The mole explained step by step for IGCSE Chemistry
The mole trips up many students: it means counting things you cannot see with an enormous number. This guide goes step by step, from what a mole is and how to find Mr to the formulas and reacting-mass calculations, with interactives to drag the numbers yourself.
The short answer
The mole (mol) is the unit of amount of substance. One mole contains 6.02 × 10²³ particles; this number is the Avogadro constant.
The key formula is amount (mol) = mass (g) ÷ molar mass (g/mol), and one mole of any gas occupies 24 dm³ at room temperature and pressure (r.t.p.).
In Cambridge 0620 (2026–2028), mole calculations are Supplement content, so they are examined on the Extended papers (Paper 2 and Paper 4). Core candidates calculate reacting masses in simple proportions, without the mole.
Why count in moles
We buy eggs by the dozen (12) and paper by the ream (500) because counting one by one is impractical. Atoms are far smaller: a single drop of water holds around 10²¹ molecules. So chemists use a very large counting unit: 1 mole = 6.02 × 10²³ particles.
We cannot count atoms, but we can weigh things. The mole is the bridge between the mass we weigh and the number of particles that actually react.
Always find Mr first
The relative atomic mass (Ar) of each element is read from the Periodic Table printed with the exam: H = 1, C = 12, O = 16, Na = 23, Cl = 35.5.
The relative molecular mass (Mr) is the sum of the Ar of every atom in the formula: H₂O = 2 × 1 + 16 = 18. For ionic compounds it is called the relative formula mass, still written Mr: NaCl = 23 + 35.5 = 58.5.
The molar mass has the same value as Mr in g/mol, so one mole of water weighs 18 g.
The formulas, to drag yourself
amount (mol) = mass (g) ÷ molar mass (g/mol)
number of particles = amount (mol) × 6.02 × 10²³
gas volume (dm³) = amount (mol) × 24, only for gases at r.t.p. Given cm³? Divide by 1000 first (1 dm³ = 1000 cm³).
concentration (mol/dm³) = amount (mol) ÷ volume of solution (dm³)
Pick a substance and drag the mass below to see the moles, the particles and the gas volume change together.
H₂O = water · Relative molecular mass (Mr) = 2 × H (1) + 1 × O (16) = 18
In the jar, 0.1 mol = 1 molecule · now 1 mol = 10 molecules
Mass
18 g
Amount
1 mol
Number of particles
6.02 × 10²³
Not a gas at r.t.p., so 24 dm³ does not apply
Reacting masses in three steps
Example: 16 g of methane (CH₄) burns as CH₄ + 2O₂ → CO₂ + 2H₂O. What mass of water forms?
Step 1, moles of what you know: 16 g ÷ 16 g/mol = 1 mol. Step 2, the ratio from the equation: CH₄ : H₂O = 1 : 2, so 2 mol of water. Step 3, back to mass: 2 mol × 18 g/mol = 36 g.
Change the mass and the equation below and reveal one step at a time.
Step 1 · Find the moles of CH₄
n = 16 g ÷ 16 g/mol = 1 mol
Common mistakes
Mixing up O and O₂. Oxygen gas is diatomic: Mr of O₂ is 32, not 16.
Forgetting the ratio from the equation and carrying the moles of the reactant straight to the product. Check the numbers in front of the formulae in the balanced equation every time.
Using 24 dm³ for a solid, a liquid or a solution. It only applies to gases at r.t.p.
Forgetting to convert cm³ to dm³ before working out a concentration or a gas volume.
Rounding in the first step. Keep the full value and round only the final answer.
Keep going on Boom-Tutor
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