How to Calculate Moles: Every Formula, With Worked Examples
There is no single mole formula. There are four, and which one you use depends entirely on what the question gives you.
| If you are given | Use | Where |
|---|---|---|
| A mass in grams | n = m / Mr | Solids, and anything weighed |
| A volume of gas | n = V / Vm | Gases. Vm is 22.7 or 24, see below |
| A concentration and volume | n = c × V | Solutions and titrations |
| A number of particles | n = N / NA | Atom and molecule counting |
Students rarely get the arithmetic wrong here. What goes wrong is picking the wrong formula, or using the wrong number for Vm — and that last one is board-dependent in a way almost no revision page warns you about. It is covered properly further down.
The mole is the hinge of the whole of quantitative chemistry. Every reacting-mass question, every titration, every percentage yield and every empirical formula goes through it. Get fluent here and a large part of the subject stops being difficult.
What a mole actually is
A mole is a counting word, like “dozen” or “pair”. A dozen is 12 of something. A mole is 6.02 × 1023 of something.
One mole = 6.02 × 1023 particles. That number is called the Avogadro constant.
Why such an absurd number? Because atoms are absurdly small. Chemists needed a counting unit that turns the invisible into something you can put on a balance, and 6.02 × 1023 is the number that does it. Count out that many carbon-12 atoms and they weigh exactly 12 grams. Count out that many water molecules and they weigh 18 grams — about a tablespoon and a half.
That is the whole trick, and it is worth saying plainly:
The relative mass
Water has Mr = 18.
A bare number. No units.
The mass of one mole
One mole of water weighs 18 g.
Same number, now in grams.
So you never have to look up “the mass of a mole” of anything. It is simply the relative formula mass with “grams” written after it. If you are shaky on working out Mr, that is the step before this one and it is covered in how to calculate relative atomic mass.
From a mass: n = m / Mr
The one you will use most. Divide the mass in grams by the relative formula mass.
n is moles (mol), m is mass in grams, and Mr is the relative formula mass, which has no units.
How many moles are there in 36 g of water, H2O?
Mᵣ of H₂O = (2 × 1.0) + 16.0 = 18.0
n = 36 ÷ 18.0
n = 2.0 mol
What is the mass of 0.25 mol of calcium carbonate, CaCO3?
Rearrange: m = n × Mᵣ
Mᵣ of CaCO₃ = 40.0 + 12.0 + (3 × 16.0) = 100.0
m = 0.25 × 100.0
m = 25 g
A word about the mole triangle
You may have been taught a triangle with m on top and n and Mr underneath. It works, and it is a fine crutch while you are learning. But it only works for this one formula, and there are four. Students who rely on triangles come unstuck the moment a question involves gas volume or concentration, because there is no triangle for those in their head.
Learn to rearrange instead. It is the same skill every time, it works for all four formulas, and it is the skill the exam is actually testing.
Mr is grams per mole, so m must be in grams. A mass given in kilograms must be multiplied by 1000 first, and a mass in milligrams divided by 1000. This is the single most common silent error in mole calculations, because the answer still looks like a normal number.
From a gas volume, and the 22.4 / 22.7 / 24 problem
Gases are easier than solids, because one mole of any gas takes up the same volume at the same temperature and pressure. It does not matter whether it is hydrogen or carbon dioxide.
V is the gas volume in dm3 and Vm is the molar volume. And here is the problem nobody warns you about: Vm is not one number. It depends on the temperature and pressure your exam board defines, and the three common values are all different.
| Conditions | Means | Vm | Used by |
|---|---|---|---|
| STP | 0 °C (273 K) and 100 kPa | 22.7 dm3 mol−1 | The IB |
| Old STP | 0 °C (273 K) and 1 atm | 22.4 dm3 mol−1 | Older textbooks, many US courses |
| RTP | About 20–25 °C and 1 atm | 24 dm3 mol−1 | UK GCSE and most A-level |
Using 24 when the question wanted 22.7 does not give a slightly wrong answer — it gives an answer about 6% out, which is comfortably outside any mark scheme. The three values are not interchangeable and they are not rounding differences. They describe genuinely different conditions.
Find out which one your course uses, once, and then always use it. If the question states the conditions, follow the question. If your data booklet prints a value, that is the one.
How many moles are there in 480 cm3 of oxygen gas at room temperature and pressure? (Vm = 24 dm3 mol−1)
First convert cm³ to dm³. Divide by 1000
480 ÷ 1000 = 0.480 dm³
Now divide by the molar volume
n = 0.480 ÷ 24
n = 0.020 mol
Skipping the cm3 conversion gives 20 mol, which is a thousand times too big — about half a kilogram of oxygen in a container the size of a drinks can. Sense-checking the size of your answer catches this instantly.
What volume does 0.50 mol of carbon dioxide occupy at STP, as the IB defines it?
IB STP means Vᵐ = 22.7 dm³ mol⁻¹
Rearrange: V = n × Vᵐ
V = 0.50 × 22.7
V = 11.35 dm³, so 11 dm³ to 2 s.f.
Notice that the identity of the gas never entered the calculation. Half a mole of any gas occupies the same volume under the same conditions.
From a solution: n = c × V
For anything dissolved in water, moles come from the concentration multiplied by the volume. This is the formula every titration question is built on.
c is concentration in mol dm−3 and V is volume in dm3. Volumes in titrations are almost always given in cm3, so dividing by 1000 is nearly always the first thing you do.
How many moles of HCl are there in 25.0 cm3 of 0.100 mol dm−3 hydrochloric acid?
Convert the volume to dm³
25.0 ÷ 1000 = 0.0250 dm³
n = 0.100 × 0.0250
n = 2.50 × 10⁻³ mol
4.0 g of sodium hydroxide is dissolved to make 250 cm3 of solution. What is the concentration? (Mr NaOH = 40.0)
Two steps. First the moles, from the mass
n = 4.0 ÷ 40.0 = 0.10 mol
Then the concentration. Rearrange: c = n ÷ V
V = 250 ÷ 1000 = 0.250 dm³
c = 0.10 ÷ 0.250
c = 0.40 mol dm⁻³
This is the shape of most solution questions: get to moles first, then go wherever the question wants. Moles is always the middle of the journey.
You will also see concentration written in g dm−3. To convert, multiply or divide by Mr: g dm−3 = mol dm−3 × Mr. The NaOH solution above is 0.40 × 40.0 = 16 g dm−3.
From a number of particles
This is the mole in its original sense: a count.
N is the number of particles and NA is the Avogadro constant, 6.02 × 1023 mol−1.
How many water molecules are there in 2.0 mol of water?
Rearrange: N = n × Nₐ
N = 2.0 × 6.02 × 10²³
N = 1.2 × 10²⁴ molecules
Read the question carefully. 2.0 mol of water contains 1.2 × 1024 molecules, but each molecule has three atoms in it, so it contains 3.6 × 1024 atoms. Questions ask for both, and the difference is a factor of three.
Using the mole ratio in a reaction
On its own, a number of moles is not very interesting. What makes the mole powerful is that a balanced equation is a recipe in moles, so once you know the moles of one substance you know the moles of everything else.
The big numbers in a balanced equation are mole ratios. Nothing else.
Every reacting-mass question follows the same three steps, and they are worth learning as a shape rather than as individual questions.
- Go from what you were given to moles, using whichever of the four formulas fits.
- Use the balanced equation to change species, multiplying or dividing by the mole ratio.
- Go from moles back to what the question wants — a mass, a volume, a concentration.
What mass of magnesium oxide is made when 4.8 g of magnesium burns completely in oxygen?
2Mg + O2 → 2MgO (Mr: Mg 24.0, MgO 40.0)
Step 1 — mass to moles
n(Mg) = 4.8 ÷ 24.0 = 0.20 mol
Step 2 — the ratio Mg : MgO is 2 : 2, which is 1 : 1
n(MgO) = 0.20 mol
Step 3 — moles back to mass
m = 0.20 × 40.0
m = 8.0 g
Notice you cannot shortcut this by comparing 4.8 g directly with the answer. Magnesium oxide is heavier than the magnesium it came from, because oxygen joined on. Masses do not follow the equation’s ratios — only moles do.
When a question gives you two starting amounts, one of them runs out first and sets the limit. Finding that limiting reactant, and turning the answer into a percentage yield, is covered in the percentage yield formula.
Practice questions
Eight questions, one for each route and then some that combine them. The answers are hidden — do each one on paper first, then tap to check.
Use these where you need them: H = 1.0, C = 12.0, O = 16.0, Na = 23.0, Cl = 35.5, Ca = 40.0, NA = 6.02 × 1023.
1. How many moles are there in 88 g of carbon dioxide, CO2?
Mᵣ of CO₂ = 12.0 + (2 × 16.0) = 44.0
n = 88 ÷ 44.0
n = 2.0 mol
2. What is the mass of 0.40 mol of sodium chloride, NaCl?
Mᵣ of NaCl = 23.0 + 35.5 = 58.5
m = 0.40 × 58.5
m = 23.4 g
3. How many moles are there in 1.2 dm3 of nitrogen gas at RTP? (Vm = 24)
Already in dm³, so no conversion needed
n = 1.2 ÷ 24
n = 0.050 mol
4. How many moles are in 50.0 cm3 of 0.200 mol dm−3 sodium hydroxide?
Convert cm³ to dm³ first
50.0 ÷ 1000 = 0.0500 dm³
n = 0.200 × 0.0500
n = 0.0100 mol
5. How many moles is 3.01 × 1023 particles?
n = 3.01 × 10²³ ÷ 6.02 × 10²³
n = 0.500 mol
Exactly half the Avogadro constant, so exactly half a mole. Spotting that saves you using the calculator at all.
6. 2.0 g of sodium hydroxide is dissolved to make 500 cm3 of solution. What is the concentration in mol dm−3? (Mr NaOH = 40.0)
Mass to moles first
n = 2.0 ÷ 40.0 = 0.050 mol
Then concentration. V = 500 ÷ 1000 = 0.500 dm³
c = 0.050 ÷ 0.500
c = 0.10 mol dm⁻³
7. 25.0 g of calcium carbonate is heated and decomposes completely: CaCO3 → CaO + CO2. What volume of carbon dioxide is made at RTP? (Vm = 24)
Step 1 — mass to moles. Mᵣ CaCO₃ = 40.0 + 12.0 + 48.0 = 100.0
n(CaCO₃) = 25.0 ÷ 100.0 = 0.250 mol
Step 2 — the ratio is 1 : 1
n(CO₂) = 0.250 mol
Step 3 — moles to gas volume
V = 0.250 × 24
V = 6.0 dm³
Three steps, three different formulas, moles in the middle. This is the shape of nearly every harder mole question.
8. What volume does 0.25 mol of a gas occupy at STP as the IB defines it?
IB STP: Vᵐ = 22.7 dm³ mol⁻¹, NOT 24
V = 0.25 × 22.7
V = 5.7 dm³ to 2 s.f.
Using 24 here gives 6.0 dm3, which is about 6% out and would be marked wrong. Check which value your course uses.
Where the marks get lost
- Not converting cm3 to dm3. Divide by 1000. This is the most frequent error in the whole topic, and it makes the answer a thousand times too big.
- Using kilograms instead of grams. Mr is grams per mole, so the mass must be in grams.
- Using the wrong molar volume. 22.7 for the IB, 24 for most UK A-level and GCSE, 22.4 in older books. They are not interchangeable.
- Comparing masses instead of moles. Only moles follow the ratios in a balanced equation. Masses never do.
- Forgetting the mole ratio entirely when it is not 1 : 1. Read the big numbers in the equation before you move between substances.
- Using an unbalanced equation. The ratio is then wrong and everything after it fails.
- Confusing molecules with atoms when counting particles.
- Rounding part-way through. Keep the full value in the calculator and round once, at the end.
Does the size of the answer make sense? A few grams of something is usually a fraction of a mole, not hundreds. A test tube of gas is milli-moles, not kilo-moles. Answers like 20 mol or 0.000003 mol in a school question almost always mean a factor of 1000 went the wrong way, which means a cm3 or a kilogram.
The whole thing on one card
Photograph this before the paper.
Moles — quick reference
- One mole = 6.02 × 1023 particles. It is a counting word, like “dozen”.
- The mass of one mole is Mr in grams. Nothing to look up.
- Vm is 22.7 for the IB and 24 for most UK A-level and GCSE. Check yours once, then stick to it.
- cm3 → dm3: divide by 1000. The commonest error in the topic.
- Mass must be in grams, never kilograms.
- The big numbers in an equation are mole ratios. Masses never follow them.
- Every reacting question is three steps: to moles, across with the ratio, back out again.
- Molecules are not atoms. One mole of H2O has three moles of atoms in it.
- Sense check: a few grams is a fraction of a mole, not hundreds.
Moles sits between two topics on this site. Before it comes relative atomic mass, which is where Mr comes from. After it comes percentage yield, which is what you do with the answer. Our IB Chemistry revision notes carry the whole sequence at SL and HL.
Frequently asked questions
How do you calculate moles?
What is a mole in chemistry?
What is the formula for moles from mass?
Is the molar volume 22.4, 22.7 or 24?
How do you convert cm³ to dm³?
What is the Avogadro constant?
How do you find moles from concentration?
Why can’t I just compare the masses in a reaction?
Should I use the mole triangle?
How many atoms are in a mole of water?
What units are moles measured in?
What is the difference between molar mass and Mr?
Moles is the topic worth getting fluent at
Almost everything else in quantitative chemistry runs through it, so time spent here pays back across the whole paper. Our tutors teach IB and A-level chemistry and will work through your own past papers rather than a generic worksheet.
Book a free 30-minute meetingRevising the sequence? Start with relative atomic mass, then this page, then percentage yield. Writing a chemistry investigation? See our guide to IB internal assessments, or read about IB Diploma tutoring.