This mini-lesson walks you through the whole of Cambridge IGCSE Topic 3 — Stoichiometry: writing and balancing equations, relative masses Ar and Mr, the mole, and every calculation — reacting masses, concentration, gas volumes, % yield & purity, empirical formulae and titrations.
Work through each screen, answer the questions as you go (lots are calculations — get your Ar values right!) and collect ⭐ stars. Press Start when you're ready.
A chemical formula shows the type and number of atoms in a substance. The little subscript number multiplies only the symbol it follows:
The molecular formula is the number and type of different atoms in one molecule. You can deduce a simple compound's formula from a diagram of its atoms.
Watch out: a big number in front (a coefficient) multiplies the whole formula. 2 H₂O = 4 H + 2 O, but it does not change what one water molecule is.
In a reaction, atoms are only rearranged — none are made or lost. So a balanced equation must have the same number of each type of atom on both sides. You balance by changing the big numbers in front (coefficients), never the subscripts.
State symbols show the physical state: (s) solid, (l) liquid, (g) gas, (aq) aqueous (dissolved in water).
CH₄ + ?O₂ → CO₂ + ?H₂O
Balance H (4 on left → 2H₂O), then O (right now has 2+2 = 4 O → 2O₂):
CH₄ + 2O₂ → CO₂ + 2H₂O ✓
Extended only. An ionic equation shows only the ions and species that actually take part — the spectator ions (those unchanged on both sides) are cancelled out.
Full equation for a neutralisation:
Na⁺ and Cl⁻ are spectators, so the ionic equation is simply:
Tip: for a precipitation such as silver chloride forming, the ionic equation is Ag⁺(aq) + Cl⁻(aq) → AgCl(s). Charges must balance too, not just atoms.
Relative atomic mass, Ar, is the average mass of an element's isotopes compared to 1⁄12 of the mass of a carbon-12 atom.
Relative molecular mass (or relative formula mass), Mr, is just the sum of all the Ar values in the formula. For ionic compounds we say relative formula mass.
(2 × H) + S + (4 × O) = (2×1) + 32 + (4×16)
= 2 + 32 + 64 = 98
Common error: Ar and Mr are ratios, so they have no units. Don't write "98 g" for an Mr — it is just 98.
Extended only. The mole (mol) is the unit of amount of substance. One mole always contains the same number of particles — the Avogadro constant:
The link between mass and moles uses the mole triangle. The molar mass (in g/mol) is numerically the same as the Mr:
How many moles are in 80 g of NaOH? (Mr = 23 + 16 + 1 = 40)
moles = 80 ÷ 40 = 2 mol (= 2 × 6.02 × 10²³ = 1.204 × 10²⁴ formula units)
Extended only. The big numbers in a balanced equation give the mole ratio in which substances react. To find a reacting mass:
Mg + 2HCl → MgCl₂ + H₂. What mass of H₂ forms from 12 g of Mg? (Ar: Mg = 24, H = 1)
moles Mg = 12 ÷ 24 = 0.5 mol → ratio Mg : H₂ is 1 : 1 → 0.5 mol H₂
mass H₂ = 0.5 × 2 = 1 g
Concentration measures how much solute is dissolved in a given volume. It can be given in g/dm³ (Core) or mol/dm³ (the calculation is Supplement). Volumes are in dm³, so remember 1 dm³ = 1000 cm³.
0.5 mol of NaCl is dissolved to make 250 cm³ of solution. Find the concentration in mol/dm³.
250 cm³ = 0.25 dm³ → conc = 0.5 ÷ 0.25 = 2 mol/dm³
Don't mix them up: g/dm³ uses mass; mol/dm³ uses moles. Convert between them with g/dm³ = mol/dm³ × Mr.
Extended only. At room temperature and pressure (r.t.p.), one mole of any gas occupies the same volume:
What volume does 0.5 mol of CO₂ occupy at r.t.p.?
volume = 0.5 × 24 = 12 dm³
Key idea: it does not matter which gas — 24 dm³ per mole works for H₂, CO₂, NH₃ alike, because equal volumes of gases contain equal numbers of molecules.
Extended only. Reactions rarely give 100% of the expected product. Two different ideas:
A reaction could make 8 g of product but only 6 g is collected.
% yield = (6 ÷ 8) × 100 = 75%
Don't confuse them: % yield compares product made to the theoretical maximum; % purity compares the pure part of a sample to the whole sample. Different questions, different formulae.
Extended only. The empirical formula is the simplest whole-number ratio of atoms in a compound. The molecular formula is the real number in one molecule (a whole-number multiple of the empirical one).
A compound is 4.8 g C and 1.2 g H. (Ar: C = 12, H = 1)
C: 4.8 ÷ 12 = 0.4 | H: 1.2 ÷ 1 = 1.2
Divide by smallest (0.4): C = 1, H = 3 → empirical formula CH₃
Remember: the empirical formula is the simplest ratio, so H₂O₂'s empirical formula is HO and C₆H₁₂O₆'s is CH₂O — they are not the same as the molecular formula.
Extended only. A titration finds an unknown concentration by reacting it with a solution of known concentration. The route:
25.0 cm³ of NaOH is neutralised by 20.0 cm³ of 0.10 mol/dm³ HCl. NaOH + HCl → NaCl + H₂O (ratio 1:1). Find the NaOH concentration.
moles HCl = 0.10 × (20.0 ÷ 1000) = 0.0020 mol → moles NaOH = 0.0020 mol
conc NaOH = 0.0020 ÷ (25.0 ÷ 1000) = 0.0020 ÷ 0.025 = 0.08 mol/dm³
Watch the units: always convert cm³ to dm³ (÷ 1000) before dividing, or your concentration will be 1000× wrong.
Tap the correct ending for each stoichiometry skill.
Relative formula mass: Mr = sum of all Ar values (no units)
Moles: amount (mol) = mass (g) ÷ molar mass [S]
Particles: number = moles × 6.02 × 10²³ [S]
Concentration: conc = amount ÷ volume (dm³), in g/dm³ or mol/dm³
Gas volume: volume = moles × 24 dm³ at r.t.p. [S]
% yield: (actual ÷ theoretical) × 100 [S]
% purity: (pure mass ÷ sample mass) × 100 [S]
Empirical formula: simplest whole-number atom ratio [S]
You've covered all of Cambridge IGCSE Topic 3 — formulae & equations, relative masses, the mole, and every calculation. [S] marks Supplement (Extended-only) content. Press Finish to see your score.
You've worked through Stoichiometry for Cambridge IGCSE Chemistry. 🎉
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