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AS · Practice questions · Empirical and molecular formulae

Find the ratio.

Six original Cambridge-style questions on empirical and molecular formulae, from simple percentage data to combustion analysis and hydrated salts.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
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01
Calculation
[2 marks]

A compound contains 32.4% sodium, 22.6% sulfur and 45.0% oxygen by mass. Find its empirical formula. (Ar: O 16.0, Na 23.0, S 32.1)

  • Na: 32.4 ÷ 23.0 = 1.41; S: 22.6 ÷ 32.1 = 0.704; O: 45.0 ÷ 16.0 = 2.81. ✓
  • Dividing by 0.704 gives 2.00 : 1.00 : 3.99, so Na2SO4. ✓
02
Calculation
[3 marks]

A compound contains 54.5% carbon, 9.1% hydrogen and 36.4% oxygen by mass. Its Mr is 88.0. Find its empirical and molecular formulae. (Ar: H 1.0, C 12.0, O 16.0)

  • C: 54.5 ÷ 12.0 = 4.54; H: 9.1 ÷ 1.0 = 9.1; O: 36.4 ÷ 16.0 = 2.275. ✓
  • Dividing by 2.275 gives 2 : 4 : 1, so the empirical formula is C2H4O (mass 44.0). ✓
  • n = 88.0 ÷ 44.0 = 2, so the molecular formula is C4H8O2. ✓
03
Calculation
[2 marks]

An oxide of iron contains 70.0% iron by mass. Find its empirical formula. (Ar: O 16.0, Fe 55.8)

  • Fe: 70.0 ÷ 55.8 = 1.25; O: 30.0 ÷ 16.0 = 1.875. ✓
  • Ratio 1 : 1.50; doubling gives 2 : 3, so Fe2O3. ✓
04
Calculation
[3 marks]

0.230 g of a compound containing carbon, hydrogen and oxygen burns completely to give 0.440 g of CO2 and 0.270 g of H2O. Find its empirical formula. (Ar: H 1.0, C 12.0, O 16.0)

  • Mass of C = 0.440 × 12.0/44.0 = 0.120 g; mass of H = 0.270 × 2.0/18.0 = 0.030 g. ✓
  • Mass of O = 0.230 − 0.120 − 0.030 = 0.080 g. ✓
  • Moles: C 0.0100, H 0.030, O 0.0050, a ratio of 2 : 6 : 1, so C2H6O. ✓
05
Calculation
[2 marks]

2.86 g of hydrated sodium carbonate, Na2CO3·xH2O, is heated to constant mass. The anhydrous residue has a mass of 1.06 g. Find x. (Mr: Na2CO3 106.0, H2O 18.0)

  • Water lost = 2.86 − 1.06 = 1.80 g = 0.100 mol; Na2CO3 = 1.06 ÷ 106.0 = 0.0100 mol. ✓
  • x = 0.100 ÷ 0.0100 = 10. ✓
06
Explain
[2 marks]

In the experiment above, explain why the solid is heated to constant mass, and state the effect on the value of x if it is not.

  • Heating to constant mass shows that all the water of crystallisation has been driven off. ✓
  • If some water remains, the mass lost is too small, so the calculated x is too low. ✓

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.