Moles in chemistry explained: formula and examples

TutorAura guide · Updated 26 October 2026

A mole is a counting unit for particles: one mole contains $6.02 \times 10^{23}$ of them (the Avogadro constant). The useful part is that one mole of any substance has a mass in grams equal to its $M_r$ . That gives the key equation:

\text{moles} = \frac{\text{mass (g)}}{M_r}

Practise with the moles calculator.

Key takeaways

Moles = mass in grams ÷ Mr.
One mole contains 6.02 × 10²³ particles.
Balanced equations give the ratio of moles that react.
At room temperature and pressure, one mole of any gas takes up 24 dm³.

Basic calculations

Moles from mass: 36 g of water ( $M_r = 18$ ): $36 \div 18 = 2$ mol.
Mass from moles: 0.5 mol of NaCl ( $M_r = 58.5$ ): $0.5 \times 58.5 = 29.25$ g.
Number of particles: 2 mol of water contains $2 \times 6.02 \times 10^{23} = 1.204 \times 10^{24}$ molecules.

Reacting masses

What mass of calcium oxide is made by heating 50 g of calcium carbonate?

\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2

Moles of CaCO₃: $50 \div 100 = 0.5$ mol.
Ratio from the equation: 1 CaCO₃ : 1 CaO, so 0.5 mol of CaO.
Mass of CaO: $0.5 \times 56 = 28$ g.

The pattern is always: mass to moles, use the ratio, moles to mass.

Limiting reactants

When one reactant runs out first, it limits how much product forms; the other is in excess. To find the limiting reactant, work out the moles of each and compare with the ratio in the equation. For $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ , 0.4 mol of Mg needs 0.2 mol of O₂. If 0.3 mol of O₂ is available, magnesium is limiting and oxygen is in excess.

Concentration

\text{concentration (mol/dm}^3\text{)} = \frac{\text{moles}}{\text{volume (dm}^3\text{)}}

Remember $1 \text{ dm}^3 = 1{,}000 \text{ cm}^3$ . Dissolving 0.2 mol in 500 cm³ (0.5 dm³) gives $0.2 \div 0.5 = 0.4$ mol/dm³. Concentration can also be given in g/dm³: mass ÷ volume.

Gas volumes

At room temperature and pressure, one mole of any gas occupies 24 dm³ (24,000 cm³). So 0.25 mol of carbon dioxide takes up $0.25 \times 24 = 6$ dm³. Check whether your course and tier include this: it's in Cambridge IGCSE and AQA's separate chemistry Higher tier.

Common mistakes

Using kilograms instead of grams.
Ignoring the ratio in the balanced equation.
Forgetting to convert cm³ to dm³ for concentration.

Need $M_r$ first? See relative formula mass.

Empirical formulae

Some courses, including Cambridge IGCSE, ask you to find an empirical formula: the simplest whole-number ratio of atoms. Divide each element's mass by its $A_r$ , then divide by the smallest answer.

Example: a compound contains 2.4 g of magnesium and 1.6 g of oxygen. Mg: $2.4 \div 24 = 0.1$ ; O: $1.6 \div 16 = 0.1$ . The ratio is 1 : 1, so the empirical formula is MgO.

Practice questions

How many moles are in 8 g of oxygen gas, O₂?
What is the mass of 0.1 mol of calcium carbonate, CaCO₃?
In 2H₂ + O₂ → 2H₂O, how many moles of oxygen react with 2 mol of hydrogen?

Answers: $8 \div 32 = 0.25$ mol; 10 g; 1 mol.

Frequently asked questions

What is the volume of one mole of gas?

24 dm³ (24,000 cm³) at room temperature and pressure, for any gas.

Why do chemists use moles?

Atoms are far too small to count, so chemists count them in moles. Using moles lets you work out exactly how much of each reactant to use and how much product to expect.

How do you convert cm³ to dm³?

Divide by 1,000. For example, 250 cm³ is 0.25 dm³.

What is a limiting reactant?

The reactant that runs out first. It limits how much product can form; the other reactants are in excess.

What is a mole in chemistry?

An amount of substance containing 6.02 × 10²³ particles. One mole has a mass in grams equal to the substance's Mr.

How do you calculate moles from mass?

Divide the mass in grams by the relative formula mass (Mr).

What is the Avogadro constant?

6.02 × 10²³ per mole: the number of particles in one mole.

Sources

GCSE Chemistry (8462) specification: subject content. AQA.
Cambridge IGCSE Chemistry 0620 syllabus for 2026, 2027 and 2028. Cambridge International.

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