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Cambridge IGCSE Chemistry (0620) · Topic 3 — Stoichiometry
Mini-Lesson

Stoichiometry

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.

Make a guess: how do you fix the missing oxygen atom?

  1. Mission 1Why can't you change water's formula?
  2. Mission 2How do you count atoms by weighing?
  3. Mission 3Why does a mole of any gas fill the same space?

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.

Formulae · 3.1

Formulae of elements & compounds

A chemical formula shows the type and number of atoms in a substance. The little subscript number multiplies only the symbol it follows:

  • H₂O — 2 hydrogen atoms + 1 oxygen atom.
  • CaCO₃ — 1 Ca, 1 C, 3 O.
  • Ca(OH)₂ — the bracket means 2 of everything inside: 1 Ca, 2 O, 2 H.

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.

Quick check

Counting atoms

?How many oxygen atoms are there in total in one formula unit of magnesium nitrate, Mg(NO₃)₂?
Equations · 3.1

Writing & balancing equations

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.

2H₂ + O₂ → 2H₂O4 H and 2 O on each side — balanced

State symbols show the physical state: (s) solid, (l) liquid, (g) gas, (aq) aqueous (dissolved in water).

reactants products equal mass
Mass is conserved: the pans balance because no atoms are created or destroyed.
Worked example — balance the burning of methane

CH₄ + ?O₂ → CO₂ + ?H₂O

Balance H (4 on left → 2H₂O), then O (right now has 2+2 = 4 O → 2O₂):

CH₄ + 2O₂ → CO₂ + 2H₂O ✓

Calculate

Your turn — balance it

1Balance the formation of ammonia: N₂ + ___ H₂ → 2NH₃. What is the missing coefficient in front of H₂?
× H₂
Hint: 2NH₃ contains 6 H atoms. How many H₂ molecules give 6 H?
Supplement · 3.1

Ionic equations

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:

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Na⁺ and Cl⁻ are spectators, so the ionic equation is simply:

H⁺(aq) + OH⁻(aq) → H₂O(l)this single reaction is what all acid–alkali neutralisations share

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.

Quick check · Supplement

Spot the spectators

?When silver nitrate solution is added to sodium chloride solution, a white precipitate of AgCl forms. Which is the correct ionic equation?
Relative masses · 3.2

Relative atomic & molecular mass

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.

Worked example — Mr of sulfuric acid, H₂SO₄

(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.

Calculate

Your turn — relative formula mass

2Calculate the relative formula mass Mr of calcium carbonate, CaCO₃. (Ar: Ca = 40, C = 12, O = 16)
Hint: 40 + 12 + (3 × 16). No units!
Mission 1 check

Why can't you change water's formula?

?A student 'balances' H₂ + O₂ → H₂O by changing it to H₂ + O₂ → H₂O₂. What should the student have written?
Mission 2 of 3

How do you count atoms by weighing?

Make a guess: how can you count atoms just by weighing?

Supplement · 3.3

The mole & the Avogadro constant

Extended only. The mole (mol) is the unit of amount of substance. One mole always contains the same number of particles — the Avogadro constant:

6.02 × 10²³ particles per moleatoms, ions or molecules — e.g. 1 mol of water = 6.02 × 10²³ H₂O molecules

The link between mass and moles uses the mole triangle. The molar mass (in g/mol) is numerically the same as the Mr:

mass (g) mol (amount) M molar mass
Cover what you want: moles = mass ÷ M, mass = mol × M, M = mass ÷ mol.
amount (mol) = mass (g) ÷ molar mass (g/mol)rearrange to find mass, molar mass, or number of particles (× 6.02 × 10²³)
Worked example

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)

Calculate · Supplement

Your turn — moles from mass

3How many moles are there in 36 g of water, H₂O? (Mr of H₂O = 18)
mol
Hint: moles = mass ÷ Mr = 36 ÷ 18.
Mission 2 check

How do you count atoms by weighing?

?A chemist weighs out 24 g of magnesium (Aᵣ = 24) and 12 g of carbon (Aᵣ = 12). Which statement about the number of atoms in each pile is correct?
Supplement · 3.3

Reacting masses & the mole ratio

Extended only. The big numbers in a balanced equation give the mole ratio in which substances react. To find a reacting mass:

  • 1. Write the balanced equation.
  • 2. Convert the known mass to moles.
  • 3. Use the mole ratio to get moles of the unknown.
  • 4. Convert back to mass (mass = mol × Mr).
mass of A (known, g) moles of A moles of B mass of B (answer, g) ÷ Mr of A × mole ratio × Mr of B
The mole ratio comes from the balanced equation's big numbers.
Worked example

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

Limiting reactant: if one reactant is used up first, it is the limiting reactant (the other is in excess). Compare the moles of each with the mole ratio, then use the moles of the limiting reactant to work out the product.

Calculate · Supplement

Your turn — reacting masses

4Calcium carbonate decomposes: CaCO₃ → CaO + CO₂. What mass of calcium oxide (CaO) is made from 50 g of CaCO₃? (Mr: CaCO₃ = 100, CaO = 56)
g
Hint: moles CaCO₃ = 50 ÷ 100 = 0.5. Ratio 1 : 1, so 0.5 mol CaO. Mass = 0.5 × 56.
Concentration · 3.3

Concentration of solutions

Concentration measures how much solute is dissolved in a given volume. It can be measured in g/dm³ or mol/dm³ (Core: know both units); calculating a concentration in either unit is Supplement. Volumes are in dm³, so remember 1 dm³ = 1000 cm³.

solute dissolved concentration = amount of solute (g or mol) volume of solution (dm³)
concentration = amount of solute ÷ volume in dm³
conc (g/dm³) = mass (g) ÷ volume (dm³)conc (mol/dm³) = moles ÷ volume (dm³) — both calculations are Supplement
Worked example

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.

Calculate · Supplement

Your turn — concentration in g/dm³

520 g of copper(II) sulfate is dissolved to make 500 cm³ of solution. What is the concentration in g/dm³?
g/dm³
Hint: 500 cm³ = 0.5 dm³. conc = mass ÷ volume = 20 ÷ 0.5.
Calculate · Supplement

Your turn — concentration in mol/dm³

60.5 mol of HCl is dissolved to make 250 cm³ of solution. What is the concentration in mol/dm³?
mol/dm³
Hint: 250 cm³ = 0.25 dm³. conc = moles ÷ volume = 0.5 ÷ 0.25.
Mission 3 of 3

Why does a mole of any gas fill the same space?

Make a guess: why does a mole of any gas fill the same space?

Supplement · 3.3

Molar gas volume

Extended only. At room temperature and pressure (r.t.p.), one mole of any gas occupies the same volume:

1 mol of gas = 24 dm³ at r.t.p.volume (dm³) = moles × 24  ·  moles = volume ÷ 24
Worked example

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 at the same temperature and pressure contain equal numbers of molecules.

Calculate · Supplement

Your turn — gas volume

7What volume, in dm³, does 0.25 mol of hydrogen gas occupy at r.t.p.? (molar gas volume = 24 dm³/mol)
dm³
Hint: volume = moles × 24 = 0.25 × 24.
Mission 3 check

Why does a mole of any gas fill the same space?

?At room temperature and pressure, a flask holds 24 dm³ of oxygen. It is emptied and filled with 24 dm³ of chlorine, which has much heavier molecules. Which statement is correct?
Supplement · 3.3

Percentage yield & purity

Extended only. Reactions rarely give 100% of the expected product. Two different ideas:

% yield = (actual ÷ theoretical) × 100how much product you got vs the maximum possible
% purity = (mass of pure substance ÷ mass of impure sample) × 100how much of a sample is the substance you want
Worked example — yield

A reaction could make 8 g of product but only 6 g is collected.

% yield = (6 ÷ 8) × 100 = 75%

% composition by mass of an element = (Ar × number of its atoms ÷ Mr) × 100 — e.g. carbon in CO₂ = (12 ÷ 44) × 100 = 27.3%.

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.

Calculate · Supplement

Your turn — percentage yield

8The theoretical (maximum) yield of a reaction is 40 g, but only 30 g is actually obtained. Calculate the percentage yield.
%
Hint: (actual ÷ theoretical) × 100 = (30 ÷ 40) × 100.
Supplement · 3.1 / 3.3

Empirical & molecular 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).

  • 1. Divide each element's mass (or %) by its Ar → gives moles.
  • 2. Divide all by the smallest result → gives the ratio.
  • 3. Round to whole numbers.
Worked example

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.

Quick check · Supplement

Your turn — empirical formula

?A compound contains 4.6 g of sodium and 1.6 g of oxygen. (Ar: Na = 23, O = 16). What is its empirical formula?
Supplement · 3.3

Titration calculations

Extended only. A titration finds an unknown concentration by reacting it with a solution of known concentration. The route:

  • 1. moles of known = conc × volume (in dm³).
  • 2. Use the mole ratio from the equation → moles of unknown.
  • 3. conc of unknown = moles ÷ its volume (in dm³).
Worked example

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.

Calculate · Supplement

Your turn — titration

9In a titration, 25.0 cm³ of HCl is exactly neutralised by 0.0020 mol of NaOH (ratio 1:1). What is the concentration of the HCl in mol/dm³?
mol/dm³
Hint: moles HCl = 0.0020 (1:1 ratio). 25.0 cm³ = 0.025 dm³. conc = 0.0020 ÷ 0.025.
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Tap the correct ending for each stoichiometry skill.

Recap

The formulae to know

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  [S]

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 the key ideas in 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.

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