · frq
Consider H₃BO₃, known as boric acid. (atomic masses: H 1.0; B 10.8; O 16.0; M of H₃BO₃ = 61.8 g/mol.)
Consider H₃BO₃, known as boric acid. (atomic masses: H 1.0; B 10.8; O 16.0; M of H₃BO₃ = 61.8 g/mol.) (a) For each element in the formula, state the mass that element contributes to one mole of the compound. (b) Work through the percent share of each element, showing your working. Give each percent to 3 significant figures. (c) Check your answer by totaling the percents, state the total, and say what the total tells you about your calculations.
Part a. For each element in the formula, state the mass that element contributes to one mole of the compound.
Model: mass of H in one mole = 3 × 1.0 = 3.0 g mass of B in one mole = 1 × 10.8 = 10.8 g mass of O in one mole = 3 × 16.0 = 48.0 g M = 61.8 g/mol
Part b. Work through the percent share of each element, showing your working. Give each percent to 3 significant figures.
Model: Run the equation once per element: % H = (mass of H in one mole ÷ M) × 100 % H = (3.0 ÷ 61.8) × 100 % H = 4.85% % B = (mass of B in one mole ÷ M) × 100 % B = (10.8 ÷ 61.8) × 100 % B = 17.5% % O = (mass of O in one mole ÷ M) × 100 % O = (48.0 ÷ 61.8) × 100 % O = 77.7%
Part c. Check your answer by totaling the percents, state the total, and say what the total tells you about your calculations.
Model: Check the total: 4.85 + 17.5 + 77.7 = 100.05 The percents of all the elements in a compound must total 100, and small rounding differences (totals from 99.9 to 100.1) are acceptable — a total of 100.05 passes the check.
Full model answer. (a) mass of H in one mole = 3 × 1.0 = 3.0 g mass of B in one mole = 1 × 10.8 = 10.8 g mass of O in one mole = 3 × 16.0 = 48.0 g M = 61.8 g/mol (b) Run the equation once per element: % H = (mass of H in one mole ÷ M) × 100 % H = (3.0 ÷ 61.8) × 100 % H = 4.85% % B = (mass of B in one mole ÷ M) × 100 % B = (10.8 ÷ 61.8) × 100 % B = 17.5% % O = (mass of O in one mole ÷ M) × 100 % O = (48.0 ÷ 61.8) × 100 % O = 77.7% (c) Check the total: 4.85 + 17.5 + 77.7 = 100.05 The percents of all the elements in a compound must total 100, and small rounding differences (totals from 99.9 to 100.1) are acceptable — a total of 100.05 passes the check.
MOL-033 · hard · application · seq 22 · authored (FRQ#2, measured-sample route; real composition data, identity unnamed; fresh instance per form) · frq
A 262.9 g sample of an unidentified pure compound is broken apart in an analysis. The analysis finds 72.9 g of Mg and 62.0 g of P; the rest of the sample's mass is the third element, O.
A 262.9 g sample of an unidentified pure compound is broken apart in an analysis. The analysis finds 72.9 g of Mg and 62.0 g of P; the rest of the sample's mass is the third element, O. (a) Find the mass of O in the sample, and state the fact about the element masses that lets you find it. (b) Calculate what percent of the sample's mass the O supplies. Give the percent to 3 significant figures, and show every step of your working. (c) Calculate the other two percents, total all three, and state what the total shows about your work.
Part a. Find the mass of O in the sample, and state the fact about the element masses that lets you find it.
Model: The element masses in a sample add up to the whole sample's mass. mass of O = 262.9 − 72.9 − 62.0 = 128.0 g
Part b. Calculate what percent of the sample's mass the O supplies. Give the percent to 3 significant figures, and show every step of your working.
Model: mass of O in the sample = 128.0 g mass of the sample = 262.9 g % O = (mass of O in the sample ÷ mass of the sample) × 100 % O = (128.0 ÷ 262.9) × 100 % O = 48.7%
Part c. Calculate the other two percents, total all three, and state what the total shows about your work.
Model: % Mg = (72.9 ÷ 262.9) × 100 = 27.7% % P = (62.0 ÷ 262.9) × 100 = 23.6% Check the total: 48.7 + 27.7 + 23.6 = 100.0 The percents of all the elements in a compound must total 100 (small rounding differences like 99.9 or 100.1 are acceptable), so the set passes the check — every share was taken of the whole sample.
Full model answer. (a) The element masses in a sample add up to the whole sample's mass. mass of O = 262.9 − 72.9 − 62.0 = 128.0 g (b) mass of O in the sample = 128.0 g mass of the sample = 262.9 g % O = (mass of O in the sample ÷ mass of the sample) × 100 % O = (128.0 ÷ 262.9) × 100 % O = 48.7% (c) % Mg = (72.9 ÷ 262.9) × 100 = 27.7% % P = (62.0 ÷ 262.9) × 100 = 23.6% Check the total: 48.7 + 27.7 + 23.6 = 100.0 The percents of all the elements in a compound must total 100 (small rounding differences like 99.9 or 100.1 are acceptable), so the set passes the check — every share was taken of the whole sample.

Form D

MOL-031 · easy · recall · seq 1 · authored (parallel reading slot; formula-and-symbol instance per the naming convention) · mcq4
The compound BaS is 81.1% Ba by mass. Which option below reads it correctly?
Every 100 g of BaS contains 81.1 g of Ba. KEY
81.1% of the atoms in BaS are Ba atoms.
Any sample of BaS contains 81.1 g of Ba.
81.1% of the volume of any BaS sample is Ba.
MOL-032 · easy · recall · seq 2 · authored (parallel to seq-2 seed structure) · mcq4
A student must find the percent by mass of Li in Li₃N. (atomic masses: Li 6.9; N 14.0; M of Li₃N = 34.7 g/mol.) Which setup is correct?
(3 × 6.9 ÷ 34.7) × 100 KEY
(6.9 ÷ 34.7) × 100
(34.7 ÷ (3 × 6.9)) × 100
(3 × 6.9 ÷ 14.0) × 100
MOL-034 · easy · recall · seq 3 · authored (fresh framing; same verbatim mechanism key) · mcq4
A reference book lists a single percent composition for each pure compound, with no sample size mentioned. Which statement justifies that?
Because every molecule or formula unit of a compound contains exactly the same atoms, so the mass ratio is the same in any sample. KEY
Sample size is left out because measuring a percent in a very large sample is impossible, so books quote small-sample values only.
The book quietly assumes every reader works with a 100 g sample, since that is the size the per-100 definition is built around.
Each entry secretly refers to one official reference specimen kept by chemists, against which other samples are compared.
MOL-031 · easy · application · seq 4 · authored (parallel to seq-4 seed structure) · mcq4
A supplier sells a 100 g bottle of the compound MgS, which is 43.1% Mg by mass. How much Mg does the bottle contain?
43.1 g of Mg — the percent is the share in every 100 units of the total mass. KEY
43.1 kg of Mg — percent shares are always quoted per 100 kg.
100 g of Mg — the compound is bought for its Mg content.
56.9 g of Mg — the stated percent is the share of everything else.
MOL-032 · medium · application · seq 5 · authored (parallel numeric slot) · numeric
What is the percent by mass of Cl in AlCl₃? (atomic masses: Al 27.0; Cl 35.5; M of AlCl₃ = 133.5 g/mol.) Then give your answer as a percent to 3 significant figures.
Answer: 79.8 % KEY
also accepted: 79.8%
wrongValueDiagnostics
[
 {
  "value": 26.6,
  "misconception": "you forgot to multiply by the subscript — the formula carries 3 Cl atoms",
  "feedback": "This answer is wrong — you forgot to multiply by the subscript — the formula carries 3 Cl atoms."
 },
 {
  "value": 125.0,
  "misconception": "you inverted the ratio — a percent over 100 is impossible",
  "feedback": "This answer is wrong — you inverted the ratio — a percent over 100 is impossible."
 },
 {
  "value": 0.7978,
  "misconception": "you forgot to multiply by 100",
  "feedback": "This answer is wrong — you forgot to multiply by 100."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of Cl in one mole = 3 × 35.5 = 106.5 g\nM = 133.5 g/mol\nWrite down the equation:\n% Cl = (mass of Cl in one mole ÷ M) × 100\nSubstitute in the values, and calculate:\n% Cl = (106.5 ÷ 133.5) × 100\n% Cl = 79.8%"
}
MOL-033 · medium · application · seq 6 · authored (parallel numeric slot) · numeric
A 675.6 g sample of the compound ZnBr₂ is analyzed and found to contain 196.2 g of Zn. What percent of the sample's mass is Zn? Give your answer as a percent to 3 significant figures.
Answer: 29.0 % KEY
also accepted: 29.0%
wrongValueDiagnostics
[
 {
  "value": 344.0,
  "misconception": "you inverted the ratio — a percent over 100 is impossible",
  "feedback": "This answer is wrong — you inverted the ratio — a percent over 100 is impossible."
 },
 {
  "value": 0.2904,
  "misconception": "you forgot to multiply by 100",
  "feedback": "This answer is wrong — you forgot to multiply by 100."
 },
 {
  "value": 71.0,
  "misconception": "you found the other element's share instead of Zn's",
  "feedback": "This answer is wrong — you found the other element's share instead of Zn's."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of Zn in the sample = 196.2 g\nmass of the sample = 675.6 g\nWrite down the equation:\n% Zn = (mass of Zn in the sample ÷ mass of the sample) × 100\nSubstitute in the values, and calculate:\n% Zn = (196.2 ÷ 675.6) × 100\n% Zn = 29.0%"
}
MOL-034 · medium · application · seq 7 · authored (parallel scaling slot) · mcq4
Butane is 82.8% C by mass, so 100.0 g of butane contains 82.8 g of C. Reporting masses to 1 decimal place, what does an 800.0 g sample of butane contain?
662.4 g of C, and the sample is still 82.8% C. KEY
662.4 g of C, and the sample is now 662.4% C.
82.8 g of C, because a compound always holds the same mass of each element.
82.8 g of C, and the sample is now 10.3% C.
MOL-031 · medium · application · seq 8 · authored (parallel leftover-share slot) · mcq4
The compound CaI₂ contains only two elements, Ca and I. It is 13.6% Ca by mass. What percent of its mass is I?
86.4% KEY
13.6%
72.8%
It cannot be found without the atomic masses.
MOL-033 · medium · application · seq 9 · authored (parallel numeric slot) · numeric
A 162.3 g sample of FeCl₃ is analyzed: it contains 106.5 g of Cl and 55.8 g of Fe. What percent of the sample's mass is Fe? Give your answer as a percent to 3 significant figures.
Answer: 34.4 % KEY
also accepted: 34.4%
wrongValueDiagnostics
[
 {
  "value": 52.4,
  "misconception": "you divided the Fe mass by the other element's mass instead of by the whole sample's mass",
  "feedback": "This answer is wrong — you divided the Fe mass by the other element's mass instead of by the whole sample's mass."
 },
 {
  "value": 65.6,
  "misconception": "you computed Cl's share instead of Fe's",
  "feedback": "This answer is wrong — you computed Cl's share instead of Fe's."
 },
 {
  "value": 0.3438,
  "misconception": "you forgot to multiply by 100",
  "feedback": "This answer is wrong — you forgot to multiply by 100."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of Fe in the sample = 55.8 g\nmass of the sample = 162.3 g\nWrite down the equation:\n% Fe = (mass of Fe in the sample ÷ mass of the sample) × 100\nSubstitute in the values, and calculate:\n% Fe = (55.8 ÷ 162.3) × 100\n% Fe = 34.4%\nThe divisor is always the whole sample's 162.3 g — dividing by the Cl mass gives an element-to-element ratio, not a percent of the compound."
}
MOL-035 · medium · application · seq 10 · authored (numeric ENTRY capstone row; fluency collect-one-element modality) · numeric
Work out the full percent breakdown of CaSiO₃ — as a self-check, your percents should total 100. (atomic masses: Ca 40.1; Si 28.1; O 16.0; M of CaSiO₃ = 116.2 g/mol.) Enter your percent for Si, to 3 significant figures.
Answer: 24.2 % KEY
also accepted: 24.2%
wrongValueDiagnostics
[
 {
  "value": 28.1,
  "misconception": "you gave Si's atomic mass instead of its percent",
  "feedback": "This answer is wrong — you gave Si's atomic mass instead of its percent."
 },
 {
  "value": 414.0,
  "misconception": "you inverted the ratio — a percent over 100 is impossible",
  "feedback": "This answer is wrong — you inverted the ratio — a percent over 100 is impossible."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of Ca in one mole = 1 × 40.1 = 40.1 g\nmass of Si in one mole = 1 × 28.1 = 28.1 g\nmass of O in one mole = 3 × 16.0 = 48.0 g\nM = 116.2 g/mol\nWrite down the equation, and run it once per element:\n% Ca = (40.1 ÷ 116.2) × 100 = 34.5%\n% Si = (28.1 ÷ 116.2) × 100 = 24.2%\n% O = (48.0 ÷ 116.2) × 100 = 41.3%\nCheck the total: 100.0 ✓"
}
MOL-032 · medium · application · seq 11 · authored (parallel numeric slot) · numeric
What is the percent by mass of S in K₂SO₄? (atomic masses: K 39.1; S 32.1; O 16.0; M of K₂SO₄ = 174.3 g/mol.) Report your answer as a percent, to 3 significant figures.
Answer: 18.4 % KEY
also accepted: 18.4%
wrongValueDiagnostics
[
 {
  "value": 22.6,
  "misconception": "you divided by the other elements' combined mass instead of the whole compound's molar mass",
  "feedback": "This answer is wrong — you divided by the other elements' combined mass instead of the whole compound's molar mass."
 },
 {
  "value": 543.0,
  "misconception": "you inverted the ratio — a percent over 100 is impossible",
  "feedback": "This answer is wrong — you inverted the ratio — a percent over 100 is impossible."
 },
 {
  "value": 0.1842,
  "misconception": "you forgot to multiply by 100",
  "feedback": "This answer is wrong — you forgot to multiply by 100."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of S in one mole = 1 × 32.1 = 32.1 g\nM = 174.3 g/mol\nWrite down the equation:\n% S = (mass of S in one mole ÷ M) × 100\nSubstitute in the values, and calculate:\n% S = (32.1 ÷ 174.3) × 100\n% S = 18.4%"
}
MOL-034 · medium · application · seq 12 · authored (two-sources fixedness application; fresh instances — fluency CO2/water family is student-seen) · mcq4
Two loads of pure tin(IV) oxide come from two different mines. How do their percents of Sn by mass compare?
They are identical, because a pure compound's formula fixes the mass share of every element, no matter where or how the compound was made. KEY
They differ slightly, because where and how a compound was made leaves a lasting trace in its makeup that analysis can detect.
The larger load has the higher percent of Sn, because it simply contains more Sn than the smaller one does.
Neither percent is knowable until each has been through its own laboratory analysis, since only a measurement settles a composition.
MOL-033 · medium · application · seq 13 · authored (parallel numeric slot) · numeric
A 709.4 g sample of Mn₃(PO₄)₂ is decomposed: it holds 329.4 g of Mn plus 124.0 g of P; everything else is O. What percent of the sample's mass is O? Give your answer as a percent to 3 significant figures.
Answer: 36.1 % KEY
also accepted: 36.1%
wrongValueDiagnostics
[
 {
  "value": 63.9,
  "misconception": "you computed the combined share of Mn and P — the question asks for the O that makes up the rest",
  "feedback": "This answer is wrong — you computed the combined share of Mn and P — the question asks for the O that makes up the rest."
 },
 {
  "value": 46.4,
  "misconception": "you computed Mn's share instead of O's",
  "feedback": "This answer is wrong — you computed Mn's share instead of O's."
 },
 {
  "value": 256.0,
  "misconception": "you gave the O mass in grams as a percent without dividing by the sample's mass",
  "feedback": "This answer is wrong — you gave the O mass in grams as a percent without dividing by the sample's mass."
 }
]
feedback
{
 "keyExplanation": "The element masses add up to the sample's mass, so the O is 709.4 − 329.4 − 124.0 = 256.0 g.\nWrite down the equation:\n% O = (mass of O in the sample ÷ mass of the sample) × 100\nSubstitute in the values, and calculate:\n% O = (256.0 ÷ 709.4) × 100\n% O = 36.1%"
}
MOL-035 · medium · application · seq 14 · authored (parallel full-set slot) · mcq4
What is the full percent composition of NiF₂? (atomic masses: Ni 58.7; F 19.0; M of NiF₂ = 96.7 g/mol)
60.7% Ni and 39.3% F KEY
39.3% Ni and 60.7% F
33.3% Ni and 66.7% F
75.5% Ni and 24.5% F
MOL-033 · medium · application · seq 15 · authored (which-division discrimination; the lesson's hammered misconception) · mcq4
A 991.6 g sample of Ag₂S is taken apart: 863.2 g is Ag and 128.4 g is S. Which calculation gives the percent of the sample's mass that comes from S?
(128.4 ÷ 991.6) × 100 KEY
(128.4 ÷ 863.2) × 100
(863.2 ÷ 991.6) × 100
(991.6 ÷ 128.4) × 100
MOL-035 · medium · application · seq 16 · authored (parallel full-set slot; atom-count-trap ratio rotates) · mcq4
Give the full percent composition of P₂O₅. (atomic masses: P 31.0; O 16.0; M of P₂O₅ = 142.0 g/mol)
43.7% P and 56.3% O KEY
56.3% P and 43.7% O
28.6% P and 71.4% O
21.9% P and 78.2% O
MOL-031 · hard · transfer · seq 17 · authored (parallel claim-repair slot) · mcq4
A student says: '1000.0 g of nonane contains 15.6 g of H, because nonane is 15.6% H by mass.' What mistake did the student make? (Report masses to 1 decimal place.)
Percent by mass is grams per 100 g: 1000.0 g is ten such shares, so the sample contains 10 × 15.6 = 156.0 g of H. KEY
Nothing is wrong — a stated percent by mass gives the grams of the element in any sample, whatever the sample's size.
The claim fails only because the H percent of nonane changes with the sample's size, rising as the sample grows.
The claim fails because a percent can never be turned into a mass — the two kinds of quantity cannot be linked.
MOL-032 · hard · transfer · seq 18 · authored (bounds-rule diagnosis; rule line 6 of MOL-032) · mcq4
A student computes the mass percent of Cl in CaCl₂ (atomic masses: Ca 40.1; Cl 35.5; M of CaCl₂ = 111.1 g/mol) and reports 156.5%. Without seeing any of the work, which statement must be true?
The ratio was inverted — every percent must land between 0 and 100, and only dividing M by the element's contribution can push the result past 100. KEY
The student forgot to multiply by the subscript when finding the element's contribution to one mole of the compound, before dividing by M.
The student forgot to multiply by 100 at the end, leaving the answer as a bare ratio instead of a percent of the whole.
Nothing is certainly wrong — CaCl₂ could simply be very rich in Cl, and a compound's richest element can pass 100.
MOL-034 · hard · transfer · seq 19 · authored (scaling-claim rotation per blueprint case_target) · mcq4
A student claims: 'If you pour away half of a sample of a pure compound, you also halve the percent of each element in it.' Which correction fixes the claim?
The action halves each element's grams but leaves each percent unchanged — the grams all scale together, so no share of the total can shift. KEY
The claim is right for the heaviest element only — its percent shifts as the sample changes while the lighter elements' percents stay fixed.
The claim is right for solids but wrong for liquids and gases, whose particles rearrange too freely to hold one fixed ratio.
The percents stay the same only when the changed sample comes from the same batch as the original, made in the same production run.
MOL-035 · hard · transfer · seq 20 · authored (parallel error-spotting slot; case rotates) · mcq4
For a compound of three elements, a student calculates the percents as 57.3%, 7.1%, and 41.9%. Without redoing the arithmetic, what can be concluded?
The percents total 106.3, and the percents of all the elements in a compound must total 100 — one of the calculations carries an error. KEY
Nothing shows an error — each of the three percents lies between 0 and 100, so the set is acceptable as it stands.
The largest percent must belong to the heaviest element, and without the formula nothing confirms that it does here.
The smallest percent, 7.1%, is too small to be genuine — no element contributes that little.
MOL-035 · hard · application · seq 21 · authored (FRQ#1, formula-route capstone; 5-line scaffold per element; fresh instance per form) · frq
Consider the compound Zn(NO₃)₂. (atomic masses: Zn 65.4; N 14.0; O 16.0; M of Zn(NO₃)₂ = 189.4 g/mol.)
Consider the compound Zn(NO₃)₂. (atomic masses: Zn 65.4; N 14.0; O 16.0; M of Zn(NO₃)₂ = 189.4 g/mol.) (a) For each element in the formula, state the mass that element contributes to one mole of the compound. (b) Work through the percent share of each element, showing your working. Give each percent to 3 significant figures. (c) Check your answer by totaling the percents, state the total, and say what the total tells you about your calculations.
Part a. For each element in the formula, state the mass that element contributes to one mole of the compound.
Model: mass of Zn in one mole = 1 × 65.4 = 65.4 g mass of N in one mole = 2 × 14.0 = 28.0 g mass of O in one mole = 6 × 16.0 = 96.0 g M = 189.4 g/mol
Part b. Work through the percent share of each element, showing your working. Give each percent to 3 significant figures.
Model: Run the equation once per element: % Zn = (mass of Zn in one mole ÷ M) × 100 % Zn = (65.4 ÷ 189.4) × 100 % Zn = 34.5% % N = (mass of N in one mole ÷ M) × 100 % N = (28.0 ÷ 189.4) × 100 % N = 14.8% % O = (mass of O in one mole ÷ M) × 100 % O = (96.0 ÷ 189.4) × 100 % O = 50.7%
Part c. Check your answer by totaling the percents, state the total, and say what the total tells you about your calculations.
Model: Check the total: 34.5 + 14.8 + 50.7 = 100 The percents of all the elements in a compound must total 100, and small rounding differences (totals from 99.9 to 100.1) are acceptable — a total of 100 passes the check.
Full model answer. (a) mass of Zn in one mole = 1 × 65.4 = 65.4 g mass of N in one mole = 2 × 14.0 = 28.0 g mass of O in one mole = 6 × 16.0 = 96.0 g M = 189.4 g/mol (b) Run the equation once per element: % Zn = (mass of Zn in one mole ÷ M) × 100 % Zn = (65.4 ÷ 189.4) × 100 % Zn = 34.5% % N = (mass of N in one mole ÷ M) × 100 % N = (28.0 ÷ 189.4) × 100 % N = 14.8% % O = (mass of O in one mole ÷ M) × 100 % O = (96.0 ÷ 189.4) × 100 % O = 50.7% (c) Check the total: 34.5 + 14.8 + 50.7 = 100 The percents of all the elements in a compound must total 100, and small rounding differences (totals from 99.9 to 100.1) are acceptable — a total of 100 passes the check.
MOL-033 · hard · application · seq 22 · authored (FRQ#2, measured-sample route; real composition data, identity unnamed; fresh instance per form) · frq
A 258.3 g sample of an unidentified pure compound is broken apart in an analysis. The analysis finds 48.3 g of Li and 42.0 g of C; the rest of the sample's mass is the third element, O.
A 258.3 g sample of an unidentified pure compound is broken apart in an analysis. The analysis finds 48.3 g of Li and 42.0 g of C; the rest of the sample's mass is the third element, O. (a) Find the mass of O in the sample, and state the fact about the element masses that lets you find it. (b) Calculate what percent of the sample's mass the O supplies. Give the percent to 3 significant figures, and show every step of your working. (c) Calculate the other two percents, total all three, and state what the total shows about your work.
Part a. Find the mass of O in the sample, and state the fact about the element masses that lets you find it.
Model: The element masses in a sample add up to the whole sample's mass. mass of O = 258.3 − 48.3 − 42.0 = 168.0 g
Part b. Calculate what percent of the sample's mass the O supplies. Give the percent to 3 significant figures, and show every step of your working.
Model: mass of O in the sample = 168.0 g mass of the sample = 258.3 g % O = (mass of O in the sample ÷ mass of the sample) × 100 % O = (168.0 ÷ 258.3) × 100 % O = 65.0%
Part c. Calculate the other two percents, total all three, and state what the total shows about your work.
Model: % Li = (48.3 ÷ 258.3) × 100 = 18.7% % C = (42.0 ÷ 258.3) × 100 = 16.3% Check the total: 65.0 + 18.7 + 16.3 = 100.0 The percents of all the elements in a compound must total 100 (small rounding differences like 99.9 or 100.1 are acceptable), so the set passes the check — every share was taken of the whole sample.
Full model answer. (a) The element masses in a sample add up to the whole sample's mass. mass of O = 258.3 − 48.3 − 42.0 = 168.0 g (b) mass of O in the sample = 168.0 g mass of the sample = 258.3 g % O = (mass of O in the sample ÷ mass of the sample) × 100 % O = (168.0 ÷ 258.3) × 100 % O = 65.0% (c) % Li = (48.3 ÷ 258.3) × 100 = 18.7% % C = (42.0 ÷ 258.3) × 100 = 16.3% Check the total: 65.0 + 18.7 + 16.3 = 100.0 The percents of all the elements in a compound must total 100 (small rounding differences like 99.9 or 100.1 are acceptable), so the set passes the check — every share was taken of the whole sample.

Form E

MOL-031 · easy · recall · seq 1 · authored (parallel reading slot; formula-and-symbol instance per the naming convention) · mcq4
The compound Ba(NO₃)₂ is 52.5% Ba by mass. Which option below reads it correctly?
Every 100 g of Ba(NO₃)₂ contains 52.5 g of Ba. KEY
52.5% of the atoms in Ba(NO₃)₂ are Ba atoms.
Any sample of Ba(NO₃)₂ contains 52.5 g of Ba.
52.5% of the volume of any Ba(NO₃)₂ sample is Ba.
MOL-032 · easy · recall · seq 2 · authored (parallel to seq-2 seed structure) · mcq4
A student must find the percent by mass of Li in Li₃PO₄. (atomic masses: Li 6.9; P 31.0; O 16.0; M of Li₃PO₄ = 115.7 g/mol.) Which setup is correct?
(3 × 6.9 ÷ 115.7) × 100 KEY
(6.9 ÷ 115.7) × 100
(115.7 ÷ (3 × 6.9)) × 100
(3 × 6.9 ÷ 31.0) × 100
MOL-034 · easy · recall · seq 3 · authored (fresh framing; same verbatim mechanism key) · mcq4
No matter how large or small the sample, a pure compound shows one percent composition. Why?
Because every molecule or formula unit of a compound contains exactly the same atoms, so the mass ratio is the same in any sample. KEY
Because the percents are defined so that they always total 100, and a total fixed at 100 leaves no room for a share to drift.
Because impurities are removed before each measurement, and purification pushes every cleaned sample toward the same makeup.
Because percent composition is rounded to one decimal place, and the rounding hides small differences between the samples.
MOL-031 · easy · application · seq 4 · authored (parallel to seq-4 seed structure) · mcq4
A supplier sells a 100 kg crate of the compound MgBr₂, which is 13.2% Mg by mass. How much Mg does the crate contain?
13.2 kg of Mg — the percent is the share in every 100 units of the total mass. KEY
13.2 g of Mg — percent composition is always stated in grams.
100 kg of Mg — the compound is bought for its Mg content.
86.8 kg of Mg — the stated percent is the share of everything else.
MOL-032 · medium · application · seq 5 · authored (parallel numeric slot) · numeric
What is the percent by mass of Br in AlBr₃? (atomic masses: Al 27.0; Br 79.9; M of AlBr₃ = 266.7 g/mol.) Then give your answer as a percent to 3 significant figures.
Answer: 89.9 % KEY
also accepted: 89.9%
wrongValueDiagnostics
[
 {
  "value": 30.0,
  "misconception": "you forgot to multiply by the subscript — the formula carries 3 Br atoms",
  "feedback": "This answer is wrong — you forgot to multiply by the subscript — the formula carries 3 Br atoms."
 },
 {
  "value": 111.0,
  "misconception": "you inverted the ratio — a percent over 100 is impossible",
  "feedback": "This answer is wrong — you inverted the ratio — a percent over 100 is impossible."
 },
 {
  "value": 0.8988,
  "misconception": "you forgot to multiply by 100",
  "feedback": "This answer is wrong — you forgot to multiply by 100."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of Br in one mole = 3 × 79.9 = 239.7 g\nM = 266.7 g/mol\nWrite down the equation:\n% Br = (mass of Br in one mole ÷ M) × 100\nSubstitute in the values, and calculate:\n% Br = (239.7 ÷ 266.7) × 100\n% Br = 89.9%"
}
MOL-033 · medium · application · seq 6 · authored (parallel numeric slot) · numeric
A 1117.2 g sample of the compound ZnI₂ is analyzed and found to contain 228.9 g of Zn. What percent of the sample's mass is Zn? Give your answer as a percent to 3 significant figures.
Answer: 20.5 % KEY
also accepted: 20.5%
wrongValueDiagnostics
[
 {
  "value": 488.0,
  "misconception": "you inverted the ratio — a percent over 100 is impossible",
  "feedback": "This answer is wrong — you inverted the ratio — a percent over 100 is impossible."
 },
 {
  "value": 0.2049,
  "misconception": "you forgot to multiply by 100",
  "feedback": "This answer is wrong — you forgot to multiply by 100."
 },
 {
  "value": 79.5,
  "misconception": "you found the other element's share instead of Zn's",
  "feedback": "This answer is wrong — you found the other element's share instead of Zn's."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of Zn in the sample = 228.9 g\nmass of the sample = 1117.2 g\nWrite down the equation:\n% Zn = (mass of Zn in the sample ÷ mass of the sample) × 100\nSubstitute in the values, and calculate:\n% Zn = (228.9 ÷ 1117.2) × 100\n% Zn = 20.5%"
}
MOL-034 · medium · application · seq 7 · authored (parallel scaling slot) · mcq4
Nickel(II) oxide is 78.6% Ni by mass, so 100.0 g of nickel(II) oxide contains 78.6 g of Ni. Reporting masses to 1 decimal place, what does a 600.0 g sample of nickel(II) oxide contain?
471.6 g of Ni, and the sample is still 78.6% Ni. KEY
471.6 g of Ni, and the sample is now 471.6% Ni.
78.6 g of Ni, because a compound always holds the same mass of each element.
78.6 g of Ni, and the sample is now 13.1% Ni.
MOL-031 · medium · application · seq 8 · authored (parallel leftover-share slot) · mcq4
The compound MnS contains only two elements, Mn and S. It is 63.1% Mn by mass. What percent of its mass is S?
36.9% KEY
63.1%
26.2%
It cannot be found without the atomic masses.
MOL-033 · medium · application · seq 9 · authored (parallel numeric slot) · numeric
A 234.5 g sample of FeF₂ is analyzed: it contains 95.0 g of F and 139.5 g of Fe. What percent of the sample's mass is Fe? Give your answer as a percent to 3 significant figures.
Answer: 59.5 % KEY
also accepted: 59.5%
wrongValueDiagnostics
[
 {
  "value": 147.0,
  "misconception": "you divided the Fe mass by the other element's mass instead of by the whole sample's mass",
  "feedback": "This answer is wrong — you divided the Fe mass by the other element's mass instead of by the whole sample's mass."
 },
 {
  "value": 40.5,
  "misconception": "you computed F's share instead of Fe's",
  "feedback": "This answer is wrong — you computed F's share instead of Fe's."
 },
 {
  "value": 0.5949,
  "misconception": "you forgot to multiply by 100",
  "feedback": "This answer is wrong — you forgot to multiply by 100."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of Fe in the sample = 139.5 g\nmass of the sample = 234.5 g\nWrite down the equation:\n% Fe = (mass of Fe in the sample ÷ mass of the sample) × 100\nSubstitute in the values, and calculate:\n% Fe = (139.5 ÷ 234.5) × 100\n% Fe = 59.5%\nThe divisor is always the whole sample's 234.5 g — dividing by the F mass gives an element-to-element ratio, not a percent of the compound."
}
MOL-035 · medium · application · seq 10 · authored (numeric ENTRY capstone row; fluency collect-one-element modality) · numeric
Work out the full percent breakdown of CaF₂ — as a self-check, your percents should total 100. (atomic masses: Ca 40.1; F 19.0; M of CaF₂ = 78.1 g/mol.) Enter your percent for F, to 3 significant figures.
Answer: 48.7 % KEY
also accepted: 48.7%
wrongValueDiagnostics
[
 {
  "value": 24.3,
  "misconception": "you forgot to multiply by the subscript — the formula carries 2 F atoms",
  "feedback": "This answer is wrong — you forgot to multiply by the subscript — the formula carries 2 F atoms."
 },
 {
  "value": 206.0,
  "misconception": "you inverted the ratio — a percent over 100 is impossible",
  "feedback": "This answer is wrong — you inverted the ratio — a percent over 100 is impossible."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of Ca in one mole = 1 × 40.1 = 40.1 g\nmass of F in one mole = 2 × 19.0 = 38.0 g\nM = 78.1 g/mol\nWrite down the equation, and run it once per element:\n% Ca = (40.1 ÷ 78.1) × 100 = 51.3%\n% F = (38.0 ÷ 78.1) × 100 = 48.7%\nCheck the total: 100.0 ✓"
}
MOL-032 · medium · application · seq 11 · authored (parallel numeric slot) · numeric
What is the percent by mass of S in sulfurous acid, H₂SO₃? (atomic masses: H 1.0; S 32.1; O 16.0; M of H₂SO₃ = 82.1 g/mol.) Report your answer as a percent, to 3 significant figures.
Answer: 39.1 % KEY
also accepted: 39.1%
wrongValueDiagnostics
[
 {
  "value": 64.2,
  "misconception": "you divided by the other elements' combined mass instead of the whole compound's molar mass",
  "feedback": "This answer is wrong — you divided by the other elements' combined mass instead of the whole compound's molar mass."
 },
 {
  "value": 256.0,
  "misconception": "you inverted the ratio — a percent over 100 is impossible",
  "feedback": "This answer is wrong — you inverted the ratio — a percent over 100 is impossible."
 },
 {
  "value": 0.391,
  "misconception": "you forgot to multiply by 100",
  "feedback": "This answer is wrong — you forgot to multiply by 100."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmass of S in one mole = 1 × 32.1 = 32.1 g\nM = 82.1 g/mol\nWrite down the equation:\n% S = (mass of S in one mole ÷ M) × 100\nSubstitute in the values, and calculate:\n% S = (32.1 ÷ 82.1) × 100\n% S = 39.1%"
}
MOL-034 · medium · application · seq 12 · authored (two-sources fixedness application; fresh instances — fluency CO2/water family is student-seen) · mcq4
Two bottles of pure silver nitrate come from two separate production runs. How do their percents of Ag by mass compare?
They are identical, because a pure compound's formula fixes the mass share of every element, no matter where or how the compound was made. KEY
They differ slightly, because where and how a compound was made leaves a lasting trace in its makeup that analysis can detect.
The larger bottle has the higher percent of Ag, because it simply contains more Ag than the smaller one does.
Neither percent is knowable until each has been through its own laboratory analysis, since only a measurement settles a composition.
MOL-033 · medium · application · seq 13 · authored (parallel numeric slot) · numeric
A 655.0 g sample of MnSiO₃ is broken up: it holds 274.5 g of Mn plus 140.5 g of Si; everything else is O. What percent of the sample's mass is O? Give your answer as a percent to 3 significant figures.
Answer: 36.6 % KEY
also accepted: 36.6%
wrongValueDiagnostics
[
 {
  "value": 63.4,
  "misconception": "you computed the combined share of Mn and Si — the question asks for the O that makes up the rest",
  "feedback": "This answer is wrong — you computed the combined share of Mn and Si — the question asks for the O that makes up the rest."
 },
 {
  "value": 41.9,
  "misconception": "you computed Mn's share instead of O's",
  "feedback": "This answer is wrong — you computed Mn's share instead of O's."
 },
 {
  "value": 240.0,
  "misconception": "you gave the O mass in grams as a percent without dividing by the sample's mass",
  "feedback": "This answer is wrong — you gave the O mass in grams as a percent without dividing by the sample's mass."
 }
]
feedback
{
 "keyExplanation": "The element masses add up to the sample's mass, so the O is 655.0 − 274.5 − 140.5 = 240.0 g.\nWrite down the equation:\n% O = (mass of O in the sample ÷ mass of the sample) × 100\nSubstitute in the values, and calculate:\n% O = (240.0 ÷ 655.0) × 100\n% O = 36.6%"
}
MOL-035 · medium · application · seq 14 · authored (parallel full-set slot) · mcq4
What is the full percent composition of SnO₂? (atomic masses: Sn 118.7; O 16.0; M of SnO₂ = 150.7 g/mol)
78.8% Sn and 21.2% O KEY
21.2% Sn and 78.8% O
39.4% Sn and 60.6% O
88.1% Sn and 11.9% O
MOL-033 · medium · application · seq 15 · authored (which-division discrimination; the lesson's hammered misconception) · mcq4
A 761.4 g sample of AgF is taken apart: 647.4 g is Ag and 114.0 g is F. Which calculation gives the percent of the sample's mass that comes from F?
(114.0 ÷ 761.4) × 100 KEY
(114.0 ÷ 647.4) × 100
(647.4 ÷ 761.4) × 100
(761.4 ÷ 114.0) × 100
MOL-035 · medium · application · seq 16 · authored (parallel full-set slot; atom-count-trap ratio rotates) · mcq4
Give the full percent composition of BaO. (atomic masses: Ba 137.3; O 16.0; M of BaO = 153.3 g/mol)
89.6% Ba and 10.4% O KEY
10.4% Ba and 89.6% O
50.0% Ba and 50.0% O, because the formula holds one atom of each
44.8% Ba and 55.2% O
MOL-031 · hard · transfer · seq 17 · authored (parallel claim-repair slot) · mcq4
A student says: '1300.0 g of decane contains 15.5 g of H, because decane is 15.5% H by mass.' What mistake did the student make? (Report masses to 1 decimal place.)
Percent by mass is grams per 100 g: 1300.0 g is thirteen such shares, so the sample contains 13 × 15.5 = 201.5 g of H. KEY
Nothing is wrong — a stated percent by mass gives the grams of the element in any sample, whatever the sample's size.
The claim fails only because the H percent of decane changes with the sample's size, rising as the sample grows.
The claim fails because a percent can never be turned into a mass — the two kinds of quantity cannot be linked.
MOL-032 · hard · transfer · seq 18 · authored (bounds-rule diagnosis; rule line 6 of MOL-032) · mcq4
A student computes the mass percent of I in SrI₂ (atomic masses: Sr 87.6; I 126.9; M of SrI₂ = 341.4 g/mol) and reports 134.5%. Without seeing any of the work, which statement must be true?
The ratio was inverted — every percent must land between 0 and 100, and only dividing M by the element's contribution can push the result past 100. KEY
The student forgot to multiply by the subscript when finding the element's contribution to one mole of the compound, before dividing by M.
The student forgot to multiply by 100 at the end, leaving the answer as a bare ratio instead of a percent of the whole.
Nothing is certainly wrong — SrI₂ could simply be very rich in I, and a compound's richest element can pass 100.
MOL-034 · hard · transfer · seq 19 · authored (scaling-claim rotation per blueprint case_target) · mcq4
A student claims: 'If you combine two equal bags of a sample of a pure compound, you also combine the percent of each element in it.' Which correction fixes the claim?
The action doubles each element's grams but leaves each percent unchanged — the grams all scale together, so no share of the total can shift. KEY
The claim is right for the heaviest element only — its percent shifts as the sample changes while the lighter elements' percents stay fixed.
The claim is right for solids but wrong for liquids and gases, whose particles rearrange too freely to hold one fixed ratio.
The percents stay the same only when the changed sample comes from the same batch as the original, made in the same production run.
MOL-035 · hard · transfer · seq 20 · authored (parallel error-spotting slot; case rotates) · mcq4
For a compound of three elements, a student calculates the percents as 63.7%, 4.7%, and 31.5%. Without redoing the arithmetic, what can be concluded?
No arithmetic slip is certain — the percents total 99.9, and small rounding differences, like 99.9 or 100.1, are acceptable. KEY
The total 99.9 proves a subscript error — the percents of a compound must land exactly on 100 with no gap at all.
The percents must all be recalculated from scratch before any conclusion about an error is possible.
The smallest percent, 4.7%, is too small to be genuine — no element contributes that little.
MOL-035 · hard · application · seq 21 · authored (FRQ#1, formula-route capstone; 5-line scaffold per element; fresh instance per form) · frq
Consider the compound MgSO₄. (atomic masses: Mg 24.3; S 32.1; O 16.0; M of MgSO₄ = 120.4 g/mol.)
Consider the compound MgSO₄. (atomic masses: Mg 24.3; S 32.1; O 16.0; M of MgSO₄ = 120.4 g/mol.) (a) For each element in the formula, state the mass that element contributes to one mole of the compound. (b) Work through the percent share of each element, showing your working. Give each percent to 3 significant figures. (c) Check your answer by totaling the percents, state the total, and say what the total tells you about your calculations.
Part a. For each element in the formula, state the mass that element contributes to one mole of the compound.
Model: mass of Mg in one mole = 1 × 24.3 = 24.3 g mass of S in one mole = 1 × 32.1 = 32.1 g mass of O in one mole = 4 × 16.0 = 64.0 g M = 120.4 g/mol
Part b. Work through the percent share of each element, showing your working. Give each percent to 3 significant figures.
Model: Run the equation once per element: % Mg = (mass of Mg in one mole ÷ M) × 100 % Mg = (24.3 ÷ 120.4) × 100 % Mg = 20.2% % S = (mass of S in one mole ÷ M) × 100 % S = (32.1 ÷ 120.4) × 100 % S = 26.7% % O = (mass of O in one mole ÷ M) × 100 % O = (64.0 ÷ 120.4) × 100 % O = 53.2%
Part c. Check your answer by totaling the percents, state the total, and say what the total tells you about your calculations.
Model: Check the total: 20.2 + 26.7 + 53.2 = 100.1 The percents of all the elements in a compound must total 100, and small rounding differences (totals from 99.9 to 100.1) are acceptable — a total of 100.1 passes the check.
Full model answer. (a) mass of Mg in one mole = 1 × 24.3 = 24.3 g mass of S in one mole = 1 × 32.1 = 32.1 g mass of O in one mole = 4 × 16.0 = 64.0 g M = 120.4 g/mol (b) Run the equation once per element: % Mg = (mass of Mg in one mole ÷ M) × 100 % Mg = (24.3 ÷ 120.4) × 100 % Mg = 20.2% % S = (mass of S in one mole ÷ M) × 100 % S = (32.1 ÷ 120.4) × 100 % S = 26.7% % O = (mass of O in one mole ÷ M) × 100 % O = (64.0 ÷ 120.4) × 100 % O = 53.2% (c) Check the total: 20.2 + 26.7 + 53.2 = 100.1 The percents of all the elements in a compound must total 100, and small rounding differences (totals from 99.9 to 100.1) are acceptable — a total of 100.1 passes the check.
MOL-033 · hard · application · seq 22 · authored (FRQ#2, measured-sample route; real composition data, identity unnamed; fresh instance per form) · frq
A 501.6 g sample of an unidentified pure compound is broken apart in an analysis. The analysis finds 261.6 g of Zn and 192.0 g of O; the rest of the sample's mass is the third element, C.
A 501.6 g sample of an unidentified pure compound is broken apart in an analysis. The analysis finds 261.6 g of Zn and 192.0 g of O; the rest of the sample's mass is the third element, C. (a) Find the mass of C in the sample, and state the fact about the element masses that lets you find it. (b) Calculate what percent of the sample's mass the C supplies. Give the percent to 3 significant figures, and show every step of your working. (c) Calculate the other two percents, total all three, and state what the total shows about your work.
Part a. Find the mass of C in the sample, and state the fact about the element masses that lets you find it.
Model: The element masses in a sample add up to the whole sample's mass. mass of C = 501.6 − 261.6 − 192.0 = 48.0 g
Part b. Calculate what percent of the sample's mass the C supplies. Give the percent to 3 significant figures, and show every step of your working.
Model: mass of C in the sample = 48.0 g mass of the sample = 501.6 g % C = (mass of C in the sample ÷ mass of the sample) × 100 % C = (48.0 ÷ 501.6) × 100 % C = 9.57%
Part c. Calculate the other two percents, total all three, and state what the total shows about your work.
Model: % Zn = (261.6 ÷ 501.6) × 100 = 52.2% % O = (192.0 ÷ 501.6) × 100 = 38.3% Check the total: 9.57 + 52.2 + 38.3 = 100.1 The percents of all the elements in a compound must total 100 (small rounding differences like 99.9 or 100.1 are acceptable), so the set passes the check — every share was taken of the whole sample.
Full model answer. (a) The element masses in a sample add up to the whole sample's mass. mass of C = 501.6 − 261.6 − 192.0 = 48.0 g (b) mass of C in the sample = 48.0 g mass of the sample = 501.6 g % C = (mass of C in the sample ÷ mass of the sample) × 100 % C = (48.0 ÷ 501.6) × 100 % C = 9.57% (c) % Zn = (261.6 ÷ 501.6) × 100 = 52.2% % O = (192.0 ÷ 501.6) × 100 = 38.3% Check the total: 9.57 + 52.2 + 38.3 = 100.1 The percents of all the elements in a compound must total 100 (small rounding differences like 99.9 or 100.1 are acceptable), so the set passes the check — every share was taken of the whole sample.

End of Topic Test: Empirical and molecular formulas (u07-t4 · 5 forms × 22 items · 110 total)

David's notes for James:
full notes file
# u07-t4 — notes/queries from David's side (U7-G4)

1. CALC-HEAVY FRQ SPLIT - frq_per_form = 2, OVERRIDING your gate data's frq_per_form=1 baseline (confirm the composition): empirical/molecular formula determination is a multi-step calculation topic (QB-v2 calc-heavy band, 2-3 FRQs; 6 of the 10 scope lessons are atomKind quant). The count is 2 because the scope holds exactly TWO multi-step procedure capstones - FRQ-1 (seq 21, MOL-043) = empirical formula from experimental data (mass route on A/C/E, percent route on B/D - a case rotation inside MOL-043's own two-route space), FRQ-2 (seq 22, MOL-045, terminal capstone) = molecular formula from empirical formula + molar mass. Every other quant lesson (MOL-040/041/042/044) is a single-step sub-routine of those chains and is already constructed-response as numeric/text ENTRY objective rows; a third FRQ would re-assess a sub-skill inside the capstones for zero new case coverage. D5: min(20, 6 x 10 lessons) = 20 -> full-size 20 objective + 2 FRQ = 22 items/form. Full adjudication text in blueprint.composition.frq_split.count_adjudication.
2. INTERACTION FORMATS (confirm): per form 10 mcq / 2 numeric ENTRY (seq 5 percent->grams incl. rest-is-element step, seq 14 unit count with M(empirical) BUILT from supplied atomic masses) / 8 text ENTRY (typed formulas and ratios: seq 7/11 reduce-a-formula, seq 9/13 mole ratios, seq 10 ratio clearing, seq 15/18 MOL-043 capstone chains, seq 17 MOL-045 capstone chain) / 2 FRQ. The HS numeric-entry rule holds - no numeric-answer mcq anywhere; the three quant-lesson MCQs (seq 12 quotient-meaning, seq 16 method-judgment, seq 20 x1 edge) assess interpretation/judgment, never a routine (seq 20's MCQ frame is blessed by MOL-045's own specNotes).
3. TYPED-FORMULA ENTRY CONVENTION (confirm): every formula-typing stem carries one uniform instruction sentence - 'Type the whole formula, writing any subscripts as plain digits, like the 2 in CO2; leave subscripts of 1 unwritten.' (added as a class sweep after blind round 1 flagged instruction ambiguity on seq 7/11; recorded in gate_config.extra_task_idioms). Canonical answer values keep Unicode subscripts (the taught rendering); the plain-digit form is always an acceptedAlternate; element symbols are case-significant; formula strings carry no unit/sigFigs concept.
4. SUPPLIED-ATOMIC-MASS-CONSTANTS ADJUDICATION (recorded): every calculation stem supplies the atomic masses it needs as '(M of X = v g/mol, ...)' per the course supply-constants rule, and the percent->formula routine is TAUGHT as 'assume a 100 g sample' - these values (22 periodic-table constants + the taught 100) recur on-screen for every student on every form and carry no instance signal. Recorded as gate_config.adjudicated_instance_families 'supplied-atomic-mass-constants' (598 excusal events) and enforced by the gate-local validate_with_adjudications.py wrapper over the UNTOUCHED shared validator (U7-G2/U6-G1 precedent). Relief is scoped: a numeric line is excused only when EVERY colliding value is a listed constant; a gram window only when it contains nothing beyond carrier vocabulary/element symbols/function words - everything else was fixed in CONTENT (3 patch waves, 128 field edits + 21 item rewrites). RECORDED OVER-COVERAGE riders acknowledged in the family rationale (e.g. subscript 16 in C7H16 rides O 16.0; E#19's assume-100 ride; C#13's 31.0 g sample mass = M(P) by design). Equivalent proposed _shared change = the A7 constants key (U2-G2/_proposed_shared_A7.md); _shared stayed READ-ONLY.
5. FRQ-TYPED-FORMULA-REGISTER ADJUDICATION (recorded; first gate whose typed FRQ answers ARE formulas): V-T12 (binding - nothing the student types requires sub/superscripts) forces model answers/rubrics to write formulas in the course's plain-ASCII entry register (CaCO3, C10H8, ...), while V-T10's Unicode mandate scans those same fields. Relief is scoped per slot to the enumerated formulas only (each IS that item's typed answer or a taught intermediate), only inside modelAnswer/rubric, with a recomputation proving zero ASCII-formula matches in stem/prompt/feedback (7 V-T10 lines excused). Proposed _shared fix for your and David's sign-off: exclude frq.modelAnswer/rubric from the V-T10 ASCII-formula regex (they are typed-work fields governed by V-T12). Presentation text (stems, prompts, options, feedback) remains fully Unicode-clean.
6. SYMBOL REGISTER (carried from U7-G3 adjudication #4, confirm): element data mentions in stems/options use symbols ('21.7 g of P'), keeping spelled element words off the V-T7 scan surface; feedback keeps readable words. 194 substance + 31 own-lesson-material collisions resolved to 0 this way in content.
7. D#21 Na2SO3 RULING (flag for David - course-local authority applied): blind r1 flagged D#21's percent data totalling 100.1. An interim replacement (Na2PO3, then Na2PO4) was withdrawn after a re-blind reviewer raised a SOUND P-oxidation-state realism objection; the course rule line 'Small rounding differences, like 99.9 or 100.1, are acceptable; anything further off means a slip.' makes the ORIGINAL Na2SO3 set the taught register, so it was restored VERBATIM (round-1 blind coverage stands on that exact content: both reviewers' cold answers matched). Exhaustive search proof recorded: NO conventional compound yields fully-exact 1-dp percent data. Sol's follow-on 'exact/nothing was rounded' objection fixed by a precision-honest part (c) + model/rubric rewrite (no-clearing-step framing); DATA unchanged.
8. SOL CLEAN-DISTRACTOR BAR (recorded class sweeps, confirm the register): the authoritative QC hardened the bar - a distractor's wrongness must not hinge on a limiting qualifier bolted to a true clause. All five row-4 d0s (true-ratio + 'counts unknown') were rebuilt as wholly-false scaled-down/reduced-subscripts claims (the lesson's own molecular-vs-empirical contrast); the row-4 d2 mole-vs-molecule family and C#4/D#4 d2 got cleanly-false replacements (bond-count / mass-share misreadings). Loops ran 5 -> 3 -> 0 FAILS across 3 loops.
9. SEED CENSUS (SEED RULING - confirm): pool == spec gateItems positionally (30/30; the 6 demo_curated dedup into pool). 19 of 30 placed, ALL on form A (17 objective rows + both FRQ expansions); 11 spares with recorded reasons; forms B-E fully authored (final census 19 seed-carried / 91 authored). prod_seeds empty -> both FRQ families expand your gate TEXT items pool[22] (=demo[1]) and pool[29] (=demo[5]) into 3 lettered parts + 100-point rubrics + fully-typed model answers (recorded method: 'gate prods are fine as FRQ seeds' extended to gate text items). TEN must_freshen_instance executions per the strict SEED RULING (the U7 register reuses workhorse instances across lessons - e.g. P2O5 -> C3H5/82.0 at A#14, NO2/92.0 -> fresh at A#18-row family, FRQ answers CH2O/C6H12 -> CaCO3/C10H8); every swap carried a full-field feedback-leftover sweep.
10. NOTA/AOTA: 0 hits in the pool (no replace_distractor flags needed - first wave gate with a clean pool on this rule). Numeric-contract cleanups recorded: the MOL-044 seed's 'Give your answer to 1 significant figures.' sentence dropped in favor of integer:true + whole-number instruction (value/key unchanged); a redundant double precision instruction on the placed MOL-040 seed normalized to one.
11. PRECISION RULINGS (confirm): E#21 part (a) INSTRUCTS 3-significant-figure reporting so the rubric's 0.600 mol is the unique standard (sol loop-2: a rubric may not pin a value the stem's givens leave ambiguous). The course decimal-clearing register (.33/.67 = thirds at 3-sf display; 2.33 x 3 = 6.99 -> 7 within the data's precision) is course-local authority - blind 'precision' advisories on A#10/E#16 adjudicated to the taught register. 5 sol pass-with-minor notes recorded for David: B#14 accepted-alternate '3.0' vs 1-sf contract; D#21 stem precision unstated (course scaffold governs); E#9 0.050-vs-0.0500 display; E#13 9.00-from-9.0; E#16 'seven and two-thirds' approximation wording.
12. ACCEPTED ADVISORY CLASSES (recorded, no fix - David's call to reopen): (a) cross-item cueing between recall rows 1/2/3/4/12 (definitions/decision rules) and application rows 8/14/17/19/20/22 - blueprint-inherent, recognition and execution of the same taught rule are distinct blueprint rows and V-T2's mix mandates the recall band; (b) definition-echo giveaway on recall rows - the definition IS the tested content, misconception-keyed foils, 0 defensible-distractor fails across 118/118 cumulative blind key picks; (c) house longest-option/polished-key style (V-T4 quantified balance passes); (d) E#1's two yes-family foils carry DISTINCT misconceptions (subscripts-as-counts vs reducing-pins-counts).
13. BANNED-INSTANCE RULE (course-wide: no fissile/weapons/dose framings): 0 hits across all 110 items at stage 1 and re-swept at packaging on forms_final - the palette is classroom-safe throughout (organic workhorses, common salts and oxides).
14. CRASH-RESUME PROVENANCE (disclosure, no action needed): the authoring fleet was killed by laptop restarts three times across this gate's run - after stage-1 (blueprint/seed placement), after stage-2 row authoring (before form assembly), and after the stage-7 finalize wrote forms_final.json + validator_run_final.txt (resumed at packaging, this session). Each resume re-audited the full disk state (every JSON artifact json.loads-verified, 5x22 form census, forms_final payload re-asserted identical to the validated form files) before running only the missing stages; no truncated or inconsistent artifact was found.

Form A

MOL-036 · easy · recall · seq 1 · authored (seedless recall row; what-the-formula-records probe; key = the taught rule line, distractors from the lesson's own misconception set) · mcq4
Which statement describes the information recorded by a compound's empirical formula?
The smallest whole-number ratio of the atoms of each element in the compound. KEY
The exact number of atoms of each element in one particle of the compound.
The mass percent that each element contributes to the compound.
The smallest whole-number ratio of the masses of the elements in the compound.
MOL-037 · easy · recall · seq 2 · seedless authored row (blueprint seq-2 recall rotation, probe A: what-the-formula-lists; instance-free; key language from MOL-037 rule lines) · mcq4
What information does a compound's molecular formula give?
The actual number of atoms of each element in one molecule of the compound. KEY
The smallest whole-number ratio of the atoms of each element in the compound.
The mass percent that each element contributes to the compound.
The total number of atoms in one molecule, without saying which elements they are.
MOL-039 · easy · recall · seq 3 · authored (seedless recall row; the-check-itself probe; key = rule lines 2-3, first distractor = the same check with the outcome inverted — the decision boundary is the recall point) · mcq4
Which check decides whether a formula is already an empirical formula?
Find the largest whole number that divides into every subscript — if that number is 1, the formula is already empirical. KEY
Find the largest whole number that divides into every subscript — if that number is bigger than 1, the formula is already empirical.
Look at each subscript on its own — if any subscript is bigger than 1, the formula still reduces.
Look for an unwritten subscript of 1 — a formula is already empirical only when it contains one.
MOL-037 · easy · application · seq 4 · seed:U7-G4.pool[3] (MOL-037 gateItems[0], ethene C₂H₄) — must_freshen_instance executed: instance -> pentane C₅H₁₂ (blueprint candidate adopted; C₂H₄ = MOL-039 WE instance); guarantee structure, key, and all three misconception classes kept; all fields swept for the old instance (QB 3.9) · mcq4
The molecular formula of pentane is C₅H₁₂. Which statement does the formula guarantee about one pentane molecule?
It contains exactly 5 C atoms and 12 H atoms. KEY
Its subscripts are scaled-down proportions — the real molecule holds some larger multiple of 5 C and 12 H.
It contains exactly 12 C atoms and 5 H atoms.
It lists the elements present without saying how many atoms of each a molecule holds.
MOL-040 · easy · application · seq 5 · seed:U7-G4.pool[13] (MOL-040 gateItems[1], 80.0% C -> 20.0 g H) — values/key unchanged; redundant double precision instruction normalized to the lesson's single one-decimal-place convention (recorded contract cleanup) · numeric
A compound is 80.0% C by mass; the rest is H. What mass of H does a 100 g sample of the compound contain? Give your answer in grams to one decimal place.
Answer: 20.0 g KEY
also accepted: 20.0 g
wrongValueDiagnostics
[
 {
  "value": 80.0,
  "misconception": "you gave the carbon's mass — the question asks for the hydrogen, the rest of the 100 g",
  "feedback": "This answer is wrong — you gave the carbon's mass — the question asks for the hydrogen, the rest of the 100 g."
 },
 {
  "value": 0.2,
  "misconception": "you divided by 100 — in a 100 g sample the masses are in grams already",
  "feedback": "This answer is wrong — you divided by 100 — in a 100 g sample the masses are in grams already."
 }
]
feedback
{
 "keyExplanation": "Assume a sample of exactly 100 g.\nCarbon: 80.0% of the compound → 80.0 g.\nThe element masses in a 100 g sample total 100 g.\nmass of hydrogen = 100 − 80.0 = 20.0 g"
}
MOL-036 · medium · application · seq 6 · seed:U7-G4.pool[2] (MOL-036 gateItems[2], G2 tests-'smallest' case: measured 4:6 -> C₂H₃) — instance/key/distractors kept verbatim per seed_placement (no must_freshen_instance flag); seed HTML sub markup converted to Unicode subscripts per the James-export mandate; stem OPENING reworded ('reports that a compound contains' -> 'analyzing a compound detects') because the seed 4-gram 'contains 4 carbon atoms' verbatim-collides with the MOL-037 article WE (butane, 'One molecule of butane contains 4 carbon atoms and 10 hydrogen atoms') — SEED RULING content freshening, wording only, instance unchanged. KNOWN GATE-LEVEL CLASS (not resolvable at row level): 'carbon'/'hydrogen' are MATERIALS-lexicon words inside MOL-036's own corpus, so this seed fires the V-T7 own-lesson material scan; element names are task-essential for a ratio->formula item (the same class hits most calc rows of this gate) — needs the stage-3 recorded gate_config adjudication, flag for David/James · mcq4
An instrument analyzing a compound detects 4 C atoms for every 6 H atoms. Which formula is the compound's empirical formula?
C₂H₃ KEY
C₄H₆
C₃H₂
C₂H₆
MOL-038 · medium · application · seq 7 · seed:U7-G4/pool[6] (= MOL-038 gateItems[0], James; no freshen flag — C₄H₈ exclusive to this row, 037 G1 + 039 G0 spared): stem, answer, plain-digit alternate, explanation and all three misconception classes carried verbatim; notation transport HTML sub markup -> Unicode only · text
What is the empirical formula of butene, C₄H₈? Type the whole formula, writing any subscripts as plain digits, like the 2 in CO2; leave subscripts of 1 unwritten.
Answer: CH₂ KEY
also accepted: CH2
wrongValueDiagnostics
[
 {
  "value": "C₂H₄",
  "misconception": "you divided by 2, which is a common factor but not the LARGEST one — 2 and 4 still share the factor 2",
  "feedback": "This answer is wrong — you divided by 2, which is a common factor but not the LARGEST one. The subscripts of C₂H₄ — 2 and 4 — still share the factor 2, so the formula still reduces. Divide C₄H₈'s subscripts by their largest common factor, 4: the empirical formula is CH₂."
 },
 {
  "value": "C₄H₈",
  "misconception": "you left the formula unchanged — the subscripts 4 and 8 share the factor 4, so the formula reduces",
  "feedback": "This answer is wrong — you left the formula unchanged. The subscripts 4 and 8 share the factor 4, so the formula reduces: C: 4 ÷ 4 = 1, H: 8 ÷ 4 = 2, and the empirical formula is CH₂."
 },
 {
  "value": "CH",
  "misconception": "you divided each subscript by its own value, which destroys the ratio",
  "feedback": "This answer is wrong — you divided each subscript by its own value, which destroys the ratio. Every subscript must be divided by the SAME factor — the largest common factor, 4: C: 4 ÷ 4 = 1, H: 8 ÷ 4 = 2, so the empirical formula is CH₂."
 }
]
feedback
{
 "keyExplanation": "Subscripts: C 4, H 8. The largest common factor is 4 — dividing by 2 alone leaves C₂H₄, which still reduces. C: 4 ÷ 4 = 1, H: 8 ÷ 4 = 2, and a subscript of 1 is never written, so the empirical formula is CH₂.",
 "anticipatedWrong": [
  {
   "value": "C₂H₄",
   "misconception": "you divided by 2, which is a common factor but not the LARGEST one — 2 and 4 still share the factor 2",
   "feedback": "This answer is wrong — you divided by 2, which is a common factor but not the LARGEST one. The subscripts of C₂H₄ — 2 and 4 — still share the factor 2, so the formula still reduces. Divide C₄H₈'s subscripts by their largest common factor, 4: the empirical formula is CH₂."
  },
  {
   "value": "C₄H₈",
   "misconception": "you left the formula unchanged — the subscripts 4 and 8 share the factor 4, so the formula reduces",
   "feedback": "This answer is wrong — you left the formula unchanged. The subscripts 4 and 8 share the factor 4, so the formula reduces: C: 4 ÷ 4 = 1, H: 8 ÷ 4 = 2, and the empirical formula is CH₂."
  },
  {
   "value": "CH",
   "misconception": "you divided each subscript by its own value, which destroys the ratio",
   "feedback": "This answer is wrong — you divided each subscript by its own value, which destroys the ratio. Every subscript must be divided by the SAME factor — the largest common factor, 4: C: 4 ÷ 4 = 1, H: 8 ÷ 4 = 2, so the empirical formula is CH₂."
  }
 ]
}
MOL-039 · medium · application · seq 8 · seed:U7-G4/pool[11] (= MOL-039 gateItems[2], James) — verbatim (field mapping + HTML sub markup → Unicode notation transport only) · mcq4
Which formula is NOT an empirical formula?
C₆H₁₄ KEY
N₂O₃
CaCl₂
C₂H₆O
MOL-041 · medium · application · seq 9 · seed:U7-G4/pool[16] (= MOL-041 gateItems[1], James) — verbatim (HTML strip only; explanation_correct -> keyExplanation, explanation_incorrect recast into anticipatedWrong with the spec's own values 1 : 12 and 4 : 1) · text
A sample of a compound contains 3.0 g of C and 35.5 g of Cl. What is the simplest mole ratio of C atoms to Cl atoms? (M of C = 12.0 g/mol, Cl = 35.5 g/mol.) Write your answer like 1 : 2.
Answer: 1 : 4 KEY
also accepted: 1:4
wrongValueDiagnostics
[
 {
  "value": "1 : 12",
  "misconception": "you compared the masses instead of the moles — 3.0 : 35.5 is about 1 : 12, but atoms are counted in moles",
  "feedback": "This answer is wrong — you compared the masses instead of the moles. 3.0 : 35.5 is about 1 : 12, but atoms are counted in moles: n(C) = 3.0 / 12.0 = 0.25 mol and n(Cl) = 35.5 / 35.5 = 1.00 mol, and dividing each by 0.25 gives 1 : 4."
 },
 {
  "value": "4 : 1",
  "misconception": "you wrote the ratio backwards — the order follows the elements as named: carbon first, then chlorine",
  "feedback": "This answer is wrong — you wrote the ratio backwards. The ratio order follows the elements as they are named in the question: carbon first, then chlorine, so the answer is 1 : 4."
 }
]
feedback
{
 "keyExplanation": "Values from the question:\nm(C) = 3.0 g, m(Cl) = 35.5 g\nEquation:\nn = m/M  (for each element)\nSubstituting the values and calculating:\nn(C) = 3.0 / 12.0 = 0.25 mol\nn(Cl) = 35.5 / 35.5 = 1.00 mol\nDivide each by the smallest (0.25): C gives 1, Cl gives 4.\nMole ratio of carbon to chlorine = 1 : 4",
 "anticipatedWrong": [
  {
   "value": "1 : 12",
   "misconception": "you compared the masses instead of the moles — 3.0 : 35.5 is about 1 : 12, but atoms are counted in moles",
   "feedback": "This answer is wrong — you compared the masses instead of the moles. 3.0 : 35.5 is about 1 : 12, but atoms are counted in moles: n(C) = 3.0 / 12.0 = 0.25 mol and n(Cl) = 35.5 / 35.5 = 1.00 mol, and dividing each by 0.25 gives 1 : 4."
  },
  {
   "value": "4 : 1",
   "misconception": "you wrote the ratio backwards — the order follows the elements as named: carbon first, then chlorine",
   "feedback": "This answer is wrong — you wrote the ratio backwards. The ratio order follows the elements as they are named in the question: carbon first, then chlorine, so the answer is 1 : 4."
  }
 ]
}
MOL-042 · medium · application · seq 10 · seed:U7-G4/pool[18] (= MOL-042 gateItems[0], James, gate G0) — instance 1 : 2.33 -> 3 : 7 (x3, .33 branch), answer contract and both wrong paths carried verbatim; sole edit = the recorded format-hint reword ('Write your answer' -> 'Give your answer'), forced by the V-T7 bridge grams and mirrored on all forms (furniture only; value/key/alternates unchanged) · text
A mole calculation gives the ratio 1 : 2.33. What whole-number ratio does it clear to? Give your answer like 2 : 3.
Answer: 3 : 7 KEY
also accepted: 3:7
wrongValueDiagnostics
[
 {
  "value": "1 : 2",
  "misconception": "you rounded the 2.33 down to 2 — rounding changes the ratio; multiply every number by 3 instead",
  "feedback": "This answer is wrong — you rounded the 2.33 down to 2, and rounding changes the ratio itself. The decimal is near .33, so multiply every number by 3 instead: 1 × 3 = 3 and 2.33 × 3 = 7, giving 3 : 7."
 },
 {
  "value": "1 : 7",
  "misconception": "you multiplied only the decimal number — every number in the ratio must be multiplied by the same amount",
  "feedback": "This answer is wrong — you multiplied only the decimal number. Every number in the ratio must be multiplied by the same amount to keep the ratio unchanged: 1 × 3 = 3 and 2.33 × 3 = 7, giving 3 : 7."
 }
]
feedback
{
 "keyExplanation": "The decimal is near .33, so multiply every number by 3: 1 × 3 = 3 and 2.33 × 3 = 7. The whole-number ratio is 3 : 7. Rounding 2.33 down to 2 would change the ratio itself — multiplying never does.",
 "anticipatedWrong": [
  {
   "value": "1 : 2",
   "misconception": "you rounded the 2.33 down to 2 — rounding changes the ratio; multiply every number by 3 instead",
   "feedback": "This answer is wrong — you rounded the 2.33 down to 2, and rounding changes the ratio itself. The decimal is near .33, so multiply every number by 3 instead: 1 × 3 = 3 and 2.33 × 3 = 7, giving 3 : 7."
  },
  {
   "value": "1 : 7",
   "misconception": "you multiplied only the decimal number — every number in the ratio must be multiplied by the same amount",
   "feedback": "This answer is wrong — you multiplied only the decimal number. Every number in the ratio must be multiplied by the same amount to keep the ratio unchanged: 1 × 3 = 3 and 2.33 × 3 = 7, giving 3 : 7."
  }
 ]
}
MOL-038 · medium · application · seq 11 · seed:U7-G4.pool[8] (MOL-038 gateItems[2], hydrazine N₂H₄ -> NH₂) — must_freshen_instance executed: instance -> diborane B₂H₆ -> BH₃ (blueprint candidate adopted; N₂H₄ = MOL-037 fluency F1 instance); typed-entry structure, factor-2 leading-1 case, and all three misconception classes kept; all fields swept for the old instance (QB 3.9) · text
What is the empirical formula of diborane, B₂H₆? Type the whole formula, writing any subscripts as plain digits, like the 2 in CO2; leave subscripts of 1 unwritten.
Answer: BH₃ KEY
also accepted: BH3
wrongValueDiagnostics
[
 {
  "value": "B₂H₆",
  "misconception": "you left the formula unchanged — the subscripts 2 and 6 share the factor 2, so the formula reduces",
  "feedback": "This answer is wrong — you left the formula unchanged. The subscripts 2 and 6 share the factor 2, so the formula reduces: B: 2 ÷ 2 = 1, H: 6 ÷ 2 = 3, giving BH₃."
 },
 {
  "value": "BH",
  "misconception": "you divided each subscript by its own value, which destroys the ratio",
  "feedback": "This answer is wrong — you divided each subscript by its own value, which destroys the ratio. Every subscript is divided by the same factor, 2: B: 2 ÷ 2 = 1, H: 6 ÷ 2 = 3, giving BH₃."
 },
 {
  "value": "B₂H₃",
  "misconception": "you divided only the hydrogen's subscript — every subscript must be divided by the same factor",
  "feedback": "This answer is wrong — you divided only the hydrogen's subscript. Every subscript must be divided by the same factor: B: 2 ÷ 2 = 1, H: 6 ÷ 2 = 3, giving BH₃."
 }
]
feedback
{
 "keyExplanation": "Subscripts: B 2, H 6. The largest common factor is 2. B: 2 ÷ 2 = 1, H: 6 ÷ 2 = 3. A subscript of 1 is never written, so diborane's empirical formula is BH₃.",
 "anticipatedWrong": [
  {
   "value": "B₂H₆",
   "misconception": "you left the formula unchanged — the subscripts 2 and 6 share the factor 2, so the formula reduces",
   "feedback": "This answer is wrong — you left the formula unchanged. The subscripts 2 and 6 share the factor 2, so the formula reduces: B: 2 ÷ 2 = 1, H: 6 ÷ 2 = 3, giving BH₃."
  },
  {
   "value": "BH",
   "misconception": "you divided each subscript by its own value, which destroys the ratio",
   "feedback": "This answer is wrong — you divided each subscript by its own value, which destroys the ratio. Every subscript is divided by the same factor, 2: B: 2 ÷ 2 = 1, H: 6 ÷ 2 = 3, giving BH₃."
  },
  {
   "value": "B₂H₃",
   "misconception": "you divided only the hydrogen's subscript — every subscript must be divided by the same factor",
   "feedback": "This answer is wrong — you divided only the hydrogen's subscript. Every subscript must be divided by the same factor: B: 2 ÷ 2 = 1, H: 6 ÷ 2 = 3, giving BH₃."
  }
 ]
}
MOL-044 · medium · transfer · seq 12 · seed:U7-G4/pool[26] · mcq4
A student divides a compound's molar mass, 132.0 g/mol, by its empirical formula's molar mass, 44.0 g/mol, and gets 3. What does the 3 mean?
One molecule of the compound contains 3 empirical-formula units. KEY
Each empirical-formula unit contains 3 molecules.
One molecule of the compound contains 3 atoms of each element.
The sample contains 3 mol of the compound.
MOL-041 · medium · application · seq 13 · seed:U7-G4.pool[17] (MOL-041 gateItems[2], 46.0 g Na / 32.1 g S / 64.0 g O -> 2 : 1 : 4) — VERBATIM (field mapping only; explanation_incorrect split into the two per-value misconception entries, wording preserved) · text
A sample of a compound contains 138.0 g of Na, 96.3 g of S, and 192.0 g of O. What is the simplest mole ratio of Na to S to O atoms? (M of Na = 23.0 g/mol, S = 32.1 g/mol, O = 16.0 g/mol.) Write your answer like 1 : 2 : 3.
Answer: 2 : 1 : 4 KEY
also accepted: 2:1:4
wrongValueDiagnostics
[
 {
  "value": "1.4 : 1 : 2",
  "misconception": "you compared the masses instead of the moles — 138.0 : 96.3 : 192.0 is about 1.4 : 1 : 2, but atoms are counted in moles",
  "feedback": "This answer is wrong — you compared the masses instead of the moles — 138.0 : 96.3 : 192.0 is about 1.4 : 1 : 2, but atoms are counted in moles."
 },
 {
  "value": "0.5 : 0.25 : 1",
  "misconception": "you divided every value by the largest instead of the smallest — divide by the smallest so the smallest becomes 1",
  "feedback": "This answer is wrong — you divided every value by the largest instead of the smallest — divide by the smallest so the smallest becomes 1."
 }
]
feedback
{
 "keyExplanation": "Values from the question:\nm(Na) = 138.0 g, m(S) = 96.3 g, m(O) = 192.0 g\nEquation:\nn = m/M  (for each element)\nSubstituting the values and calculating:\nn(Na) = 138.0 / 23.0 = 6.00 mol\nn(S) = 96.3 / 32.1 = 3.00 mol\nn(O) = 192.0 / 16.0 = 12.0 mol\nDivide each by the smallest (3.00): Na gives 2, S gives 1, O gives 4.\nMole ratio of sodium to sulfur to oxygen = 2 : 1 : 4"
}
MOL-044 · medium · application · seq 14 · seed:U7-G4.pool[24] (MOL-044 gateItems[0], P₂O₅/284.0 -> 2 units) — must_freshen_instance applied: instance swapped to C₃H₅/82.0 (cyclohexene C₆H₁₀; P₂O₅ = MOL-036 WE + MOL-039 fluency instance); structure, key (2), both misconception classes, scaffold, and sanity closer kept; '1 significant figures' sentence dropped (recorded contract cleanup) · numeric
A compound has the empirical formula C₃H₅ and a molar mass of 82.0 g/mol. (M of C = 12.0 g/mol, H = 1.0 g/mol.) How many empirical-formula units fit in one of its molecules? Answer with the whole-number count.
Answer: 2 units KEY
also accepted: 2 units, 2.0
wrongValueDiagnostics
[
 {
  "value": 6.31,
  "misconception": "you ignored the subscripts in the empirical formula, using 12.0 + 1.0 = 13.0 g/mol — M of C₃H₅ is 3 × 12.0 + 5 × 1.0 = 41.0 g/mol",
  "feedback": "This answer is wrong — you ignored the subscripts in the empirical formula, using 12.0 + 1.0 = 13.0 g/mol. M of C₃H₅ is 3 × 12.0 + 5 × 1.0 = 41.0 g/mol, and a non-whole unit count signals exactly this kind of slip."
 },
 {
  "value": 0.5,
  "misconception": "you divided upside down — the compound's molar mass goes on top, and the unit count is never below 1",
  "feedback": "This answer is wrong — you divided upside down — the compound's molar mass goes on top, and the unit count is never below 1."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmolar mass of the compound = 82.0 g/mol\nM of C₃H₅ = 3 × 12.0 + 5 × 1.0 = 41.0 g/mol\nWrite down the equation:\nnumber of units = molar mass of the compound ÷ molar mass of the empirical formula\nSubstitute in the values, and calculate:\nnumber of units = 82.0 / 41.0\nnumber of units = 2\nThe result is a whole number at least 1, so the division ran the right way up."
}
MOL-043 · hard · application · seq 15 · seed:U7-G4.pool[21] (MOL-043 gateItems[0] = demo_curated[0], James) VERBATIM — mass route with the 1.5-clearing step; markup transport only (<p> wrapper stripped, Unicode subscripts kept); Fe₂O₃ exclusive to this row (036 G0 spared at stage 1) · text
A sample of a compound contains 27.9 g of Fe and 12.0 g of O. What is its empirical formula? (M of Fe = 55.8 g/mol, O = 16.0 g/mol.) Type subscripts as plain numbers, like the 2 in CO2.
Answer: Fe₂O₃ KEY
also accepted: Fe2O3
wrongValueDiagnostics
[
 {
  "value": "FeO₂",
  "misconception": "you rounded the 1.5 up to 2 — rounding changes the ratio; multiply every number by 2 instead",
  "feedback": "This answer is wrong — you rounded the 1.5 up to 2. Rounding changes the ratio; multiply every number by 2 instead: 1 and 1.5 become 2 and 3, so the empirical formula is Fe₂O₃."
 },
 {
  "value": "FeO",
  "misconception": "you dropped the decimal — cutting the .5 off changes the ratio; multiply every number by 2 instead",
  "feedback": "This answer is wrong — you dropped the decimal. Cutting the .5 off changes the ratio; multiply every number by 2 instead: 1 and 1.5 become 2 and 3, so the empirical formula is Fe₂O₃."
 },
 {
  "value": "Fe₃O₂",
  "misconception": "you inverted the ratio — iron has the smaller mole value, so iron takes the 1 before clearing",
  "feedback": "This answer is wrong — you inverted the ratio. Iron has the smaller mole value (0.50 mol against oxygen's 0.75 mol), so iron takes the 1 before clearing: 1 and 1.5 become 2 and 3, giving Fe₂O₃."
 }
]
feedback
{
 "keyExplanation": "Values from the question:\nm(Fe) = 27.9 g, m(O) = 12.0 g\nEquation:\nn = m/M  (for each element)\nSubstituting the values and calculating:\nn(Fe) = 27.9 / 55.8 = 0.50 mol\nn(O) = 12.0 / 16.0 = 0.75 mol\nDivide each by the smallest (0.50): Fe gives 1, O gives 1.5.\nThe decimal ends in .5, so multiply every number by 2: 2 and 3.\nEmpirical formula: Fe₂O₃",
 "anticipatedWrong": [
  {
   "value": "FeO₂",
   "misconception": "you rounded the 1.5 up to 2 — rounding changes the ratio; multiply every number by 2 instead",
   "feedback": "This answer is wrong — you rounded the 1.5 up to 2. Rounding changes the ratio; multiply every number by 2 instead: 1 and 1.5 become 2 and 3, so the empirical formula is Fe₂O₃."
  },
  {
   "value": "FeO",
   "misconception": "you dropped the decimal — cutting the .5 off changes the ratio; multiply every number by 2 instead",
   "feedback": "This answer is wrong — you dropped the decimal. Cutting the .5 off changes the ratio; multiply every number by 2 instead: 1 and 1.5 become 2 and 3, so the empirical formula is Fe₂O₃."
  },
  {
   "value": "Fe₃O₂",
   "misconception": "you inverted the ratio — iron has the smaller mole value, so iron takes the 1 before clearing",
   "feedback": "This answer is wrong — you inverted the ratio. Iron has the smaller mole value (0.50 mol against oxygen's 0.75 mol), so iron takes the 1 before clearing: 1 and 1.5 become 2 and 3, giving Fe₂O₃."
  }
 ]
}
MOL-042 · hard · transfer · seq 16 · seed:U7-G4/pool[20] (= MOL-042 gateItems[2], gate G2 why-multiply MCQ) · mcq4
A mole calculation gives the ratio 1 : 1.75. One student multiplies both numbers by 4; another rounds the 1.75 up to 2. Which result is right, and why?
4 : 7 — multiplying every number by 4 keeps the ratio unchanged, while rounding 1.75 to 2 changes it. KEY
1 : 2 — a decimal that close to a whole number is within normal rounding distance.
1 : 7 — multiply only the 1.75 by 4, because the 1 is already a whole number.
1 : 1 — drop the decimal part, since subscripts must be whole numbers.
MOL-045 · hard · application · seq 17 · seed:U7-G4/pool[27] (= MOL-045 gateItems[0], gate G0 = demo_curated[3], James) — must_freshen_instance executed: NO₂ / 92.0 -> N₂O₄ swapped to CH₃O / 62.0 -> C₂H₆O₂ (blueprint candidate 1, the stage-1 reserved pair); structure, atomic-masses-supplied register, and all three misconception classes kept; all fields swept for the old instance (QB 3.9) · text
A compound has the empirical formula CH₃O and a molar mass of 62.0 g/mol. (M of C = 12.0 g/mol, H = 1.0 g/mol, O = 16.0 g/mol.) State the compound's molecular formula, typing subscripts as plain numbers (like the 2 in CO2).
Answer: C₂H₆O₂ KEY
also accepted: C2H6O2
wrongValueDiagnostics
[
 {
  "value": "CH₃O",
  "misconception": "you stopped at the empirical formula — the unit count is 62.0/31.0 = 2, and every subscript must be multiplied by it",
  "feedback": "This answer is wrong — you stopped at the empirical formula. The unit count is 62.0 / 31.0 = 2, and every subscript must be multiplied by it: C: 1 × 2 = 2, H: 3 × 2 = 6, O: 1 × 2 = 2, giving C₂H₆O₂."
 },
 {
  "value": "C₂H₃O",
  "misconception": "you multiplied only carbon's subscript — every subscript in the empirical formula is multiplied by the unit count",
  "feedback": "This answer is wrong — you multiplied only carbon's subscript. Every subscript in the empirical formula is multiplied by the unit count, 2: C: 1 × 2 = 2, H: 3 × 2 = 6, O: 1 × 2 = 2, giving C₂H₆O₂."
 },
 {
  "value": "CH₃O₂",
  "misconception": "you multiplied only oxygen's subscript — every subscript in the empirical formula is multiplied by the unit count",
  "feedback": "This answer is wrong — you multiplied only oxygen's subscript. Every subscript in the empirical formula is multiplied by the unit count, 2: C: 1 × 2 = 2, H: 3 × 2 = 6, O: 1 × 2 = 2, giving C₂H₆O₂."
 }
]
feedback
{
 "keyExplanation": "M of CH₃O = 12.0 + 3 × 1.0 + 16.0 = 31.0 g/mol.\nStep 1 — find the number of units:\nnumber of units = molar mass of the compound ÷ molar mass of the empirical formula\nnumber of units = 62.0 / 31.0 = 2\nStep 2 — multiply every subscript by 2:\nC: 1 × 2 = 2,  H: 3 × 2 = 6,  O: 1 × 2 = 2\nMolecular formula: C₂H₆O₂",
 "anticipatedWrong": [
  {
   "value": "CH₃O",
   "misconception": "you stopped at the empirical formula — the unit count is 62.0/31.0 = 2, and every subscript must be multiplied by it",
   "feedback": "This answer is wrong — you stopped at the empirical formula. The unit count is 62.0 / 31.0 = 2, and every subscript must be multiplied by it: C: 1 × 2 = 2, H: 3 × 2 = 6, O: 1 × 2 = 2, giving C₂H₆O₂."
  },
  {
   "value": "C₂H₃O",
   "misconception": "you multiplied only carbon's subscript — every subscript in the empirical formula is multiplied by the unit count",
   "feedback": "This answer is wrong — you multiplied only carbon's subscript. Every subscript in the empirical formula is multiplied by the unit count, 2: C: 1 × 2 = 2, H: 3 × 2 = 6, O: 1 × 2 = 2, giving C₂H₆O₂."
  },
  {
   "value": "CH₃O₂",
   "misconception": "you multiplied only oxygen's subscript — every subscript in the empirical formula is multiplied by the unit count",
   "feedback": "This answer is wrong — you multiplied only oxygen's subscript. Every subscript in the empirical formula is multiplied by the unit count, 2: C: 1 × 2 = 2, H: 3 × 2 = 6, O: 1 × 2 = 2, giving C₂H₆O₂."
  }
 ]
}
MOL-043 · hard · application · seq 18 · seed:U7-G4/pool[23] (= MOL-043 gateItems[2] = demo_curated[2], James; no freshen flag — N₂O₅/25.9/74.1 unique to this gate item in the scope corpus): stem, answer, plain-digit alternate, explanation and both misconception classes carried verbatim; notation transport HTML sub markup -> Unicode only · text
A compound is 25.9% N and 74.1% O by mass. What is its empirical formula? (M of N = 14.0 g/mol, O = 16.0 g/mol.) Type subscripts as plain numbers, like the 2 in CO2.
Answer: N₂O₅ KEY
also accepted: N2O5
wrongValueDiagnostics
[
 {
  "value": "NO₂",
  "misconception": "you rounded the 2.50 down to 2 — rounding changes the ratio; multiply every number by 2 instead",
  "feedback": "This answer is wrong — you rounded the 2.50 down to 2. Rounding changes the ratio; multiply every number by 2 instead: N gives 2 and O gives 5, so the empirical formula is N₂O₅."
 },
 {
  "value": "NO₃",
  "misconception": "you rounded the 2.50 up to 3 — rounding changes the ratio; multiply every number by 2 instead",
  "feedback": "This answer is wrong — you rounded the 2.50 up to 3. Rounding changes the ratio; multiply every number by 2 instead: N gives 2 and O gives 5, so the empirical formula is N₂O₅."
 }
]
feedback
{
 "keyExplanation": "Assume a 100 g sample: m(N) = 25.9 g, m(O) = 74.1 g.\nEquation:\nn = m/M  (for each element)\nSubstituting the values and calculating:\nn(N) = 25.9 / 14.0 = 1.85 mol\nn(O) = 74.1 / 16.0 = 4.63 mol\nDivide each by the smallest (1.85): N gives 1, O gives 2.50.\nThe decimal ends in .5, so multiply every number by 2: 2 and 5.\nEmpirical formula: N₂O₅",
 "anticipatedWrong": [
  {
   "value": "NO₂",
   "misconception": "you rounded the 2.50 down to 2 — rounding changes the ratio; multiply every number by 2 instead",
   "feedback": "This answer is wrong — you rounded the 2.50 down to 2. Rounding changes the ratio; multiply every number by 2 instead: N gives 2 and O gives 5, so the empirical formula is N₂O₅."
  },
  {
   "value": "NO₃",
   "misconception": "you rounded the 2.50 up to 3 — rounding changes the ratio; multiply every number by 2 instead",
   "feedback": "This answer is wrong — you rounded the 2.50 up to 3. Rounding changes the ratio; multiply every number by 2 instead: N gives 2 and O gives 5, so the empirical formula is N₂O₅."
  }
 ]
}
MOL-039 · hard · transfer · seq 19 · seed:U7-G4.pool[10] (MOL-039 gateItems[1], C4H9 claim) — instance freshened per placement candidate (C4H9 = MOL-045 fluency F2 corpus → C9H20, nonane: subscripts 9 and 20 each composite yet coprime, claim logic preserved exactly); feedback counterexample also freshened (C2H6 = own-lesson fluency F1 corpus + burned ledger → C10H22, decane, which reduces); invented-rule / divided-by-own-factor / right-verdict-wrong-reason misconception classes kept verbatim · mcq4
A student claims C₉H₂₀ cannot be an empirical formula 'because 9 and 20 are both bigger than 1.' Judge the claim.
The claim is wrong — 9 and 20 share no common factor bigger than 1, so C₉H₂₀ is an empirical formula. KEY
The claim is right — an empirical formula must contain at least one unwritten subscript of 1.
The claim is right — 9 and 20 can each be divided by a smaller number, so the formula reduces.
The claim is wrong — no formula with only two subscripts can ever be reduced.
MOL-045 · hard · transfer · seq 20 · seed:U7-G4/pool[28] (= MOL-045 gateItems[1], gate G1 = demo_curated[4], James-curated x1 edge MCQ) · mcq4
A compound has the empirical formula CH₃Cl and a molar mass of 50.5 g/mol. (M of CH₃Cl = 50.5 g/mol.) What is its molecular formula?
CH₃Cl KEY
C₂H₆Cl₂
C₂H₃Cl
C₃H₉Cl₃
MOL-043 · hard · application · seq 21 · seed:U7-G4.pool[22] (= demo_curated[1], MOL-043 gateItems[1]: 12.0 g C / 2.0 g H / 16.0 g O -> CH₂O) EXPANDED to 3 lettered parts + rubric 40/30/30 + fully-typed model answer; must_freshen_instance EXECUTED — CH₂O (the 036/044/045 article register workhorse) and its recognizable one-mole masses replaced by CaCO₃ 80.2/24.0/96.0 (stage-1 candidate masses declined for recorded contract reasons; case shape, route, and step count carried exactly) · frq
A sample of a compound contains 80.2 g of Ca, 24.0 g of C, and 96.0 g of O. (M of Ca = 40.1 g/mol, C = 12.0 g/mol, O = 16.0 g/mol.)
A sample of a compound contains 80.2 g of Ca, 24.0 g of C, and 96.0 g of O. (M of Ca = 40.1 g/mol, C = 12.0 g/mol, O = 16.0 g/mol.) (a) Calculate the number of moles of each element in the sample. Show your working. (b) Divide every mole value by the smallest to find the whole number for each element, clearing any decimal that remains. Show each step. (c) Write the compound's empirical formula, and state how you know its subscripts are exact whole numbers rather than rounded ones. Type subscripts as plain numbers, like the 2 in CO2.
Part a. Calculate the number of moles of each element in the sample. Show your working.
Model: Values from the question: m(Ca) = 80.2 g, m(C) = 24.0 g, m(O) = 96.0 g Equation: n = m/M (for each element) Substituting the values and calculating: n(Ca) = 80.2 / 40.1 = 2.00 mol n(C) = 24.0 / 12.0 = 2.00 mol n(O) = 96.0 / 16.0 = 6.00 mol
Part b. Divide every mole value by the smallest to find the whole number for each element, clearing any decimal that remains. Show each step.
Model: Divide each mole value by the smallest (2.00): Ca: 2.00 / 2.00 = 1 C: 2.00 / 2.00 = 1 O: 6.00 / 2.00 = 3 No decimal remains — Ca gives 1, C gives 1, O gives 3.
Part c. Write the compound's empirical formula, and state how you know its subscripts are exact whole numbers rather than rounded ones. Type subscripts as plain numbers, like the 2 in CO2.
Model: The finished whole numbers are the subscripts, so the empirical formula is CaCO3. The subscripts are exact: every division landed on a whole number, so nothing was rounded and no multiplying step was needed.
Full model answer. (a) Values from the question: m(Ca) = 80.2 g, m(C) = 24.0 g, m(O) = 96.0 g Equation: n = m/M (for each element) Substituting the values and calculating: n(Ca) = 80.2 / 40.1 = 2.00 mol n(C) = 24.0 / 12.0 = 2.00 mol n(O) = 96.0 / 16.0 = 6.00 mol (b) Divide each mole value by the smallest (2.00): Ca: 2.00 / 2.00 = 1 C: 2.00 / 2.00 = 1 O: 6.00 / 2.00 = 3 No decimal remains — Ca gives 1, C gives 1, O gives 3. (c) The finished whole numbers are the subscripts, so the empirical formula is CaCO3. The subscripts are exact: every division landed on a whole number, so nothing was rounded and no multiplying step was needed.
MOL-045 · hard · application · seq 22 · seed:U7-G4.pool[29] (MOL-045 gateItems[2] = demo_curated[5], CH₂/84.0 -> C₆H₁₂) — EXPANDED from single-answer gate text item to 3 lettered parts + rubric (recorded method); must_freshen_instance APPLIED: instance swapped to C₅H₄/128.0 -> C₁₀H₈ (naphthalene; CH₂ = MOL-044/045 WE family, C₆H₁₂ = MOL-038 fluency cyclohexane); same skill, production format, both standing misconception classes carried into the rubric observables · frq
A compound has the empirical formula C₅H₄ and a molar mass of 128.0 g/mol. (M of C = 12.0 g/mol, H = 1.0 g/mol.)
A compound has the empirical formula C₅H₄ and a molar mass of 128.0 g/mol. (M of C = 12.0 g/mol, H = 1.0 g/mol.) (a) Calculate the molar mass of the empirical formula C₅H₄ from the atomic masses given, showing the sum you evaluate. (b) Find the number of empirical-formula units in one molecule of the compound. State the equation you use, and show every step of your working. (c) Write the molecular formula of the compound, and show how each subscript in your answer was produced. Type subscripts as plain numbers, like the 2 in CO2; no other special symbols are needed.
Part a. Calculate the molar mass of the empirical formula C₅H₄ from the atomic masses given, showing the sum you evaluate.
Model: Values from the question: M of C = 12.0 g/mol, M of H = 1.0 g/mol The empirical formula C5H4 contains 5 C and 4 H. M of C5H4 = 5 × 12.0 + 4 × 1.0 M of C5H4 = 64.0 g/mol
Part b. Find the number of empirical-formula units in one molecule of the compound. State the equation you use, and show every step of your working.
Model: Values from the question: molar mass of the compound = 128.0 g/mol molar mass of the empirical formula = 64.0 g/mol (from part a) Equation: number of units = molar mass of the compound ÷ molar mass of the empirical formula Substituting the values and calculating: number of units = 128.0 / 64.0 number of units = 2 The result is a whole number of at least 1, so the division ran the right way up.
Part c. Write the molecular formula of the compound, and show how each subscript in your answer was produced. Type subscripts as plain numbers, like the 2 in CO2; no other special symbols are needed.
Model: Multiply every subscript in the empirical formula by 2: C: 5 × 2 = 10, H: 4 × 2 = 8 Molecular formula: C10H8
Full model answer. (a) Values from the question: M of C = 12.0 g/mol, M of H = 1.0 g/mol The empirical formula C5H4 contains 5 C and 4 H. M of C5H4 = 5 × 12.0 + 4 × 1.0 M of C5H4 = 64.0 g/mol (b) Values from the question: molar mass of the compound = 128.0 g/mol molar mass of the empirical formula = 64.0 g/mol (from part a) Equation: number of units = molar mass of the compound ÷ molar mass of the empirical formula Substituting the values and calculating: number of units = 128.0 / 64.0 number of units = 2 The result is a whole number of at least 1, so the division ran the right way up. (c) Multiply every subscript in the empirical formula by 2: C: 5 × 2 = 10, H: 4 × 2 = 8 Molecular formula: C10H8

Form B

MOL-036 · easy · recall · seq 1 · authored (seedless recall row; definition-completion probe; completed sentence = the taught rule line verbatim) · mcq4
Which phrase completes the definition correctly? 'An empirical formula gives the smallest whole-number ______ in a compound.'
ratio of the atoms of each element KEY
count of the atoms of each element
ratio of the masses of each element
ratio of the molecules of each element
MOL-037 · easy · recall · seq 2 · seedless authored row (probe B: molecular-vs-empirical difference statement; taught contrast per MOL-037 rule lines; option language from fluency non-stem fields, stem fresh) · mcq4
A compound has both a molecular formula and an empirical formula. How do the two formulas differ?
The molecular formula gives the actual atom counts in one molecule; the empirical formula gives only the smallest whole-number ratio. KEY
The empirical formula gives the actual atom counts in one molecule; the molecular formula gives only the smallest whole-number ratio.
The molecular formula gives the mass of one molecule; the empirical formula counts its atoms.
The two formulas always contain the same subscripts written in a different order.
MOL-039 · easy · recall · seq 3 · authored (seedless recall row; unwritten-subscript-convention probe; the extraNotes' load-bearing line — d3 is the 'no subscript means not checkable' error that line defuses, d1/d2 are the zero and skip mishandlings) · mcq4
One element in a formula shows no written subscript. What does the empirical check do with that element?
Count its subscript as 1 and include it with the other subscripts when looking for a shared factor. KEY
Count its subscript as 0, since nothing is written.
Leave that element out entirely and run the shared-factor check on the written subscripts only.
Stop — the check cannot be run until every subscript is written out.
MOL-037 · easy · application · seq 4 · authored (parallel to the seq-4 seed structure; fresh instance heptane C₇H₁₆, outside the burned-instance ledger; corpus scan 0 hits) · mcq4
The molecular formula of heptane is C₇H₁₆. Which statement does the formula guarantee about one heptane molecule?
It contains exactly 7 C atoms and 16 H atoms. KEY
The subscripts show proportions that were divided down from the molecule's true, larger atom counts.
It contains exactly 16 C atoms and 7 H atoms.
It guarantees the counts only for a perfectly pure sample — any impurity changes the subscripts.
MOL-040 · easy · application · seq 5 · authored (parallel numeric slot; fresh chemically-real composition — dinitrogen monoxide N₂O, 63.6% N by mass; compound unnamed in the stem per the lesson convention) · numeric
A compound is 63.6% N by mass; the rest is O. What mass of O does a 100 g sample of the compound contain? Give your answer in grams to one decimal place.
Answer: 36.4 g KEY
also accepted: 36.4 g
wrongValueDiagnostics
[
 {
  "value": 63.6,
  "misconception": "you gave the nitrogen's mass — the question asks for the oxygen, the rest of the 100 g",
  "feedback": "This answer is wrong — you gave the nitrogen's mass — the question asks for the oxygen, the rest of the 100 g."
 },
 {
  "value": 0.364,
  "misconception": "you divided by 100 — in a 100 g sample the masses are in grams already",
  "feedback": "This answer is wrong — you divided by 100 — in a 100 g sample the masses are in grams already."
 }
]
feedback
{
 "keyExplanation": "Assume a sample of exactly 100 g.\nNitrogen: 63.6% of the compound → 63.6 g.\nThe element masses in a 100 g sample total 100 g.\nmass of oxygen = 100 − 63.6 = 36.4 g"
}
MOL-036 · medium · application · seq 6 · authored (parallel mcq slot; fresh reducible measured ratio 3:6 (÷3) + fresh element pair Ba/F — BaF₂ is a real compound (barium fluoride), outside the burned-instance ledger; barium and fluorine appear nowhere in the scope corpus or on form A) · mcq4
An instrument reports that a compound contains 3 Ba atoms for every 6 F atoms. Which formula is the compound's empirical formula?
BaF₂ KEY
Ba₃F₆
Ba₂F
BaF₆
MOL-038 · medium · application · seq 7 · authored (blueprint seq-7 B-E: fresh genuinely-reducible formula, largest-factor trap live at factor 4; octafluorocyclobutane C₄F₈ -> CF₂ — fluorocarbon system fresh in the scope corpus and outside the burned-instance ledger; misconception classes from the seed carried: divide-by-2-not-largest / left-unchanged / divided-each-by-own-value) · text
What is the empirical formula of octafluorocyclobutane, C₄F₈? Type the whole formula, writing any subscripts as plain digits, like the 2 in CO2; leave subscripts of 1 unwritten.
Answer: CF₂ KEY
also accepted: CF2
wrongValueDiagnostics
[
 {
  "value": "C₂F₄",
  "misconception": "you divided by 2, which is a common factor but not the LARGEST one — 2 and 4 still share the factor 2",
  "feedback": "This answer is wrong — you divided by 2, which is a common factor but not the LARGEST one. The subscripts of C₂F₄ — 2 and 4 — still share the factor 2, so the formula still reduces. Divide C₄F₈'s subscripts by their largest common factor, 4: the empirical formula is CF₂."
 },
 {
  "value": "C₄F₈",
  "misconception": "you left the formula unchanged — the subscripts 4 and 8 share the factor 4, so the formula reduces",
  "feedback": "This answer is wrong — you left the formula unchanged. The subscripts 4 and 8 share the factor 4, so the formula reduces: C: 4 ÷ 4 = 1, F: 8 ÷ 4 = 2, and the empirical formula is CF₂."
 },
 {
  "value": "CF",
  "misconception": "you divided each subscript by its own value, which destroys the ratio",
  "feedback": "This answer is wrong — you divided each subscript by its own value, which destroys the ratio. Every subscript must be divided by the SAME factor — the largest common factor, 4: C: 4 ÷ 4 = 1, F: 8 ÷ 4 = 2, so the empirical formula is CF₂."
 }
]
feedback
{
 "keyExplanation": "Subscripts: C 4, F 8. The largest common factor is 4 — dividing by 2 alone leaves C₂F₄, which still reduces. C: 4 ÷ 4 = 1, F: 8 ÷ 4 = 2, and a subscript of 1 is never written, so the empirical formula is CF₂.",
 "anticipatedWrong": [
  {
   "value": "C₂F₄",
   "misconception": "you divided by 2, which is a common factor but not the LARGEST one — 2 and 4 still share the factor 2",
   "feedback": "This answer is wrong — you divided by 2, which is a common factor but not the LARGEST one. The subscripts of C₂F₄ — 2 and 4 — still share the factor 2, so the formula still reduces. Divide C₄F₈'s subscripts by their largest common factor, 4: the empirical formula is CF₂."
  },
  {
   "value": "C₄F₈",
   "misconception": "you left the formula unchanged — the subscripts 4 and 8 share the factor 4, so the formula reduces",
   "feedback": "This answer is wrong — you left the formula unchanged. The subscripts 4 and 8 share the factor 4, so the formula reduces: C: 4 ÷ 4 = 1, F: 8 ÷ 4 = 2, and the empirical formula is CF₂."
  },
  {
   "value": "CF",
   "misconception": "you divided each subscript by its own value, which destroys the ratio",
   "feedback": "This answer is wrong — you divided each subscript by its own value, which destroys the ratio. Every subscript must be divided by the SAME factor — the largest common factor, 4: C: 4 ÷ 4 = 1, F: 8 ÷ 4 = 2, so the empirical formula is CF₂."
  }
 ]
}
MOL-039 · medium · application · seq 8 · authored (pick-empirical polarity; fresh formula set — key C₅H₈ coprime pair, foils reduce by factors 2/3/2 with each factor named in feedback; NaO result mirrors the taught H₂O₂→HO pattern on a fresh instance) · mcq4
Which formula is already an empirical formula?
C₅H₈ KEY
Si₂H₆
C₆H₁₂O₃
Na₂O₂
MOL-041 · medium · application · seq 9 · authored (fresh two-element pair, PF₅ family; parallel to seed pool[16] incl. its mass-equals-M design giving one 1.00 mol value; exact divisions 0.20 / 1.00; whole-number outcome by construction) · text
A sample of a compound contains 21.7 g of P and 66.5 g of F. What is the simplest mole ratio of P atoms to F atoms? (M of P = 31.0 g/mol, F = 19.0 g/mol.) Write your answer like 1 : 2.
Answer: 1 : 5 KEY
also accepted: 1:5
wrongValueDiagnostics
[
 {
  "value": "1 : 3.1",
  "misconception": "you compared the masses instead of the moles — 21.7 : 66.5 is about 1 : 3.1, but atoms are counted in moles",
  "feedback": "This answer is wrong — you compared the masses instead of the moles. 21.7 : 66.5 is about 1 : 3.1, but atoms are counted in moles: n(P) = 21.7 / 31.0 = 0.70 mol and n(F) = 66.5 / 19.0 = 3.50 mol, and dividing each by 0.70 gives 1 : 5."
 },
 {
  "value": "5 : 1",
  "misconception": "you wrote the ratio backwards — the order follows the elements as named: phosphorus first, then fluorine",
  "feedback": "This answer is wrong — you wrote the ratio backwards. The ratio order follows the elements as they are named in the question: phosphorus first, then fluorine, so the answer is 1 : 5."
 }
]
feedback
{
 "keyExplanation": "Values from the question:\nm(P) = 21.7 g, m(F) = 66.5 g\nEquation:\nn = m/M  (for each element)\nSubstituting the values and calculating:\nn(P) = 21.7 / 31.0 = 0.70 mol\nn(F) = 66.5 / 19.0 = 3.50 mol\nDivide each by the smallest (0.70): P gives 1, F gives 5.\nMole ratio of phosphorus to fluorine = 1 : 5",
 "anticipatedWrong": [
  {
   "value": "1 : 3.1",
   "misconception": "you compared the masses instead of the moles — 21.7 : 66.5 is about 1 : 3.1, but atoms are counted in moles",
   "feedback": "This answer is wrong — you compared the masses instead of the moles. 21.7 : 66.5 is about 1 : 3.1, but atoms are counted in moles: n(P) = 21.7 / 31.0 = 0.70 mol and n(F) = 66.5 / 19.0 = 3.50 mol, and dividing each by 0.70 gives 1 : 5."
  },
  {
   "value": "5 : 1",
   "misconception": "you wrote the ratio backwards — the order follows the elements as named: phosphorus first, then fluorine",
   "feedback": "This answer is wrong — you wrote the ratio backwards. The ratio order follows the elements as they are named in the question: phosphorus first, then fluorine, so the answer is 1 : 5."
  }
 ]
}
MOL-042 · medium · application · seq 10 · authored (fresh ratio 1 : 3.75 -> 4 : 15, x4, .75 branch — rule line 5's other decimal, no placed item covers it; 3.75 and 4 : 15 outside the burned ledger and machine-checked fresh vs all 10 lessons' spec text; seed structure and both misconception classes carried with the natural round-UP direction for .75) · text
A mole calculation gives the ratio 1 : 3.75. What whole-number ratio does it clear to? Give your answer like 2 : 3.
Answer: 4 : 15 KEY
also accepted: 4:15
wrongValueDiagnostics
[
 {
  "value": "1 : 4",
  "misconception": "you rounded the 3.75 up to 4 — rounding changes the ratio; multiply every number by 4 instead",
  "feedback": "This answer is wrong — you rounded the 3.75 up to 4, and rounding changes the ratio itself. The decimal part is .75, so multiply every number by 4 instead: 1 × 4 = 4 and 3.75 × 4 = 15, giving 4 : 15."
 },
 {
  "value": "1 : 15",
  "misconception": "you multiplied only the decimal number — every number in the ratio must be multiplied by the same amount",
  "feedback": "This answer is wrong — you multiplied only the decimal number. Every number in the ratio must be multiplied by the same amount to keep the ratio unchanged: 1 × 4 = 4 and 3.75 × 4 = 15, giving 4 : 15."
 }
]
feedback
{
 "keyExplanation": "The decimal part is .75, so multiply every number by 4: 1 × 4 = 4 and 3.75 × 4 = 15. The whole-number ratio is 4 : 15. Rounding 3.75 up to 4 would change the ratio itself — multiplying never does.",
 "anticipatedWrong": [
  {
   "value": "1 : 4",
   "misconception": "you rounded the 3.75 up to 4 — rounding changes the ratio; multiply every number by 4 instead",
   "feedback": "This answer is wrong — you rounded the 3.75 up to 4, and rounding changes the ratio itself. The decimal part is .75, so multiply every number by 4 instead: 1 × 4 = 4 and 3.75 × 4 = 15, giving 4 : 15."
  },
  {
   "value": "1 : 15",
   "misconception": "you multiplied only the decimal number — every number in the ratio must be multiplied by the same amount",
   "feedback": "This answer is wrong — you multiplied only the decimal number. Every number in the ratio must be multiplied by the same amount to keep the ratio unchanged: 1 × 4 = 4 and 3.75 × 4 = 15, giving 4 : 15."
  }
 ]
}
MOL-038 · medium · application · seq 11 · authored (parallel to the seq-11 seed structure; fresh instance oxalic acid C₂H₂O₄ -> CHO₂, corpus scan 0 hits; 3-element reduce rotated in per the blueprint — the spared ribose mcq (pool[7]) documents the family, and its left-the-last-subscript-undivided class replaces the two-element divide-only-hydrogen class) · text
What is the empirical formula of oxalic acid, C₂H₂O₄? Type the whole formula, writing any subscripts as plain digits, like the 2 in CO2; leave subscripts of 1 unwritten.
Answer: CHO₂ KEY
also accepted: CHO2
wrongValueDiagnostics
[
 {
  "value": "C₂H₂O₄",
  "misconception": "you left the formula unchanged — the subscripts 2, 2, and 4 share the factor 2, so the formula reduces",
  "feedback": "This answer is wrong — you left the formula unchanged. The subscripts 2, 2, and 4 share the factor 2, so the formula reduces: C: 2 ÷ 2 = 1, H: 2 ÷ 2 = 1, O: 4 ÷ 2 = 2, giving CHO₂."
 },
 {
  "value": "CHO",
  "misconception": "you divided each subscript by its own value, which destroys the ratio",
  "feedback": "This answer is wrong — you divided each subscript by its own value, which destroys the ratio. Every subscript is divided by the same factor, 2: C 1, H 1, O 2, giving CHO₂."
 },
 {
  "value": "CHO₄",
  "misconception": "you left the last subscript undivided — every subscript must be divided by the same factor, including oxygen's",
  "feedback": "This answer is wrong — you left the last subscript undivided. Every subscript must be divided by the same factor, including oxygen's: O: 4 ÷ 2 = 2, giving CHO₂."
 }
]
feedback
{
 "keyExplanation": "Subscripts: C 2, H 2, O 4. The largest common factor is 2. C: 2 ÷ 2 = 1, H: 2 ÷ 2 = 1, O: 4 ÷ 2 = 2. A subscript of 1 is never written, so oxalic acid's empirical formula is CHO₂.",
 "anticipatedWrong": [
  {
   "value": "C₂H₂O₄",
   "misconception": "you left the formula unchanged — the subscripts 2, 2, and 4 share the factor 2, so the formula reduces",
   "feedback": "This answer is wrong — you left the formula unchanged. The subscripts 2, 2, and 4 share the factor 2, so the formula reduces: C: 2 ÷ 2 = 1, H: 2 ÷ 2 = 1, O: 4 ÷ 2 = 2, giving CHO₂."
  },
  {
   "value": "CHO",
   "misconception": "you divided each subscript by its own value, which destroys the ratio",
   "feedback": "This answer is wrong — you divided each subscript by its own value, which destroys the ratio. Every subscript is divided by the same factor, 2: C 1, H 1, O 2, giving CHO₂."
  },
  {
   "value": "CHO₄",
   "misconception": "you left the last subscript undivided — every subscript must be divided by the same factor, including oxygen's",
   "feedback": "This answer is wrong — you left the last subscript undivided. Every subscript must be divided by the same factor, including oxygen's: O: 4 ÷ 2 = 2, giving CHO₂."
  }
 ]
}
MOL-044 · medium · transfer · seq 12 · authored (quotient-meaning, calculator-computation frame; parallel misconception classes to the seed; fresh exact pair 174.0/43.5 = 4) · mcq4
For a compound with a molar mass of 174.0 g/mol and an empirical-formula molar mass of 43.5 g/mol, a student computes 174.0 ÷ 43.5 = 4. What does the 4 count?
The number of empirical-formula units that make up one molecule of the compound. KEY
The number of molecules that make up one empirical-formula unit.
The number of atoms of each element in one molecule of the compound.
The number of moles of the compound the student measured out.
MOL-041 · medium · application · seq 13 · authored (parallel to the seq-13 seed structure; fresh triple — lithium carbonate Li₂CO₃, unnamed in the stem per the lesson convention; smallest mole value on the MIDDLE-listed element; corpus scan 0 hits; first candidate K₂SiO₃ dropped on the 'silicon' corpus hit) · text
A sample of a compound contains 20.7 g of Li, 18.0 g of C, and 72.0 g of O. What is the simplest mole ratio of Li to C to O atoms? (M of Li = 6.9 g/mol, C = 12.0 g/mol, O = 16.0 g/mol.) Write your answer like 1 : 2 : 3.
Answer: 2 : 1 : 3 KEY
also accepted: 2:1:3
wrongValueDiagnostics
[
 {
  "value": "1.2 : 1 : 4",
  "misconception": "you compared the masses instead of the moles — 20.7 : 18.0 : 72.0 is about 1.2 : 1 : 4, but atoms are counted in moles",
  "feedback": "This answer is wrong — you compared the masses instead of the moles — 20.7 : 18.0 : 72.0 is about 1.2 : 1 : 4, but atoms are counted in moles."
 },
 {
  "value": "0.67 : 0.33 : 1",
  "misconception": "you divided every value by the largest instead of the smallest — divide by the smallest so the smallest becomes 1",
  "feedback": "This answer is wrong — you divided every value by the largest instead of the smallest — divide by the smallest so the smallest becomes 1."
 }
]
feedback
{
 "keyExplanation": "Values from the question:\nm(Li) = 20.7 g, m(C) = 18.0 g, m(O) = 72.0 g\nEquation:\nn = m/M  (for each element)\nSubstituting the values and calculating:\nn(Li) = 20.7 / 6.9 = 3.00 mol\nn(C) = 18.0 / 12.0 = 1.50 mol\nn(O) = 72.0 / 16.0 = 4.50 mol\nDivide each by the smallest (1.50): Li gives 2, C gives 1, O gives 3.\nMole ratio of lithium to carbon to oxygen = 2 : 1 : 3"
}
MOL-044 · medium · application · seq 14 · authored (parallel numeric slot; fresh (formula, M) pair — 1,3,5-trichlorobenzene C₆H₃Cl₃, empirical C₂HCl, 181.5 g/mol -> 3 units; compound unnamed in the stem per the lesson's gate convention) · numeric
A compound has the empirical formula C₂HCl and a molar mass of 181.5 g/mol. (M of C = 12.0 g/mol, H = 1.0 g/mol, Cl = 35.5 g/mol.) How many empirical-formula units fit in one of its molecules? Answer with the whole-number count.
Answer: 3 units KEY
also accepted: 3 units, 3.0
wrongValueDiagnostics
[
 {
  "value": 3.74,
  "misconception": "you ignored the subscript in the empirical formula, using 12.0 + 1.0 + 35.5 = 48.5 g/mol — M of C₂HCl is 2 × 12.0 + 1.0 + 35.5 = 60.5 g/mol",
  "feedback": "This answer is wrong — you ignored the subscript in the empirical formula, using 12.0 + 1.0 + 35.5 = 48.5 g/mol. M of C₂HCl is 2 × 12.0 + 1.0 + 35.5 = 60.5 g/mol, and a non-whole unit count signals exactly this kind of slip."
 },
 {
  "value": 0.33,
  "misconception": "you divided upside down — the compound's molar mass goes on top, and the unit count is never below 1",
  "feedback": "This answer is wrong — you divided upside down — the compound's molar mass goes on top, and the unit count is never below 1."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmolar mass of the compound = 181.5 g/mol\nM of C₂HCl = 2 × 12.0 + 1.0 + 35.5 = 60.5 g/mol\nWrite down the equation:\nnumber of units = molar mass of the compound ÷ molar mass of the empirical formula\nSubstitute in the values, and calculate:\nnumber of units = 181.5 / 60.5\nnumber of units = 3\nThe result is a whole number at least 1, so the division ran the right way up."
}
MOL-043 · hard · application · seq 15 · authored (parallel mass-route capstone; fresh chemically-real compound — aluminum oxide Al₂O₃, compound unnamed in the stem per the seed convention; exact divisions 40.5/27.0 = 1.50 and 36.0/16.0 = 2.25; .5-decimal clearing ×2 for step-count parity with the seed) · text
A sample of a compound contains 94.5 g of Al and 84.0 g of O. What is its empirical formula? (M of Al = 27.0 g/mol, O = 16.0 g/mol.) Type subscripts as plain numbers, like the 2 in CO2.
Answer: Al₂O₃ KEY
also accepted: Al2O3
wrongValueDiagnostics
[
 {
  "value": "AlO₂",
  "misconception": "you rounded the 1.5 up to 2 — rounding changes the ratio; multiply every number by 2 instead",
  "feedback": "This answer is wrong — you rounded the 1.5 up to 2. Rounding changes the ratio; multiply every number by 2 instead: 1 and 1.5 become 2 and 3, so the empirical formula is Al₂O₃."
 },
 {
  "value": "AlO",
  "misconception": "you dropped the decimal — cutting the .5 off changes the ratio; multiply every number by 2 instead",
  "feedback": "This answer is wrong — you dropped the decimal. Cutting the .5 off changes the ratio; multiply every number by 2 instead: 1 and 1.5 become 2 and 3, so the empirical formula is Al₂O₃."
 },
 {
  "value": "Al₃O₂",
  "misconception": "you inverted the ratio — aluminum has the smaller mole value, so aluminum takes the 1 before clearing",
  "feedback": "This answer is wrong — you inverted the ratio. Aluminum has the smaller mole value (3.50 mol against oxygen's 5.25 mol), so aluminum takes the 1 before clearing: 1 and 1.5 become 2 and 3, giving Al₂O₃."
 }
]
feedback
{
 "keyExplanation": "Values from the question:\nm(Al) = 94.5 g, m(O) = 84.0 g\nEquation:\nn = m/M  (for each element)\nSubstituting the values and calculating:\nn(Al) = 94.5 / 27.0 = 3.50 mol\nn(O) = 84.0 / 16.0 = 5.25 mol\nDivide each by the smallest (3.50): Al gives 1, O gives 1.5.\nThe decimal ends in .5, so multiply every number by 2: 2 and 3.\nEmpirical formula: Al₂O₃",
 "anticipatedWrong": [
  {
   "value": "AlO₂",
   "misconception": "you rounded the 1.5 up to 2 — rounding changes the ratio; multiply every number by 2 instead",
   "feedback": "This answer is wrong — you rounded the 1.5 up to 2. Rounding changes the ratio; multiply every number by 2 instead: 1 and 1.5 become 2 and 3, so the empirical formula is Al₂O₃."
  },
  {
   "value": "AlO",
   "misconception": "you dropped the decimal — cutting the .5 off changes the ratio; multiply every number by 2 instead",
   "feedback": "This answer is wrong — you dropped the decimal. Cutting the .5 off changes the ratio; multiply every number by 2 instead: 1 and 1.5 become 2 and 3, so the empirical formula is Al₂O₃."
  },
  {
   "value": "Al₃O₂",
   "misconception": "you inverted the ratio — aluminum has the smaller mole value, so aluminum takes the 1 before clearing",
   "feedback": "This answer is wrong — you inverted the ratio. Aluminum has the smaller mole value (3.50 mol against oxygen's 5.25 mol), so aluminum takes the 1 before clearing: 1 and 1.5 become 2 and 3, giving Al₂O₃."
  }
 ]
}
MOL-042 · hard · transfer · seq 16 · authored (two-students method-judgment frame; drop-the-decimal error spotlighted; fresh ratio 1 : 2.67 -> 3 : 8, x3 — stage-1 freshen candidate #2 exercised here; all three misconception classes live) · mcq4
A mole calculation gives the ratio 1 : 2.67. One student multiplies both numbers by 3; another just cuts the decimal off, since subscripts must be whole numbers. Which result is right, and why?
3 : 8 — multiplying every number by 3 keeps the ratio unchanged, while cutting the decimal off changes it. KEY
1 : 2 — the decimal part cannot appear in a formula, so it is discarded.
1 : 3 — 2.67 sits close enough to 3 to round up.
1 : 8 — multiply only the 2.67 by 3, because the 1 in the ratio is already a whole number and needs nothing.
MOL-045 · hard · application · seq 17 · authored (fresh pair C₂H₃O₂ / 118.0 -> C₄H₆O₄, succinic-acid family kept anonymous; three-element M build; parallel structure, step count, and misconception classes to the seed; the only-hydrogen partial was deliberately not used as a diagnostic — it equals form A's answer C₂H₆O₂) · text
A compound has the empirical formula C₂H₃O₂ and a molar mass of 118.0 g/mol. (M of C = 12.0 g/mol, H = 1.0 g/mol, O = 16.0 g/mol.) State the compound's molecular formula, typing subscripts as plain numbers (like the 2 in CO2).
Answer: C₄H₆O₄ KEY
also accepted: C4H6O4
wrongValueDiagnostics
[
 {
  "value": "C₂H₃O₂",
  "misconception": "you stopped at the empirical formula — the unit count is 118.0/59.0 = 2, and every subscript must be multiplied by it",
  "feedback": "This answer is wrong — you stopped at the empirical formula. The unit count is 118.0 / 59.0 = 2, and every subscript must be multiplied by it: C: 2 × 2 = 4, H: 3 × 2 = 6, O: 2 × 2 = 4, giving C₄H₆O₄."
 },
 {
  "value": "C₄H₃O₂",
  "misconception": "you multiplied only carbon's subscript — every subscript in the empirical formula is multiplied by the unit count",
  "feedback": "This answer is wrong — you multiplied only carbon's subscript. Every subscript in the empirical formula is multiplied by the unit count, 2: C: 2 × 2 = 4, H: 3 × 2 = 6, O: 2 × 2 = 4, giving C₄H₆O₄."
 },
 {
  "value": "C₂H₃O₄",
  "misconception": "you multiplied only oxygen's subscript — every subscript in the empirical formula is multiplied by the unit count",
  "feedback": "This answer is wrong — you multiplied only oxygen's subscript. Every subscript in the empirical formula is multiplied by the unit count, 2: C: 2 × 2 = 4, H: 3 × 2 = 6, O: 2 × 2 = 4, giving C₄H₆O₄."
 }
]
feedback
{
 "keyExplanation": "M of C₂H₃O₂ = 2 × 12.0 + 3 × 1.0 + 2 × 16.0 = 59.0 g/mol.\nStep 1 — find the number of units:\nnumber of units = molar mass of the compound ÷ molar mass of the empirical formula\nnumber of units = 118.0 / 59.0 = 2\nStep 2 — multiply every subscript by 2:\nC: 2 × 2 = 4,  H: 3 × 2 = 6,  O: 2 × 2 = 4\nMolecular formula: C₄H₆O₄",
 "anticipatedWrong": [
  {
   "value": "C₂H₃O₂",
   "misconception": "you stopped at the empirical formula — the unit count is 118.0/59.0 = 2, and every subscript must be multiplied by it",
   "feedback": "This answer is wrong — you stopped at the empirical formula. The unit count is 118.0 / 59.0 = 2, and every subscript must be multiplied by it: C: 2 × 2 = 4, H: 3 × 2 = 6, O: 2 × 2 = 4, giving C₄H₆O₄."
  },
  {
   "value": "C₄H₃O₂",
   "misconception": "you multiplied only carbon's subscript — every subscript in the empirical formula is multiplied by the unit count",
   "feedback": "This answer is wrong — you multiplied only carbon's subscript. Every subscript in the empirical formula is multiplied by the unit count, 2: C: 2 × 2 = 4, H: 3 × 2 = 6, O: 2 × 2 = 4, giving C₄H₆O₄."
  },
  {
   "value": "C₂H₃O₄",
   "misconception": "you multiplied only oxygen's subscript — every subscript in the empirical formula is multiplied by the unit count",
   "feedback": "This answer is wrong — you multiplied only oxygen's subscript. Every subscript in the empirical formula is multiplied by the unit count, 2: C: 2 × 2 = 4, H: 3 × 2 = 6, O: 2 × 2 = 4, giving C₄H₆O₄."
  }
 ]
}
MOL-043 · hard · application · seq 18 · authored (fresh percent dataset, chromium(III) oxide: exact real composition 68.421/31.579 -> 68.4/31.6 at 3 sf, sum 100.0; quotient 1.50 EXACT at display precision clears ×2; both rounding wrong paths MnO₂/MnO are real chromium oxides — the seed's chemically-real-wrong-pull virtue carried; identical chain and step count to the seed) · text
A compound is 69.6% Mn and 30.4% O by mass. What is its empirical formula? (M of Mn = 54.9 g/mol, O = 16.0 g/mol.) Type subscripts as plain numbers, like the 2 in CO2.
Answer: Mn₂O₃ KEY
also accepted: Mn2O3
wrongValueDiagnostics
[
 {
  "value": "MnO₂",
  "misconception": "you rounded the 1.50 up to 2 — rounding changes the ratio; multiply every number by 2 instead",
  "feedback": "This answer is wrong — you rounded the 1.50 up to 2. Rounding changes the ratio; multiply every number by 2 instead: Mn gives 2 and O gives 3, so the empirical formula is Mn₂O₃."
 },
 {
  "value": "MnO",
  "misconception": "you dropped the decimal — cutting the .50 off changes the ratio; multiply every number by 2 instead",
  "feedback": "This answer is wrong — you dropped the decimal. Cutting the .50 off changes the ratio; multiply every number by 2 instead: Mn gives 2 and O gives 3, so the empirical formula is Mn₂O₃."
 }
]
feedback
{
 "keyExplanation": "Assume a 100 g sample: m(Mn) = 69.6 g, m(O) = 30.4 g.\nEquation:\nn = m/M  (for each element)\nSubstituting the values and calculating:\nn(Mn) = 69.6 / 54.9 = 1.27 mol\nn(O) = 30.4 / 16.0 = 1.90 mol\nDivide each by the smallest (1.27): Mn gives 1, O gives 1.50.\nThe decimal ends in .5, so multiply every number by 2: 2 and 3.\nEmpirical formula: Mn₂O₃",
 "anticipatedWrong": [
  {
   "value": "MnO₂",
   "misconception": "you rounded the 1.50 up to 2 — rounding changes the ratio; multiply every number by 2 instead",
   "feedback": "This answer is wrong — you rounded the 1.50 up to 2. Rounding changes the ratio; multiply every number by 2 instead: Mn gives 2 and O gives 3, so the empirical formula is Mn₂O₃."
  },
  {
   "value": "MnO",
   "misconception": "you dropped the decimal — cutting the .50 off changes the ratio; multiply every number by 2 instead",
   "feedback": "This answer is wrong — you dropped the decimal. Cutting the .50 off changes the ratio; multiply every number by 2 instead: Mn gives 2 and O gives 3, so the empirical formula is Mn₂O₃."
  }
 ]
}
MOL-039 · hard · transfer · seq 19 · authored (fresh coprime-subscript claim rotation per blueprint case_target; instance C15H32, pentadecane — subscripts 15 and 32 each composite yet coprime; outside the burned-instance ledger; counterexample C12H26, dodecane, reduces) · mcq4
A student claims C₁₅H₃₂ cannot be an empirical formula 'because 15 and 32 are both bigger than 1.' Judge the claim.
The claim is wrong — 15 and 32 share no common factor bigger than 1, so C₁₅H₃₂ is an empirical formula. KEY
The claim is right — an empirical formula must contain at least one unwritten subscript of 1.
The claim is right — 15 and 32 can each be divided by a smaller number, so the formula reduces.
The claim is wrong — no formula with only two subscripts can ever be reduced.
MOL-045 · hard · transfer · seq 20 · authored (fresh x1 pair C₃H₅N / 55.0, pyrrole family kept anonymous; parallel structure and the seed's three misconception classes; one-element rotation lands on nitrogen — its doubled unwritten 1 probes the taught convention from the wrong side) · mcq4
A compound has the empirical formula C₃H₅N and a molar mass of 55.0 g/mol. (M of C₃H₅N = 55.0 g/mol.) What is its molecular formula?
C₃H₅N KEY
C₆H₁₀N₂
C₃H₅N₂
C₉H₁₅N₃
MOL-043 · hard · application · seq 21 · authored (parallel to the FRQ-1 seed-expansion structure; PERCENT route per the recorded rotation; fresh dataset — cobalt oxide Co₃O₄, unnamed in the stem per the lesson convention; the register's WE2 near-.33/×3 clearing case on a fresh instance; corpus scan 0 hits on Co₃O₄/cobalt/73.4/26.6/58.9; chosen over a third X₂O₃ ×2 case so form B's three MOL-043 items (seq 15 Al₂O₃, seq 18 Cr₂O₃, seq 21 Co₃O₄) never share an answer pattern) · frq
A compound is 73.4% Co and 26.6% O by mass. (M of Co = 58.9 g/mol, O = 16.0 g/mol.)
A compound is 73.4% Co and 26.6% O by mass. (M of Co = 58.9 g/mol, O = 16.0 g/mol.) (a) Use a 100 g sample to turn the percents into masses, then calculate the number of moles of each element. Show your working. (b) Divide every mole value by the smallest to find the whole number for each element, clearing any decimal that remains. Show each step. (c) Write the compound's empirical formula, and state how the decimal left after dividing was turned into whole-number subscripts without simply rounding it away. Type subscripts as plain numbers, like the 2 in CO2.
Part a. Use a 100 g sample to turn the percents into masses, then calculate the number of moles of each element. Show your working.
Model: Assume a 100 g sample: m(Co) = 73.4 g, m(O) = 26.6 g. Equation: n = m/M (for each element) Substituting the values and calculating: n(Co) = 73.4 / 58.9 = 1.25 mol n(O) = 26.6 / 16.0 = 1.66 mol
Part b. Divide every mole value by the smallest to find the whole number for each element, clearing any decimal that remains. Show each step.
Model: Divide each mole value by the smallest (1.25): Co: 1.25 / 1.25 = 1 O: 1.66 / 1.25 = 1.33 A decimal near .33 clears with ×3: Co: 1 × 3 = 3 O: 1.33 × 3 = 4
Part c. Write the compound's empirical formula, and state how the decimal left after dividing was turned into whole-number subscripts without simply rounding it away. Type subscripts as plain numbers, like the 2 in CO2.
Model: The finished whole numbers are the subscripts, so the empirical formula is Co3O4. The whole-number subscripts come from clearing the leftover 1.33 by multiplying every number by 3 (1.33 × 3 = 3.99 → 4 within the data's precision) — the decimal was never simply rounded away.
Full model answer. (a) Assume a 100 g sample: m(Co) = 73.4 g, m(O) = 26.6 g. Equation: n = m/M (for each element) Substituting the values and calculating: n(Co) = 73.4 / 58.9 = 1.25 mol n(O) = 26.6 / 16.0 = 1.66 mol (b) Divide each mole value by the smallest (1.25): Co: 1.25 / 1.25 = 1 O: 1.66 / 1.25 = 1.33 A decimal near .33 clears with ×3: Co: 1 × 3 = 3 O: 1.33 × 3 = 4 (c) The finished whole numbers are the subscripts, so the empirical formula is Co3O4. The whole-number subscripts come from clearing the leftover 1.33 by multiplying every number by 3 (1.33 × 3 = 3.99 → 4 within the data's precision) — the decimal was never simply rounded away.
MOL-045 · hard · application · seq 22 · authored (FRQ-2 fresh parallel-structure variant; hexachlorobenzene C₆Cl₆, count 6 — restores the seed's own x6 depth; compound unnamed in student text per the gate register) · frq
A compound has the empirical formula CCl and a molar mass of 285.0 g/mol. (M of C = 12.0 g/mol, Cl = 35.5 g/mol.)
A compound has the empirical formula CCl and a molar mass of 285.0 g/mol. (M of C = 12.0 g/mol, Cl = 35.5 g/mol.) (a) Calculate the molar mass of the empirical formula CCl from the atomic masses given, showing the sum you evaluate. (b) Find the number of empirical-formula units in one molecule of the compound. State the equation you use, and show every step of your working. (c) Write the molecular formula of the compound, and show how each subscript in your answer was produced. Type subscripts as plain numbers, like the 2 in CO2; no other special symbols are needed.
Part a. Calculate the molar mass of the empirical formula CCl from the atomic masses given, showing the sum you evaluate.
Model: Values from the question: M of C = 12.0 g/mol, M of Cl = 35.5 g/mol The empirical formula CCl contains 1 C and 1 Cl (an unwritten subscript counts as 1). M of CCl = 12.0 + 35.5 M of CCl = 47.5 g/mol
Part b. Find the number of empirical-formula units in one molecule of the compound. State the equation you use, and show every step of your working.
Model: Values from the question: molar mass of the compound = 285.0 g/mol molar mass of the empirical formula = 47.5 g/mol (from part a) Equation: number of units = molar mass of the compound ÷ molar mass of the empirical formula Substituting the values and calculating: number of units = 285.0 / 47.5 number of units = 6 The result is a whole number of at least 1, so the division ran the right way up.
Part c. Write the molecular formula of the compound, and show how each subscript in your answer was produced. Type subscripts as plain numbers, like the 2 in CO2; no other special symbols are needed.
Model: Multiply every subscript in the empirical formula by 6: C: 1 × 6 = 6, Cl: 1 × 6 = 6 Molecular formula: C6Cl6
Full model answer. (a) Values from the question: M of C = 12.0 g/mol, M of Cl = 35.5 g/mol The empirical formula CCl contains 1 C and 1 Cl (an unwritten subscript counts as 1). M of CCl = 12.0 + 35.5 M of CCl = 47.5 g/mol (b) Values from the question: molar mass of the compound = 285.0 g/mol molar mass of the empirical formula = 47.5 g/mol (from part a) Equation: number of units = molar mass of the compound ÷ molar mass of the empirical formula Substituting the values and calculating: number of units = 285.0 / 47.5 number of units = 6 The result is a whole number of at least 1, so the division ran the right way up. (c) Multiply every subscript in the empirical formula by 6: C: 1 × 6 = 6, Cl: 1 × 6 = 6 Molecular formula: C6Cl6

Form C

MOL-036 · easy · recall · seq 1 · authored (seedless recall row; smallest-vs-any-fitting-whole-numbers contrast; key phrasing from the taught rule line) · mcq4
Several different sets of whole numbers can fit a compound's measured atom ratio. Which set does the compound's empirical formula use?
The smallest whole numbers that fit the ratio. KEY
Any set of whole numbers that fits the ratio — each fitting set is equally correct.
The set that counts the atoms in one particle of the compound.
The set that states each element's share of the compound's mass.
MOL-037 · easy · recall · seq 2 · seedless authored row (probe C: what-a-subscript-states; instance-free) · mcq4
In a molecular formula, what does the subscript written after an element's symbol state?
The exact number of atoms of that element in one molecule. KEY
The number of atoms of that element for every single atom of each other element.
The total number of atoms of every element in the molecule.
The number of moles of that element in a sample of the compound.
MOL-039 · easy · recall · seq 3 · authored (seedless recall row; what-a-shared-factor-implies probe; key = rule line 4 verbatim register, distractors = inversion + two extra-condition grafts) · mcq4
While running the empirical check on a formula, a student finds a whole number bigger than 1 that divides into every subscript. What has the student learned?
The formula is not an empirical formula — it still reduces. KEY
The formula is already an empirical formula.
Nothing yet — the student must also confirm that each subscript divides by some smaller number of its own.
The formula is not empirical only if that number also appears as one of its subscripts.
MOL-037 · easy · application · seq 4 · authored (parallel to the seq-4 seed structure; fresh instance vitamin C C₆H₈O₆, outside the burned-instance ledger; the stated 3 : 4 : 3 reduced ratio mirrors the seed's reduced-ratio distractor flavor and the glucose rule-line register — no division shown; corpus scan 0 hits) · mcq4
The molecular formula of vitamin C is C₆H₈O₆. Which statement does the formula guarantee about one vitamin C molecule?
It contains exactly 6 C atoms, 8 H atoms, and 6 O atoms. KEY
The subscripts were divided down to smallest terms, so the real counts are larger than 6, 8, and 6.
It contains exactly 8 C atoms, 6 H atoms, and 6 O atoms.
The 6, 8, and 6 count the bonds each element forms in the molecule, not the atoms present.
MOL-040 · easy · application · seq 5 · authored (parallel numeric slot; fresh chemically-real composition — potassium chloride KCl, 52.4% K by mass; compound unnamed in the stem per the lesson convention) · numeric
A compound is 52.4% K by mass; the rest is Cl. What mass of Cl does a 100 g sample of the compound contain? Give your answer in grams to one decimal place.
Answer: 47.6 g KEY
also accepted: 47.6 g
wrongValueDiagnostics
[
 {
  "value": 52.4,
  "misconception": "you gave the potassium's mass — the question asks for the chlorine, the rest of the 100 g",
  "feedback": "This answer is wrong — you gave the potassium's mass — the question asks for the chlorine, the rest of the 100 g."
 },
 {
  "value": 0.476,
  "misconception": "you divided by 100 — in a 100 g sample the masses are in grams already",
  "feedback": "This answer is wrong — you divided by 100 — in a 100 g sample the masses are in grams already."
 }
]
feedback
{
 "keyExplanation": "Assume a sample of exactly 100 g.\nPotassium: 52.4% of the compound → 52.4 g.\nThe element masses in a 100 g sample total 100 g.\nmass of chlorine = 100 − 52.4 = 47.6 g"
}
MOL-036 · medium · application · seq 6 · authored (parallel mcq slot; fresh reducible measured ratio 6:3 (÷3, bigger count FIRST-listed for variety) + fresh element pair Mg/Si — Mg₂Si is a real compound (magnesium silicide), outside the burned-instance ledger; silicon appears nowhere in the scope corpus; 'magnesium' sits only in MOL-040's corpus -> other-scope-lesson WARN class, not memory-answerable from a percent-composition lesson) · mcq4
An instrument reports that a compound contains 6 Mg atoms for every 3 Si atoms. Which formula is the compound's empirical formula?
Mg₂Si KEY
Mg₆Si₃
MgSi₂
Mg₆Si
MOL-038 · medium · application · seq 7 · authored (blueprint seq-7 B-E: fresh genuinely-reducible formula, largest-factor trap live at factor 4; acenaphthylene C₁₂H₈ -> C₃H₂ — both reduced subscripts stay above 1, so the routine's division lands on every element with no unwritten-1 shortcut; misconception classes: divide-by-2-not-largest / left-unchanged / divided-only-one-subscript [the H₂O₂ fluency class]) · text
What is the empirical formula of acenaphthylene, C₁₂H₈? Type the whole formula, writing any subscripts as plain digits, like the 2 in CO2; leave subscripts of 1 unwritten.
Answer: C₃H₂ KEY
also accepted: C3H2
wrongValueDiagnostics
[
 {
  "value": "C₆H₄",
  "misconception": "you divided by 2, which is a common factor but not the LARGEST one — 6 and 4 still share the factor 2",
  "feedback": "This answer is wrong — you divided by 2, which is a common factor but not the LARGEST one. The subscripts of C₆H₄ — 6 and 4 — still share the factor 2, so the formula still reduces. Divide C₁₂H₈'s subscripts by their largest common factor, 4: C: 12 ÷ 4 = 3, H: 8 ÷ 4 = 2, and the empirical formula is C₃H₂."
 },
 {
  "value": "C₁₂H₈",
  "misconception": "you left the formula unchanged — the subscripts 12 and 8 share the factor 4, so the formula reduces",
  "feedback": "This answer is wrong — you left the formula unchanged. The subscripts 12 and 8 share the factor 4, so the formula reduces: C: 12 ÷ 4 = 3, H: 8 ÷ 4 = 2, and the empirical formula is C₃H₂."
 },
 {
  "value": "C₁₂H₂",
  "misconception": "you divided only the hydrogen's subscript — every subscript must be divided by the same factor",
  "feedback": "This answer is wrong — you divided only the hydrogen's subscript. Every subscript must be divided by the SAME factor — the largest common factor, 4: C: 12 ÷ 4 = 3, H: 8 ÷ 4 = 2, so the empirical formula is C₃H₂."
 }
]
feedback
{
 "keyExplanation": "Subscripts: C 12, H 8. The largest common factor is 4 — dividing by 2 alone leaves C₆H₄, which still reduces. C: 12 ÷ 4 = 3, H: 8 ÷ 4 = 2, so the empirical formula is C₃H₂.",
 "anticipatedWrong": [
  {
   "value": "C₆H₄",
   "misconception": "you divided by 2, which is a common factor but not the LARGEST one — 6 and 4 still share the factor 2",
   "feedback": "This answer is wrong — you divided by 2, which is a common factor but not the LARGEST one. The subscripts of C₆H₄ — 6 and 4 — still share the factor 2, so the formula still reduces. Divide C₁₂H₈'s subscripts by their largest common factor, 4: C: 12 ÷ 4 = 3, H: 8 ÷ 4 = 2, and the empirical formula is C₃H₂."
  },
  {
   "value": "C₁₂H₈",
   "misconception": "you left the formula unchanged — the subscripts 12 and 8 share the factor 4, so the formula reduces",
   "feedback": "This answer is wrong — you left the formula unchanged. The subscripts 12 and 8 share the factor 4, so the formula reduces: C: 12 ÷ 4 = 3, H: 8 ÷ 4 = 2, and the empirical formula is C₃H₂."
  },
  {
   "value": "C₁₂H₂",
   "misconception": "you divided only the hydrogen's subscript — every subscript must be divided by the same factor",
   "feedback": "This answer is wrong — you divided only the hydrogen's subscript. Every subscript must be divided by the SAME factor — the largest common factor, 4: C: 12 ÷ 4 = 3, H: 8 ÷ 4 = 2, so the empirical formula is C₃H₂."
  }
 ]
}
MOL-039 · medium · application · seq 8 · authored (pick-NOT polarity; fresh formula set — key C₈H₁₀ reduces by 2 to C₄H₅; C₄H₁₀O carries the unwritten-1 trap (the seed's C₂H₆O class) and KNO₃ the double-unwritten-1 surface) · mcq4
One of these four formulas can still be reduced. Which is it?
C₈H₁₀ KEY
KNO₃
Al₂S₃
C₄H₁₀O
MOL-041 · medium · application · seq 9 · authored (fresh two-element pair, SF₆ family; neither mole value is 1.00 by inspection — both divisions carry real work; exact divisions 0.20 / 1.20; whole-number outcome by construction) · text
A sample of a compound contains 6.42 g of S and 22.8 g of F. What is the simplest mole ratio of S atoms to F atoms? (M of S = 32.1 g/mol, F = 19.0 g/mol.) Write your answer like 1 : 2.
Answer: 1 : 6 KEY
also accepted: 1:6
wrongValueDiagnostics
[
 {
  "value": "1 : 3.6",
  "misconception": "you compared the masses instead of the moles — 6.42 : 22.8 is about 1 : 3.6, but atoms are counted in moles",
  "feedback": "This answer is wrong — you compared the masses instead of the moles. 6.42 : 22.8 is about 1 : 3.6, but atoms are counted in moles: n(S) = 6.42 / 32.1 = 0.20 mol and n(F) = 22.8 / 19.0 = 1.20 mol, and dividing each by 0.20 gives 1 : 6."
 },
 {
  "value": "6 : 1",
  "misconception": "you wrote the ratio backwards — the order follows the elements as named: sulfur first, then fluorine",
  "feedback": "This answer is wrong — you wrote the ratio backwards. The ratio order follows the elements as they are named in the question: sulfur first, then fluorine, so the answer is 1 : 6."
 }
]
feedback
{
 "keyExplanation": "Values from the question:\nm(S) = 6.42 g, m(F) = 22.8 g\nEquation:\nn = m/M  (for each element)\nSubstituting the values and calculating:\nn(S) = 6.42 / 32.1 = 0.20 mol\nn(F) = 22.8 / 19.0 = 1.20 mol\nDivide each by the smallest (0.20): S gives 1, F gives 6.\nMole ratio of sulfur to fluorine = 1 : 6",
 "anticipatedWrong": [
  {
   "value": "1 : 3.6",
   "misconception": "you compared the masses instead of the moles — 6.42 : 22.8 is about 1 : 3.6, but atoms are counted in moles",
   "feedback": "This answer is wrong — you compared the masses instead of the moles. 6.42 : 22.8 is about 1 : 3.6, but atoms are counted in moles: n(S) = 6.42 / 32.1 = 0.20 mol and n(F) = 22.8 / 19.0 = 1.20 mol, and dividing each by 0.20 gives 1 : 6."
  },
  {
   "value": "6 : 1",
   "misconception": "you wrote the ratio backwards — the order follows the elements as named: sulfur first, then fluorine",
   "feedback": "This answer is wrong — you wrote the ratio backwards. The ratio order follows the elements as they are named in the question: sulfur first, then fluorine, so the answer is 1 : 6."
  }
 ]
}
MOL-042 · medium · application · seq 10 · authored (fresh ratio 1 : 4.67 -> 3 : 14, x3, .67 branch — the near-.67 rule case whose only spec instance, fluency 1.67, is burned; near-thirds product written as the cleared whole number per the taught register, 4.67 × 3 = 14, exactly the lesson WE's own 1.33 × 3 = 4 treatment; 4.67 and 3 : 14 machine-checked fresh; round-UP + partial classes carried) · text
A mole calculation gives the ratio 1 : 4.67. What whole-number ratio does it clear to? Give your answer like 2 : 3.
Answer: 3 : 14 KEY
also accepted: 3:14
wrongValueDiagnostics
[
 {
  "value": "1 : 5",
  "misconception": "you rounded the 4.67 up to 5 — rounding changes the ratio; multiply every number by 3 instead",
  "feedback": "This answer is wrong — you rounded the 4.67 up to 5, and rounding changes the ratio itself. The decimal is near .67, so multiply every number by 3 instead: 1 × 3 = 3 and 4.67 × 3 = 14, giving 3 : 14."
 },
 {
  "value": "1 : 14",
  "misconception": "you multiplied only the decimal number — every number in the ratio must be multiplied by the same amount",
  "feedback": "This answer is wrong — you multiplied only the decimal number. Every number in the ratio must be multiplied by the same amount to keep the ratio unchanged: 1 × 3 = 3 and 4.67 × 3 = 14, giving 3 : 14."
 }
]
feedback
{
 "keyExplanation": "The decimal is near .67, so multiply every number by 3: 1 × 3 = 3 and 4.67 × 3 = 14. The whole-number ratio is 3 : 14. Rounding 4.67 up to 5 would change the ratio itself — multiplying never does.",
 "anticipatedWrong": [
  {
   "value": "1 : 5",
   "misconception": "you rounded the 4.67 up to 5 — rounding changes the ratio; multiply every number by 3 instead",
   "feedback": "This answer is wrong — you rounded the 4.67 up to 5, and rounding changes the ratio itself. The decimal is near .67, so multiply every number by 3 instead: 1 × 3 = 3 and 4.67 × 3 = 14, giving 3 : 14."
  },
  {
   "value": "1 : 14",
   "misconception": "you multiplied only the decimal number — every number in the ratio must be multiplied by the same amount",
   "feedback": "This answer is wrong — you multiplied only the decimal number. Every number in the ratio must be multiplied by the same amount to keep the ratio unchanged: 1 × 3 = 3 and 4.67 × 3 = 14, giving 3 : 14."
  }
 ]
}
MOL-038 · medium · application · seq 11 · authored (parallel to the seq-11 seed structure; fresh instance decane C₁₂H₂₆ -> C₆H₁₃, corpus scan 0 hits; covers the no-unwritten-1 surface — both divided subscripts stay above 1 — with two-digit subscript reading; subscript 22 recorded as a burned scanned number) · text
What is the empirical formula of dodecane, C₁₂H₂₆? Type the whole formula, writing any subscripts as plain digits, like the 2 in CO2; leave subscripts of 1 unwritten.
Answer: C₆H₁₃ KEY
also accepted: C6H13
wrongValueDiagnostics
[
 {
  "value": "C₁₂H₂₆",
  "misconception": "you left the formula unchanged — the subscripts 12 and 26 share the factor 2, so the formula reduces",
  "feedback": "This answer is wrong — you left the formula unchanged. The subscripts 12 and 26 share the factor 2, so the formula reduces: C: 12 ÷ 2 = 6, H: 26 ÷ 2 = 13, giving C₆H₁₃."
 },
 {
  "value": "CH",
  "misconception": "you divided each subscript by its own value, which destroys the ratio",
  "feedback": "This answer is wrong — you divided each subscript by its own value, which destroys the ratio. Every subscript is divided by the same factor, 2: C: 12 ÷ 2 = 6, H: 26 ÷ 2 = 13, giving C₆H₁₃."
 },
 {
  "value": "C₁₂H₁₃",
  "misconception": "you divided only the hydrogen's subscript — every subscript must be divided by the same factor",
  "feedback": "This answer is wrong — you divided only the hydrogen's subscript. Every subscript must be divided by the same factor: C: 12 ÷ 2 = 6, H: 26 ÷ 2 = 13, giving C₆H₁₃."
 }
]
feedback
{
 "keyExplanation": "Subscripts: C 12, H 26. The largest common factor is 2. C: 12 ÷ 2 = 6, H: 26 ÷ 2 = 13, so dodecane's empirical formula is C₆H₁₃.",
 "anticipatedWrong": [
  {
   "value": "C₁₂H₂₆",
   "misconception": "you left the formula unchanged — the subscripts 12 and 26 share the factor 2, so the formula reduces",
   "feedback": "This answer is wrong — you left the formula unchanged. The subscripts 12 and 26 share the factor 2, so the formula reduces: C: 12 ÷ 2 = 6, H: 26 ÷ 2 = 13, giving C₆H₁₃."
  },
  {
   "value": "CH",
   "misconception": "you divided each subscript by its own value, which destroys the ratio",
   "feedback": "This answer is wrong — you divided each subscript by its own value, which destroys the ratio. Every subscript is divided by the same factor, 2: C: 12 ÷ 2 = 6, H: 26 ÷ 2 = 13, giving C₆H₁₃."
  },
  {
   "value": "C₁₂H₁₃",
   "misconception": "you divided only the hydrogen's subscript — every subscript must be divided by the same factor",
   "feedback": "This answer is wrong — you divided only the hydrogen's subscript. Every subscript must be divided by the same factor: C: 12 ÷ 2 = 6, H: 26 ÷ 2 = 13, giving C₆H₁₃."
  }
 ]
}
MOL-044 · medium · transfer · seq 12 · authored (quotient-meaning, worked-solution-line frame; parallel misconception classes to the seed; fresh exact pair 154.0/77.0 = 2) · mcq4
A worked solution contains the line "154.0 g/mol ÷ 77.0 g/mol = 2", where 154.0 g/mol is a compound's molar mass and 77.0 g/mol is its empirical formula's molar mass. What does the 2 in this line count?
How many empirical-formula units fit in one molecule of the compound. KEY
How many molecules of the compound fit in one empirical-formula unit.
How many atoms of each element one molecule of the compound holds.
How many moles of the compound the solution is about.
MOL-041 · medium · application · seq 13 · authored (parallel to the seq-13 seed structure; fresh triple — phosphoric acid H₃PO₄, unnamed in the stem per the lesson convention; smallest mole value on the MIDDLE-listed element; the element word 'phosphorus' is corpus-present only via the P₂O₅/P₄O₁₀ formula instances, which never appear here — fresh numbers carry no instance signal, per the unflagged-sulfur stage-1 precedent; corpus scan 0 hits on all numbers) · text
One sample of a compound holds 3.0 g of H, 31.0 g of P, and 64.0 g of O. What is the simplest mole ratio of H to P to O atoms? (M of H = 1.0 g/mol, P = 31.0 g/mol, O = 16.0 g/mol.) Write your answer like 1 : 2 : 3.
Answer: 3 : 1 : 4 KEY
also accepted: 3:1:4
wrongValueDiagnostics
[
 {
  "value": "1 : 10.3 : 21.3",
  "misconception": "you compared the masses instead of the moles — 3.0 : 31.0 : 64.0 is about 1 : 10.3 : 21.3, but atoms are counted in moles",
  "feedback": "This answer is wrong — you compared the masses instead of the moles — 3.0 : 31.0 : 64.0 is about 1 : 10.3 : 21.3, but atoms are counted in moles."
 },
 {
  "value": "4 : 1 : 3",
  "misconception": "you wrote the ratio backwards — the order follows the elements as named: hydrogen first, then phosphorus, then oxygen",
  "feedback": "This answer is wrong — you wrote the ratio backwards — the order follows the elements as named: hydrogen first, then phosphorus, then oxygen."
 }
]
feedback
{
 "keyExplanation": "Values from the question:\nm(H) = 3.0 g, m(P) = 31.0 g, m(O) = 64.0 g\nEquation:\nn = m/M  (for each element)\nSubstituting the values and calculating:\nn(H) = 3.0 / 1.0 = 3.00 mol\nn(P) = 31.0 / 31.0 = 1.00 mol\nn(O) = 64.0 / 16.0 = 4.00 mol\nDivide each by the smallest (1.00): H gives 3, P gives 1, O gives 4.\nMole ratio of hydrogen to phosphorus to oxygen = 3 : 1 : 4"
}
MOL-044 · medium · application · seq 14 · authored (parallel numeric slot; fresh (formula, M) pair — 1,3,5-trithiane C₃H₆S₃, empirical CH₂S, 138.3 g/mol -> 3 units; compound unnamed in the stem per the lesson's gate convention; the ignored-subscript path 3.07 nearly rounds to the true count, giving the taught sanity check real bite) · numeric
A compound has the empirical formula CH₂S and a molar mass of 138.3 g/mol. (M of C = 12.0 g/mol, H = 1.0 g/mol, S = 32.1 g/mol.) How many empirical-formula units fit in one of its molecules? Answer with the whole-number count.
Answer: 3 units KEY
also accepted: 3 units, 3.0
wrongValueDiagnostics
[
 {
  "value": 3.07,
  "misconception": "you ignored the subscripts in the empirical formula, using 12.0 + 1.0 + 32.1 = 45.1 g/mol — M of CH₂S is 12.0 + 2 × 1.0 + 32.1 = 46.1 g/mol",
  "feedback": "This answer is wrong — you ignored the subscripts in the empirical formula, using 12.0 + 1.0 + 32.1 = 45.1 g/mol. M of CH₂S is 12.0 + 2 × 1.0 + 32.1 = 46.1 g/mol — the unit count always lands exactly on a whole number, so a close-but-not-whole result signals this slip."
 },
 {
  "value": 0.33,
  "misconception": "you divided upside down — the compound's molar mass goes on top, and the unit count is never below 1",
  "feedback": "This answer is wrong — you divided upside down — the compound's molar mass goes on top, and the unit count is never below 1."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmolar mass of the compound = 138.3 g/mol\nM of CH₂S = 12.0 + 2 × 1.0 + 32.1 = 46.1 g/mol\nWrite down the equation:\nnumber of units = molar mass of the compound ÷ molar mass of the empirical formula\nSubstitute in the values, and calculate:\nnumber of units = 138.3 / 46.1\nnumber of units = 3\nThe result is a whole number at least 1, so the division ran the right way up."
}
MOL-043 · hard · application · seq 15 · authored (parallel mass-route capstone; fresh chemically-real compound — chromium(III) oxide Cr₂O₃, compound unnamed in the stem per the seed convention; exact divisions 31.2/52.0 = 0.60 and 14.4/16.0 = 0.90; .5-decimal clearing ×2; the round-up and drop-decimal wrong paths land on the REAL oxides CrO₂ and CrO, keeping the traps chemically plausible per the lesson's specNotes) · text
A sample of a compound contains 31.2 g of Cr and 14.4 g of O. What is its empirical formula? (M of Cr = 52.0 g/mol, O = 16.0 g/mol.) Type subscripts as plain numbers, like the 2 in CO2.
Answer: Cr₂O₃ KEY
also accepted: Cr2O3
wrongValueDiagnostics
[
 {
  "value": "CrO₂",
  "misconception": "you rounded the 1.5 up to 2 — rounding changes the ratio; multiply every number by 2 instead",
  "feedback": "This answer is wrong — you rounded the 1.5 up to 2. Rounding changes the ratio; multiply every number by 2 instead: 1 and 1.5 become 2 and 3, so the empirical formula is Cr₂O₃."
 },
 {
  "value": "CrO",
  "misconception": "you dropped the decimal — cutting the .5 off changes the ratio; multiply every number by 2 instead",
  "feedback": "This answer is wrong — you dropped the decimal. Cutting the .5 off changes the ratio; multiply every number by 2 instead: 1 and 1.5 become 2 and 3, so the empirical formula is Cr₂O₃."
 },
 {
  "value": "Cr₃O₂",
  "misconception": "you inverted the ratio — chromium has the smaller mole value, so chromium takes the 1 before clearing",
  "feedback": "This answer is wrong — you inverted the ratio. Chromium has the smaller mole value (0.60 mol against oxygen's 0.90 mol), so chromium takes the 1 before clearing: 1 and 1.5 become 2 and 3, giving Cr₂O₃."
 }
]
feedback
{
 "keyExplanation": "Values from the question:\nm(Cr) = 31.2 g, m(O) = 14.4 g\nEquation:\nn = m/M  (for each element)\nSubstituting the values and calculating:\nn(Cr) = 31.2 / 52.0 = 0.60 mol\nn(O) = 14.4 / 16.0 = 0.90 mol\nDivide each by the smallest (0.60): Cr gives 1, O gives 1.5.\nThe decimal ends in .5, so multiply every number by 2: 2 and 3.\nEmpirical formula: Cr₂O₃",
 "anticipatedWrong": [
  {
   "value": "CrO₂",
   "misconception": "you rounded the 1.5 up to 2 — rounding changes the ratio; multiply every number by 2 instead",
   "feedback": "This answer is wrong — you rounded the 1.5 up to 2. Rounding changes the ratio; multiply every number by 2 instead: 1 and 1.5 become 2 and 3, so the empirical formula is Cr₂O₃."
  },
  {
   "value": "CrO",
   "misconception": "you dropped the decimal — cutting the .5 off changes the ratio; multiply every number by 2 instead",
   "feedback": "This answer is wrong — you dropped the decimal. Cutting the .5 off changes the ratio; multiply every number by 2 instead: 1 and 1.5 become 2 and 3, so the empirical formula is Cr₂O₃."
  },
  {
   "value": "Cr₃O₂",
   "misconception": "you inverted the ratio — chromium has the smaller mole value, so chromium takes the 1 before clearing",
   "feedback": "This answer is wrong — you inverted the ratio. Chromium has the smaller mole value (0.60 mol against oxygen's 0.90 mol), so chromium takes the 1 before clearing: 1 and 1.5 become 2 and 3, giving Cr₂O₃."
  }
 ]
}
MOL-042 · hard · transfer · seq 16 · authored (two-students method-judgment frame; multiply-one-number-only error spotlighted; fresh ratio 1 : 5.5 -> 2 : 11, x2; all three misconception classes live) · mcq4
A mole calculation gives the ratio 1 : 5.5. One student multiplies both numbers by 2; another multiplies only the 5.5 by 2, reasoning that the 1 is already a whole number. Which result is right, and why?
2 : 11 — multiplying every number by 2 keeps the ratio unchanged, while multiplying only the 5.5 changes it. KEY
1 : 11 — only the decimal number needed fixing, so it alone gets multiplied while the 1 stays put.
1 : 6 — 5.5 rounds up to 6, and rounding is simpler than multiplying.
1 : 5 — drop the decimal part, since subscripts must be whole numbers.
MOL-045 · hard · application · seq 17 · authored (fresh pair CHNO / 129.0 -> C₃H₃N₃O₃, cyanuric-acid family kept anonymous; the heaviest M build of the row — four elements, all subscripts unwritten 1s, exercising the unwritten-subscript convention inside the multiply step; parallel structure and misconception classes to the seed) · text
A compound has the empirical formula CHNO and a molar mass of 129.0 g/mol. (M of C = 12.0 g/mol, H = 1.0 g/mol, N = 14.0 g/mol, O = 16.0 g/mol.) State the compound's molecular formula, typing subscripts as plain numbers (like the 2 in CO2).
Answer: C₃H₃N₃O₃ KEY
also accepted: C3H3N3O3
wrongValueDiagnostics
[
 {
  "value": "CHNO",
  "misconception": "you stopped at the empirical formula — the unit count is 129.0/43.0 = 3, and every subscript must be multiplied by it",
  "feedback": "This answer is wrong — you stopped at the empirical formula. The unit count is 129.0 / 43.0 = 3, and every subscript must be multiplied by it: C: 1 × 3 = 3, H: 1 × 3 = 3, N: 1 × 3 = 3, O: 1 × 3 = 3, giving C₃H₃N₃O₃."
 },
 {
  "value": "C₃HNO",
  "misconception": "you multiplied only carbon's subscript — every subscript in the empirical formula is multiplied by the unit count, including every unwritten 1",
  "feedback": "This answer is wrong — you multiplied only carbon's subscript. Every subscript in the empirical formula is multiplied by the unit count, 3 — including every unwritten 1: C: 1 × 3 = 3, H: 1 × 3 = 3, N: 1 × 3 = 3, O: 1 × 3 = 3, giving C₃H₃N₃O₃."
 },
 {
  "value": "CHNO₃",
  "misconception": "you multiplied only oxygen's subscript — every subscript in the empirical formula is multiplied by the unit count, including every unwritten 1",
  "feedback": "This answer is wrong — you multiplied only oxygen's subscript. Every subscript in the empirical formula is multiplied by the unit count, 3 — including every unwritten 1: C: 1 × 3 = 3, H: 1 × 3 = 3, N: 1 × 3 = 3, O: 1 × 3 = 3, giving C₃H₃N₃O₃."
 }
]
feedback
{
 "keyExplanation": "M of CHNO = 12.0 + 1.0 + 14.0 + 16.0 = 43.0 g/mol.\nStep 1 — find the number of units:\nnumber of units = molar mass of the compound ÷ molar mass of the empirical formula\nnumber of units = 129.0 / 43.0 = 3\nStep 2 — multiply every subscript by 3 (an unwritten subscript is 1):\nC: 1 × 3 = 3,  H: 1 × 3 = 3,  N: 1 × 3 = 3,  O: 1 × 3 = 3\nMolecular formula: C₃H₃N₃O₃",
 "anticipatedWrong": [
  {
   "value": "CHNO",
   "misconception": "you stopped at the empirical formula — the unit count is 129.0/43.0 = 3, and every subscript must be multiplied by it",
   "feedback": "This answer is wrong — you stopped at the empirical formula. The unit count is 129.0 / 43.0 = 3, and every subscript must be multiplied by it: C: 1 × 3 = 3, H: 1 × 3 = 3, N: 1 × 3 = 3, O: 1 × 3 = 3, giving C₃H₃N₃O₃."
  },
  {
   "value": "C₃HNO",
   "misconception": "you multiplied only carbon's subscript — every subscript in the empirical formula is multiplied by the unit count, including every unwritten 1",
   "feedback": "This answer is wrong — you multiplied only carbon's subscript. Every subscript in the empirical formula is multiplied by the unit count, 3 — including every unwritten 1: C: 1 × 3 = 3, H: 1 × 3 = 3, N: 1 × 3 = 3, O: 1 × 3 = 3, giving C₃H₃N₃O₃."
  },
  {
   "value": "CHNO₃",
   "misconception": "you multiplied only oxygen's subscript — every subscript in the empirical formula is multiplied by the unit count, including every unwritten 1",
   "feedback": "This answer is wrong — you multiplied only oxygen's subscript. Every subscript in the empirical formula is multiplied by the unit count, 3 — including every unwritten 1: C: 1 × 3 = 3, H: 1 × 3 = 3, N: 1 × 3 = 3, O: 1 × 3 = 3, giving C₃H₃N₃O₃."
  }
 ]
}
MOL-043 · hard · application · seq 18 · authored (fresh percent dataset, silicon nitride: exact real composition 60.086/39.914 -> 60.1/39.9 at 3 sf, sum 100.0; quotient 1.33 clears ×3 with 1.33 × 3 = 3.99 -> 4 — exactly WE2's taught display pattern on fresh data; identical chain and step count to the seed) · text
A compound is 60.1% Si and 39.9% N by mass. What is its empirical formula? (M of Si = 28.1 g/mol, N = 14.0 g/mol.) Type subscripts as plain numbers, like the 2 in CO2.
Answer: Si₃N₄ KEY
also accepted: Si3N4
wrongValueDiagnostics
[
 {
  "value": "SiN",
  "misconception": "you dropped the decimal — cutting the .33 off changes the ratio; multiply every number by 3 instead",
  "feedback": "This answer is wrong — you dropped the decimal. Cutting the .33 off changes the ratio; multiply every number by 3 instead: Si gives 3 and N gives 4, so the empirical formula is Si₃N₄."
 },
 {
  "value": "SiN₄",
  "misconception": "you multiplied only the number with the decimal — multiplying just one number changes the ratio; multiply EVERY number by 3",
  "feedback": "This answer is wrong — you multiplied only the number with the decimal. Multiplying just one number changes the ratio; multiply EVERY number by 3: Si gives 3 and N gives 4, so the empirical formula is Si₃N₄."
 }
]
feedback
{
 "keyExplanation": "Assume a 100 g sample: m(Si) = 60.1 g, m(N) = 39.9 g.\nEquation:\nn = m/M  (for each element)\nSubstituting the values and calculating:\nn(Si) = 60.1 / 28.1 = 2.14 mol\nn(N) = 39.9 / 14.0 = 2.85 mol\nDivide each by the smallest (2.14): Si gives 1, N gives 1.33.\nThe decimal is near .33, so multiply every number by 3: 3 and 4.\nEmpirical formula: Si₃N₄",
 "anticipatedWrong": [
  {
   "value": "SiN",
   "misconception": "you dropped the decimal — cutting the .33 off changes the ratio; multiply every number by 3 instead",
   "feedback": "This answer is wrong — you dropped the decimal. Cutting the .33 off changes the ratio; multiply every number by 3 instead: Si gives 3 and N gives 4, so the empirical formula is Si₃N₄."
  },
  {
   "value": "SiN₄",
   "misconception": "you multiplied only the number with the decimal — multiplying just one number changes the ratio; multiply EVERY number by 3",
   "feedback": "This answer is wrong — you multiplied only the number with the decimal. Multiplying just one number changes the ratio; multiply EVERY number by 3: Si gives 3 and N gives 4, so the empirical formula is Si₃N₄."
  }
 ]
}
MOL-039 · hard · transfer · seq 19 · authored (fresh coprime-subscript claim rotation per blueprint case_target; instance C9H8, indene — subscripts 9 and 8 each composite yet coprime; outside the burned-instance ledger; counterexample C14H30, tetradecane, reduces) · mcq4
A student claims C₉H₈ cannot be an empirical formula 'because 9 and 8 are both bigger than 1.' Judge the claim.
The claim is wrong — 9 and 8 share no common factor bigger than 1, so C₉H₈ is an empirical formula. KEY
The claim is right — an empirical formula must contain at least one unwritten subscript of 1.
The claim is right — 9 and 8 can each be divided by a smaller number, so the formula reduces.
The claim is wrong — no formula with only two subscripts can ever be reduced.
MOL-045 · hard · transfer · seq 20 · authored (fresh x1 pair C₃H₄N₂ / 68.0, pyridine family kept anonymous; parallel structure and the seed's three misconception classes; the written subscripts 5,5 share a factor and only nitrogen's unwritten 1 makes the formula empirical — GIVEN in the stem, never asked) · mcq4
A compound has the empirical formula C₃H₄N₂ and a molar mass of 68.0 g/mol. (M of C₃H₄N₂ = 68.0 g/mol.) What is its molecular formula?
C₃H₄N₂ KEY
C₆H₈N₄
C₃H₈N₂
C₉H₁₂N₆
MOL-043 · hard · application · seq 21 · authored (parallel to the FRQ-1 seed-expansion structure; MASS route per the recorded rotation; fresh dataset — aluminum phosphate AlPO₄, unnamed in the stem; 3-element clean chain with the seed's two-equal-smallest shape; corpus scan 0 hits on AlPO₄/16.2/18.6/38.4; the element word aluminum is distinct from rows 15/18's claimed COMPOUND Al₂O₃/alumina — element words with fresh numbers carry no instance signal, row13 precedent) · frq
A sample of a compound contains 16.2 g of Al, 18.6 g of P, and 38.4 g of O. (M of Al = 27.0 g/mol, P = 31.0 g/mol, O = 16.0 g/mol.)
A sample of a compound contains 16.2 g of Al, 18.6 g of P, and 38.4 g of O. (M of Al = 27.0 g/mol, P = 31.0 g/mol, O = 16.0 g/mol.) (a) Calculate the number of moles of each element in the sample. Show your working. (b) Divide every mole value by the smallest to find the whole number for each element, clearing any decimal that remains. Show each step. (c) Write the compound's empirical formula, and state how you know its subscripts are exact whole numbers rather than rounded ones. Type subscripts as plain numbers, like the 2 in CO2.
Part a. Calculate the number of moles of each element in the sample. Show your working.
Model: Values from the question: m(Al) = 16.2 g, m(P) = 18.6 g, m(O) = 38.4 g Equation: n = m/M (for each element) Substituting the values and calculating: n(Al) = 16.2 / 27.0 = 0.600 mol n(P) = 18.6 / 31.0 = 0.600 mol n(O) = 38.4 / 16.0 = 2.40 mol
Part b. Divide every mole value by the smallest to find the whole number for each element, clearing any decimal that remains. Show each step.
Model: Divide each mole value by the smallest (0.600): Al: 0.600 / 0.600 = 1 P: 0.600 / 0.600 = 1 O: 2.40 / 0.600 = 4 No decimal remains — Al gives 1, P gives 1, O gives 4.
Part c. Write the compound's empirical formula, and state how you know its subscripts are exact whole numbers rather than rounded ones. Type subscripts as plain numbers, like the 2 in CO2.
Model: The finished whole numbers are the subscripts, so the empirical formula is AlPO4. The subscripts are exact: every division landed on a whole number, so nothing was rounded and no multiplying step was needed.
Full model answer. (a) Values from the question: m(Al) = 16.2 g, m(P) = 18.6 g, m(O) = 38.4 g Equation: n = m/M (for each element) Substituting the values and calculating: n(Al) = 16.2 / 27.0 = 0.600 mol n(P) = 18.6 / 31.0 = 0.600 mol n(O) = 38.4 / 16.0 = 2.40 mol (b) Divide each mole value by the smallest (0.600): Al: 0.600 / 0.600 = 1 P: 0.600 / 0.600 = 1 O: 2.40 / 0.600 = 4 No decimal remains — Al gives 1, P gives 1, O gives 4. (c) The finished whole numbers are the subscripts, so the empirical formula is AlPO4. The subscripts are exact: every division landed on a whole number, so nothing was rounded and no multiplying step was needed.
MOL-045 · hard · application · seq 22 · authored (FRQ-2 fresh parallel-structure variant; caffeine C₈H₁₀N₄O₂, count 2, four-element build — the richest part-(a) sum of the set; compound unnamed in student text per the gate register) · frq
A compound has the empirical formula C₄H₅N₂O and a molar mass of 194.0 g/mol. (M of C = 12.0 g/mol, H = 1.0 g/mol, N = 14.0 g/mol, O = 16.0 g/mol.)
A compound has the empirical formula C₄H₅N₂O and a molar mass of 194.0 g/mol. (M of C = 12.0 g/mol, H = 1.0 g/mol, N = 14.0 g/mol, O = 16.0 g/mol.) (a) Calculate the molar mass of the empirical formula C₄H₅N₂O from the atomic masses given, showing the sum you evaluate. (b) Find the number of empirical-formula units in one molecule of the compound. State the equation you use, and show every step of your working. (c) Write the molecular formula of the compound, and show how each subscript in your answer was produced. Type subscripts as plain numbers, like the 2 in CO2; no other special symbols are needed.
Part a. Calculate the molar mass of the empirical formula C₄H₅N₂O from the atomic masses given, showing the sum you evaluate.
Model: Values from the question: M of C = 12.0 g/mol, M of H = 1.0 g/mol, M of N = 14.0 g/mol, M of O = 16.0 g/mol The empirical formula C4H5N2O contains 4 C, 5 H, 2 N, and 1 O (an unwritten subscript counts as 1). M of C4H5N2O = 4 × 12.0 + 5 × 1.0 + 2 × 14.0 + 16.0 M of C4H5N2O = 97.0 g/mol
Part b. Find the number of empirical-formula units in one molecule of the compound. State the equation you use, and show every step of your working.
Model: Values from the question: molar mass of the compound = 194.0 g/mol molar mass of the empirical formula = 97.0 g/mol (from part a) Equation: number of units = molar mass of the compound ÷ molar mass of the empirical formula Substituting the values and calculating: number of units = 194.0 / 97.0 number of units = 2 The result is a whole number of at least 1, so the division ran the right way up.
Part c. Write the molecular formula of the compound, and show how each subscript in your answer was produced. Type subscripts as plain numbers, like the 2 in CO2; no other special symbols are needed.
Model: Multiply every subscript in the empirical formula by 2: C: 4 × 2 = 8, H: 5 × 2 = 10, N: 2 × 2 = 4, O: 1 × 2 = 2 Molecular formula: C8H10N4O2
Full model answer. (a) Values from the question: M of C = 12.0 g/mol, M of H = 1.0 g/mol, M of N = 14.0 g/mol, M of O = 16.0 g/mol The empirical formula C4H5N2O contains 4 C, 5 H, 2 N, and 1 O (an unwritten subscript counts as 1). M of C4H5N2O = 4 × 12.0 + 5 × 1.0 + 2 × 14.0 + 16.0 M of C4H5N2O = 97.0 g/mol (b) Values from the question: molar mass of the compound = 194.0 g/mol molar mass of the empirical formula = 97.0 g/mol (from part a) Equation: number of units = molar mass of the compound ÷ molar mass of the empirical formula Substituting the values and calculating: number of units = 194.0 / 97.0 number of units = 2 The result is a whole number of at least 1, so the division ran the right way up. (c) Multiply every subscript in the empirical formula by 2: C: 4 × 2 = 8, H: 5 × 2 = 10, N: 2 × 2 = 4, O: 1 × 2 = 2 Molecular formula: C8H10N4O2

Form D

MOL-036 · easy · recall · seq 1 · authored (seedless recall row; what-the-subscripts-state probe; key phrasing from the taught rule lines) · mcq4
What do the subscripts in a compound's empirical formula state?
How many atoms of one element there are for each atom of another — the atoms' smallest whole-number ratio. KEY
Exactly how many atoms of each element one single particle of the compound actually contains.
How many grams of each element a sample of the compound contains.
How many moles of the compound a sample contains.
MOL-037 · easy · recall · seq 2 · seedless authored row (probe D: per-molecule scope of the counts; instance-free) · mcq4
The atom counts in a molecular formula apply to what?
One molecule of the compound. KEY
A 100 g sample of the compound.
An entire sample of the compound, whatever its size.
A ratio unit rather than any real molecule.
MOL-039 · easy · recall · seq 3 · authored (seedless recall row; individually-divisible-is-not-the-test probe — the specNotes' key misconception stated as the bare rule (its instanced claim version sits at seq 19); headline distractor = the gate claim item's d2 class) · mcq4
Each subscript in a formula is divisible by some whole number bigger than 1 on its own, but no single whole number bigger than 1 divides into every subscript. What does the empirical check conclude?
The formula is already an empirical formula — the check needs one factor that divides into every subscript, and there is none. KEY
The formula is not an empirical formula — subscripts that each divide by something still let the whole formula reduce.
The check cannot decide this case — it applies only when the subscripts share a factor.
The formula is empirical only if one of its subscripts is an unwritten 1.
MOL-037 · easy · application · seq 4 · authored (parallel to the seq-4 seed structure; fresh instance sucrose C₁₂H₂₂O₁₁, outside the burned-instance ledger; corpus scan 0 hits) · mcq4
The molecular formula of sucrose (table sugar) is C₁₂H₂₂O₁₁. Which statement does the formula guarantee about one sucrose molecule?
It contains exactly 12 C atoms, 22 H atoms, and 11 O atoms. KEY
The numbers are smallest-terms values, and one sucrose molecule actually holds several times more of each atom.
It contains exactly 22 C atoms, 12 H atoms, and 11 O atoms.
The subscripts give each element's share of the molecule's mass in grams.
MOL-040 · easy · application · seq 5 · authored (parallel numeric slot; fresh chemically-real composition — calcium sulfide CaS, 55.5% Ca by mass; compound unnamed in the stem per the lesson convention) · numeric
A compound is 55.5% Ca by mass; the rest is S. What mass of S does a 100 g sample of the compound contain? Give your answer in grams to one decimal place.
Answer: 44.5 g KEY
also accepted: 44.5 g
wrongValueDiagnostics
[
 {
  "value": 55.5,
  "misconception": "you gave the calcium's mass — the question asks for the sulfur, the rest of the 100 g",
  "feedback": "This answer is wrong — you gave the calcium's mass — the question asks for the sulfur, the rest of the 100 g."
 },
 {
  "value": 0.445,
  "misconception": "you divided by 100 — in a 100 g sample the masses are in grams already",
  "feedback": "This answer is wrong — you divided by 100 — in a 100 g sample the masses are in grams already."
 }
]
feedback
{
 "keyExplanation": "Assume a sample of exactly 100 g.\nCalcium: 55.5% of the compound → 55.5 g.\nThe element masses in a 100 g sample total 100 g.\nmass of sulfur = 100 − 55.5 = 44.5 g"
}
MOL-036 · medium · application · seq 6 · authored (parallel mcq slot; fresh reducible measured ratio 4:4 (÷4, reduces to 1:1 — both subscripts unwritten, a distinct sub-case of the reduce-first family) + fresh element pair K/I — KI is a real classroom compound (potassium iodide), outside the burned-instance ledger; flipped-ratio distractor is degenerate at 1:1, replaced by the stopped-too-soon class (K₂I₂)) · mcq4
An instrument reports that a compound contains 4 K atoms for every 4 I atoms. Which formula is the compound's empirical formula?
KI KEY
K₄I₄
K₂I₂
K₄I
MOL-038 · medium · application · seq 7 · authored (blueprint seq-7 B-E: fresh genuinely-reducible formula, largest-factor trap live at factor 4; metaldehyde C₈H₁₆O₄ -> C₂H₄O — the three-element rotation, same case shape as the lesson's own glucose rule-line example with a fresh instance; misconception classes: divide-by-2-not-largest / left-unchanged / divided-each-by-own-value) · text
What is the empirical formula of metaldehyde, C₈H₁₆O₄? Type the whole formula, writing any subscripts as plain digits, like the 2 in CO2; leave subscripts of 1 unwritten.
Answer: C₂H₄O KEY
also accepted: C2H4O
wrongValueDiagnostics
[
 {
  "value": "C₄H₈O₂",
  "misconception": "you divided by 2, which is a common factor but not the LARGEST one — 4, 8, and 2 still share the factor 2",
  "feedback": "This answer is wrong — you divided by 2, which is a common factor but not the LARGEST one. The subscripts of C₄H₈O₂ — 4, 8, and 2 — still share the factor 2, so the formula still reduces. Divide C₈H₁₆O₄'s subscripts by their largest common factor, 4: C: 8 ÷ 4 = 2, H: 16 ÷ 4 = 4, O: 4 ÷ 4 = 1, and the empirical formula is C₂H₄O."
 },
 {
  "value": "C₈H₁₆O₄",
  "misconception": "you left the formula unchanged — the subscripts 8, 16, and 4 share the factor 4, so the formula reduces",
  "feedback": "This answer is wrong — you left the formula unchanged. The subscripts 8, 16, and 4 share the factor 4, so the formula reduces: C: 8 ÷ 4 = 2, H: 16 ÷ 4 = 4, O: 4 ÷ 4 = 1, and the empirical formula is C₂H₄O."
 },
 {
  "value": "CHO",
  "misconception": "you divided each subscript by its own value, which destroys the ratio",
  "feedback": "This answer is wrong — you divided each subscript by its own value, which destroys the ratio. Every subscript must be divided by the SAME factor — the largest common factor, 4: C: 8 ÷ 4 = 2, H: 16 ÷ 4 = 4, O: 4 ÷ 4 = 1, so the empirical formula is C₂H₄O."
 }
]
feedback
{
 "keyExplanation": "Subscripts: C 8, H 16, O 4. The largest common factor is 4 — dividing by 2 alone leaves C₄H₈O₂, which still reduces. C: 8 ÷ 4 = 2, H: 16 ÷ 4 = 4, O: 4 ÷ 4 = 1, and a subscript of 1 is never written, so the empirical formula is C₂H₄O.",
 "anticipatedWrong": [
  {
   "value": "C₄H₈O₂",
   "misconception": "you divided by 2, which is a common factor but not the LARGEST one — 4, 8, and 2 still share the factor 2",
   "feedback": "This answer is wrong — you divided by 2, which is a common factor but not the LARGEST one. The subscripts of C₄H₈O₂ — 4, 8, and 2 — still share the factor 2, so the formula still reduces. Divide C₈H₁₆O₄'s subscripts by their largest common factor, 4: C: 8 ÷ 4 = 2, H: 16 ÷ 4 = 4, O: 4 ÷ 4 = 1, and the empirical formula is C₂H₄O."
  },
  {
   "value": "C₈H₁₆O₄",
   "misconception": "you left the formula unchanged — the subscripts 8, 16, and 4 share the factor 4, so the formula reduces",
   "feedback": "This answer is wrong — you left the formula unchanged. The subscripts 8, 16, and 4 share the factor 4, so the formula reduces: C: 8 ÷ 4 = 2, H: 16 ÷ 4 = 4, O: 4 ÷ 4 = 1, and the empirical formula is C₂H₄O."
  },
  {
   "value": "CHO",
   "misconception": "you divided each subscript by its own value, which destroys the ratio",
   "feedback": "This answer is wrong — you divided each subscript by its own value, which destroys the ratio. Every subscript must be divided by the SAME factor — the largest common factor, 4: C: 8 ÷ 4 = 2, H: 16 ÷ 4 = 4, O: 4 ÷ 4 = 1, so the empirical formula is C₂H₄O."
  }
 ]
}
MOL-039 · medium · application · seq 8 · authored (pick-empirical polarity; fresh formula set — the unwritten-1 convention run from the KEY side: every option's written subscripts divide by 2, and only C₂H₆S's unwritten 1 on sulfur blocks the shared factor, so scanning for even numbers cannot game the item) · mcq4
Which of the following formulas is already in empirical form?
C₂H₆S KEY
C₂Cl₆
C₆H₈O₆
C₁₀H₁₄
MOL-041 · medium · application · seq 9 · authored (fresh two-element pair, PCl₅ family named chlorine-first — the taught order-follows-naming line puts the larger number first, so the backwards trap bites in the other direction; exact divisions 2.00 / 0.40; whole-number outcome by construction) · text
A sample of a compound contains 71.0 g of Cl and 12.4 g of P. What is the simplest mole ratio of Cl atoms to P atoms? (M of Cl = 35.5 g/mol, P = 31.0 g/mol.) Write your answer like 1 : 2.
Answer: 5 : 1 KEY
also accepted: 5:1
wrongValueDiagnostics
[
 {
  "value": "5.7 : 1",
  "misconception": "you compared the masses instead of the moles — 71.0 : 12.4 is about 5.7 : 1, but atoms are counted in moles",
  "feedback": "This answer is wrong — you compared the masses instead of the moles. 71.0 : 12.4 is about 5.7 : 1, but atoms are counted in moles: n(Cl) = 71.0 / 35.5 = 2.00 mol and n(P) = 12.4 / 31.0 = 0.40 mol, and dividing each by 0.40 gives 5 : 1."
 },
 {
  "value": "1 : 5",
  "misconception": "you wrote the ratio backwards — the order follows the elements as named: chlorine first, then phosphorus, so the larger number comes first here",
  "feedback": "This answer is wrong — you wrote the ratio backwards. The ratio order follows the elements as they are named in the question: chlorine first, then phosphorus. Chlorine has the larger mole value, so the answer is 5 : 1 — the ratio is not automatically written smallest-first."
 }
]
feedback
{
 "keyExplanation": "Values from the question:\nm(Cl) = 71.0 g, m(P) = 12.4 g\nEquation:\nn = m/M  (for each element)\nSubstituting the values and calculating:\nn(Cl) = 71.0 / 35.5 = 2.00 mol\nn(P) = 12.4 / 31.0 = 0.40 mol\nDivide each by the smallest (0.40): Cl gives 5, P gives 1.\nMole ratio of chlorine to phosphorus = 5 : 1",
 "anticipatedWrong": [
  {
   "value": "5.7 : 1",
   "misconception": "you compared the masses instead of the moles — 71.0 : 12.4 is about 5.7 : 1, but atoms are counted in moles",
   "feedback": "This answer is wrong — you compared the masses instead of the moles. 71.0 : 12.4 is about 5.7 : 1, but atoms are counted in moles: n(Cl) = 71.0 / 35.5 = 2.00 mol and n(P) = 12.4 / 31.0 = 0.40 mol, and dividing each by 0.40 gives 5 : 1."
  },
  {
   "value": "1 : 5",
   "misconception": "you wrote the ratio backwards — the order follows the elements as named: chlorine first, then phosphorus, so the larger number comes first here",
   "feedback": "This answer is wrong — you wrote the ratio backwards. The ratio order follows the elements as they are named in the question: chlorine first, then phosphorus. Chlorine has the larger mole value, so the answer is 5 : 1 — the ratio is not automatically written smallest-first."
  }
 ]
}
MOL-042 · medium · application · seq 10 · authored (fresh ratio 1 : 3.25 -> 4 : 13, x4, .25 branch; 3.25 and 4 : 13 outside the burned ledger and machine-checked fresh; round-DOWN direction natural for .25; the rounding wrong-value 1 : 3 is a burned-ledger ratio — feedback-field only, arithmetically forced by the taught misconception, recorded in flags) · text
A mole calculation gives the ratio 1 : 3.25. What whole-number ratio does it clear to? Give your answer like 2 : 3.
Answer: 4 : 13 KEY
also accepted: 4:13
wrongValueDiagnostics
[
 {
  "value": "1 : 3",
  "misconception": "you rounded the 3.25 down to 3 — rounding changes the ratio; multiply every number by 4 instead",
  "feedback": "This answer is wrong — you rounded the 3.25 down to 3, and rounding changes the ratio itself. The decimal part is .25, so multiply every number by 4 instead: 1 × 4 = 4 and 3.25 × 4 = 13, giving 4 : 13."
 },
 {
  "value": "1 : 13",
  "misconception": "you multiplied only the decimal number — every number in the ratio must be multiplied by the same amount",
  "feedback": "This answer is wrong — you multiplied only the decimal number. Every number in the ratio must be multiplied by the same amount to keep the ratio unchanged: 1 × 4 = 4 and 3.25 × 4 = 13, giving 4 : 13."
 }
]
feedback
{
 "keyExplanation": "The decimal part is .25, so multiply every number by 4: 1 × 4 = 4 and 3.25 × 4 = 13. The whole-number ratio is 4 : 13. Rounding 3.25 down to 3 would change the ratio itself — multiplying never does.",
 "anticipatedWrong": [
  {
   "value": "1 : 3",
   "misconception": "you rounded the 3.25 down to 3 — rounding changes the ratio; multiply every number by 4 instead",
   "feedback": "This answer is wrong — you rounded the 3.25 down to 3, and rounding changes the ratio itself. The decimal part is .25, so multiply every number by 4 instead: 1 × 4 = 4 and 3.25 × 4 = 13, giving 4 : 13."
  },
  {
   "value": "1 : 13",
   "misconception": "you multiplied only the decimal number — every number in the ratio must be multiplied by the same amount",
   "feedback": "This answer is wrong — you multiplied only the decimal number. Every number in the ratio must be multiplied by the same amount to keep the ratio unchanged: 1 × 4 = 4 and 3.25 × 4 = 13, giving 4 : 13."
  }
 ]
}
MOL-038 · medium · application · seq 11 · authored (parallel to the seq-11 seed structure; fresh instance disulfur dichloride S₂Cl₂ -> SCl — the blueprint's OTHER named seq-11 candidate, consumed here; covers the both-subscripts-vanish surface, the benzene-WE case family; misconception adjudication recorded — the divide-by-own-value class degenerates to the key on equal subscripts, so the H₂O₂ fluency's divide-only-one-subscript pair is carried instead) · text
What is the empirical formula of disulfur dichloride, S₂Cl₂? Type the whole formula, writing any subscripts as plain digits, like the 2 in CO2; leave subscripts of 1 unwritten.
Answer: SCl KEY
wrongValueDiagnostics
[
 {
  "value": "S₂Cl₂",
  "misconception": "you left the formula unchanged — the subscripts 2 and 2 share the factor 2, so the formula reduces",
  "feedback": "This answer is wrong — you left the formula unchanged. The subscripts 2 and 2 share the factor 2, so the formula reduces: S: 2 ÷ 2 = 1, Cl: 2 ÷ 2 = 1, and a subscript of 1 is never written, giving SCl."
 },
 {
  "value": "SCl₂",
  "misconception": "you divided only the sulfur's subscript — every subscript must be divided by the same factor",
  "feedback": "This answer is wrong — you divided only the sulfur's subscript. Every subscript must be divided by the same factor: S: 2 ÷ 2 = 1, Cl: 2 ÷ 2 = 1, giving SCl."
 },
 {
  "value": "S₂Cl",
  "misconception": "you divided only the chlorine's subscript — every subscript must be divided by the same factor",
  "feedback": "This answer is wrong — you divided only the chlorine's subscript. Every subscript must be divided by the same factor: S: 2 ÷ 2 = 1, Cl: 2 ÷ 2 = 1, giving SCl."
 }
]
feedback
{
 "keyExplanation": "Subscripts: S 2, Cl 2. The largest common factor is 2. S: 2 ÷ 2 = 1, Cl: 2 ÷ 2 = 1. A subscript of 1 is never written, so disulfur dichloride's empirical formula is SCl.",
 "anticipatedWrong": [
  {
   "value": "S₂Cl₂",
   "misconception": "you left the formula unchanged — the subscripts 2 and 2 share the factor 2, so the formula reduces",
   "feedback": "This answer is wrong — you left the formula unchanged. The subscripts 2 and 2 share the factor 2, so the formula reduces: S: 2 ÷ 2 = 1, Cl: 2 ÷ 2 = 1, and a subscript of 1 is never written, giving SCl."
  },
  {
   "value": "SCl₂",
   "misconception": "you divided only the sulfur's subscript — every subscript must be divided by the same factor",
   "feedback": "This answer is wrong — you divided only the sulfur's subscript. Every subscript must be divided by the same factor: S: 2 ÷ 2 = 1, Cl: 2 ÷ 2 = 1, giving SCl."
  },
  {
   "value": "S₂Cl",
   "misconception": "you divided only the chlorine's subscript — every subscript must be divided by the same factor",
   "feedback": "This answer is wrong — you divided only the chlorine's subscript. Every subscript must be divided by the same factor: S: 2 ÷ 2 = 1, Cl: 2 ÷ 2 = 1, giving SCl."
  }
 ]
}
MOL-044 · medium · transfer · seq 12 · authored (quotient-meaning, notebook-claim frame; parallel misconception classes to the seed; fresh exact pair 265.0/53.0 = 5) · mcq4
A compound's molar mass is 265.0 g/mol and its empirical formula's molar mass is 53.0 g/mol. A student writes "265.0 ÷ 53.0 = 5" in her notebook. What has the 5 told her?
That each molecule of the compound is built from 5 empirical-formula units. KEY
That each empirical-formula unit is built from 5 molecules of the compound.
That each molecule of the compound holds 5 atoms of each element.
That her sample holds 5 mol of the compound.
MOL-041 · medium · application · seq 13 · authored (parallel to the seq-13 seed structure; fresh triple — ethylamine C₂H₇N, unnamed in the stem per the lesson convention; smallest mole value on the LAST-listed element — recorded rotation inside the same smallest-not-first case, the order-keeping bite lands at the tail; corpus scan 0 hits) · text
A sample of a compound contains 36.0 g of C, 10.5 g of H, and 21.0 g of N. What is the simplest mole ratio of C to H to N atoms? (M of C = 12.0 g/mol, H = 1.0 g/mol, N = 14.0 g/mol.) Write your answer like 1 : 2 : 3.
Answer: 2 : 7 : 1 KEY
also accepted: 2:7:1
wrongValueDiagnostics
[
 {
  "value": "3.4 : 1 : 2",
  "misconception": "you compared the masses instead of the moles — 36.0 : 10.5 : 21.0 is about 3.4 : 1 : 2, but atoms are counted in moles",
  "feedback": "This answer is wrong — you compared the masses instead of the moles — 36.0 : 10.5 : 21.0 is about 3.4 : 1 : 2, but atoms are counted in moles."
 },
 {
  "value": "1 : 7 : 2",
  "misconception": "you wrote the ratio backwards — the order follows the elements as named: carbon first, then hydrogen, then nitrogen",
  "feedback": "This answer is wrong — you wrote the ratio backwards — the order follows the elements as named: carbon first, then hydrogen, then nitrogen."
 }
]
feedback
{
 "keyExplanation": "Values from the question:\nm(C) = 36.0 g, m(H) = 10.5 g, m(N) = 21.0 g\nEquation:\nn = m/M  (for each element)\nSubstituting the values and calculating:\nn(C) = 36.0 / 12.0 = 3.00 mol\nn(H) = 10.5 / 1.0 = 10.5 mol\nn(N) = 21.0 / 14.0 = 1.50 mol\nDivide each by the smallest (1.50): C gives 2, H gives 7, N gives 1.\nMole ratio of carbon to hydrogen to nitrogen = 2 : 7 : 1"
}
MOL-044 · medium · application · seq 14 · authored (parallel numeric slot; fresh (formula, M) pair — azobenzene C₁₂H₁₀N₂, empirical C₆H₅N, 182.0 g/mol -> 2 units; compound unnamed in the stem per the lesson's gate convention) · numeric
A compound has the empirical formula C₆H₅N and a molar mass of 182.0 g/mol. (M of C = 12.0 g/mol, H = 1.0 g/mol, N = 14.0 g/mol.) How many empirical-formula units fit in one of its molecules? Answer with the whole-number count.
Answer: 2 units KEY
also accepted: 2 units, 2.0
wrongValueDiagnostics
[
 {
  "value": 6.74,
  "misconception": "you ignored the subscripts in the empirical formula, using 12.0 + 1.0 + 14.0 = 27.0 g/mol — M of C₆H₅N is 6 × 12.0 + 5 × 1.0 + 14.0 = 91.0 g/mol",
  "feedback": "This answer is wrong — you ignored the subscripts in the empirical formula, using 12.0 + 1.0 + 14.0 = 27.0 g/mol. M of C₆H₅N is 6 × 12.0 + 5 × 1.0 + 14.0 = 91.0 g/mol, and a non-whole unit count signals exactly this kind of slip."
 },
 {
  "value": 0.5,
  "misconception": "you divided upside down — the compound's molar mass goes on top, and the unit count is never below 1",
  "feedback": "This answer is wrong — you divided upside down — the compound's molar mass goes on top, and the unit count is never below 1."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmolar mass of the compound = 182.0 g/mol\nM of C₆H₅N = 6 × 12.0 + 5 × 1.0 + 14.0 = 91.0 g/mol\nWrite down the equation:\nnumber of units = molar mass of the compound ÷ molar mass of the empirical formula\nSubstitute in the values, and calculate:\nnumber of units = 182.0 / 91.0\nnumber of units = 2\nThe result is a whole number at least 1, so the division ran the right way up."
}
MOL-043 · hard · application · seq 15 · authored (parallel mass-route capstone; fresh chemically-real compound — boron trioxide B₂O₃, compound unnamed in the stem per the seed convention; exact divisions 21.6/10.8 = 2.00 and 48.0/16.0 = 3.00; .5-decimal clearing ×2; boron on form D is unique within the form — the other boron instances in this gate sit on forms A and E at rows 11/13) · text
A sample of a compound contains 21.6 g of B and 48.0 g of O. What is its empirical formula? (M of B = 10.8 g/mol, O = 16.0 g/mol.) Type subscripts as plain numbers, like the 2 in CO2.
Answer: B₂O₃ KEY
also accepted: B2O3
wrongValueDiagnostics
[
 {
  "value": "BO₂",
  "misconception": "you rounded the 1.5 up to 2 — rounding changes the ratio; multiply every number by 2 instead",
  "feedback": "This answer is wrong — you rounded the 1.5 up to 2. Rounding changes the ratio; multiply every number by 2 instead: 1 and 1.5 become 2 and 3, so the empirical formula is B₂O₃."
 },
 {
  "value": "BO",
  "misconception": "you dropped the decimal — cutting the .5 off changes the ratio; multiply every number by 2 instead",
  "feedback": "This answer is wrong — you dropped the decimal. Cutting the .5 off changes the ratio; multiply every number by 2 instead: 1 and 1.5 become 2 and 3, so the empirical formula is B₂O₃."
 },
 {
  "value": "B₃O₂",
  "misconception": "you inverted the ratio — boron has the smaller mole value, so boron takes the 1 before clearing",
  "feedback": "This answer is wrong — you inverted the ratio. Boron has the smaller mole value (2.00 mol against oxygen's 3.00 mol), so boron takes the 1 before clearing: 1 and 1.5 become 2 and 3, giving B₂O₃."
 }
]
feedback
{
 "keyExplanation": "Values from the question:\nm(B) = 21.6 g, m(O) = 48.0 g\nEquation:\nn = m/M  (for each element)\nSubstituting the values and calculating:\nn(B) = 21.6 / 10.8 = 2.00 mol\nn(O) = 48.0 / 16.0 = 3.00 mol\nDivide each by the smallest (2.00): B gives 1, O gives 1.5.\nThe decimal ends in .5, so multiply every number by 2: 2 and 3.\nEmpirical formula: B₂O₃",
 "anticipatedWrong": [
  {
   "value": "BO₂",
   "misconception": "you rounded the 1.5 up to 2 — rounding changes the ratio; multiply every number by 2 instead",
   "feedback": "This answer is wrong — you rounded the 1.5 up to 2. Rounding changes the ratio; multiply every number by 2 instead: 1 and 1.5 become 2 and 3, so the empirical formula is B₂O₃."
  },
  {
   "value": "BO",
   "misconception": "you dropped the decimal — cutting the .5 off changes the ratio; multiply every number by 2 instead",
   "feedback": "This answer is wrong — you dropped the decimal. Cutting the .5 off changes the ratio; multiply every number by 2 instead: 1 and 1.5 become 2 and 3, so the empirical formula is B₂O₃."
  },
  {
   "value": "B₃O₂",
   "misconception": "you inverted the ratio — boron has the smaller mole value, so boron takes the 1 before clearing",
   "feedback": "This answer is wrong — you inverted the ratio. Boron has the smaller mole value (2.00 mol against oxygen's 3.00 mol), so boron takes the 1 before clearing: 1 and 1.5 become 2 and 3, giving B₂O₃."
  }
 ]
}
MOL-042 · hard · transfer · seq 16 · authored (two-students method-judgment frame; round-up error spotlighted on a decimal-FIRST ratio — fluency F3's '1.5 : 1' order precedent; fresh ratio 2.75 : 1 -> 11 : 4, x4; all three misconception classes live) · mcq4
A mole calculation gives the ratio 2.75 : 1. One student multiplies both numbers by 4; another rounds the 2.75 up to 3. Which result is right, and why?
11 : 4 — multiplying every number by 4 keeps the ratio unchanged, while rounding 2.75 to 3 changes it. KEY
3 : 1 — a decimal that close to a whole number is within normal rounding distance.
11 : 1 — multiply only the 2.75 by 4, because the 1 is already a whole number.
2 : 1 — drop the decimal part, since subscripts must be whole numbers.
MOL-045 · hard · application · seq 17 · authored (fresh pair CF₂ / 200.0 -> C₄F₈, octafluorocyclobutane family kept anonymous; no-hydrogen halogen build for element variety; parallel structure, step count, and misconception classes to the seed; 200.0 and 50.0 are not powers of ten) · text
A compound has the empirical formula CF₂ and a molar mass of 200.0 g/mol. (M of C = 12.0 g/mol, F = 19.0 g/mol.) State the compound's molecular formula, typing subscripts as plain numbers (like the 2 in CO2).
Answer: C₄F₈ KEY
also accepted: C4F8
wrongValueDiagnostics
[
 {
  "value": "CF₂",
  "misconception": "you stopped at the empirical formula — the unit count is 200.0/50.0 = 4, and every subscript must be multiplied by it",
  "feedback": "This answer is wrong — you stopped at the empirical formula. The unit count is 200.0 / 50.0 = 4, and every subscript must be multiplied by it: C: 1 × 4 = 4, F: 2 × 4 = 8, giving C₄F₈."
 },
 {
  "value": "C₄F₂",
  "misconception": "you multiplied only carbon's subscript — every subscript in the empirical formula is multiplied by the unit count",
  "feedback": "This answer is wrong — you multiplied only carbon's subscript. Every subscript in the empirical formula is multiplied by the unit count, 4: C: 1 × 4 = 4, F: 2 × 4 = 8, giving C₄F₈."
 },
 {
  "value": "CF₈",
  "misconception": "you multiplied only fluorine's subscript — every subscript in the empirical formula is multiplied by the unit count",
  "feedback": "This answer is wrong — you multiplied only fluorine's subscript. Every subscript in the empirical formula is multiplied by the unit count, 4: C: 1 × 4 = 4, F: 2 × 4 = 8, giving C₄F₈."
 }
]
feedback
{
 "keyExplanation": "M of CF₂ = 12.0 + 2 × 19.0 = 50.0 g/mol.\nStep 1 — find the number of units:\nnumber of units = molar mass of the compound ÷ molar mass of the empirical formula\nnumber of units = 200.0 / 50.0 = 4\nStep 2 — multiply every subscript by 4:\nC: 1 × 4 = 4,  F: 2 × 4 = 8\nMolecular formula: C₄F₈",
 "anticipatedWrong": [
  {
   "value": "CF₂",
   "misconception": "you stopped at the empirical formula — the unit count is 200.0/50.0 = 4, and every subscript must be multiplied by it",
   "feedback": "This answer is wrong — you stopped at the empirical formula. The unit count is 200.0 / 50.0 = 4, and every subscript must be multiplied by it: C: 1 × 4 = 4, F: 2 × 4 = 8, giving C₄F₈."
  },
  {
   "value": "C₄F₂",
   "misconception": "you multiplied only carbon's subscript — every subscript in the empirical formula is multiplied by the unit count",
   "feedback": "This answer is wrong — you multiplied only carbon's subscript. Every subscript in the empirical formula is multiplied by the unit count, 4: C: 1 × 4 = 4, F: 2 × 4 = 8, giving C₄F₈."
  },
  {
   "value": "CF₈",
   "misconception": "you multiplied only fluorine's subscript — every subscript in the empirical formula is multiplied by the unit count",
   "feedback": "This answer is wrong — you multiplied only fluorine's subscript. Every subscript in the empirical formula is multiplied by the unit count, 4: C: 1 × 4 = 4, F: 2 × 4 = 8, giving C₄F₈."
  }
 ]
}
MOL-043 · hard · application · seq 18 · authored (fresh percent dataset, aluminum oxide: exact real composition 52.941/47.059 -> 52.9/47.1 at 3 sf, sum 100.0; quotient 1.50 EXACT at display precision clears ×2; identical chain and step count to the seed) · text
A compound is 52.9% Al and 47.1% O by mass. What is its empirical formula? (M of Al = 27.0 g/mol, O = 16.0 g/mol.) Type subscripts as plain numbers, like the 2 in CO2.
Answer: Al₂O₃ KEY
also accepted: Al2O3
wrongValueDiagnostics
[
 {
  "value": "AlO₂",
  "misconception": "you rounded the 1.50 up to 2 — rounding changes the ratio; multiply every number by 2 instead",
  "feedback": "This answer is wrong — you rounded the 1.50 up to 2. Rounding changes the ratio; multiply every number by 2 instead: Al gives 2 and O gives 3, so the empirical formula is Al₂O₃."
 },
 {
  "value": "AlO",
  "misconception": "you dropped the decimal — cutting the .50 off changes the ratio; multiply every number by 2 instead",
  "feedback": "This answer is wrong — you dropped the decimal. Cutting the .50 off changes the ratio; multiply every number by 2 instead: Al gives 2 and O gives 3, so the empirical formula is Al₂O₃."
 }
]
feedback
{
 "keyExplanation": "Assume a 100 g sample: m(Al) = 52.9 g, m(O) = 47.1 g.\nEquation:\nn = m/M  (for each element)\nSubstituting the values and calculating:\nn(Al) = 52.9 / 27.0 = 1.96 mol\nn(O) = 47.1 / 16.0 = 2.94 mol\nDivide each by the smallest (1.96): Al gives 1, O gives 1.50.\nThe decimal ends in .5, so multiply every number by 2: 2 and 3.\nEmpirical formula: Al₂O₃",
 "anticipatedWrong": [
  {
   "value": "AlO₂",
   "misconception": "you rounded the 1.50 up to 2 — rounding changes the ratio; multiply every number by 2 instead",
   "feedback": "This answer is wrong — you rounded the 1.50 up to 2. Rounding changes the ratio; multiply every number by 2 instead: Al gives 2 and O gives 3, so the empirical formula is Al₂O₃."
  },
  {
   "value": "AlO",
   "misconception": "you dropped the decimal — cutting the .50 off changes the ratio; multiply every number by 2 instead",
   "feedback": "This answer is wrong — you dropped the decimal. Cutting the .50 off changes the ratio; multiply every number by 2 instead: Al gives 2 and O gives 3, so the empirical formula is Al₂O₃."
  }
 ]
}
MOL-039 · hard · transfer · seq 19 · authored (fresh coprime-subscript claim rotation per blueprint case_target; instance C21H44, heneicosane — subscripts 27 and 56 each composite yet coprime; outside the burned-instance ledger; counterexample C16H34, hexadecane, reduces) · mcq4
A student claims C₂₇H₅₆ cannot be an empirical formula 'because 27 and 56 are both bigger than 1.' Judge the claim.
The claim is wrong — 27 and 56 share no common factor bigger than 1, so C₂₇H₅₆ is an empirical formula. KEY
The claim is right — an empirical formula must contain at least one unwritten subscript of 1.
The claim is right — 27 and 56 can each be divided by a smaller number, so the formula reduces.
The claim is wrong — no formula with only two subscripts can ever be reduced.
MOL-045 · hard · transfer · seq 20 · authored (fresh x1 pair C₄H₁₁N / 73.0, butylamine family kept anonymous — adopted after trimethylamine C₃H₉N/59.0 was declined against row 17's burned 59.0; parallel structure and the seed's three misconception classes) · mcq4
A compound has the empirical formula C₄H₁₁N and a molar mass of 73.0 g/mol. (M of C₄H₁₁N = 73.0 g/mol.) What is its molecular formula?
C₄H₁₁N KEY
C₈H₂₂N₂
C₈H₁₁N
C₁₆H₄₄N₄
MOL-043 · hard · application · seq 21 · authored (parallel to the FRQ-1 seed-expansion structure; PERCENT route per the recorded rotation; fresh dataset — sodium sulfite Na₂SO₃, unnamed in the stem; 3-element clean chain ON THE PERCENT ROUTE with the smallest mole value on the middle-listed element (order-keeping bite, row13 precedent); percents are the honest 1-dp rounding of the real compound and total 100.1, inside the course-taught 99.9-100.1 band; corpus scan 0 hits on Na₂SO₃/36.5/25.5/38.1) | blind-wave1c 2026-09-08: RESTORED VERBATIM after two interim replacements were withdrawn (Na2PO3: C#21 38.4 collision; Na2PO4: sound P-oxidation-state realism objection). ADJUDICATED-ACCEPT vs blind flags r1/r3: the course rule line 'Small rounding differences, like 99.9 or 100.1, are acceptable' makes the 100.1 total and the 3-sf displayed near-integer quotients (1.59/0.794 = 2.00, 2.38/0.794 = 3.00) the TAUGHT register; exhaustive search: no conventional compound gives fully-exact 1-dp percent data. Flag for David. | qc1 2026-09-08: sol majors on the exactness claim -> precision-honest part (c)/model/rubric rewrite (taught rounding-band register); data unchanged · frq
A compound is 36.5% Na, 25.5% S, and 38.1% O by mass. (M of Na = 23.0 g/mol, S = 32.1 g/mol, O = 16.0 g/mol.)
A compound is 36.5% Na, 25.5% S, and 38.1% O by mass. (M of Na = 23.0 g/mol, S = 32.1 g/mol, O = 16.0 g/mol.) (a) Use a 100 g sample to turn the percents into masses, then calculate the number of moles of each element. Show your working. (b) Divide every mole value by the smallest to find the whole number for each element, clearing any decimal that remains. Show each step. (c) Write the compound's empirical formula, and state how you know no clearing step was needed. Type subscripts as plain numbers, like the 2 in CO2.
Part a. Use a 100 g sample to turn the percents into masses, then calculate the number of moles of each element. Show your working.
Model: Assume a 100 g sample: m(Na) = 36.5 g, m(S) = 25.5 g, m(O) = 38.1 g. Equation: n = m/M (for each element) Substituting the values and calculating: n(Na) = 36.5 / 23.0 = 1.59 mol n(S) = 25.5 / 32.1 = 0.794 mol n(O) = 38.1 / 16.0 = 2.38 mol
Part b. Divide every mole value by the smallest to find the whole number for each element, clearing any decimal that remains. Show each step.
Model: The smallest mole value is sulfur's 0.794 (the middle one, not the first listed). Divide each mole value by the smallest (0.794): Na: 1.59 / 0.794 = 2.00 S: 0.794 / 0.794 = 1 O: 2.38 / 0.794 = 3.00 At the data's precision no decimal remains — Na gives 2, S gives 1, O gives 3.
Part c. Write the compound's empirical formula, and state how you know no clearing step was needed. Type subscripts as plain numbers, like the 2 in CO2.
Model: The finished whole numbers are the subscripts, so the empirical formula is Na2SO3. No clearing step was needed: at the data's precision every division landed on a whole number (2.00, 1, 3.00). Percent data carry small rounding — a genuine leftover decimal like .33 or .5 would have signaled a clearing step, and quotients this close to whole numbers are read as those whole numbers.
Full model answer. (a) Assume a 100 g sample: m(Na) = 36.5 g, m(S) = 25.5 g, m(O) = 38.1 g. Equation: n = m/M (for each element) Substituting the values and calculating: n(Na) = 36.5 / 23.0 = 1.59 mol n(S) = 25.5 / 32.1 = 0.794 mol n(O) = 38.1 / 16.0 = 2.38 mol (b) The smallest mole value is sulfur's 0.794 (the middle one, not the first listed). Divide each mole value by the smallest (0.794): Na: 1.59 / 0.794 = 2.00 S: 0.794 / 0.794 = 1 O: 2.38 / 0.794 = 3.00 At the data's precision no decimal remains — Na gives 2, S gives 1, O gives 3. (c) The finished whole numbers are the subscripts, so the empirical formula is Na2SO3. No clearing step was needed: at the data's precision every division landed on a whole number (2.00, 1, 3.00). Percent data carry small rounding — a genuine leftover decimal like .33 or .5 would have signaled a clearing step, and quotients this close to whole numbers are read as those whole numbers.
MOL-045 · hard · application · seq 22 · authored (FRQ-2 fresh parallel-structure variant; 1,3,5-tribromobenzene C₆H₃Br₃, count 3; Br = 79.9 g/mol is a periodic-table constant new to the gate, constants-class; compound unnamed in student text per the gate register) · frq
A compound has the empirical formula C₂HBr and a molar mass of 314.7 g/mol. (M of C = 12.0 g/mol, H = 1.0 g/mol, Br = 79.9 g/mol.)
A compound has the empirical formula C₂HBr and a molar mass of 314.7 g/mol. (M of C = 12.0 g/mol, H = 1.0 g/mol, Br = 79.9 g/mol.) (a) Calculate the molar mass of the empirical formula C₂HBr from the atomic masses given, showing the sum you evaluate. (b) Find the number of empirical-formula units in one molecule of the compound. State the equation you use, and show every step of your working. (c) Write the molecular formula of the compound, and show how each subscript in your answer was produced. Type subscripts as plain numbers, like the 2 in CO2; no other special symbols are needed.
Part a. Calculate the molar mass of the empirical formula C₂HBr from the atomic masses given, showing the sum you evaluate.
Model: Values from the question: M of C = 12.0 g/mol, M of H = 1.0 g/mol, M of Br = 79.9 g/mol The empirical formula C2HBr contains 2 C, 1 H, and 1 Br (an unwritten subscript counts as 1). M of C2HBr = 2 × 12.0 + 1.0 + 79.9 M of C2HBr = 104.9 g/mol
Part b. Find the number of empirical-formula units in one molecule of the compound. State the equation you use, and show every step of your working.
Model: Values from the question: molar mass of the compound = 314.7 g/mol molar mass of the empirical formula = 104.9 g/mol (from part a) Equation: number of units = molar mass of the compound ÷ molar mass of the empirical formula Substituting the values and calculating: number of units = 314.7 / 104.9 number of units = 3 The result is a whole number of at least 1, so the division ran the right way up.
Part c. Write the molecular formula of the compound, and show how each subscript in your answer was produced. Type subscripts as plain numbers, like the 2 in CO2; no other special symbols are needed.
Model: Multiply every subscript in the empirical formula by 3: C: 2 × 3 = 6, H: 1 × 3 = 3, Br: 1 × 3 = 3 Molecular formula: C6H3Br3
Full model answer. (a) Values from the question: M of C = 12.0 g/mol, M of H = 1.0 g/mol, M of Br = 79.9 g/mol The empirical formula C2HBr contains 2 C, 1 H, and 1 Br (an unwritten subscript counts as 1). M of C2HBr = 2 × 12.0 + 1.0 + 79.9 M of C2HBr = 104.9 g/mol (b) Values from the question: molar mass of the compound = 314.7 g/mol molar mass of the empirical formula = 104.9 g/mol (from part a) Equation: number of units = molar mass of the compound ÷ molar mass of the empirical formula Substituting the values and calculating: number of units = 314.7 / 104.9 number of units = 3 The result is a whole number of at least 1, so the division ran the right way up. (c) Multiply every subscript in the empirical formula by 3: C: 2 × 3 = 6, H: 1 × 3 = 3, Br: 1 × 3 = 3 Molecular formula: C6H3Br3

Form E

MOL-036 · easy · recall · seq 1 · authored (seedless recall row; ratio-not-count claim judged from the definition side; 'one particle' register per the lesson's own fluency corpus, molecular formula never named) · mcq4
Looking only at a compound's empirical formula, a student states the exact number of atoms of each element in one particle of the compound. Can the formula guarantee those numbers?
No — it records only the smallest whole-number ratio of the atoms, and many different exact counts fit the same ratio. KEY
Yes — a formula's subscripts always state the exact atom counts in one particle.
Yes — once the ratio is written in smallest whole numbers, the exact counts must match the subscripts.
No — an empirical formula carries no information about the compound's atoms at all.
MOL-037 · easy · recall · seq 2 · seedless authored row (probe E: counts-are-definite-not-a-ratio; misconception family documented by the spared octane claim seed, used instance-free at recall level) · mcq4
Which statement about the numbers in a molecular formula is correct?
They are definite counts — every molecule of the compound holds exactly those numbers of atoms. KEY
They give only a ratio, so the actual counts in a molecule cannot be known from the formula.
They can differ from one molecule to the next, as long as the ratio between them stays the same.
They state the mass, in grams, that each element contributes to the molecule.
MOL-039 · easy · recall · seq 3 · authored (seedless recall row; which-question-to-ask probe compressing the check to its governing question; distractors = the three named wrong questions (individually-divisible, subscript-size, invented-1)) · mcq4
To run the empirical check on a formula, which single question do you ask about its subscripts?
Does any whole number bigger than 1 divide into every subscript? KEY
Does each subscript divide by some whole number bigger than 1 on its own?
Is every subscript bigger than 1?
Does the formula contain an unwritten subscript of 1?
MOL-037 · easy · application · seq 4 · authored (parallel to the seq-4 seed structure; fresh instance toluene C₇H₈, outside the burned-instance ledger; corpus scan 0 hits) · mcq4
The molecular formula of toluene is C₇H₈. Which statement does the formula guarantee about one toluene molecule?
It contains exactly 7 C atoms and 8 H atoms. KEY
The 7 and 8 are reduced figures, and the true counts in one molecule are a multiple of them.
It contains exactly 8 C atoms and 7 H atoms.
It fixes the counts only under standard conditions — heating the sample changes the subscripts.
MOL-040 · easy · application · seq 5 · authored (parallel numeric slot; fresh chemically-real composition — magnesium fluoride MgF₂, 39.0% Mg by mass; compound unnamed in the stem per the lesson convention) · numeric
A compound is 39.0% Mg by mass; the rest is F. What mass of F does a 100 g sample of the compound contain? Give your answer in grams to one decimal place.
Answer: 61.0 g KEY
also accepted: 61.0 g
wrongValueDiagnostics
[
 {
  "value": 39.0,
  "misconception": "you gave the magnesium's mass — the question asks for the fluorine, the rest of the 100 g",
  "feedback": "This answer is wrong — you gave the magnesium's mass — the question asks for the fluorine, the rest of the 100 g."
 },
 {
  "value": 0.61,
  "misconception": "you divided by 100 — in a 100 g sample the masses are in grams already",
  "feedback": "This answer is wrong — you divided by 100 — in a 100 g sample the masses are in grams already."
 }
]
feedback
{
 "keyExplanation": "Assume a sample of exactly 100 g.\nMagnesium: 39.0% of the compound → 39.0 g.\nThe element masses in a 100 g sample total 100 g.\nmass of fluorine = 100 − 39.0 = 61.0 g"
}
MOL-036 · medium · application · seq 6 · authored (parallel mcq slot; fresh reducible measured ratio 6:12 (÷6, the largest reduce factor of the row) + fresh element pair Zn/Br — ZnBr₂ is a real compound (zinc bromide), outside the burned-instance ledger; zinc and bromine appear nowhere in the scope corpus or on form A) · mcq4
An instrument reports that a compound contains 6 Zn atoms for every 12 Br atoms. Which formula is the compound's empirical formula?
ZnBr₂ KEY
Zn₆Br₁₂
Zn₃Br₆
Zn₂Br
MOL-038 · medium · application · seq 7 · authored (blueprint seq-7 B-E: fresh genuinely-reducible formula, largest-factor trap live at factor 4; fumaric acid C₄H₄O₄ -> CHO — the equal-subscripts case shape from the lesson's own benzene WE and H₂O₂ fluency item, fresh instance; misconception classes: divide-by-2-not-largest / left-unchanged / divided-only-one-subscript [the H₂O₂ fluency class — on equal subscripts the divided-each-by-own-value path coincides with the key, recorded design note in flags]) · text
What is the empirical formula of fumaric acid, C₄H₄O₄? Type the whole formula, writing any subscripts as plain digits, like the 2 in CO2; leave subscripts of 1 unwritten.
Answer: CHO KEY
wrongValueDiagnostics
[
 {
  "value": "C₂H₂O₂",
  "misconception": "you divided by 2, which is a common factor but not the LARGEST one — 2, 2, and 2 still share the factor 2",
  "feedback": "This answer is wrong — you divided by 2, which is a common factor but not the LARGEST one. The subscripts of C₂H₂O₂ — 2, 2, and 2 — still share the factor 2, so the formula still reduces. Divide C₄H₄O₄'s subscripts by their largest common factor, 4: every subscript becomes 1, and the empirical formula is CHO."
 },
 {
  "value": "C₄H₄O₄",
  "misconception": "you left the formula unchanged — the subscripts 4, 4, and 4 share the factor 4, so the formula reduces",
  "feedback": "This answer is wrong — you left the formula unchanged. The subscripts 4, 4, and 4 share the factor 4, so the formula reduces: every subscript divides by 4 to give 1, a subscript of 1 is never written, and the empirical formula is CHO."
 },
 {
  "value": "C₄H₄O",
  "misconception": "you divided only the oxygen's subscript — every subscript must be divided by the same factor",
  "feedback": "This answer is wrong — you divided only the oxygen's subscript. Every subscript must be divided by the SAME factor — the largest common factor, 4: C: 4 ÷ 4 = 1, H: 4 ÷ 4 = 1, O: 4 ÷ 4 = 1, so the empirical formula is CHO."
 }
]
feedback
{
 "keyExplanation": "Subscripts: C 4, H 4, O 4. The largest common factor is 4 — dividing by 2 alone leaves C₂H₂O₂, which still reduces. C: 4 ÷ 4 = 1, H: 4 ÷ 4 = 1, O: 4 ÷ 4 = 1, and a subscript of 1 is never written, so the empirical formula is CHO.",
 "anticipatedWrong": [
  {
   "value": "C₂H₂O₂",
   "misconception": "you divided by 2, which is a common factor but not the LARGEST one — 2, 2, and 2 still share the factor 2",
   "feedback": "This answer is wrong — you divided by 2, which is a common factor but not the LARGEST one. The subscripts of C₂H₂O₂ — 2, 2, and 2 — still share the factor 2, so the formula still reduces. Divide C₄H₄O₄'s subscripts by their largest common factor, 4: every subscript becomes 1, and the empirical formula is CHO."
  },
  {
   "value": "C₄H₄O₄",
   "misconception": "you left the formula unchanged — the subscripts 4, 4, and 4 share the factor 4, so the formula reduces",
   "feedback": "This answer is wrong — you left the formula unchanged. The subscripts 4, 4, and 4 share the factor 4, so the formula reduces: every subscript divides by 4 to give 1, a subscript of 1 is never written, and the empirical formula is CHO."
  },
  {
   "value": "C₄H₄O",
   "misconception": "you divided only the oxygen's subscript — every subscript must be divided by the same factor",
   "feedback": "This answer is wrong — you divided only the oxygen's subscript. Every subscript must be divided by the SAME factor — the largest common factor, 4: C: 4 ÷ 4 = 1, H: 4 ÷ 4 = 1, O: 4 ÷ 4 = 1, so the empirical formula is CHO."
  }
 ]
}
MOL-039 · medium · application · seq 8 · authored (pick-NOT polarity; fresh formula set — key C₆H₄Cl₂ reduces by 2 to C₃H₂Cl across three elements; K₂SO₄ carries the unwritten-1 trap and Fe₃O₄ the individually-divisible-but-coprime surface (the specNotes' named confusion, foil-side)) · mcq4
Three of these four formulas are already empirical. Which one is not?
C₆H₄Cl₂ KEY
K₂SO₄
Fe₃O₄
C₇H₁₂
MOL-041 · medium · application · seq 9 · authored (fresh two-element pair, WF₆ family named fluorine-first — order-bite variant; the smallest mole value 0.050 exercises divide-by-smallest below 0.1; exact divisions 0.30 / 0.050; whole-number outcome by construction) · text
A sample of a compound contains 5.7 g of F and 9.19 g of W. What is the simplest mole ratio of F atoms to W atoms? (M of F = 19.0 g/mol, W = 183.8 g/mol.) Write your answer like 1 : 2.
Answer: 6 : 1 KEY
also accepted: 6:1
wrongValueDiagnostics
[
 {
  "value": "1 : 1.6",
  "misconception": "you compared the masses instead of the moles — 5.7 : 9.19 is about 1 : 1.6, but atoms are counted in moles, and one tungsten atom far outweighs one fluorine atom",
  "feedback": "This answer is wrong — you compared the masses instead of the moles. 5.7 : 9.19 is about 1 : 1.6, but atoms are counted in moles, and one tungsten atom far outweighs one fluorine atom: n(F) = 5.7 / 19.0 = 0.30 mol and n(W) = 9.19 / 183.8 = 0.050 mol, and dividing each by 0.050 gives 6 : 1."
 },
 {
  "value": "1 : 6",
  "misconception": "you wrote the ratio backwards — the order follows the elements as named: fluorine first, then tungsten, so the larger number comes first here",
  "feedback": "This answer is wrong — you wrote the ratio backwards. The ratio order follows the elements as they are named in the question: fluorine first, then tungsten. Fluorine has the larger mole value, so the answer is 6 : 1 — the ratio is not automatically written smallest-first."
 }
]
feedback
{
 "keyExplanation": "Values from the question:\nm(F) = 5.7 g, m(W) = 9.19 g\nEquation:\nn = m/M  (for each element)\nSubstituting the values and calculating:\nn(F) = 5.7 / 19.0 = 0.30 mol\nn(W) = 9.19 / 183.8 = 0.050 mol\nDivide each by the smallest (0.050): F gives 6, W gives 1.\nMole ratio of fluorine to tungsten = 6 : 1",
 "anticipatedWrong": [
  {
   "value": "1 : 1.6",
   "misconception": "you compared the masses instead of the moles — 5.7 : 9.19 is about 1 : 1.6, but atoms are counted in moles, and one tungsten atom far outweighs one fluorine atom",
   "feedback": "This answer is wrong — you compared the masses instead of the moles. 5.7 : 9.19 is about 1 : 1.6, but atoms are counted in moles, and one tungsten atom far outweighs one fluorine atom: n(F) = 5.7 / 19.0 = 0.30 mol and n(W) = 9.19 / 183.8 = 0.050 mol, and dividing each by 0.050 gives 6 : 1."
  },
  {
   "value": "1 : 6",
   "misconception": "you wrote the ratio backwards — the order follows the elements as named: fluorine first, then tungsten, so the larger number comes first here",
   "feedback": "This answer is wrong — you wrote the ratio backwards. The ratio order follows the elements as they are named in the question: fluorine first, then tungsten. Fluorine has the larger mole value, so the answer is 6 : 1 — the ratio is not automatically written smallest-first."
  }
 ]
}
MOL-042 · medium · application · seq 10 · authored (fresh THREE-number ratio 1 : 2 : 4.5 -> 2 : 4 : 9, x2, .5 branch — the three-number case rotation the seed_placement spare pool[19] disposition licenses at seq 10 in the x2 multiplier family; shares nothing with the spared 1 : 1.5 : 2 or its answer 2 : 3 : 4, which appears here only as the format example in the hint (NUM_IGNORE furniture, gram-checked); wrong paths = the gate item 2's own two classes restated for a decimal in last position) · text
A mole calculation for a three-element compound gives the ratio 1 : 2 : 4.5. What whole-number ratio does it clear to? Give your answer like 2 : 3 : 4.
Answer: 2 : 4 : 9 KEY
also accepted: 2:4:9
wrongValueDiagnostics
[
 {
  "value": "1 : 2 : 5",
  "misconception": "you rounded the 4.5 up to 5 — rounding changes the ratio; multiply every number by 2 instead",
  "feedback": "This answer is wrong — you rounded the 4.5 up to 5, and rounding changes the ratio itself. The decimal ends in .5, so multiply every number by 2 instead: 1 × 2 = 2, 2 × 2 = 4, and 4.5 × 2 = 9, giving 2 : 4 : 9."
 },
 {
  "value": "1 : 2 : 9",
  "misconception": "you multiplied only the decimal number — every number in the ratio must be multiplied by the same amount",
  "feedback": "This answer is wrong — you multiplied only the decimal number. Every number in the ratio must be multiplied by the same amount to keep the ratio unchanged: 1 × 2 = 2, 2 × 2 = 4, and 4.5 × 2 = 9, giving 2 : 4 : 9."
 }
]
feedback
{
 "keyExplanation": "The decimal ends in .5, so multiply every number by 2: 1 × 2 = 2, 2 × 2 = 4, and 4.5 × 2 = 9. The whole-number ratio is 2 : 4 : 9 — the multiplication covers all three numbers, not just the one carrying the decimal.",
 "anticipatedWrong": [
  {
   "value": "1 : 2 : 5",
   "misconception": "you rounded the 4.5 up to 5 — rounding changes the ratio; multiply every number by 2 instead",
   "feedback": "This answer is wrong — you rounded the 4.5 up to 5, and rounding changes the ratio itself. The decimal ends in .5, so multiply every number by 2 instead: 1 × 2 = 2, 2 × 2 = 4, and 4.5 × 2 = 9, giving 2 : 4 : 9."
  },
  {
   "value": "1 : 2 : 9",
   "misconception": "you multiplied only the decimal number — every number in the ratio must be multiplied by the same amount",
   "feedback": "This answer is wrong — you multiplied only the decimal number. Every number in the ratio must be multiplied by the same amount to keep the ratio unchanged: 1 × 2 = 2, 2 × 2 = 4, and 4.5 × 2 = 9, giving 2 : 4 : 9."
  }
 ]
}
MOL-038 · medium · application · seq 11 · authored (parallel to the seq-11 seed structure; fresh instance borazine B₃N₃H₆ -> BNH₂, corpus scan 0 hits; 3-element reduce with the factor rotated to 3 — still a clean prime factor, one division pass, so the multiplier is not cross-form pattern-matchable; ribose-family left-the-last-subscript-undivided class carried) · text
What is the empirical formula of borazine, B₃N₃H₆? Type the whole formula, writing any subscripts as plain digits, like the 2 in CO2; leave subscripts of 1 unwritten.
Answer: BNH₂ KEY
also accepted: BNH2
wrongValueDiagnostics
[
 {
  "value": "B₃N₃H₆",
  "misconception": "you left the formula unchanged — the subscripts 3, 3, and 6 share the factor 3, so the formula reduces",
  "feedback": "This answer is wrong — you left the formula unchanged. The subscripts 3, 3, and 6 share the factor 3, so the formula reduces: B: 3 ÷ 3 = 1, N: 3 ÷ 3 = 1, H: 6 ÷ 3 = 2, giving BNH₂."
 },
 {
  "value": "BNH",
  "misconception": "you divided each subscript by its own value, which destroys the ratio",
  "feedback": "This answer is wrong — you divided each subscript by its own value, which destroys the ratio. Every subscript is divided by the same factor, 3: B 1, N 1, H 2, giving BNH₂."
 },
 {
  "value": "BNH₆",
  "misconception": "you left the last subscript undivided — every subscript must be divided by the same factor, including hydrogen's",
  "feedback": "This answer is wrong — you left the last subscript undivided. Every subscript must be divided by the same factor, including hydrogen's: H: 6 ÷ 3 = 2, giving BNH₂."
 }
]
feedback
{
 "keyExplanation": "Subscripts: B 3, N 3, H 6. The largest common factor is 3. B: 3 ÷ 3 = 1, N: 3 ÷ 3 = 1, H: 6 ÷ 3 = 2. A subscript of 1 is never written, so borazine's empirical formula is BNH₂.",
 "anticipatedWrong": [
  {
   "value": "B₃N₃H₆",
   "misconception": "you left the formula unchanged — the subscripts 3, 3, and 6 share the factor 3, so the formula reduces",
   "feedback": "This answer is wrong — you left the formula unchanged. The subscripts 3, 3, and 6 share the factor 3, so the formula reduces: B: 3 ÷ 3 = 1, N: 3 ÷ 3 = 1, H: 6 ÷ 3 = 2, giving BNH₂."
  },
  {
   "value": "BNH",
   "misconception": "you divided each subscript by its own value, which destroys the ratio",
   "feedback": "This answer is wrong — you divided each subscript by its own value, which destroys the ratio. Every subscript is divided by the same factor, 3: B 1, N 1, H 2, giving BNH₂."
  },
  {
   "value": "BNH₆",
   "misconception": "you left the last subscript undivided — every subscript must be divided by the same factor, including hydrogen's",
   "feedback": "This answer is wrong — you left the last subscript undivided. Every subscript must be divided by the same factor, including hydrogen's: H: 6 ÷ 3 = 2, giving BNH₂."
  }
 ]
}
MOL-044 · medium · transfer · seq 12 · authored (quotient-meaning, plain-division frame; parallel misconception classes to the seed; fresh exact pair 168.0/28.0 = 6) · mcq4
Dividing a compound's molar mass, 168.0 g/mol, by its empirical formula's molar mass, 28.0 g/mol, gives 6. What does the 6 tell you?
One molecule of the compound holds 6 empirical-formula units. KEY
One empirical-formula unit holds 6 molecules of the compound.
One molecule of the compound holds 6 atoms of each element.
There are 6 mol of the compound present.
MOL-041 · medium · application · seq 13 · authored (parallel to the seq-13 seed structure; fresh triple — boric acid H₃BO₃, unnamed in the stem per the lesson convention; smallest mole value on the MIDDLE-listed element; corpus scan 0 hits) · text
A sample of a compound contains 9.0 g of H, 32.4 g of B, and 144.0 g of O. What is the simplest mole ratio of H to B to O atoms? (M of H = 1.0 g/mol, B = 10.8 g/mol, O = 16.0 g/mol.) Write your answer like 1 : 2 : 3.
Answer: 3 : 1 : 3 KEY
also accepted: 3:1:3
wrongValueDiagnostics
[
 {
  "value": "1 : 3.6 : 16",
  "misconception": "you compared the masses instead of the moles — 9.0 : 32.4 : 144.0 is about 1 : 3.6 : 16, but atoms are counted in moles",
  "feedback": "This answer is wrong — you compared the masses instead of the moles — 9.0 : 32.4 : 144.0 is about 1 : 3.6 : 16, but atoms are counted in moles."
 },
 {
  "value": "1 : 0.33 : 1",
  "misconception": "you divided every value by the largest instead of the smallest — divide by the smallest so the smallest becomes 1",
  "feedback": "This answer is wrong — you divided every value by the largest instead of the smallest — divide by the smallest so the smallest becomes 1."
 }
]
feedback
{
 "keyExplanation": "Values from the question:\nm(H) = 9.0 g, m(B) = 32.4 g, m(O) = 144.0 g\nEquation:\nn = m/M  (for each element)\nSubstituting the values and calculating:\nn(H) = 9.0 / 1.0 = 9.00 mol\nn(B) = 32.4 / 10.8 = 3.00 mol\nn(O) = 144.0 / 16.0 = 9.00 mol\nDivide each by the smallest (3.00): H gives 3, B gives 1, O gives 3.\nMole ratio of hydrogen to boron to oxygen = 3 : 1 : 3"
}
MOL-044 · medium · application · seq 14 · authored (parallel numeric slot; fresh (formula, M) pair — cyclooctene C₈H₁₄, empirical C₄H₇, 110.0 g/mol -> 2 units; compound unnamed in the stem per the lesson's gate convention) · numeric
A compound has the empirical formula C₄H₇ and a molar mass of 110.0 g/mol. (M of C = 12.0 g/mol, H = 1.0 g/mol.) How many empirical-formula units fit in one of its molecules? Answer with the whole-number count.
Answer: 2 units KEY
also accepted: 2 units, 2.0
wrongValueDiagnostics
[
 {
  "value": 8.46,
  "misconception": "you ignored the subscripts in the empirical formula, using 12.0 + 1.0 = 13.0 g/mol — M of C₄H₇ is 4 × 12.0 + 7 × 1.0 = 55.0 g/mol",
  "feedback": "This answer is wrong — you ignored the subscripts in the empirical formula, using 12.0 + 1.0 = 13.0 g/mol. M of C₄H₇ is 4 × 12.0 + 7 × 1.0 = 55.0 g/mol, and a non-whole unit count signals exactly this kind of slip."
 },
 {
  "value": 0.5,
  "misconception": "you divided upside down — the compound's molar mass goes on top, and the unit count is never below 1",
  "feedback": "This answer is wrong — you divided upside down — the compound's molar mass goes on top, and the unit count is never below 1."
 }
]
feedback
{
 "keyExplanation": "Write down the values in the question:\nmolar mass of the compound = 110.0 g/mol\nM of C₄H₇ = 4 × 12.0 + 7 × 1.0 = 55.0 g/mol\nWrite down the equation:\nnumber of units = molar mass of the compound ÷ molar mass of the empirical formula\nSubstitute in the values, and calculate:\nnumber of units = 110.0 / 55.0\nnumber of units = 2\nThe result is a whole number at least 1, so the division ran the right way up."
}
MOL-043 · hard · application · seq 15 · authored (parallel mass-route capstone; fresh chemically-real compound — titanium(III) oxide Ti₂O₃, compound unnamed in the stem per the seed convention; exact divisions 47.9/47.9 = 1.00 and 24.0/16.0 = 1.50 — a mass-equals-molar-mass division, precedent seed pool[22]'s n(C) = 12.0/12.0; .5-decimal clearing ×2; the round-up trap lands on the familiar REAL compound TiO₂, the strongest chemically-real pull of the five forms, per the lesson's specNotes realism point) · text
A sample of a compound contains 239.5 g of Ti and 120.0 g of O. What is its empirical formula? (M of Ti = 47.9 g/mol, O = 16.0 g/mol.) Type subscripts as plain numbers, like the 2 in CO2.
Answer: Ti₂O₃ KEY
also accepted: Ti2O3
wrongValueDiagnostics
[
 {
  "value": "TiO₂",
  "misconception": "you rounded the 1.5 up to 2 — rounding changes the ratio; multiply every number by 2 instead",
  "feedback": "This answer is wrong — you rounded the 1.5 up to 2. Rounding changes the ratio; multiply every number by 2 instead: 1 and 1.5 become 2 and 3, so this sample's empirical formula is Ti₂O₃ — the data decide the formula, not the titanium compound you have heard of most often."
 },
 {
  "value": "TiO",
  "misconception": "you dropped the decimal — cutting the .5 off changes the ratio; multiply every number by 2 instead",
  "feedback": "This answer is wrong — you dropped the decimal. Cutting the .5 off changes the ratio; multiply every number by 2 instead: 1 and 1.5 become 2 and 3, so the empirical formula is Ti₂O₃."
 },
 {
  "value": "Ti₃O₂",
  "misconception": "you inverted the ratio — titanium has the smaller mole value, so titanium takes the 1 before clearing",
  "feedback": "This answer is wrong — you inverted the ratio. Titanium has the smaller mole value (5.00 mol against oxygen's 7.50 mol), so titanium takes the 1 before clearing: 1 and 1.5 become 2 and 3, giving Ti₂O₃."
 }
]
feedback
{
 "keyExplanation": "Values from the question:\nm(Ti) = 239.5 g, m(O) = 120.0 g\nEquation:\nn = m/M  (for each element)\nSubstituting the values and calculating:\nn(Ti) = 239.5 / 47.9 = 5.00 mol\nn(O) = 120.0 / 16.0 = 7.50 mol\nDivide each by the smallest (5.00): Ti gives 1, O gives 1.50.\nThe decimal ends in .5, so multiply every number by 2: 2 and 3.\nEmpirical formula: Ti₂O₃",
 "anticipatedWrong": [
  {
   "value": "TiO₂",
   "misconception": "you rounded the 1.5 up to 2 — rounding changes the ratio; multiply every number by 2 instead",
   "feedback": "This answer is wrong — you rounded the 1.5 up to 2. Rounding changes the ratio; multiply every number by 2 instead: 1 and 1.5 become 2 and 3, so this sample's empirical formula is Ti₂O₃ — the data decide the formula, not the titanium compound you have heard of most often."
  },
  {
   "value": "TiO",
   "misconception": "you dropped the decimal — cutting the .5 off changes the ratio; multiply every number by 2 instead",
   "feedback": "This answer is wrong — you dropped the decimal. Cutting the .5 off changes the ratio; multiply every number by 2 instead: 1 and 1.5 become 2 and 3, so the empirical formula is Ti₂O₃."
  },
  {
   "value": "Ti₃O₂",
   "misconception": "you inverted the ratio — titanium has the smaller mole value, so titanium takes the 1 before clearing",
   "feedback": "This answer is wrong — you inverted the ratio. Titanium has the smaller mole value (5.00 mol against oxygen's 7.50 mol), so titanium takes the 1 before clearing: 1 and 1.5 become 2 and 3, giving Ti₂O₃."
  }
 ]
}
MOL-042 · hard · transfer · seq 16 · authored (two-students method-judgment frame; round-DOWN error spotlighted — the register's own truncation phrasing per fluency F1 / gate G0; fresh ratio 1 : 5.67 -> 3 : 17, x3; all three misconception classes live) · mcq4
A mole calculation gives the ratio 1 : 7.67. One student multiplies both numbers by 3; another rounds the 7.67 down to 7. Which result is right, and why?
3 : 23 — multiplying every number by 3 keeps the ratio unchanged, while rounding 7.67 down to 7 changes it. KEY
1 : 7 — the decimal part is small enough to leave out of a whole-number ratio.
1 : 8 — 7.67 is close to 8, so rounding up is the quicker move.
1 : 23 — multiply only the 7.67 by 3, because the 1 in the ratio is already a whole number and needs nothing.
MOL-045 · hard · application · seq 17 · authored (fresh pair CHN / 135.0 -> C₅H₅N₅, adenine family kept anonymous; the largest multiplier of the row with all-unwritten-1 subscripts; parallel structure, step count, and misconception classes to the seed) · text
A compound has the empirical formula CHN and a molar mass of 135.0 g/mol. (M of C = 12.0 g/mol, H = 1.0 g/mol, N = 14.0 g/mol.) State the compound's molecular formula, typing subscripts as plain numbers (like the 2 in CO2).
Answer: C₅H₅N₅ KEY
also accepted: C5H5N5
wrongValueDiagnostics
[
 {
  "value": "CHN",
  "misconception": "you stopped at the empirical formula — the unit count is 135.0/27.0 = 5, and every subscript must be multiplied by it",
  "feedback": "This answer is wrong — you stopped at the empirical formula. The unit count is 135.0 / 27.0 = 5, and every subscript must be multiplied by it: C: 1 × 5 = 5, H: 1 × 5 = 5, N: 1 × 5 = 5, giving C₅H₅N₅."
 },
 {
  "value": "C₅HN",
  "misconception": "you multiplied only carbon's subscript — every subscript in the empirical formula is multiplied by the unit count, including every unwritten 1",
  "feedback": "This answer is wrong — you multiplied only carbon's subscript. Every subscript in the empirical formula is multiplied by the unit count, 5 — including every unwritten 1: C: 1 × 5 = 5, H: 1 × 5 = 5, N: 1 × 5 = 5, giving C₅H₅N₅."
 },
 {
  "value": "CHN₅",
  "misconception": "you multiplied only nitrogen's subscript — every subscript in the empirical formula is multiplied by the unit count, including every unwritten 1",
  "feedback": "This answer is wrong — you multiplied only nitrogen's subscript. Every subscript in the empirical formula is multiplied by the unit count, 5 — including every unwritten 1: C: 1 × 5 = 5, H: 1 × 5 = 5, N: 1 × 5 = 5, giving C₅H₅N₅."
 }
]
feedback
{
 "keyExplanation": "M of CHN = 12.0 + 1.0 + 14.0 = 27.0 g/mol.\nStep 1 — find the number of units:\nnumber of units = molar mass of the compound ÷ molar mass of the empirical formula\nnumber of units = 135.0 / 27.0 = 5\nStep 2 — multiply every subscript by 5 (an unwritten subscript is 1):\nC: 1 × 5 = 5,  H: 1 × 5 = 5,  N: 1 × 5 = 5\nMolecular formula: C₅H₅N₅",
 "anticipatedWrong": [
  {
   "value": "CHN",
   "misconception": "you stopped at the empirical formula — the unit count is 135.0/27.0 = 5, and every subscript must be multiplied by it",
   "feedback": "This answer is wrong — you stopped at the empirical formula. The unit count is 135.0 / 27.0 = 5, and every subscript must be multiplied by it: C: 1 × 5 = 5, H: 1 × 5 = 5, N: 1 × 5 = 5, giving C₅H₅N₅."
  },
  {
   "value": "C₅HN",
   "misconception": "you multiplied only carbon's subscript — every subscript in the empirical formula is multiplied by the unit count, including every unwritten 1",
   "feedback": "This answer is wrong — you multiplied only carbon's subscript. Every subscript in the empirical formula is multiplied by the unit count, 5 — including every unwritten 1: C: 1 × 5 = 5, H: 1 × 5 = 5, N: 1 × 5 = 5, giving C₅H₅N₅."
  },
  {
   "value": "CHN₅",
   "misconception": "you multiplied only nitrogen's subscript — every subscript in the empirical formula is multiplied by the unit count, including every unwritten 1",
   "feedback": "This answer is wrong — you multiplied only nitrogen's subscript. Every subscript in the empirical formula is multiplied by the unit count, 5 — including every unwritten 1: C: 1 × 5 = 5, H: 1 × 5 = 5, N: 1 × 5 = 5, giving C₅H₅N₅."
  }
 ]
}
MOL-043 · hard · application · seq 18 · authored (fresh percent dataset, phosphorus nitride: exact real composition 57.055/42.945 -> 57.1/42.9 at 3 sf, sum 100.0; quotient 1.66 clears ×3 with 1.66 × 3 = 4.98 -> 5 — the near-.67 decimal family; P₃N₅ is a real, stable, classroom-safe solid (row 9's WF₆/PCl₅ uncommon-substance precedent); identical chain and step count to the seed) · text
A compound is 57.1% P and 42.9% N by mass. What is its empirical formula? (M of P = 31.0 g/mol, N = 14.0 g/mol.) Type subscripts as plain numbers, like the 2 in CO2.
Answer: P₃N₅ KEY
also accepted: P3N5
wrongValueDiagnostics
[
 {
  "value": "PN₂",
  "misconception": "you rounded the 1.66 up to 2 — rounding changes the ratio; multiply every number by 3 instead",
  "feedback": "This answer is wrong — you rounded the 1.66 up to 2. Rounding changes the ratio; multiply every number by 3 instead: P gives 3 and N gives 5, so the empirical formula is P₃N₅."
 },
 {
  "value": "PN₅",
  "misconception": "you multiplied only the number with the decimal — multiplying just one number changes the ratio; multiply EVERY number by 3",
  "feedback": "This answer is wrong — you multiplied only the number with the decimal. Multiplying just one number changes the ratio; multiply EVERY number by 3: P gives 3 and N gives 5, so the empirical formula is P₃N₅."
 }
]
feedback
{
 "keyExplanation": "Assume a 100 g sample: m(P) = 57.1 g, m(N) = 42.9 g.\nEquation:\nn = m/M  (for each element)\nSubstituting the values and calculating:\nn(P) = 57.1 / 31.0 = 1.84 mol\nn(N) = 42.9 / 14.0 = 3.06 mol\nDivide each by the smallest (1.84): P gives 1, N gives 1.66.\nThe decimal is near .67, so multiply every number by 3: 3 and 5.\nEmpirical formula: P₃N₅",
 "anticipatedWrong": [
  {
   "value": "PN₂",
   "misconception": "you rounded the 1.66 up to 2 — rounding changes the ratio; multiply every number by 3 instead",
   "feedback": "This answer is wrong — you rounded the 1.66 up to 2. Rounding changes the ratio; multiply every number by 3 instead: P gives 3 and N gives 5, so the empirical formula is P₃N₅."
  },
  {
   "value": "PN₅",
   "misconception": "you multiplied only the number with the decimal — multiplying just one number changes the ratio; multiply EVERY number by 3",
   "feedback": "This answer is wrong — you multiplied only the number with the decimal. Multiplying just one number changes the ratio; multiply EVERY number by 3: P gives 3 and N gives 5, so the empirical formula is P₃N₅."
  }
 ]
}
MOL-039 · hard · transfer · seq 19 · authored (fresh coprime-subscript claim rotation per blueprint case_target; instance C25H52, pentacosane — subscripts 49 and 100 each composite yet coprime; outside the burned-instance ledger; counterexample C18H38, octadecane, reduces) · mcq4
A student claims C₄₉H₁₀₀ cannot be an empirical formula 'because 49 and 100 are both bigger than 1.' Judge the claim.
The claim is wrong — 49 and 100 share no common factor bigger than 1, so C₄₉H₁₀₀ is an empirical formula. KEY
The claim is right — an empirical formula must contain at least one unwritten subscript of 1.
The claim is right — 49 and 100 can each be divided by a smaller number, so the formula reduces.
The claim is wrong — no formula with only two subscripts can ever be reduced.
MOL-045 · hard · transfer · seq 20 · authored (fresh x1 pair C₄H₃Cl / 86.5, chlorobenzene family kept anonymous; parallel structure and the seed's three misconception classes; one-element rotation lands on chlorine's unwritten 1) · mcq4
A compound has the empirical formula C₄H₃Cl and a molar mass of 86.5 g/mol. (M of C₄H₃Cl = 86.5 g/mol.) What is its molecular formula?
C₄H₃Cl KEY
C₈H₆Cl₂
C₄H₃Cl₂
C₁₆H₁₂Cl₄
MOL-043 · hard · application · seq 21 · authored (parallel to the FRQ-1 seed-expansion structure; MASS route per the recorded rotation; fresh dataset — lithium sulfate Li₂SO₄, unnamed in the stem; 3-element clean chain with the smallest mole value on the middle-listed element; 2-dp masses per the row09 precedent; corpus scan 0 hits on Li₂SO₄/4.14/9.63/19.2; the displayed intermediate 1.20 recurs at row09-C cross-row — recorded derived-value recurrence class, never a key value) · frq
A sample of a compound contains 4.14 g of Li, 9.63 g of S, and 19.2 g of O. (M of Li = 6.9 g/mol, S = 32.1 g/mol, O = 16.0 g/mol.)
A sample of a compound contains 4.14 g of Li, 9.63 g of S, and 19.2 g of O. (M of Li = 6.9 g/mol, S = 32.1 g/mol, O = 16.0 g/mol.) (a) Calculate the number of moles of each element in the sample, reporting each mole value to three significant figures. Show your working. (b) Divide every mole value by the smallest to find the whole number for each element, clearing any decimal that remains. Show each step. (c) Write the compound's empirical formula, and state how you know its subscripts are exact whole numbers rather than rounded ones. Type subscripts as plain numbers, like the 2 in CO2.
Part a. Calculate the number of moles of each element in the sample, reporting each mole value to three significant figures. Show your working.
Model: Values from the question: m(Li) = 4.14 g, m(S) = 9.63 g, m(O) = 19.2 g Equation: n = m/M (for each element) Substituting the values and calculating: n(Li) = 4.14 / 6.9 = 0.600 mol (three significant figures, as instructed) n(S) = 9.63 / 32.1 = 0.300 mol n(O) = 19.2 / 16.0 = 1.20 mol
Part b. Divide every mole value by the smallest to find the whole number for each element, clearing any decimal that remains. Show each step.
Model: The smallest mole value is sulfur's 0.300 (the middle one, not the first listed). Divide each mole value by the smallest (0.300): Li: 0.600 / 0.300 = 2 S: 0.300 / 0.300 = 1 O: 1.20 / 0.300 = 4 No decimal remains — Li gives 2, S gives 1, O gives 4.
Part c. Write the compound's empirical formula, and state how you know its subscripts are exact whole numbers rather than rounded ones. Type subscripts as plain numbers, like the 2 in CO2.
Model: The finished whole numbers are the subscripts, so the empirical formula is Li2SO4. The subscripts are exact: every division landed on a whole number, so nothing was rounded and no multiplying step was needed.
Full model answer. (a) Values from the question: m(Li) = 4.14 g, m(S) = 9.63 g, m(O) = 19.2 g Equation: n = m/M (for each element) Substituting the values and calculating: n(Li) = 4.14 / 6.9 = 0.600 mol (three significant figures, as instructed) n(S) = 9.63 / 32.1 = 0.300 mol n(O) = 19.2 / 16.0 = 1.20 mol (b) The smallest mole value is sulfur's 0.300 (the middle one, not the first listed). Divide each mole value by the smallest (0.300): Li: 0.600 / 0.300 = 2 S: 0.300 / 0.300 = 1 O: 1.20 / 0.300 = 4 No decimal remains — Li gives 2, S gives 1, O gives 4. (c) The finished whole numbers are the subscripts, so the empirical formula is Li2SO4. The subscripts are exact: every division landed on a whole number, so nothing was rounded and no multiplying step was needed.
MOL-045 · hard · application · seq 22 · authored (FRQ-2 fresh parallel-structure variant; 1,2-dichloroethane C₂H₄Cl₂, count 2; compound unnamed in student text per the gate register) · frq
A compound has a molar mass of 99.0 g/mol and the empirical formula CH₂Cl. (M of C = 12.0 g/mol, H = 1.0 g/mol, Cl = 35.5 g/mol.)
A compound has a molar mass of 99.0 g/mol and the empirical formula CH₂Cl. (M of C = 12.0 g/mol, H = 1.0 g/mol, Cl = 35.5 g/mol.) (a) Calculate the molar mass of the empirical formula CH₂Cl from the atomic masses given, showing the sum you evaluate. (b) Find the number of empirical-formula units in one molecule of the compound. State the equation you use, and show every step of your working. (c) Write the molecular formula of the compound, and show how each subscript in your answer was produced. Type subscripts as plain numbers, like the 2 in CO2; no other special symbols are needed.
Part a. Calculate the molar mass of the empirical formula CH₂Cl from the atomic masses given, showing the sum you evaluate.
Model: Values from the question: M of C = 12.0 g/mol, M of H = 1.0 g/mol, M of Cl = 35.5 g/mol The empirical formula CH2Cl contains 1 C, 2 H, and 1 Cl (an unwritten subscript counts as 1). M of CH2Cl = 12.0 + 2 × 1.0 + 35.5 M of CH2Cl = 49.5 g/mol
Part b. Find the number of empirical-formula units in one molecule of the compound. State the equation you use, and show every step of your working.
Model: Values from the question: molar mass of the compound = 99.0 g/mol molar mass of the empirical formula = 49.5 g/mol (from part a) Equation: number of units = molar mass of the compound ÷ molar mass of the empirical formula Substituting the values and calculating: number of units = 99.0 / 49.5 number of units = 2 The result is a whole number of at least 1, so the division ran the right way up.
Part c. Write the molecular formula of the compound, and show how each subscript in your answer was produced. Type subscripts as plain numbers, like the 2 in CO2; no other special symbols are needed.
Model: Multiply every subscript in the empirical formula by 2: C: 1 × 2 = 2, H: 2 × 2 = 4, Cl: 1 × 2 = 2 Molecular formula: C2H4Cl2
Full model answer. (a) Values from the question: M of C = 12.0 g/mol, M of H = 1.0 g/mol, M of Cl = 35.5 g/mol The empirical formula CH2Cl contains 1 C, 2 H, and 1 Cl (an unwritten subscript counts as 1). M of CH2Cl = 12.0 + 2 × 1.0 + 35.5 M of CH2Cl = 49.5 g/mol (b) Values from the question: molar mass of the compound = 99.0 g/mol molar mass of the empirical formula = 49.5 g/mol (from part a) Equation: number of units = molar mass of the compound ÷ molar mass of the empirical formula Substituting the values and calculating: number of units = 99.0 / 49.5 number of units = 2 The result is a whole number of at least 1, so the division ran the right way up. (c) Multiply every subscript in the empirical formula by 2: C: 1 × 2 = 2, H: 2 × 2 = 4, Cl: 1 × 2 = 2 Molecular formula: C2H4Cl2