IGCSE · Practice questions · Empirical and molecular formulae

Find the ratio.

Six original Cambridge-style questions on empirical and molecular formulae, from reacting masses to percentage composition. Extended content.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Keep these straight

Divide, divide, make whole.

Divide each mass or % by A_r, then by the smallest answer.
Double if a value ends in .5; never round it away.
Molecular formula: M_r ÷ empirical formula mass = the multiplier.
Find a missing mass by subtraction: mass of oxide − mass of metal = mass of oxygen.

Recall

[2 marks]

Define (a) empirical formula and (b) molecular formula.

(a) The simplest whole-number ratio of the different atoms or ions in a compound. ✓
(b) The number and type of each atom in one molecule of the compound. ✓

Calculation

[2 marks]

1.4 g of nitrogen combines with 3.2 g of oxygen. Find the empirical formula of the oxide. (A_r: N 14, O 16)

N: 1.4 ÷ 14 = 0.1 mol; O: 3.2 ÷ 16 = 0.2 mol. ✓
Ratio 1 : 2, so NO₂. ✓

Calculation

[3 marks]

A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its M_r is 42. Find its empirical and molecular formulae. (A_r: H 1, C 12)

C: 85.7 ÷ 12 = 7.14; H: 14.3 ÷ 1 = 14.3. ✓
Dividing by 7.14 gives 1 : 2, so the empirical formula is CH₂ (mass 14). ✓
42 ÷ 14 = 3, so the molecular formula is C₃H₆. ✓

Calculation

[3 marks]

A compound contains 52.2% carbon, 13.0% hydrogen and 34.8% oxygen by mass. Its M_r is 46. Find its empirical and molecular formulae. (A_r: H 1, C 12, O 16)

C: 52.2 ÷ 12 = 4.35; H: 13.0 ÷ 1 = 13.0; O: 34.8 ÷ 16 = 2.175. ✓
Dividing by 2.175 gives 2 : 5.98 : 1, which is 2 : 6 : 1, so C₂H₆O (mass 24 + 6 + 16 = 46). ✓
46 ÷ 46 = 1, so the molecular formula is also C₂H₆O. ✓

Analysis

[3 marks]

5.4 g of aluminium is heated in oxygen and forms 10.2 g of aluminium oxide. Find the empirical formula of the oxide. (A_r: O 16, Al 27)

Mass of oxygen = 10.2 − 5.4 = 4.8 g. ✓
Al: 5.4 ÷ 27 = 0.2 mol; O: 4.8 ÷ 16 = 0.3 mol. ✓
Ratio 1 : 1.5, doubled to 2 : 3, so Al₂O₃. ✓

Explain

[2 marks]

A student finds the ratio of iron atoms to oxygen atoms in an oxide is 1 : 1.5 and writes the formula FeO₂. Explain the mistake and give the correct formula.

The student rounded 1.5 up to 2; a ratio ending in .5 is a real ratio and must not be rounded. ✓
Doubling both numbers gives 2 : 3, so the formula is Fe₂O₃. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.