IGCSE Chemistry — Stoichiometry: Notes, Practice Questions & Worked Solutions

IGCSE Chemistry 0620 · Topic 3: Stoichiometry

Stoichiometry

This is the most calculation-heavy topic in the whole syllabus. Almost every calculation starts from the same idea: the mole, a “counting unit” for atoms and molecules, linked to mass by the relative formula mass.

1. Moles, Mass and Relative Formula Mass

moles (mol) = mass (g) / Mr

Mr (relative formula mass) is found by adding up the relative atomic masses (Ar) of every atom in the formula.

Worked example — Mr and moles Find the Mr of CaCO₃ (Ca=40, C=12, O=16), then find the number of moles in 5.6 g.
Mr = 40+12+(16×3) = 100.
moles = 5.6/100 = 0.056 mol

2. Gas Volumes and Concentration

1 mole of any gas occupies 24 dm³ at room temperature and pressure (rtp)

concentration (mol/dm³) = moles / volume (dm³)

Worked example — titration calculation 25.0 cm³ of 0.100 mol/dm³ NaOH exactly neutralises 20.0 cm³ of HCl. Find the concentration of the HCl (1:1 reacting ratio).
moles NaOH = 0.0250 × 0.100 = 0.00250 mol. Since the ratio is 1:1, moles HCl = 0.00250 mol.
concentration HCl = 0.00250 / 0.0200 = 0.125 mol/dm³

3. Empirical Formula, Yield and Purity

% yield = (actual yield / theoretical yield) × 100

% purity = (mass of pure substance / mass of sample) × 100

Worked example — empirical formula A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Find its empirical formula.
Divide each % by its Ar: C: 40.0/12=3.33, H: 6.7/1=6.7, O: 53.3/16=3.33.
Divide by the smallest (3.33): C:1, H:2, O:1 → empirical formula CH₂O

Worked example — percentage yield A reaction has a theoretical yield of 4.0 g, but only 3.4 g is actually obtained. Find the percentage yield.
% yield = (3.4/4.0) × 100 = 85%

Common mistakes to watch for

Forgetting to convert cm³ to dm³ (divide by 1000) before using the concentration formula.
Using the wrong mole ratio from the balanced equation — always check the equation’s coefficients before scaling between reactants and products.
Rounding empirical formula ratios too early — round to the nearest whole number only at the very end.

Key Vocabulary Recap

Mole — the amount of substance containing 6.02×10²³ particles (the Avogadro constant)

Empirical formula — the simplest whole-number ratio of atoms in a compound

Syllabus Reference

This note covers IGCSE Chemistry 0620 Topic 3, Stoichiometry: formulae, equations, relative masses, moles, the Avogadro constant, molar gas volume, concentration, titration calculations, empirical/molecular formulae, percentage yield, and percentage purity.

Practice questions

IGCSE Chemistry 0620 · Topic 3: Stoichiometry — Practice Question Bank (no answers)

Stoichiometry — Practice

Covers: Mr, moles, gas volume, concentration, titration, empirical formula, yield and purity. 10 questions. (Ar: H=1, C=12, O=16, Na=23, S=32, Ca=40)

Q1 Core

Calculate the relative formula mass (Mr) of CaCO₃.

Q2 Core

Calculate the number of moles in 5.6 g of CaCO₃ (Mr=100).

Q3 Core

Calculate the mass of 0.25 mol of NaOH (Mr=40).

Q4 Extended

Calculate the volume, in dm³, occupied by 0.1 mol of a gas at room temperature and pressure.

Q5 Extended

Calculate the concentration, in mol/dm³, of a solution containing 0.5 mol of solute in 250 cm³ of solution.

Q6 Extended

25.0 cm³ of 0.100 mol/dm³ NaOH exactly neutralises 20.0 cm³ of HCl (1:1 ratio). Calculate the concentration of the HCl.

Q7 Extended

A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Find its empirical formula.

Q8 Core

A reaction has a theoretical yield of 4.0 g, but only 3.4 g is obtained. Calculate the percentage yield.

Q9 Core

A 10.0 g sample of impure calcium carbonate is found to contain 9.5 g of pure calcium carbonate. Calculate the percentage purity.

Q10 Core

Balance the following equation: Fe₂O₃ + CO → Fe + CO₂.

Worked solutions

Show worked solutions

IGCSE Chemistry 0620 · Topic 3: Stoichiometry — Solutions

Stoichiometry — Practice: Solutions

Mr of CaCO₃.

Add the Ar values

40+12+(16×3)

✓Mr = 100

Moles in 5.6g CaCO₃.

moles = mass/Mr

5.6÷100

✓0.056 mol

Mass of 0.25mol NaOH.

mass = moles×Mr

0.25×40

✓10 g

Volume of 0.1mol gas at rtp.

volume = moles × 24 dm³

0.1×24

✓2.4 dm³

Concentration: 0.5mol in 250cm³.

Convert volume to dm³, then divide

250cm³=0.25dm³; 0.5÷0.25

✓2 mol/dm³

Titration: find HCl concentration.

moles NaOH = 0.0250×0.100

0.00250 mol (= moles HCl, since ratio is 1:1)

concentration = moles/volume(dm³)

0.00250÷0.0200

✓0.125 mol/dm³

Empirical formula from 40%C, 6.7%H, 53.3%O.

Divide each % by its Ar

C: 40/12=3.33; H: 6.7/1=6.7; O: 53.3/16=3.33

Divide by the smallest value (3.33)

C:1, H:2, O:1

✓CH₂O

% yield: actual3.4, theoretical4.0.

% yield = (actual/theoretical)×100

(3.4/4.0)×100

✓85%

% purity: 9.5g pure in 10.0g sample.

% purity = (pure mass/sample mass)×100

(9.5/10.0)×100

✓95%

Q10

Balance Fe₂O₃+CO→Fe+CO₂.

Balance Fe first (2 on left), then balance O and C using CO/CO₂

Fe₂O₃+3CO→2Fe+3CO₂

✓Fe₂O₃ + 3CO → 2Fe + 3CO₂ (check: Fe 2=2, O 3+3=6=3×2, C 3=3)

Book a free trial class

Tell us a little about your requirement.