Cambridge IGCSE Chemistry Notes 3: Stoichiometry

Topic 3 of Cambridge IGCSE Chemistry 0620 and 0971 turns chemical symbols into quantitative predictions. Official sections 3.1 to 3.3 move from formulae and balanced equations to relative mass, moles, particles, gases, solutions, titration and percentage calculations. The central discipline is to convert the given quantity to moles before applying an equation ratio.

Formulae name particles and proportions

A molecular formula gives the number and type of each atom in one molecule. For example, one molecule of ethene, C₂H₄, contains two carbon atoms and four hydrogen atoms.

An empirical formula gives the simplest whole-number ratio of the different atoms or ions in a compound. The molecular formula C₂H₄ therefore has empirical formula CH₂.

Ionic compounds form giant lattices rather than molecules. Their formula gives the simplest charge-balanced ratio of ions. Magnesium ions are Mg²⁺ and chloride ions are Cl⁻, so magnesium chloride is MgCl₂. Total positive and negative charge must cancel.

A model can also reveal a formula. Count the represented particles, reduce the ratio if the question asks for an empirical formula, and distinguish a finite molecule from a repeating lattice.

Write chemical equations

A word equation names reactants and products. A symbol equation replaces names with correct formulae. Balance a symbol equation by changing coefficients only. Never alter a formula to make atom counts match because that changes the substance.

For magnesium burning:

magnesium + oxygen → magnesium oxide

2Mg(s) + O₂(g) → 2MgO(s)

State symbols are (s), (l), (g) and (aq). They describe physical state or aqueous solution under the reaction conditions.

An ionic equation removes spectator ions that remain unchanged. For precipitation of silver chloride:

Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

Check both atom count and total charge. A valid ionic equation balances each.

Relative atomic, molecular and formula mass

Relative atomic mass, Ar, is the average mass of the isotopes of an element compared with one twelfth of the mass of a carbon-12 atom.

Relative molecular mass, Mr, is the sum of the relative atomic masses in a molecule. The same symbol is used for relative formula mass of an ionic compound.

For CaCO₃, using Ca = 40, C = 12 and O = 16:

Mr = 40 + 12 + (3 × 16) = 100

Ar and Mr are relative values without units. Molar mass is the mass of one mole and is written in g/mol. It has the same numerical value as the relevant Ar or Mr.

Core reacting-mass questions may use simple proportions without the mole concept. If 12 g of carbon reacts completely with 32 g of oxygen to form 44 g of carbon dioxide, 6 g of carbon forms 22 g of carbon dioxide because every mass is halved.

The mole and Avogadro constant

The mole, mol, is the unit of amount of substance. One mole contains 6.02 × 10²³ specified particles. This is the Avogadro constant.

Use:

amount (mol) = mass (g) / molar mass (g/mol)

Rearrangements give mass = amount × molar mass and molar mass = mass / amount.

Particle calculations use:

number of particles = amount × 6.02 × 10²³

Name the particle. One mole of O₂ contains one mole of oxygen molecules but two moles of oxygen atoms.

The equation ratio

Coefficients in a balanced equation give mole ratios. In:

N₂(g) + 3H₂(g) → 2NH₃(g)

one mole of nitrogen reacts with three moles of hydrogen to form two moles of ammonia. The coefficients do not give a direct gram ratio because the substances have different molar masses.

Use a consistent route:

1. Write and balance the equation.
2. Convert the given mass, gas volume, solution amount or particle count to moles.
3. Apply the coefficient ratio.
4. Convert the required moles to the requested quantity.
5. Attach a correct unit and check plausibility.

Limiting reactants

The limiting reactant is consumed first and fixes the maximum product amount. Convert every supplied reactant to moles, then compare available moles / coefficient. The smallest value identifies the limiting reactant.

Do not simply choose the smaller mass. A lower mass can represent more moles when its molar mass is much lower.

Any reactant left after the limiting reactant is consumed is in excess. Product calculation must begin from the limiting amount.

Gases at room temperature and pressure

At room temperature and pressure, r.t.p., one mole of any gas occupies 24 dm³.

Use:

amount of gas (mol) = gas volume (dm³) / 24 dm³/mol

Since 1 dm³ = 1000 cm³, one mole is also 24,000 cm³ at r.t.p. Keep volume and molar-volume units consistent before dividing.

For gaseous reactants and products at the same conditions, equation coefficients also give volume ratios. This shortcut applies only where the relevant substances are gases.

Concentration of solutions

Concentration describes solute amount per solution volume. Cambridge uses both g/dm³ and mol/dm³.

concentration (g/dm³) = solute mass (g) / solution volume (dm³)

concentration (mol/dm³) = amount (mol) / solution volume (dm³)

Convert cm³ to dm³ by dividing by 1000. To move from mol/dm³ to g/dm³, multiply by molar mass. To move from g/dm³ to mol/dm³, divide by molar mass.

The volume is the final solution volume, not automatically the solvent volume before dissolving.

Use titration data

A titration supplies an accurately measured reacting volume. The practical note owns rinsing, endpoint judgement and concordant titres; this theory note owns the stoichiometric calculation.

1. Write the balanced equation.
2. Convert the known solution volume to dm³.
3. Calculate its moles with n = cV.
4. Apply the equation ratio.
5. Divide unknown moles by unknown volume in dm³ to find concentration, or rearrange for the requested quantity.

Use the mean concordant titre specified by the data. Do not include a rough titre unless it falls within the accepted set stated in the question.

Empirical formula from composition

For masses or percentages of elements:

1. Treat percentages as masses in a 100 g sample when convenient.
2. Divide each mass by the element's Ar to obtain mole amounts.
3. Divide every amount by the smallest.
4. Multiply all ratios by the same small integer if needed to reach whole numbers.

A ratio of 1 : 1.5 becomes 2 : 3. Do not round 1.5 to 2.

To obtain a molecular formula, calculate empirical-formula mass, then use:

multiplier = molecular Mr / empirical-formula mass

Multiply every empirical-formula subscript by the whole-number multiplier.

Percentage composition, yield and purity

Percentage composition by mass is:

mass contribution of named part / Mr of compound × 100%

Percentage yield compares obtained product with the theoretical maximum:

percentage yield = actual yield / theoretical yield × 100%

Theoretical yield comes from the balanced equation and limiting reactant. Actual yield is the measured product.

Percentage purity compares desired pure substance with the total impure sample:

percentage purity = mass of pure substance / mass of impure sample × 100%

In a reaction question, use purity to find the reactive mass before converting to moles. A 10.0 g sample at 80.0% purity contains 8.00 g of the desired substance.

Worked application: link purity, limiting reagent and yield

Calcium carbonate reacts as CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. A 5.00 g limestone sample is 80.0% CaCO₃, so reactive mass is 5.00 × 0.800 = 4.00 g. With Mr(CaCO₃) = 100, this is 0.0400 mol and can form 0.0400 mol CO₂ if acid is in excess. At r.t.p. the theoretical gas volume is 0.0400 × 24 = 0.960 dm³. If 0.816 dm³ is collected, percentage yield is 0.816 / 0.960 × 100 = 85.0%. Purity changes the starting reactive amount; yield compares the observed result with the maximum from that amount.

Common misconceptions and corrections

• Changing subscripts to balance an equation. Change coefficients only.
• Leaving an equation balanced for atoms but not charge. Ionic equations require both.
• Treating spectator ions as products. Remove unchanged aqueous ions.
• Calling an empirical formula the atom count in one molecule. It is the simplest ratio.
• Using molecular terminology for an ionic lattice. Use relative formula mass.
• Giving Ar or Mr units. These are relative values.
• Giving molar mass no unit. Use g/mol.
• Using equation coefficients as gram ratios. They are mole ratios.
• Choosing the smallest reactant mass as limiting. Compare moles per coefficient.
• Multiplying mass by molar mass to find moles. Divide mass by molar mass.
• Forgetting that O₂ particles are molecules. State the requested particle type.
• Using 24 dm³ for a non-gaseous substance. Molar gas volume applies to gases at r.t.p.
• Dividing a cm³ volume directly by 24 dm³/mol. Convert units first.
• Saying 1 dm³ equals 100 cm³. It equals 1000 cm³.
• Using solvent volume as solution volume. Concentration uses total solution volume.
• Mixing g/dm³ and mol/dm³. Use molar mass to convert.
• Applying cV while V remains in cm³. Convert it to dm³.
• Averaging every titre including the rough value. Use the accepted concordant set.
• Applying a titration ratio before balancing. The equation controls the ratio.
• Rounding empirical ratios too early. Preserve values such as 1.5 and scale all ratios.
• Multiplying only one empirical subscript. Apply the molecular multiplier to all.
• Putting theoretical yield over actual yield. Actual divided by theoretical gives percentage yield.
• Using total impure mass as reactive mass. Apply percentage purity first.
• Assuming yield and purity mean the same thing. They answer different questions.
• Rounding intermediate moles heavily.

Assessment guidance

Show a transparent conversion route rather than a single calculator result. Begin with a balanced equation, label mole amounts under the relevant substances and keep units beside every conversion. For mixed-reactant questions, prove which reactant limits by comparing moles relative to coefficients. For gases, state the r.t.p. assumption and match cm³ or dm³ consistently. In titration questions, show the selected mean titre and volume conversion. For empirical formulae, display mass, Ar, moles and simplest ratio as separate rows. Finish by checking significant figures, chemical plausibility and whether a percentage lies within an expected 0% to 100% range unless the data deliberately expose an experimental problem.

Retrieval practice

Balance five equations with state symbols and reduce two to ionic equations. Calculate Mr for five substances. Practise the same balanced reaction from mass, particles, gas volume and solution concentration. Complete one limiting-reactant problem, one titration chain, two empirical-formula calculations and one molecular-formula calculation. Then calculate composition, yield and purity while explaining in one sentence what each percentage compares.

Theory and practical ownership

This theory note owns formulae, equations and stoichiometric calculation. The Chemistry practical hub owns apparatus, solution preparation, titration technique, gas collection, measurement uncertainty, hazards and method evaluation.

• Cambridge IGCSE Chemistry theory hub
• Cambridge IGCSE Chemistry practical hub

Official source

Cambridge International, Chemistry 0620 syllabus for examinations in 2026, 2027 and 2028 and Chemistry 0971 syllabus for examinations in 2026, 2027 and 2028, Topic 3 sections 3.1 Formulae, 3.2 Relative masses of atoms and molecules and 3.3 The mole and the Avogadro constant.