Here is a glass of water on a balance. The display reads 250 g. One question: how many water molecules are in the glass?
The balance settles the mass question instantly. The molecule question is another matter.
The balance reports one thing: the sample's mass — 250 g of water.
It says nothing about the number of water molecules.
Water molecules are far too small and far too many to count one by one.
No instrument you can buy reads out a particle count.
Chemical changes happen particle by particle, so a chemist often needs that count.
So the particle count of a weighed sample must be worked out from the mass — the one thing the balance does report.
The next lessons build that route, one step at a time.
Worked example 1. A spoonful of table salt sits on a balance reading 12 g. Why doesn't the reading tell you how many particles the spoonful holds?
The balance reports the mass: 12 g of salt.
The salt's particles are far too small and far too many to count one by one.
No instrument reads out a particle count.
The reading gives the mass only — the particle count must be worked out from it.
You can now explain that a balance reports a sample's mass but not the number of particles in it, so the particle count of a weighed sample must be worked out rather than read off.
A beaker of ethanol stands on a balance. Which quantity does the balance report directly?
Why must the number of particles in a weighed copper block be worked out rather than read off an instrument?
A student needs the number of molecules in a flask of acetone. What can the balance contribute?
A hardware order calls for 400 identical screws. Counting them out one by one would take all morning.
There is a faster way, and it sits on the counter.
A balance can do the counting for you.
Identical objects all have the same mass.
So a pile's total mass equals the mass of one object multiplied by the count.
Turn that around: dividing the total mass by the mass of one object gives the count.
number of objects = total mass / mass of one object
A pile of nails has a total mass of 500.0 g, and one nail has a mass of 25.0 g.
number of nails = 500.0 / 25.0 = 20.
Worked example 1. A bag of identical screws has a total mass of 96.0 g. One screw has a mass of 8.0 g. How many screws are in the bag?
Write down the values in the question
total mass = 96.0 g
mass of one screw = 8.0 g
Write down the equation
number of screws = total mass / mass of one screw
Substitute in the values, and calculate
number of screws = 96.0 / 8.0
number of screws = 12
Worked example 2. A tray of identical bolts has a total mass of 350.0 g. One bolt has a mass of 14.0 g. How many bolts are on the tray?
Write down the values in the question
total mass = 350.0 g
mass of one bolt = 14.0 g
Write down the equation
number of bolts = total mass / mass of one bolt
Substitute in the values, and calculate
number of bolts = 350.0 / 14.0
number of bolts = 25
You can now calculate how many identical objects a sample contains from the total mass and the mass of one object.
A box of identical paper clips has a total mass of 30.0 g. One paper clip has a mass of 1.5 g. How many paper clips are in the box?
A pouch of identical marbles has a total mass of 125.0 g. One marble has a mass of 5.0 g. How many marbles are in the pouch?
A tub of identical hex nuts has a total mass of 216.0 g. One hex nut has a mass of 13.5 g. How many hex nuts are in the tub?
You've already counted nails by weighing: total mass divided by the mass of one nail. The same trick can count particles — but only for some samples.
Counting nails by weighing worked because every nail had the same mass — one number to divide by.
A pile of mixed screws and nails offers no single per-piece mass, so weighing cannot count it.
Now weigh a sample of pure water instead.
Water molecules share one steady average mass.
That gives the one number the trick needs.
Weighing can count the particles of a pure substance because the particles of a pure substance share one steady average mass, so mass counts particles.
'Average' matters here: you've already seen that isotopes give atoms of one element slightly different masses, and the counting runs on the steady average.
Worked example 1. Why can weighing count the atoms in a spool of pure copper wire?
Copper atoms share one steady average mass.
Because the particles of a pure substance share one steady average mass, mass counts particles.
The spool's total mass counts its copper atoms.
Worked example 2. Why can't weighing count the pieces in a jar of mixed nuts and bolts?
A nut and a bolt have different masses.
There is no single per-piece mass to divide the total by.
Weighing cannot count a mixed sample's pieces.
You can now explain that weighing can count the particles of a pure substance because the particles of a pure substance share one steady average mass, so mass counts particles.
A bar of pure silver is weighed. Why can the bar's mass be used to count its atoms?
Weighing can count the particles or pieces of exactly one of these samples. Which one?
Why does counting by weighing fail for a box of assorted buttons?
You've already seen that a weighed sample's particle count must be worked out, and that weighing can count the particles of a pure substance. Chemists write the counts they work out in a counting unit of their own.
Everyday counting units bundle a fixed number: one pair means 2, one dozen means 12.
Chemists count particles with a unit that bundles one fixed number of particles.
Counting units
| Unit | How many it bundles |
|---|---|
| a pair | 2 |
| a dozen | 12 |
| a mole (mol) | one fixed number of particles — the same every time |
That unit is called the 'mole', written mol.
One mole means the same fixed number of particles for any substance, the way one dozen means 12 for eggs or pencils.
So 2 mol of water contains twice as many water molecules as 1 mol of water.
The count in the glass of water finally has a unit: moles.
Worked example 1. One flask holds 3 mol of carbon dioxide, CO₂, and another holds 1 mol. How many times as many CO₂ molecules does the first flask hold?
3 mol is three of the same fixed bundle that 1 mol is one of.
Three times as many molecules.
You can now state that the mole (written mol) is the unit chemists use to count particles, where one mole means a fixed number of particles in the way one dozen means 12.
One dozen means 12 eggs. In the same way, what does 1 mol of helium mean?
How does the number of molecules in 2 mol of ammonia, NH₃, compare with the number in 1 mol of ammonia?
A jar holds 1 mol of iron and a flask holds 1 mol of neon. Compare the numbers of atoms in the two containers.
You've already seen that one mole means one fixed number of particles. This lesson states the number.
One mole of any substance contains 6.022 × 10²³ particles.
Written out, 6.022 × 10²³ is 6022 followed by 20 zeros.
The number is the same for every substance.
This number is called 'Avogadro's number'.
So 1 mol of water contains 6.022 × 10²³ water molecules.
Worked example 1. How many molecules are in exactly 1 mol of oxygen gas, O₂?
One mole of any substance contains 6.022 × 10²³ particles.
6.022 × 10²³ O₂ molecules.
You can now state that one mole of any substance contains 6.022 × 10²³ particles, and that this number is named Avogadro's number.
Exactly 1 mol of neon is sealed in a flask. How many neon atoms does the flask hold? (Avogadro's number is 6.022 × 10²³.) Give your answer to 4 significant figures. Enter your answer in scientific notation.
The number of particles in one mole is 6.022 × 10²³. What is this number called?
One balloon holds exactly 1 mol of helium; a brick of lead is also exactly 1 mol. Compare the numbers of atoms in the two samples.
You've already sorted substances by formula: elements, molecular substances, and ionic compounds. A mole count is a count OF something, and the substance's kind names the something.
A mole is a fixed count of particles, so name the particle before you count.
Ask one question: what kind of substance is this?
For an element, a mole counts atoms.
For a molecular substance, a mole counts molecules.
For an ionic compound, a mole counts formula units.
A mole of copper counts copper atoms.
Which particle a mole counts
| Kind of substance | Particle counted | Example |
|---|---|---|
| element | atoms | copper |
| molecular substance | molecules | water, H₂O |
| ionic compound | formula units | table salt, NaCl |
A mole of water, H₂O, counts water molecules.
A mole of table salt, NaCl, counts formula units.
Worked example 1. Which particle does a mole of carbon dioxide, CO₂, count?
CO₂ contains only nonmetals, so it is a molecular substance — as you've already classified.
A mole of a molecular substance counts molecules.
A mole of CO₂ counts CO₂ molecules.
Worked example 2. Which particle does a mole of calcium chloride, CaCl₂, count?
CaCl₂ pairs a metal with a nonmetal — an ionic compound.
A mole of an ionic compound counts formula units.
A mole of CaCl₂ counts CaCl₂ formula units.
You can now classify which particle a mole of a given substance counts: atoms for an element, molecules for a molecular substance, or formula units for an ionic compound.
Which particle does a mole of iron, Fe, count?
Which particle does a mole of ammonia, NH₃, count?
Which particle does a mole of potassium bromide, KBr, count?
You've already seen that 1 mol of any substance is 6.022 × 10²³ particles. The counts match — but put two different moles on a balance and the readings differ.
A mole of one substance does not weigh the same as a mole of another.
Each substance has its own mass per mole.
The mass in grams of one mole of a substance is called the substance's 'molar mass'.
Molar mass carries the unit grams per mole, written g/mol.
Water's molar mass is 18.0 g/mol: 1 mol of water has a mass of 18.0 g.
Worked example 1. Carbon dioxide's molar mass is 44.0 g/mol. What is the mass of exactly 1 mol of carbon dioxide, CO₂?
Molar mass states the mass of one mole.
44.0 g.
You can now state that molar mass is the mass in grams of one mole of a substance, with the unit grams per mole (g/mol).
Helium's molar mass is 4.0 g/mol. What does that value tell you?
Which unit does molar mass carry?
Table salt, NaCl, has a molar mass of 58.5 g/mol. What is the mass, in grams, of exactly 1 mol of NaCl?
You've already read an element's average atomic mass from its periodic-table square. That same printed number now does a second job.
An element's periodic-table square prints its average atomic mass — carbon's square prints 12.0.
Read that printed number in grams per mole, and you have the element's molar mass.
Carbon's molar mass is 12.0 g/mol.
The square's other number, the atomic number, counts protons — it is not a mass and plays no part here.
Worked example 1. Use the sulfur square in the figure. What is sulfur's molar mass?
The square prints an average atomic mass of 32.1.
Read in grams per mole, that number is the molar mass.
Sulfur's molar mass is 32.1 g/mol.
Worked example 2. Use the calcium square in the figure. What is calcium's molar mass?
The square prints 40.1 — the 20 above it is the atomic number, a proton count.
Read 40.1 in grams per mole.
Calcium's molar mass is 40.1 g/mol.
You can now identify an element's molar mass in g/mol from the atomic mass printed in its periodic-table square.
The periodic-table square for aluminum is shown. State aluminum's molar mass, in g/mol.
The periodic-table square for potassium is shown. State potassium's molar mass, in g/mol.
The periodic-table square for iron is shown. State iron's molar mass, in g/mol.
You've already met two mass scales: atomic mass units (amu) for a single particle, and g/mol for a mole. Put them side by side for one element.
One oxygen atom has an average mass of 16.0 amu.
One mole of oxygen atoms has a mass of 16.0 g — a molar mass of 16.0 g/mol.
The two scales share the number 16.0.
amu is the single-particle scale: one atom or one molecule is weighed in amu.
One number, two scales
| Scale | What it weighs | Oxygen |
|---|---|---|
| amu | one particle | one O atom = 16.0 amu |
| g/mol | one mole | 1 mol of O atoms = 16.0 g |
g/mol is the weighable-sample scale: it states what one whole mole weighs.
The pattern holds for every element: read the atomic mass once, and use the number on either scale.
Worked example 1. One sodium atom has an average mass of 23.0 amu. What is sodium's molar mass?
The amu number and the g/mol number are the same number.
23.0 g/mol.
You can now state that one particle's average mass in atomic mass units (amu) and one mole's mass in grams per mole are the same number, where amu is the scale for a single particle and g/mol is the scale for a weighable sample.
One fluorine atom has an average mass of 19.0 amu. What is fluorine's molar mass?
Calcium's molar mass is 40.1 g/mol. What is the average mass of one calcium atom?
Zinc's periodic-table square prints 65.4. A single zinc atom is weighed on the atomic scale. Which mass is the atom's?
You've already seen the match: one hydrogen atom averages 1.0 amu, and 1 mol of hydrogen atoms weighs 1.0 g. Here is why the number repeats.
The match between amu and g/mol is built in, not lucky.
One gram is 6.022 × 10²³ atomic mass units.
One mole is 6.022 × 10²³ particles.
The same number sits in both facts, and that is the whole trick.
Take hydrogen atoms, 1.0 amu each.
Counting out 6.022 × 10²³ of them gives a total mass of 6.022 × 10²³ amu.
6.022 × 10²³ amu is exactly 1 g.
So 1 mol of hydrogen atoms has a mass of 1.0 g — the amu number reappears as the gram number.
A substance's molar mass in grams matches its particle's mass in amu because one gram is 6.022 × 10²³ atomic mass units, so counting out 6.022 × 10²³ particles turns a particle's mass in amu into the same number of grams.
Worked example 1. One helium atom has an average mass of 4.0 amu. Why does 1 mol of helium atoms have a mass of 4.0 g?
One helium atom averages 4.0 amu.
1 mol is 6.022 × 10²³ atoms, so the total mass is 6.022 × 10²³ × 4.0 amu.
Group it: that is 4.0 × (6.022 × 10²³ amu).
6.022 × 10²³ amu is 1 g, so the total is 4.0 × 1 g.
1 mol of helium atoms has a mass of 4.0 g — the same number as the amu value.
You can now explain that a substance's molar mass in grams matches its particle's mass in amu because one gram is 6.022 × 10²³ atomic mass units, so counting out 6.022 × 10²³ particles turns a particle's mass in amu into the same number of grams.
One nitrogen atom averages 14.0 amu, and 1 mol of nitrogen atoms has a mass of 14.0 g. Why does the number repeat?
How many atomic mass units are in one gram?
Lithium atoms average 6.9 amu. Which reasoning shows why 1 mol of lithium atoms has a mass of 6.9 g?
You've already read element molar masses from the periodic table. Some elements' particles are molecules of several atoms, and the formula tells you how many.
Some elements exist as molecules — several identical atoms bonded into each particle.
Oxygen gas has the formula O₂: each molecule contains two oxygen atoms.
Each O₂ molecule therefore weighs two oxygen atoms' worth.
So the molar mass multiplies the same way: the atom's molar mass times the subscript.
Oxygen's square prints 16.0.
molar mass of O₂ = 2 × 16.0 = 32.0 g/mol.
Worked example 1. Nitrogen gas has the formula N₂. Nitrogen's periodic-table square prints 14.0. What is the molar mass of N₂?
one nitrogen atom: 14.0 g/mol
the subscript counts the atoms in one molecule: 2
molar mass of N₂ = 28.0 g/mol
Worked example 2. Chlorine gas has the formula Cl₂. Chlorine's periodic-table square prints 35.5. What is the molar mass of Cl₂?
one chlorine atom: 35.5 g/mol
the subscript counts the atoms in one molecule: 2
molar mass of Cl₂ = 71.0 g/mol
You can now calculate the molar mass of an element that exists as molecules by multiplying the atom's molar mass by the subscript in the given formula.
Hydrogen gas has the formula H₂. Hydrogen's periodic-table square prints 1.0. Calculate the molar mass of H₂, in g/mol, to one decimal place.
Fluorine gas has the formula F₂. Fluorine's periodic-table square prints 19.0. Calculate the molar mass of F₂, in g/mol, to one decimal place.
Bromine exists as Br₂ molecules. Bromine's periodic-table square prints 79.9. Calculate the molar mass of Br₂, in g/mol, to one decimal place.
You've already doubled oxygen's value for O₂. A compound's formula mixes elements, and the same reading gives its molar mass.
A compound's formula lists every atom in one particle of the compound.
One particle's mass is the sum of its atoms' masses.
So the compound's molar mass is the sum of the molar masses of every atom the formula shows.
A subscript multiplies its own element's molar mass before the adding.
Water is H₂O: two hydrogens and one oxygen.
Hydrogen's square prints 1.0; oxygen's prints 16.0.
molar mass of H₂O = 2 × 1.0 + 16.0 = 18.0 g/mol.
The same routine covers ionic compounds — add every atom in the formula unit.
Worked example 1. Carbon dioxide has the formula CO₂. Use these molar masses: C 12.0 g/mol, O 16.0 g/mol. What is the molar mass of CO₂?
the formula counts 1 carbon and 2 oxygens
carbon: 12.0 g/mol; oxygen: 16.0 g/mol
molar mass of CO₂ = 44.0 g/mol
Worked example 2. Sodium hydroxide is the ionic compound NaOH. Use these molar masses: Na 23.0 g/mol, O 16.0 g/mol, H 1.0 g/mol. What is the molar mass of NaOH?
the formula unit counts 1 sodium, 1 oxygen, and 1 hydrogen
sodium: 23.0 g/mol; oxygen: 16.0 g/mol; hydrogen: 1.0 g/mol
molar mass of NaOH = 40.0 g/mol
You can now calculate a compound's molar mass by adding the molar mass of every atom shown in its formula.
Ammonia has the formula NH₃. Use these molar masses: N 14.0 g/mol, H 1.0 g/mol. Calculate ammonia's molar mass, in g/mol, to one decimal place.
Potassium chloride is the ionic compound KCl. Use these molar masses: K 39.1 g/mol, Cl 35.5 g/mol. Calculate KCl's molar mass, in g/mol, to one decimal place.
Magnesium chloride has the formula MgCl₂. Use these molar masses: Mg 24.3 g/mol, Cl 35.5 g/mol. Calculate MgCl₂'s molar mass, in g/mol, to one decimal place.
You've already calculated a compound's molar mass by adding the molar mass of every atom in its formula. Some formulas, such as Ca(OH)₂, use parentheses — and the parentheses change what you add.
A subscript written after a closing parenthesis multiplies everything inside the parentheses.
Ca(OH)₂ contains one calcium atom (40.1 g/mol), two oxygen atoms (16.0 g/mol each), and two hydrogen atoms (1.0 g/mol each).
To find the molar mass, first add the molar masses inside the parentheses: 16.0 + 1.0 = 17.0.
Ca(OH)₂ — what the parentheses multiply
| Atom | How many | Contribution |
|---|---|---|
| Ca | 1 | 40.1 |
| O | 2 (from the subscript outside the parentheses) | 2 × 16.0 = 32.0 |
| H | 2 (from the subscript outside the parentheses) | 2 × 1.0 = 2.0 |
| Total | 74.1 g/mol |
Then multiply that inside total by the outside subscript: 2 × 17.0 = 34.0.
Finally, add the molar masses of every atom outside the parentheses: 40.1 + 34.0 = 74.1.
The molar mass of Ca(OH)₂ is 74.1 g/mol.
Worked example 1. What is the molar mass of magnesium hydroxide, Mg(OH)₂? (molar masses: Mg 24.3, O 16.0, H 1.0 g/mol)
Inside the parentheses: 16.0 + 1.0 = 17.0
Multiply by the outside subscript: 2 × 17.0 = 34.0
Add the atoms outside the parentheses: 24.3 + 34.0
M = 58.3 g/mol
Worked example 2. What is the molar mass of aluminum nitrate, Al(NO₃)₃? (molar masses: Al 27.0, N 14.0, O 16.0 g/mol)
Inside the parentheses: 14.0 + 3 × 16.0 = 62.0
Multiply by the outside subscript: 3 × 62.0 = 186.0
Add the atoms outside the parentheses: 27.0 + 186.0
M = 213.0 g/mol
You can now calculate the molar mass of a compound whose formula uses parentheses by multiplying everything inside the parentheses by the subscript outside.
Calculate the molar mass of barium hydroxide, Ba(OH)₂. (molar masses: Ba 137.3, O 16.0, H 1.0 g/mol.) Give your answer in g/mol to one decimal place.
Calculate the molar mass of calcium nitrate, Ca(NO₃)₂. (molar masses: Ca 40.1, N 14.0, O 16.0 g/mol.) Give your answer in g/mol to one decimal place.
Calculate the molar mass of ammonium sulfide, (NH₄)₂S. (molar masses: N 14.0, H 1.0, S 32.1 g/mol.) Give your answer in g/mol to one decimal place.
You've already seen that one mole of a substance has a fixed mass — its molar mass. That link lets you predict how the number of moles changes when a sample's mass changes.
Keep the substance the same, and change only the sample's mass.
One mole of water has a mass of 18.0 g.
A 36.0 g sample of water holds two 18.0 g portions, so it holds 2 mol.
Doubling the mass doubles the number of moles.
Same substance: mass sets the moles
| Sample of water | 18.0 g portions | Moles |
|---|---|---|
| 18.0 g | 1 | 1 mol |
| 36.0 g | 2 | 2 mol |
| 54.0 g | 3 | 3 mol |
Halving the mass halves the number of moles.
Whatever factor the mass changes by, the number of moles changes by that same factor — as long as the substance stays the same.
Worked example 1. One sample of carbon dioxide has a mass of 44.0 g. A second sample has a mass of 132.0 g. How do their numbers of moles compare?
The substance is the same, so only the mass factor matters.
132.0 g is three times 44.0 g.
The 132.0 g sample holds three times as many moles as the 44.0 g sample.
Worked example 2. A 40.0 g sample of helium is poured down to 10.0 g. What happens to the number of moles?
The substance is the same, so only the mass factor matters.
10.0 g is one quarter of 40.0 g.
The number of moles falls to one quarter of what it was.
You can now predict how the number of moles of a substance changes when the sample's mass changes and the substance stays the same: doubling the mass doubles the moles.
A student weighs out two samples of sodium chloride: one of 25.0 g and one of 50.0 g. Compared with the 25.0 g sample, how many moles does the 50.0 g sample hold?
A methane sample's mass is increased from 16.0 g to 48.0 g. What happens to the number of moles in the sample?
A 60.0 g piece of aluminum is machined down to 15.0 g. Compared with the original piece, how many moles of aluminum does the machined piece hold?
You've just seen that more mass means more moles when the substance stays the same. Now hold the mass the same and change the substance.
Each mole of a substance has a mass equal to that substance's molar mass.
Water's molar mass is 18.0 g/mol; carbon dioxide's is 44.0 g/mol.
100.0 g of water splits into portions of 18.0 g each, while 100.0 g of carbon dioxide splits into portions of 44.0 g each.
The water sample holds more of its smaller portions, so it holds more moles.
Of two equal-mass samples, the substance with the larger molar mass gives fewer moles.
Of two equal-mass samples, the substance with the smaller molar mass gives more moles.
Worked example 1. One balloon holds 50.0 g of helium (molar mass 4.0 g/mol). Another holds 50.0 g of argon (molar mass 39.9 g/mol). Which balloon holds more moles of gas?
The masses are equal, so only the molar masses matter.
Helium's portions are 4.0 g each; argon's are 39.9 g each.
The same 50.0 g holds more of the smaller 4.0 g portions.
The helium balloon holds more moles — the smaller molar mass gives more moles.
Worked example 2. A cylinder holds 10.0 g of methane, CH₄ (molar mass 16.0 g/mol). Another holds 10.0 g of oxygen gas, O₂ (molar mass 32.0 g/mol). Which sample holds fewer moles?
The masses are equal, so only the molar masses matter.
Oxygen's molar mass, 32.0 g/mol, is the larger.
The oxygen sample holds fewer moles — the larger molar mass gives fewer moles.
You can now predict which of two equal-mass samples contains more moles when the substances differ and the mass stays the same: the substance with the larger molar mass gives fewer moles.
A bottle holds 20.0 g of ethanol, C₂H₆O (molar mass 46.0 g/mol). A dish holds 20.0 g of sodium chloride, NaCl (molar mass 58.5 g/mol). Which sample contains more moles?
A lab stores 100.0 g of iron (molar mass 55.8 g/mol) and 100.0 g of aluminum (molar mass 27.0 g/mol). Which sample contains more moles of atoms?
A flask holds 30.0 g of nitrogen monoxide, NO (molar mass 30.0 g/mol). Another flask holds 30.0 g of dinitrogen monoxide, N₂O (molar mass 44.0 g/mol). Which sample contains fewer moles?
You've now seen both halves of the mass–mole link: more mass means more moles, and a larger molar mass means fewer moles. One equation captures both at once.
The number of moles is the mass divided by the molar mass.
n = m / M
n stands for the amount of substance, measured in moles (mol).
| n | amount of substance (mol) |
| m | mass (g) |
| M | molar mass (g/mol) |
m stands for the mass, measured in grams (g).
M stands for the molar mass, measured in grams per mole (g/mol).
Dividing the mass by the molar mass counts how many one-mole portions the sample holds — exactly what the number of moles means.
m sits on top, so more mass gives more moles.
M sits underneath, so a larger molar mass gives fewer moles.
Worked example 1. In the equation n = m/M, what does each symbol stand for, and what unit does it carry?
Write down the values in the question
n is the amount of substance, in moles (mol).
m is the mass, in grams (g).
M is the molar mass, in grams per mole (g/mol).
Write down the equation
n = m / M
Substitute in the values, and calculate
Worked example 2. Which symbol in n = m/M stands for the mass weighed on the balance?
Answer: m — the mass, in grams.
You can now state the equation n = m/M and the meaning and unit of each symbol, where n is the amount in moles (mol), m is the mass in grams (g), and M is the molar mass in grams per mole (g/mol).
Which equation gives the number of moles in a sample?
In the equation n = m/M, what does the symbol M stand for, and in what unit?
A sample's mass is measured in grams and its molar mass is in grams per mole. What unit does the result of n = m/M carry?
You've already seen the mole equation, n = m/M. Now use it: the balance gives the mass, the molar mass is known, and one division gives the moles.
Write down the mass and the molar mass from the question.
Write down the equation, n = m/M.
Substitute the values and calculate.
| n | amount of substance (mol) |
| m | mass (g) |
| M | molar mass (g/mol) |
The result carries the unit mol.
Round the result to the correct number of significant figures, as you've already practiced.
For lab-scale samples, expect n between about 0.01 and 10 mol — an answer far outside that range signals a slip.
Worked example 1. How many moles are in 9.0 g of water? (M of water = 18.0 g/mol)
Write down the values in the question
m = 9.0 g
M = 18.0 g/mol
Write down the equation
n = m / M
Substitute in the values, and calculate
n = 9.0 / 18.0
n = 0.50 mol
Worked example 2. How many moles are in 11.0 g of carbon dioxide? (M of CO₂ = 44.0 g/mol)
Write down the values in the question
m = 11.0 g
M = 44.0 g/mol
Write down the equation
n = m / M
Substitute in the values, and calculate
n = 11.0 / 44.0
n = 0.250 mol
You can now calculate the number of moles in a sample from its mass and molar mass using n = m/M.
m/M gives g ÷ (g/mol) → mol ✓ correct
m×M gives g × (g/mol) → g²/mol ✗ you multiplied instead of dividing
M/m gives (g/mol) ÷ g → 1/mol ✗ you divided the wrong way
How many moles are in 80.0 g of sodium hydroxide, NaOH? (M of NaOH = 40.0 g/mol.) Give your answer in moles to 3 significant figures.
How many moles are in 5.85 g of sodium chloride, NaCl? (M of NaCl = 58.5 g/mol.) Give your answer in moles to 3 significant figures.
How many moles are in 12.0 g of methane, CH₄? (M of CH₄ = 16.0 g/mol.) Give your answer in moles to 3 significant figures.
Procedures often run the other way: they state the moles required, and you weigh out the matching mass.
The mole equation you've already seen still applies: n = m/M.
The question gives n and M and asks for m, so rearrange the equation.
Make m the subject by multiplying both sides by M, giving m = n × M.
Then substitute and calculate — the result carries the unit g.
| n | amount of substance (mol) |
| m | mass (g) |
| M | molar mass (g/mol) |
It is one equation, rearranged when needed — not a second equation to memorize.
Sanity check: more than 1 mol must weigh more than M, and less than 1 mol must weigh less than M.
Worked example 1. What is the mass of 0.50 mol of water? (M of water = 18.0 g/mol)
Write down the values in the question
n = 0.50 mol
M = 18.0 g/mol
Write down the equation
n = m / M
Make the unknown the subject
m = n × M
Substitute in the values, and calculate
m = 0.50 × 18.0
m = 9.0 g
Worked example 2. What is the mass of 3.00 mol of methane? (M of CH₄ = 16.0 g/mol)
Write down the values in the question
n = 3.00 mol
M = 16.0 g/mol
Write down the equation
n = m / M
Make the unknown the subject
m = n × M
Substitute in the values, and calculate
m = 3.00 × 16.0
m = 48.0 g
You can now calculate the mass of a sample from its number of moles and molar mass by rearranging n = m/M to make m the subject, giving m = n × M.
What is the mass, in grams, of 2.50 mol of sodium hydroxide, NaOH? (M of NaOH = 40.0 g/mol.) Give your answer to 3 significant figures.
What is the mass, in grams, of 0.500 mol of potassium chloride, KCl? (M of KCl = 74.6 g/mol.) Give your answer to 3 significant figures.
What is the mass, in grams, of 0.150 mol of glucose, C₆H₁₂O₆? (M of glucose = 180.0 g/mol.) Give your answer to 3 significant figures.
A table can list several samples, each with its own mass and molar mass. Which holds the most moles? The biggest mass is not automatically the answer.
Calculate n = m/M for every sample in the table, one line per sample.
Then rank the samples by their n values.
Never rank by mass alone — a large mass of a heavy-particle substance can hold fewer moles than a small mass of a light-particle substance.
In the table shown, water: n = 36.0 / 18.0 = 2.0 mol.
Helium: n = 4.0 / 4.0 = 1.0 mol.
Rank by moles, not by mass
| Sample | Mass | Molar mass | n = m/M |
|---|---|---|---|
| water | 36.0 g | 18.0 g/mol | 2.0 mol |
| helium | 4.0 g | 4.0 g/mol | 1.0 mol |
| carbon dioxide | 22.0 g | 44.0 g/mol | 0.50 mol |
Carbon dioxide: n = 22.0 / 44.0 = 0.50 mol.
Ranked from most moles to fewest: water > helium > carbon dioxide — even though carbon dioxide's 22.0 g beats helium's 4.0 g.
Worked example 1. Rank these samples from most moles to fewest: 64.0 g of oxygen gas, O₂ (M = 32.0 g/mol); 100.1 g of calcium carbonate, CaCO₃ (M = 100.1 g/mol); 20.0 g of sodium hydroxide, NaOH (M = 40.0 g/mol).
Write down the values in the question
Oxygen gas: n = 64.0 / 32.0 = 2.00 mol
Calcium carbonate: n = 100.1 / 100.1 = 1.00 mol
Sodium hydroxide: n = 20.0 / 40.0 = 0.500 mol
Write down the equation
n = m / M
Substitute in the values, and calculate
Oxygen gas > calcium carbonate > sodium hydroxide — the largest mass, 100.1 g, is not the most moles.
Worked example 2. Rank these samples from most moles to fewest: 2.0 g of hydrogen gas, H₂ (M = 2.0 g/mol); 3.2 g of methane, CH₄ (M = 16.0 g/mol); 5.85 g of sodium chloride, NaCl (M = 58.5 g/mol).
Write down the values in the question
Hydrogen gas: n = 2.0 / 2.0 = 1.0 mol
Methane: n = 3.2 / 16.0 = 0.20 mol
Sodium chloride: n = 5.85 / 58.5 = 0.100 mol
Write down the equation
n = m / M
Substitute in the values, and calculate
Hydrogen gas > methane > sodium chloride — here the largest mass holds the fewest moles.
You can now rank samples by their number of moles from a table of masses and molar masses.
Three samples: 13.8 g of lithium (M = 6.9 g/mol); 72.9 g of magnesium (M = 24.3 g/mol); 59.5 g of potassium bromide, KBr (M = 119.0 g/mol). Rank them from most moles to fewest.
Three samples: 51.0 g of ammonia, NH₃ (M = 17.0 g/mol); 200.5 g of calcium (M = 40.1 g/mol); 80.1 g of sulfur trioxide, SO₃ (M = 80.1 g/mol). Rank them from most moles to fewest.
Three samples: 142.0 g of chlorine gas, Cl₂ (M = 71.0 g/mol); 39.1 g of potassium (M = 39.1 g/mol); 81.0 g of aluminum (M = 27.0 g/mol). Rank them from most moles to fewest.
A calculation can look tidy and still be wrong. Checking a worked solution is a skill of its own — and three checks catch almost every slip.
Check the equation: moles from mass uses n = m/M, and mass from moles uses the rearrangement m = n × M.
Check the operation: multiplying where the equation divides — or dividing the wrong way around — is the most common error.
Check the size of the result: a mass smaller than the molar mass must give less than 1 mol, and more than 1 mol must weigh more than the molar mass.
Here is a supplied solution: 'Moles in 9.0 g of water (M = 18.0 g/mol): n = 9.0 × 18.0 = 162 mol.'
The solution multiplied by the molar mass instead of dividing by it.
The size check confirms the slip: 9.0 g is half of 18.0 g, so the answer must be 0.50 mol — not 162 mol.
Worked example 1. A question asks for the mass of 2.00 mol of sodium hydroxide, NaOH (M = 40.0 g/mol). A student writes: m = 40.0 / 2.00 = 20.0 g. What is the error?
The task is mass from moles, so the working should use m = n × M.
The student divided the molar mass by the moles instead of multiplying them.
Size check: 2.00 mol must weigh more than 40.0 g, and 20.0 g is less.
The error: dividing instead of multiplying — the correct working is m = 2.00 × 40.0 = 80.0 g.
Worked example 2. A question asks how many moles are in 88.0 g of carbon dioxide, CO₂ (M = 44.0 g/mol). A student writes: n = 44.0 / 88.0 = 0.50 mol. What is the error?
The equation has the right shape but is upside down — n = m/M puts the mass on top.
Size check: 88.0 g is two 44.0 g portions, so the answer must be 2.0 mol, not less than 1 mol.
The error: the division is upside down — the correct working is n = 88.0 / 44.0 = 2.0 mol.
You can now evaluate a supplied mass-to-moles or moles-to-mass worked solution and identify the error in it.
A question asks how many moles are in 10.0 g of helium (M = 4.0 g/mol). A student writes: n = 10.0 × 4.0 = 40.0 mol. What is the error?
A question asks for the mass of 0.500 mol of potassium chloride, KCl (M = 74.6 g/mol). A student writes: m = 0.500 / 74.6 = 0.00670 g. What is the error?
A question asks how many moles are in 117.0 g of sodium chloride, NaCl (M = 58.5 g/mol). A student writes: n = 58.5 / 117.0 = 0.500 mol. What is the error?
So far every mole question handed you the molar mass. A real question usually hands you only the formula — so finding the molar mass becomes the first stage of your own working.
Stage 1: calculate the molar mass from the formula, exactly as you've practiced — parentheses included.
Stage 2: apply the mole equation — n = m/M for moles from mass, or its rearrangement m = n × M for mass from moles.
Write the two stages separately, each with its own line of working.
Worked example 1. How many moles are in 10.0 g of sodium hydroxide, NaOH? (molar masses: Na 23.0, O 16.0, H 1.0 g/mol)
Write down the values in the question
Stage 1 — molar mass: M = 23.0 + 16.0 + 1.0 = 40.0 g/mol
Stage 2 — values: m = 10.0 g, M = 40.0 g/mol
Write down the equation
n = m / M
Substitute in the values, and calculate
n = 10.0 / 40.0
n = 0.250 mol
Worked example 2. What is the mass of 2.00 mol of magnesium hydroxide, Mg(OH)₂? (molar masses: Mg 24.3, O 16.0, H 1.0 g/mol)
Write down the values in the question
Stage 1 — molar mass: M = 24.3 + 2 × (16.0 + 1.0) = 58.3 g/mol
Stage 2 — values: n = 2.00 mol, M = 58.3 g/mol
Write down the equation
n = m / M
Make the unknown the subject
m = n × M
Substitute in the values, and calculate
m = 2.00 × 58.3
m = 116.6 g
You can now calculate moles from mass or mass from moles for a compound when only its formula is given, by first calculating the molar mass and then applying n = m/M or m = n × M.
How many moles are in 32.0 g of methane, CH₄? (molar masses: C 12.0, H 1.0 g/mol.) Give your answer in moles to 3 significant figures.
What is the mass, in grams, of 0.500 mol of sodium carbonate, Na₂CO₃? (molar masses: Na 23.0, C 12.0, O 16.0 g/mol.) Give your answer to 3 significant figures.
How many moles are in 328.2 g of calcium nitrate, Ca(NO₃)₂? (molar masses: Ca 40.1, N 14.0, O 16.0 g/mol.) Give your answer in moles to 3 significant figures.
End of Topic Test — five interchangeable forms, delivered separately.
You've already seen that one mole of any substance contains 6.022 × 10²³ particles. One equation turns that fact into a tool, linking the number of moles to the actual particle count.
The number of particles is the number of moles multiplied by Avogadro's number.
N = n × Nₐ
N stands for the number of particles — a plain count, with no unit.
| N | number of particles (plain count (no unit)) |
| n | amount of substance (mol) |
| Nₐ | Avogadro's number (6.022 × 10²³ particles per mole) |
n stands for the amount of substance, measured in moles (mol).
Nₐ stands for Avogadro's number: 6.022 × 10²³ particles per mole.
Capital N counts the particles one by one; small n counts them in moles — the capital letter belongs to the enormously bigger number.
Worked example 1. In the equation N = n × Nₐ, which symbol stands for the number of particles?
Answer: N — the plain count of particles.
Worked example 2. What value does Nₐ carry?
Answer: 6.022 × 10²³ particles per mole.
You can now state the equation N = n × Nₐ and the meaning of each symbol, where N is the number of particles, n is the amount in moles (mol), and Nₐ is Avogadro's number, 6.022 × 10²³ particles per mole.
Which equation links a sample's particle count to its number of moles?
In the equation N = n × Nₐ, what does the symbol n stand for, and in what unit?
In the equation N = n × Nₐ, what does Nₐ stand for, and what is its value?
You've already seen the particle equation, N = n × Nₐ. Now use it: multiply the moles by Avogadro's number, and the particle count comes out.
Write down the moles and Avogadro's number from the question.
Write down the equation, N = n × Nₐ.
Substitute the values and calculate.
| N | number of particles (plain count (no unit)) |
| n | amount of substance (mol) |
| Nₐ | Avogadro's number (6.022 × 10²³ particles per mole) |
N comes out as a plain count of particles — atoms, molecules, or formula units, whichever the substance uses.
For lab-scale samples, expect counts around 10²² to 10²⁵ — an answer smaller than 1 means the division went the wrong way.
Worked example 1. How many molecules are in 2.00 mol of water? (Nₐ = 6.022 × 10²³ mol⁻¹)
Write down the values in the question
n = 2.00 mol
Nₐ = 6.022 × 10²³ mol⁻¹
Write down the equation
N = n × Nₐ
Substitute in the values, and calculate
N = 2.00 × 6.022 × 10²³
N = 1.204 × 10²⁴ molecules
Worked example 2. How many molecules are in 0.500 mol of carbon dioxide? (Nₐ = 6.022 × 10²³ mol⁻¹)
Write down the values in the question
n = 0.500 mol
Nₐ = 6.022 × 10²³ mol⁻¹
Write down the equation
N = n × Nₐ
Substitute in the values, and calculate
N = 0.500 × 6.022 × 10²³
N = 3.011 × 10²³ molecules
You can now calculate the number of particles in a sample from its number of moles using N = n × Nₐ.
n×Nₐ gives mol × (particles/mol) → particles ✓ correct
n/Nₐ gives mol ÷ (particles/mol) → mol²/particle ✗ you divided instead of multiplying
Nₐ/n gives (particles/mol) ÷ mol → particles/mol² ✗ you divided the wrong way
How many atoms are in 5.00 mol of helium? (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.
How many molecules are in 0.250 mol of oxygen gas, O₂? (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.
How many formula units are in 1.50 mol of sodium chloride, NaCl? (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.
You can also start from a particle count and work back to the moles.
The particle equation you've already seen still applies: N = n × Nₐ.
The question gives N and asks for n, so rearrange the equation.
Make n the subject by dividing both sides by Nₐ, giving n = N / Nₐ.
Then substitute and calculate — the result carries the unit mol.
| N | number of particles (plain count (no unit)) |
| n | amount of substance (mol) |
| Nₐ | Avogadro's number (6.022 × 10²³ particles per mole) |
It is one equation, rearranged when needed — not a second equation to memorize.
Sanity check: a count above 6.022 × 10²³ means more than 1 mol, and a count below it means less than 1 mol.
Worked example 1. How many moles is 3.011 × 10²³ molecules of water? (Nₐ = 6.022 × 10²³ mol⁻¹)
Write down the values in the question
N = 3.011 × 10²³ molecules
Nₐ = 6.022 × 10²³ mol⁻¹
Write down the equation
N = n × Nₐ
Make the unknown the subject
n = N / Nₐ
Substitute in the values, and calculate
n = 3.011 × 10²³ / 6.022 × 10²³
n = 0.500 mol
Worked example 2. How many moles is 1.2044 × 10²⁴ molecules of carbon dioxide? (Nₐ = 6.022 × 10²³ mol⁻¹)
Write down the values in the question
N = 1.2044 × 10²⁴ molecules
Nₐ = 6.022 × 10²³ mol⁻¹
Write down the equation
N = n × Nₐ
Make the unknown the subject
n = N / Nₐ
Substitute in the values, and calculate
n = 1.2044 × 10²⁴ / 6.022 × 10²³
n = 2.000 mol
You can now calculate the number of moles in a sample from its number of particles by rearranging N = n × Nₐ to make n the subject, giving n = N/Nₐ.
A sample contains 9.033 × 10²³ molecules of methane, CH₄. How many moles is that? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer in moles to 3 significant figures.
A sample contains 6.022 × 10²² atoms of helium. How many moles is that? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer in moles to 3 significant figures.
A crystal contains 2.4088 × 10²⁴ formula units of sodium chloride, NaCl. How many moles is that? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer in moles to 3 significant figures.
You've already seen that a formula's subscripts count the atoms in one molecule or formula unit, and that a mole is a fixed count of particles. Put together, the subscript tells you how many moles of each element's atoms a sample of a compound holds.
Every molecule of H₂O contains 2 hydrogen atoms.
So every mole of H₂O contains 2 moles of hydrogen atoms.
The subscript scales up from single molecules to moles unchanged, because a mole is the same fixed count for molecules and for atoms alike.
To find the moles of one element's atoms, multiply the moles of the compound by that element's subscript in the formula.
A sample of 0.50 mol of H₂O therefore contains 2 × 0.50 = 1.0 mol of hydrogen atoms.
| subscript of X | number of X atoms in each molecule or formula unit of the compound (—) |
| moles of compound | amount of the compound (mol) |
| moles of X atoms | amount of element X's atoms inside the sample (mol) |
The same sample contains 1 × 0.50 = 0.50 mol of oxygen atoms, because the unwritten subscript on O is 1.
Worked example 1. How many moles of oxygen atoms are in 2.00 mol of carbon dioxide, CO₂?
Write down the values in the question
moles of CO₂ = 2.00 mol
subscript of O in CO₂ = 2
Write down the equation
moles of O atoms = subscript of O × moles of CO₂
Substitute in the values, and calculate
moles of O atoms = 2 × 2.00
moles of O atoms = 4.00 mol
Worked example 2. How many moles of hydrogen atoms are in 0.25 mol of ammonia, NH₃?
Write down the values in the question
moles of NH₃ = 0.25 mol
subscript of H in NH₃ = 3
Write down the equation
moles of H atoms = subscript of H × moles of NH₃
Substitute in the values, and calculate
moles of H atoms = 3 × 0.25
moles of H atoms = 0.75 mol
You can now calculate the moles of one element's atoms in a sample of a compound by multiplying the moles of the compound by that element's subscript in the formula.
How many moles of hydrogen atoms are in 0.50 mol of methane, CH₄? Give your answer in mol to 2 significant figures.
How many moles of carbon atoms are in 2.00 mol of propane, C₃H₈? Give your answer in mol to 3 significant figures.
How many moles of oxygen atoms are in 0.25 mol of dinitrogen tetroxide, N₂O₄? Give your answer in mol to 2 significant figures.
How many molecules are in 9.0 g of water? You have an equation that turns a mass into moles, and another that turns moles into a particle count — but no equation jumps straight from grams to molecules.
You've already used n = m/M and N = n × Nₐ on their own. This lesson is about seeing the route between them, not running the numbers.
No single taught equation links a mass directly to a particle count.
Moles sit in the middle of every route between mass and particle count.
n = m/M takes a mass to moles.
N = n × Nₐ takes moles to a particle count.
So the route from 9.0 g of water to its molecule count is two steps: first n = m/M, then N = n × Nₐ.
To travel the other way, from a particle count to a mass, run the same two equations backward: first make n the subject of N = n × Nₐ, giving n = N/Nₐ, then make m the subject of n = m/M, giving m = n × M.
The route map only names the steps — every worked problem still shows its working as equation lines.
Worked example 1. A question asks for the mass of 6.022 × 10²³ molecules of carbon dioxide, CO₂. Which two steps, in order, answer it?
The question gives a particle count, N, and asks for a mass, m.
Step 1 takes particles to moles: n = N/Nₐ.
Step 2 takes moles to mass: m = n × M.
First n = N/Nₐ, then m = n × M.
Worked example 2. A question asks for the number of molecules in 34.0 g of ammonia, NH₃. Which two steps, in order, answer it?
The question gives a mass, m, and asks for a particle count, N.
Step 1 takes mass to moles: n = m/M.
Step 2 takes moles to particles: N = n × Nₐ.
First n = m/M, then N = n × Nₐ.
You can now identify the two calculation steps between a sample's mass and its particle count, with moles as the middle quantity: n = m/M takes mass to moles, and N = n × Nₐ takes moles to particles.
A question asks for the number of molecules in 8.0 g of oxygen gas, O₂. Which two steps, in order, answer it?
A question asks for the mass of 1.204 × 10²⁴ formula units of sodium chloride, NaCl. Which two steps, in order, answer it?
Every calculation route between a sample's mass and its particle count passes through one middle quantity. Which quantity is it?
You've already seen the route: mass to moles with n = m/M, then moles to particles with N = n × Nₐ. Now run both steps and get the number out.
Write the two steps as two separate stages of working, each with its own equation line.
Stage 1: use n = m/M to turn the mass into moles.
Stage 2: use N = n × Nₐ to turn the moles into a particle count.
Write the mole value down between the stages — it is the checkpoint of the whole route.
For a lab-scale sample, that mole value should land between about 0.01 and 10 mol; a mole value in the trillions means a step was skipped.
Worked example 1. How many molecules are in 22.0 g of carbon dioxide, CO₂? (M of CO₂ = 44.0 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹)
m = 22.0 g
M = 44.0 g/mol
Nₐ = 6.022 × 10²³ mol⁻¹
N = 3.011 × 10²³ molecules
Worked example 2. How many formula units are in 80.0 g of sodium hydroxide, NaOH? (M of NaOH = 40.0 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹)
m = 80.0 g
M = 40.0 g/mol
Nₐ = 6.022 × 10²³ mol⁻¹
N = 1.204 × 10²⁴ formula units
You can now calculate the number of particles in a sample from its mass, showing two written stages: first n = m/M, then N = n × Nₐ.
How many molecules are in 8.0 g of oxygen gas, O₂? (M of O₂ = 32.0 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.
How many molecules are in 34.0 g of ammonia, NH₃? (M of NH₃ = 17.0 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.
How many atoms are in 4.0 g of helium? (M of He = 4.0 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer to 2 significant figures. Enter your answer in scientific notation, such as 1.5e23; do not type units.
You've already run the route from a mass to a particle count. This lesson runs the same route backward: a particle count in, a mass out.
The backward route uses the same two equations, each rearranged for its unknown.
Stage 1: make n the subject of N = n × Nₐ, giving n = N/Nₐ, and turn the particle count into moles.
Stage 2: make m the subject of n = m/M, giving m = n × M, and turn the moles into a mass.
Write the mole value down between the stages — it is the checkpoint of the whole route.
These are not new equations to memorize — they are N = n × Nₐ and n = m/M with a different symbol made the subject.
Worked example 1. What is the mass of 3.011 × 10²³ molecules of carbon dioxide, CO₂? (M of CO₂ = 44.0 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹)
N = 3.011 × 10²³ molecules
M = 44.0 g/mol
Nₐ = 6.022 × 10²³ mol⁻¹
m = 22.0 g
Worked example 2. What is the mass of 1.2044 × 10²⁴ molecules of oxygen gas, O₂? (M of O₂ = 32.0 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹)
N = 1.2044 × 10²⁴ molecules
M = 32.0 g/mol
Nₐ = 6.022 × 10²³ mol⁻¹
m = 64.0 g
You can now calculate the mass of a sample from its number of particles, showing two written stages: first n = N/Nₐ, then m = n × M.
What is the mass, in grams, of 3.011 × 10²³ molecules of methane, CH₄? (M of CH₄ = 16.0 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer to 2 significant figures.
What is the mass, in grams, of 6.022 × 10²² formula units of sodium chloride, NaCl? (M of NaCl = 58.5 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer to 3 significant figures.
What is the mass, in grams, of 9.033 × 10²³ molecules of ammonia, NH₃? (M of NH₃ = 17.0 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer to 3 significant figures.
You've already run mole conversions as equation lines with the mole value written between the stages. Some books and teachers write the very same steps in a different layout, and you should recognize it when you meet it.
The layout strings the conversion factors along one line, so that the units cancel as you read.
For 9.0 g of water it looks like this: 9.0 g × (1 mol / 18.0 g) = 0.50 mol.
That line carries out exactly the same step as n = 9.0/18.0 — the same division by the molar mass, written sideways.
One step, two layouts
| Equation method | Factor-label method |
|---|---|
| n = m/M | 9.0 g × (1 mol / 18.0 g) |
| n = 9.0 / 18.0 | grams cancel, moles remain |
| n = 0.50 mol | = 0.50 mol |
The layout is called the 'factor-label method', and some books call it 'dimensional analysis'.
In each factor, the unit being removed sits on the bottom and the unit being produced sits on top, so grams cancel and moles remain.
A two-factor line runs both steps of a chain at once, and the mole unit cancels in the middle — the route still passes through moles.
You won't be asked to build factor-label lines in this course; it appears here so you can recognize it as the same steps you already run with equations.
Worked example 1. A worked solution shows the line 0.50 mol × (6.022 × 10²³ molecules / 1 mol). Which equation step does the line carry out?
The factor multiplies moles by Avogadro's number, and the mol unit cancels.
The line carries out N = n × Nₐ.
You can now identify a factor-label (dimensional-analysis) setup as another way of writing the same mole conversion that the equations perform.
A worked solution for carbon dioxide, CO₂, shows the line 22.0 g × (1 mol / 44.0 g). Which equation step does the line carry out?
A worked solution for sodium hydroxide, NaOH, shows the line 2.00 mol × (40.0 g / 1 mol). Which equation step does the line carry out?
In the factor-label line 8.0 g × (1 mol / 32.0 g) for oxygen gas, O₂, which unit cancels, and which unit remains?
You've already mastered every step: mass to moles and back, particles to moles and back, and the two-stage chains. The remaining skill is choosing the route yourself, because a real question never labels its steps.
Ask two questions before any working: which quantity does the question give, and which does it ask for?
Place both quantities on the mole route: mass and particle count sit at the ends, and moles sit in the middle.
If the two quantities sit next to each other on the route, one equation finishes the job.
If they sit at opposite ends, run the two stages through moles and write the mole value between them.
Whatever the direction, use n = m/M on the mass side and N = n × Nₐ on the particle side, rearranged for the unknown when needed.
Worked example 1. How many molecules are in 4.5 g of water, H₂O? (M of H₂O = 18.0 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹)
Given a mass; asked for a particle count — opposite ends, so two stages.
m = 4.5 g
M = 18.0 g/mol
Nₐ = 6.022 × 10²³ mol⁻¹
N = 1.51 × 10²³ molecules
Worked example 2. What is the mass of 1.2044 × 10²⁴ molecules of methane, CH₄? (M of CH₄ = 16.0 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹)
Given a particle count; asked for a mass — opposite ends, so two stages.
N = 1.2044 × 10²⁴ molecules
M = 16.0 g/mol
Nₐ = 6.022 × 10²³ mol⁻¹
m = 32.0 g
Worked example 3. How many moles is 1.8066 × 10²⁴ molecules of oxygen gas, O₂? (Nₐ = 6.022 × 10²³ mol⁻¹)
Write down the values in the question
Given a particle count; asked for moles — neighbors on the route, so one equation finishes the job.
N = 1.8066 × 10²⁴ molecules
Nₐ = 6.022 × 10²³ mol⁻¹
Write down the equation
N = n × Nₐ
Make the unknown the subject
n = N/Nₐ
Substitute in the values, and calculate
n = 1.8066 × 10²⁴ / 6.022 × 10²³
n = 3.000 mol
You can now calculate any one of mass, moles, or particle count for a sample from any other one, choosing the correct route through moles.
How many molecules are in 3.4 g of ammonia, NH₃? (M of NH₃ = 17.0 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.
What is the mass, in grams, of 1.2044 × 10²⁴ formula units of sodium chloride, NaCl? (M of NaCl = 58.5 g/mol; Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer to 3 significant figures.
How many moles is 1.506 × 10²³ molecules of ethane, C₂H₆? (Nₐ = 6.022 × 10²³ mol⁻¹.) Give your answer in mol to 3 significant figures.
End of Topic Test — five interchangeable forms, delivered separately.
A farmer buys a 50 kg bag of fertilizer for its nitrogen. But the bag holds a white compound, not pure nitrogen — so how much of what the farmer paid for is actually in the bag? Answering that needs a way of saying how much of a compound's mass each element supplies.
You've already seen that a compound's formula fixes which atoms it contains. Chemists put a number on each element's share of the mass.
Water is 11.1% hydrogen by mass.
That means every 100.0 g of water contains 11.1 g of hydrogen.
The rest of the mass belongs to the oxygen: water is 88.9% oxygen by mass.
The percents of all the elements in a compound total 100.
Percent composition of water
| Element | Percent by mass | Grams in 100.0 g of water |
|---|---|---|
| hydrogen | 11.1% | 11.1 g |
| oxygen | 88.9% | 88.9 g |
| total | 100.0% | 100.0 g |
This set of shares — the percent by mass that each element contributes — is called the compound's 'percent composition'.
Worked example 1. Nitrogen dioxide, NO₂, is 30.4% nitrogen by mass. What mass of nitrogen is in 100.0 g of nitrogen dioxide?
Percent by mass means grams per 100 g of the compound.
100.0 g of NO₂ contains 30.4 g of nitrogen.
Worked example 2. Nitrogen dioxide contains only nitrogen and oxygen, and it is 30.4% nitrogen by mass. What percent of its mass is oxygen?
The percents of all the elements in a compound total 100.
100 − 30.4 = 69.6
Nitrogen dioxide is 69.6% oxygen by mass.
You can now state that a compound's percent composition is the percent by mass that each of its elements contributes to the compound.
Ammonia is 82.4% nitrogen by mass. Which statement reads that correctly?
Potassium hydroxide contains potassium, oxygen, and hydrogen. It is 69.7% potassium and 28.5% oxygen by mass. What percent of its mass is hydrogen?
Glucose is 40.0% carbon by mass. Which statement reads that correctly?
You've already seen what a percent composition means. The formula alone is enough to calculate it, because the formula fixes how much of one mole's mass each element supplies.
First find the mass that element X contributes to one mole of the compound: multiply X's atomic mass by its subscript in the formula.
The subscript step comes first, because an element that appears four times contributes four times its atomic mass.
Then divide that mass by the compound's molar mass, M, and multiply by 100.
In CH₄, hydrogen contributes 4 × 1.0 = 4.0 g of every mole's 16.0 g.
| mass of X in one mole | atomic mass of X × X's subscript in the formula (g) |
| M | molar mass of the compound (g/mol) |
| % X | percent of the compound's mass contributed by element X (%) |
So the percent of hydrogen in CH₄ is (4.0 ÷ 16.0) × 100 = 25.0%.
Every percent must land between 0 and 100 — a result over 100 means the ratio was inverted.
Worked example 1. What is the percent by mass of hydrogen in ethane, C₂H₆? (atomic mass of H = 1.0; M of C₂H₆ = 30.0 g/mol)
Write down the values in the question
mass of H in one mole = 6 × 1.0 = 6.0 g
M = 30.0 g/mol
Write down the equation
% H = (mass of H in one mole ÷ M) × 100
Substitute in the values, and calculate
% H = (6.0 ÷ 30.0) × 100
% H = 20.0%
Worked example 2. What is the percent by mass of oxygen in sodium hydroxide, NaOH? (atomic mass of O = 16.0; M of NaOH = 40.0 g/mol)
Write down the values in the question
mass of O in one mole = 1 × 16.0 = 16.0 g
M = 40.0 g/mol
Write down the equation
% O = (mass of O in one mole ÷ M) × 100
Substitute in the values, and calculate
% O = (16.0 ÷ 40.0) × 100
% O = 40.0%
You can now calculate the mass percent of one element in a compound from the formula by dividing the element's total molar mass in the formula by the compound's molar mass and multiplying by 100.
(mass of X in one mole ÷ M) × 100 gives g ÷ (g/mol) shares of one mole → % ✓ correct
(M ÷ mass of X in one mole) × 100 gives a value over 100% ✗ you inverted the ratio
(atomic mass alone ÷ M) × 100 when the subscript is 2 or more → too small ✗ you forgot to multiply by the subscript
What is the percent by mass of carbon in glucose, C₆H₁₂O₆? (atomic mass of C = 12.0; M of glucose = 180.0 g/mol.) Give your answer as a percent to 3 significant figures.
What is the percent by mass of potassium in potassium hydroxide, KOH? (atomic mass of K = 39.1; M of KOH = 56.1 g/mol.) Give your answer as a percent to 3 significant figures.
What is the percent by mass of nitrogen in ammonium nitrate, NH₄NO₃? (atomic mass of N = 14.0; M of NH₄NO₃ = 80.0 g/mol.) Give your answer as a percent to 3 significant figures.
You've already calculated percent composition from a formula. In the lab the data arrive the other way: a sample is broken apart, each element's mass is measured, and the percents come from those measurements.
Divide the element's measured mass by the whole sample's measured mass, and multiply by 100.
A 10.0 g sample that contains 8.0 g of oxygen is (8.0 ÷ 10.0) × 100 = 80.0% oxygen by mass.
The element masses in a sample add up to the whole sample's mass, so the remaining 2.0 g belongs to the other element.
Measured masses from heating 20.0 g of a copper–oxygen compound
| Part | Measured mass |
|---|---|
| copper left behind | 16.0 g |
| oxygen driven off | 4.0 g |
| whole sample | 20.0 g |
The percents from a measured sample must also total 100 — that is the same check the formula route uses.
When a table gives every element's mass, run the same division once per element.
Worked example 1. Heating 20.0 g of a copper–oxygen compound drives off the oxygen and leaves 16.0 g of copper. What percent of the compound's mass is copper?
Write down the values in the question
mass of Cu in the sample = 16.0 g
mass of the sample = 20.0 g
Write down the equation
% Cu = (mass of Cu in the sample ÷ mass of the sample) × 100
Substitute in the values, and calculate
% Cu = (16.0 ÷ 20.0) × 100
% Cu = 80.0%
Worked example 2. Splitting a 36.0 g sample of water yields 4.0 g of hydrogen and 32.0 g of oxygen. What percent of water's mass is hydrogen?
Write down the values in the question
mass of H in the sample = 4.0 g
mass of the sample = 36.0 g
Write down the equation
% H = (mass of H in the sample ÷ mass of the sample) × 100
Substitute in the values, and calculate
% H = (4.0 ÷ 36.0) × 100
% H = 11.1%
You can now calculate the mass percent of an element in a compound from measured masses by dividing the element's mass in the sample by the whole sample's mass and multiplying by 100.
A 40.0 g sample of sodium hydroxide contains 23.0 g of sodium. What percent of the sample's mass is sodium? Give your answer as a percent to 3 significant figures.
A 90.0 g sample of glucose is found to contain 36.0 g of carbon. What percent of the sample's mass is carbon? Give your answer as a percent to 3 significant figures.
Analysis of an 80.0 g sample of ammonium nitrate measures 28.0 g of nitrogen, 4.0 g of hydrogen, and 48.0 g of oxygen. What percent of the sample's mass is oxygen? Give your answer as a percent to 3 significant figures.
You've already calculated that methane is 25.0% hydrogen by mass. A fair question: was that answer only true for the sample in the question?
A 1 g sample of methane is 25.0% hydrogen by mass.
A 1000 g sample of methane is also 25.0% hydrogen by mass.
Methane samples of different sizes
| Sample of methane | Mass of hydrogen | Percent hydrogen by mass |
|---|---|---|
| 1 g | 0.25 g | 25.0% |
| 1000 g | 250 g | 25.0% |
The percent composition of a pure compound is the same in every sample, whatever its size.
It is fixed because every molecule or formula unit of a compound contains exactly the same atoms, so the mass ratio is the same in any sample.
A bigger sample is just more of the identical molecule — more hydrogen grams and more carbon grams, in the same proportion.
So percent composition is a property of the compound itself, not of any particular sample.
Worked example 1. One tank holds 5.0 g of ethane and another holds 500.0 g of ethane. Compare the percent of carbon by mass in the two tanks.
Both tanks hold pure ethane, and every molecule or formula unit of a compound contains exactly the same atoms, so the mass ratio is the same in any sample.
Ethane is 80.0% carbon by mass, so both tanks are 80.0% carbon.
The percents are identical — 80.0% carbon in both tanks.
Worked example 2. A student expects a large bag of glucose to have a higher percent of carbon than a small spoonful. What is wrong with the expectation?
The bag holds more glucose molecules, but each molecule is identical to the spoonful's molecules.
Because every molecule or formula unit of a compound contains exactly the same atoms, the mass ratio is the same in any sample.
More molecules mean more carbon grams and more of every other element's grams, in the same proportion.
Both samples are 40.0% carbon by mass — sample size cannot change a compound's percent composition.
You can now explain that a pure compound shows 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.
A drinking glass holds 250 g of pure water, and a swimming pool holds millions of grams of pure water. Compare the percent of hydrogen by mass in the two.
One gas cylinder is filled with pure carbon dioxide made by a brewery, and another with pure carbon dioxide made by burning charcoal. Compare the percent of carbon by mass in the two cylinders.
A 10.0 g sample of sodium hydroxide is 57.5% sodium by mass. What is the percent of sodium by mass in a 250.0 g sample of sodium hydroxide?
You've already calculated one element's percent from a formula. The full percent composition runs the same equation once per element — and the answers grade themselves.
Run % X = (mass of X in one mole ÷ M) × 100 once for each element in the formula.
For CH₄: % C = (12.0 ÷ 16.0) × 100 = 75.0%.
| mass of X in one mole | atomic mass of X × X's subscript in the formula (g) |
| M | molar mass of the compound (g/mol) |
| % X | percent of the compound's mass contributed by element X (%) |
For CH₄: % H = (4 × 1.0 ÷ 16.0) × 100 = 25.0%.
Then check the total: 75.0 + 25.0 = 100.0.
If the percents do not total 100, one of the calculations carries an error — recheck the subscripts first.
Small rounding differences, like 99.9 or 100.1, are acceptable; anything further off means a slip.
Worked example 1. Calculate the full percent composition of sodium hydroxide, NaOH. (atomic masses: Na 23.0, O 16.0, H 1.0; M of NaOH = 40.0 g/mol)
mass of Na in one mole = 1 × 23.0 = 23.0 g
mass of O in one mole = 1 × 16.0 = 16.0 g
mass of H in one mole = 1 × 1.0 = 1.0 g
M = 40.0 g/mol
NaOH is 57.5% sodium, 40.0% oxygen, and 2.5% hydrogen by mass.
Worked example 2. Calculate the full percent composition of ethane, C₂H₆. (atomic masses: C 12.0, H 1.0; M of C₂H₆ = 30.0 g/mol)
mass of C in one mole = 2 × 12.0 = 24.0 g
mass of H in one mole = 6 × 1.0 = 6.0 g
M = 30.0 g/mol
C₂H₆ is 80.0% carbon and 20.0% hydrogen by mass.
You can now calculate the percent composition of a compound from its formula, giving the mass percent of every element and checking that the percents total 100.
Calculate the full percent composition of propanol, C₃H₈O, and check that your percents total 100. (atomic masses: C 12.0, H 1.0, O 16.0; M of C₃H₈O = 60.0 g/mol.) Enter the percent of carbon, as a percent to 3 significant figures.
Calculate the full percent composition of glucose, C₆H₁₂O₆, and check that your percents total 100. (atomic masses: C 12.0, H 1.0, O 16.0; M of glucose = 180.0 g/mol.) Enter the percent of oxygen, as a percent to 3 significant figures.
Calculate the full percent composition of dinitrogen monoxide, N₂O, and check that your percents total 100. (atomic masses: N 14.0, O 16.0; M of N₂O = 44.0 g/mol.) Enter the percent of nitrogen, as a percent to 3 significant figures.
End of Topic Test — five interchangeable forms, delivered separately.
A laboratory instrument can report how a compound's atoms compare — one carbon atom for every two hydrogen atoms and one oxygen atom, say — without ever seeing a whole molecule. Chemists need a way to write that measured ratio down as a formula.
You've already seen that a formula's subscripts count atoms. This lesson names the formula that records a measured atom ratio.
Measurements on a compound reveal the ratio of its atoms — how many atoms of one element there are for each atom of another.
A formula can record that ratio using the smallest whole numbers that fit it.
This formula is called the compound's 'empirical formula'.
An empirical formula gives the smallest whole-number ratio of the atoms of each element in a compound.
A compound whose atoms are in a 1 : 2 : 1 ratio of carbon to hydrogen to oxygen has the empirical formula CH₂O.
Reading it back: CH₂O says 1 carbon atom for every 2 hydrogen atoms for every 1 oxygen atom — a subscript of 1 is never written.
Worked example 1. A compound contains nitrogen and oxygen atoms in a 1 : 2 ratio. What is its empirical formula?
The smallest whole numbers that fit 1 : 2 are already 1 and 2.
Answer: NO₂.
Worked example 2. What atom ratio does the empirical formula P₂O₅ give for phosphorus to oxygen?
Answer: 2 : 5 — 2 phosphorus atoms for every 5 oxygen atoms.
You can now state that an empirical formula gives the smallest whole-number ratio of the atoms of each element in a compound.
A compound contains carbon and hydrogen atoms in a 1 : 3 ratio. Which formula is its empirical formula?
What does a compound's empirical formula tell you?
A compound contains sulfur and oxygen atoms in a 1 : 3 ratio. Write its empirical formula. Type subscripts as plain numbers, like the 2 in CO2.
You've already seen that an empirical formula gives the smallest whole-number ratio of a compound's atoms. A ratio alone leaves one question open: how many atoms does one molecule actually hold?
Molecules come in definite sizes — each molecule of a compound holds an exact number of atoms.
The molecular formula — the kind of formula you have been reading since Unit 5 — lists the actual number of atoms of each element in one molecule.
The molecular formula C₆H₁₂O₆ means one glucose molecule contains exactly 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms.
Glucose's atoms are in a 1 : 2 : 1 ratio, so its empirical formula is CH₂O — the ratio; its molecular formula is C₆H₁₂O₆ — the actual counts.
The two formulas answer different questions: the empirical formula says how the atoms compare, and the molecular formula says how many each molecule holds.
Worked example 1. The molecular formula of hydrogen peroxide is H₂O₂. How many atoms of each element does one hydrogen peroxide molecule contain?
Answer: exactly 2 hydrogen atoms and 2 oxygen atoms.
Worked example 2. One molecule of butane contains 4 carbon atoms and 10 hydrogen atoms. What is butane's molecular formula?
Answer: C₄H₁₀.
You can now state that a molecular formula gives the actual number of atoms of each element in one molecule of a compound.
The molecular formula of propane is C₃H₈. What does the 8 tell you?
One molecule of hydrazine contains 2 nitrogen atoms and 4 hydrogen atoms. What is hydrazine's molecular formula?
Which statement gives the difference between a compound's molecular formula and its empirical formula?
You've already seen the two formulas side by side: the molecular formula gives the actual atom counts, and the empirical formula gives the smallest whole-number ratio. Given a molecular formula, a short routine produces the matching empirical formula.
Look at the subscripts of the molecular formula, counting an unwritten subscript as 1.
Find the largest whole number that divides into every subscript — the largest common factor.
Divide every subscript by that factor.
The divided subscripts are the empirical formula's subscripts — and a subscript of 1 is never written.
For glucose, C₆H₁₂O₆: the subscripts 6, 12, and 6 all divide by 6.
Dividing gives 1, 2, and 1, so glucose's empirical formula is CH₂O.
Molecular formula → empirical formula
| Molecular formula | Largest common factor | Empirical formula |
|---|---|---|
| C₆H₁₂O₆ | 6 | CH₂O |
Worked example 1. What is the empirical formula of tetraphosphorus decoxide, P₄O₁₀?
Write down the values in the question
Subscripts: P 4, O 10.
Largest common factor: 2.
Write down the equation
divide every subscript by the largest common factor
Substitute in the values, and calculate
P: 4 ÷ 2 = 2, O: 10 ÷ 2 = 5
Empirical formula: P₂O₅
Worked example 2. What is the empirical formula of benzene, C₆H₆?
Write down the values in the question
Subscripts: C 6, H 6.
Largest common factor: 6.
Write down the equation
divide every subscript by the largest common factor
Substitute in the values, and calculate
C: 6 ÷ 6 = 1, H: 6 ÷ 6 = 1
Empirical formula: CH
You can now identify the empirical formula that corresponds to a given molecular formula by dividing all the subscripts by their largest common factor.
What is the empirical formula of hydrogen peroxide, H₂O₂? Type subscripts as plain numbers, like the 2 in CO2.
What is the empirical formula of ethane, C₂H₆? Type subscripts as plain numbers, like the 2 in CO2.
What is the empirical formula of cyclohexane, C₆H₁₂? Type subscripts as plain numbers, like the 2 in CO2.
You've already reduced molecular formulas by dividing every subscript by the largest common factor. Some formulas have nothing to reduce — they are already empirical. One check tells you which is which.
Look at the formula's subscripts, counting an unwritten subscript as 1.
Find the largest whole number that divides into every subscript.
If that number is 1, the formula is already an empirical formula.
If that number is bigger than 1, the formula is not empirical — it still reduces.
N₂O₄ is not empirical: both subscripts divide by 2.
CO₂ is empirical: the subscripts 1 and 2 share no common factor bigger than 1.
Worked example 1. Is C₃H₈ an empirical formula?
Subscripts: C 3, H 8.
No whole number bigger than 1 divides into both 3 and 8.
Yes — C₃H₈ is already an empirical formula.
Worked example 2. Is C₂H₄ an empirical formula?
Subscripts: C 2, H 4.
Both subscripts divide by 2.
No — C₂H₄ is not empirical; it reduces to CH₂.
You can now classify a given formula as empirical or not empirical by checking whether its subscripts share a common factor.
Which formula is already an empirical formula?
Which formula is already an empirical formula?
Which formula is NOT an empirical formula?
The mole equation n = m/M works on a mass in grams — but composition data usually arrives as percents. One assumption turns every percent straight into a mass.
Percent means parts per hundred: a compound that is 75.0% carbon by mass has 75.0 g of carbon in every 100 g of the compound.
So assume a sample of exactly 100 g.
In a 100 g sample, each element's percent becomes the same number of grams.
The percents of a compound total 100, so the element masses in a 100 g sample total 100 g.
That total lets you find a missing element's mass by subtracting the known masses from 100 g.
Worked example 1. A compound is 88.9% oxygen and 11.1% hydrogen by mass. What mass of each element does a 100 g sample contain?
Assume a sample of exactly 100 g.
Oxygen: 88.9% of the compound → 88.9 g.
Hydrogen: 11.1% of the compound → 11.1 g.
88.9 g of oxygen and 11.1 g of hydrogen.
Worked example 2. A compound is 82.4% nitrogen by mass, and the rest is hydrogen. What mass of hydrogen does a 100 g sample contain?
Assume a sample of exactly 100 g.
Nitrogen: 82.4% of the compound → 82.4 g.
The element masses in a 100 g sample total 100 g.
mass of hydrogen = 100 − 82.4
mass of hydrogen = 17.6 g
You can now calculate the mass of each element in an assumed 100 g sample of a compound from its percent composition by reading each percent as that many grams.
A compound is 27.3% carbon by mass; the rest is oxygen. What mass of carbon does a 100 g sample of the compound contain? Give your answer in grams to one decimal place.
A compound is 60.3% magnesium and 39.7% oxygen by mass. What mass of oxygen does a 100 g sample of the compound contain? Give your answer in grams to one decimal place.
A compound is 71.5% calcium by mass; the rest is oxygen. What mass of oxygen does a 100 g sample of the compound contain? Give your answer in grams to one decimal place.
A sample of a compound holds 75.0 g of carbon and 25.0 g of hydrogen. Those masses cannot be compared atom for atom, because one carbon atom outweighs one hydrogen atom twelve times.
Moles count particles, so converting each element's mass to moles puts the elements on a counting scale.
Convert each element's mass to moles with n = m/M, using that element's own molar mass.
For the sample above: n(C) = 75.0/12.0 = 6.25 mol, and n(H) = 25.0/1.0 = 25.0 mol.
Divide every mole value by the smallest mole value.
Here the smallest is 6.25: carbon gives 6.25/6.25 = 1, and hydrogen gives 25.0/6.25 = 4.
The results — 1 : 4 — are the smallest ratio of the atoms counted in moles, called the elements' 'mole ratio'.
So this compound contains 1 carbon atom for every 4 hydrogen atoms.
Worked example 1. A sample of a compound contains 14.0 g of nitrogen and 32.0 g of oxygen. What is the simplest mole ratio of nitrogen atoms to oxygen atoms? (M of N = 14.0 g/mol, O = 16.0 g/mol)
Write down the values in the question
m(N) = 14.0 g
m(O) = 32.0 g
Write down the equation
n = m/M (for each element)
Substitute in the values, and calculate
n(N) = 14.0 / 14.0 = 1.00 mol n(O) = 32.0 / 16.0 = 2.00 mol Divide each by the smallest (1.00): N: 1.00 / 1.00 = 1 O: 2.00 / 1.00 = 2
Mole ratio of nitrogen to oxygen = 1 : 2
Worked example 2. A sample of a compound contains 78.2 g of potassium and 32.1 g of sulfur. What is the simplest mole ratio of potassium atoms to sulfur atoms? (M of K = 39.1 g/mol, S = 32.1 g/mol)
Write down the values in the question
m(K) = 78.2 g
m(S) = 32.1 g
Write down the equation
n = m/M (for each element)
Substitute in the values, and calculate
n(K) = 78.2 / 39.1 = 2.00 mol n(S) = 32.1 / 32.1 = 1.00 mol Divide each by the smallest (1.00): K: 2.00 / 1.00 = 2 S: 1.00 / 1.00 = 1
Mole ratio of potassium to sulfur = 2 : 1
You can now calculate the simplest mole ratio of the elements in a compound by converting each element's mass to moles and dividing every result by the smallest value.
A sample of a compound contains 12.0 g of carbon and 32.0 g of oxygen. What is the simplest mole ratio of carbon atoms to oxygen atoms? (M of C = 12.0 g/mol, O = 16.0 g/mol.) Write your answer like 1 : 2.
A sample of a compound contains 7.0 g of nitrogen and 1.5 g of hydrogen. What is the simplest mole ratio of nitrogen atoms to hydrogen atoms? (M of N = 14.0 g/mol, H = 1.0 g/mol.) Write your answer like 1 : 2.
A sample of a compound contains 13.8 g of lithium and 32.1 g of sulfur. What is the simplest mole ratio of lithium atoms to sulfur atoms? (M of Li = 6.9 g/mol, S = 32.1 g/mol.) Write your answer like 1 : 2.
Dividing by the smallest mole value does not always land on whole numbers. A ratio like 1 : 1.5 cannot become subscripts, because subscripts count atoms — and atoms come in whole numbers.
Multiply every number in the ratio by the same small whole number until every number is whole.
Multiplying every number by the same amount keeps the ratio itself unchanged.
A decimal ending in .5 clears when you multiply by 2.
A decimal near .33 or .67 clears when you multiply by 3.
Clearing a decimal from a ratio
| Decimal part | Multiply every number by |
|---|---|
| .5 | 2 |
| .33 or .67 | 3 |
| .25 or .75 | 4 |
A decimal near .25 or .75 clears when you multiply by 4.
For 1 : 1.5, multiply both numbers by 2: the ratio becomes 2 : 3.
Never round a decimal like 1.5 away — rounding changes the ratio, multiplying does not.
Worked example 1. A mole calculation gives the ratio 1 : 1.33. What whole-number ratio does it clear to?
The decimal .33 is near a third, so multiply every number by 3.
1 × 3 = 3
1.33 × 3 = 4
Whole-number ratio: 3 : 4
Worked example 2. A mole calculation gives the ratio 1 : 2.5. What whole-number ratio does it clear to?
The decimal ends in .5, so multiply every number by 2.
1 × 2 = 2
2.5 × 2 = 5
Whole-number ratio: 2 : 5
You can now calculate whole-number subscripts from a mole ratio that contains a decimal by multiplying every number in the ratio by the smallest whole number that clears the decimal.
A mole calculation gives the ratio 1 : 1.25. What whole-number ratio does it clear to? Write your answer like 2 : 3.
A mole calculation gives the ratio 1 : 1.67. What whole-number ratio does it clear to? Write your answer like 2 : 3.
A mole calculation gives the ratio 1.5 : 1. What whole-number ratio does it clear to? Write your answer like 2 : 3.
Every step of this routine is already yours: percents become masses in a 100 g sample, masses become moles with n = m/M, dividing by the smallest gives the mole ratio, and a leftover decimal clears by multiplying. This lesson chains those steps into one routine that ends in a formula.
If the data arrives as percents, assume a 100 g sample, so each percent becomes that many grams.
If the data arrives as element masses, use the masses directly.
Convert each element's mass to moles with n = m/M, using that element's own molar mass.
Divide every mole value by the smallest mole value.
If a decimal remains, multiply every number by the small whole number that clears it.
The finished whole numbers are the subscripts of the empirical formula.
For a compound of 75.0% carbon and 25.0% hydrogen: a 100 g sample holds 75.0 g of carbon and 25.0 g of hydrogen.
n(C) = 75.0/12.0 = 6.25 mol, and n(H) = 25.0/1.0 = 25.0 mol.
Composition of the compound
| Element | Percent by mass |
|---|---|
| carbon | 75.0% |
| hydrogen | 25.0% |
Dividing both by 6.25 gives 1 and 4 — no decimal remains.
So the compound's empirical formula is CH₄.
Worked example 1. A sample of a compound contains 46.0 g of sodium and 16.0 g of oxygen. What is its empirical formula? (M of Na = 23.0 g/mol, O = 16.0 g/mol)
Write down the values in the question
m(Na) = 46.0 g
m(O) = 16.0 g
Write down the equation
n = m/M (for each element)
Substitute in the values, and calculate
n(Na) = 46.0 / 23.0 = 2.00 mol n(O) = 16.0 / 16.0 = 1.00 mol Divide each by the smallest (1.00): Na: 2.00 / 1.00 = 2 O: 1.00 / 1.00 = 1
Empirical formula: Na₂O
Worked example 2. A compound is 90.0% carbon and 10.0% hydrogen by mass. What is its empirical formula? (M of C = 12.0 g/mol, H = 1.0 g/mol)
Write down the values in the question
Assume a 100 g sample:
m(C) = 90.0 g
m(H) = 10.0 g
Write down the equation
n = m/M (for each element)
Substitute in the values, and calculate
n(C) = 90.0 / 12.0 = 7.50 mol n(H) = 10.0 / 1.0 = 10.0 mol Divide each by the smallest (7.50): C: 7.50 / 7.50 = 1 H: 10.0 / 7.50 = 1.33 A decimal near .33 clears with ×3: C: 1 × 3 = 3 H: 1.33 × 3 = 4
Empirical formula: C₃H₄
You can now calculate a compound's empirical formula from percent-composition or element-mass data by converting the data to masses, then to moles, then to the simplest whole-number ratio.
A compound is 92.3% carbon and 7.7% hydrogen by mass. What is its empirical formula? (M of C = 12.0 g/mol, H = 1.0 g/mol.) Type subscripts as plain numbers, like the 2 in CO2.
A sample of a compound contains 32.1 g of sulfur and 32.0 g of oxygen. What is its empirical formula? (M of S = 32.1 g/mol, O = 16.0 g/mol.) Type subscripts as plain numbers, like the 2 in CO2.
A sample of a compound contains 6.2 g of phosphorus and 4.8 g of oxygen. What is its empirical formula? (M of P = 31.0 g/mol, O = 16.0 g/mol.) Type subscripts as plain numbers, like the 2 in CO2.
Different compounds can share one empirical formula — the ratio 1 : 2 : 1 of CH₂O fits more than one molecule. The compound's molar mass settles how big its molecule really is.
A molecule holds a whole number of empirical-formula units — 1, 2, 3, or more copies of the ratio's worth of atoms.
The empirical formula has its own molar mass, added up from the periodic table the usual way.
For CH₂O: M = 12.0 + 2 × 1.0 + 16.0 = 30.0 g/mol.
Divide the compound's molar mass by the empirical formula's molar mass.
The answer is the number of empirical-formula units in one molecule.
A compound with molar mass 180.0 g/mol built from CH₂O units holds 180.0/30.0 = 6 units per molecule.
The division lands on a whole number, and never below 1 — a result like 0.5 means the division ran upside down.
Worked example 1. A compound has the empirical formula NO₂ and a molar mass of 92.0 g/mol. How many empirical-formula units fit in one of its molecules? (M of N = 14.0 g/mol, O = 16.0 g/mol)
Write down the values in the question
molar mass of the compound = 92.0 g/mol
M of NO₂ = 14.0 + 2 × 16.0 = 46.0 g/mol
Write down the equation
number of units = molar mass of the compound ÷ molar mass of the empirical formula
Substitute in the values, and calculate
number of units = 92.0 / 46.0
number of units = 2
Worked example 2. A compound has the empirical formula CH₂ and a molar mass of 56.0 g/mol. How many empirical-formula units fit in one of its molecules? (M of C = 12.0 g/mol, H = 1.0 g/mol)
Write down the values in the question
molar mass of the compound = 56.0 g/mol
M of CH₂ = 12.0 + 2 × 1.0 = 14.0 g/mol
Write down the equation
number of units = molar mass of the compound ÷ molar mass of the empirical formula
Substitute in the values, and calculate
number of units = 56.0 / 14.0
number of units = 4
You can now calculate how many empirical-formula units fit in one molecule by dividing the compound's molar mass by the molar mass of the empirical formula.
A compound has the empirical formula HO and a molar mass of 34.0 g/mol. (M of HO = 17.0 g/mol.) How many empirical-formula units fit in one of its molecules? Enter a whole number.
A compound has the empirical formula CH and a molar mass of 78.0 g/mol. (M of CH = 13.0 g/mol.) How many empirical-formula units fit in one of its molecules? Enter a whole number.
A compound has the empirical formula C₂H₅ and a molar mass of 58.0 g/mol. (M of C₂H₅ = 29.0 g/mol.) How many empirical-formula units fit in one of its molecules? Enter a whole number.
You can already count how many empirical-formula units fit in one molecule. One more step turns that count into the molecular formula itself.
Find the number of units: divide the compound's molar mass by the empirical formula's molar mass.
Multiply every subscript in the empirical formula by that number of units.
The result is the molecular formula — the actual atom counts in one molecule.
For CH₂O with a compound molar mass of 180.0 g/mol: 180.0/30.0 = 6 units.
Multiplying the subscripts 1, 2, and 1 by 6 gives C₆H₁₂O₆ — glucose's molecular formula.
A unit count of 1 means the molecular formula and the empirical formula are the same formula.
Worked example 1. A compound has the empirical formula CH₂ and a molar mass of 42.0 g/mol. What is its molecular formula? (M of C = 12.0 g/mol, H = 1.0 g/mol)
molar mass of the compound = 42.0 g/mol
M of CH₂ = 12.0 + 2 × 1.0 = 14.0 g/mol
Molecular formula: C₃H₆
Worked example 2. A compound has the empirical formula NH₂ and a molar mass of 32.0 g/mol. What is its molecular formula? (M of N = 14.0 g/mol, H = 1.0 g/mol)
molar mass of the compound = 32.0 g/mol
M of NH₂ = 14.0 + 2 × 1.0 = 16.0 g/mol
Molecular formula: N₂H₄
You can now calculate a compound's molecular formula from its empirical formula and molar mass by finding the whole-number multiplier and multiplying every subscript by it.
A compound has the empirical formula CH and a molar mass of 26.0 g/mol. (M of CH = 13.0 g/mol.) What is its molecular formula? Type subscripts as plain numbers, like the 2 in CO2.
A compound has the empirical formula CH₂O and a molar mass of 60.0 g/mol. (M of CH₂O = 30.0 g/mol.) What is its molecular formula? Type subscripts as plain numbers, like the 2 in CO2.
A compound has the empirical formula C₄H₉ and a molar mass of 114.0 g/mol. (M of C₄H₉ = 57.0 g/mol.) What is its molecular formula? Type subscripts as plain numbers, like the 2 in CO2.
End of Topic Test — five interchangeable forms, delivered separately.