Empirical and Molecular Formulae Exam Questions
Practice IGCSE Chemistry exam questions on empirical and molecular formulae. Covers calculating empirical formula from percentage composition or mass data, finding molecular formula from empirical formula and Mr, with mark schemes and examiner notes.
Published by IGCSEChemistry.com.my
Chemistry teaching team: K. S. Tan (15+ years teaching IGCSE Chemistry) and Ms Yash (10+ years teaching IGCSE Chemistry) and Ms Kartini (15+ years teaching IGCSE Chemistry).
Mapped to Cambridge IGCSE Chemistry 0620 (2026–2028). Last updated 2026-08-19.
These questions cover all aspects of empirical and molecular formulae. Show all working clearly.
Question 1 (2 marks, Supplement)
Define empirical formula and explain how it differs from molecular formula.
Mark scheme
- The empirical formula is the simplest whole number ratio of atoms of each element in a compound [1]
- The molecular formula shows the actual number of atoms of each element in one molecule (it may be a multiple of the empirical formula) [1]
Examiner note: Example: glucose has empirical formula CH2O and molecular formula C6H12O6. The molecular formula is always a whole-number multiple of the empirical formula.
Question 2 (3 marks, Supplement)
A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.
Calculate the empirical formula of the compound. (Ar: H = 1, C = 12, O = 16)
Mark scheme
| Element | C | H | O |
|---|---|---|---|
| Mass (%) | 40.0 | 6.7 | 53.3 |
| Divide by Ar | 40.0/12 = 3.33 | 6.7/1 = 6.7 | 53.3/16 = 3.33 |
| Divide by smallest | 3.33/3.33 = 1 | 6.7/3.33 = 2 | 3.33/3.33 = 1 |
[1] for moles, [1] for correct ratio, [1] for empirical formula = CH2O
Examiner note: Always present this as a table — it makes the method clear and earns method marks even if arithmetic goes wrong. The steps are: (1) take percentages as masses, (2) divide each by Ar, (3) divide all results by the smallest.
Question 3 (4 marks, Supplement)
The compound in Question 2 has Mr = 180. Determine its molecular formula.
Mark scheme
- Empirical formula = CH2O; empirical formula mass = 12 + 2 + 16 = 30 [1]
- n = Mr / empirical formula mass = 180 / 30 = 6 [1]
- Molecular formula = (CH2O) x 6 [1]
- = C6H12O6 [1]
Examiner note: The molecular formula is always a whole-number multiple (n) of the empirical formula. If n = 1, the empirical and molecular formulae are the same. This compound is glucose.
Question 4 (4 marks, Supplement)
1.20 g of magnesium is burned in air. The magnesium oxide produced has a mass of 2.00 g.
(a) Calculate the mass of oxygen that combined with the magnesium. (1)
(b) Calculate the empirical formula of magnesium oxide. (3)
(Ar: O = 16, Mg = 24)
Mark scheme
(a) Mass of O = 2.00 - 1.20 = 0.80 g [1]
(b)
| Element | Mg | O |
|---|---|---|
| Mass | 1.20 | 0.80 |
| Moles | 1.20/24 = 0.05 | 0.80/16 = 0.05 |
| Ratio | 1 | 1 |
[1] for moles, [1] for ratio, [1] for empirical formula = MgO
Examiner note: This is a classic practical-style question. The mass of oxygen is found by subtraction: mass of product - mass of metal = mass of oxygen. Then proceed with the standard empirical formula method.
Question 5 (4 marks, Supplement)
A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its Mr is 56.
(a) Calculate the empirical formula. (2)
(b) Determine the molecular formula. (2)
(Ar: H = 1, C = 12)
Mark scheme
(a)
| Element | C | H |
|---|---|---|
| Mass (%) | 85.7 | 14.3 |
| Moles | 85.7/12 = 7.14 | 14.3/1 = 14.3 |
| Ratio | 7.14/7.14 = 1 | 14.3/7.14 = 2 |
Empirical formula = CH2 [1] [1]
(b) Empirical formula mass = 12 + 2 = 14 n = 56/14 = 4 [1] Molecular formula = C4H8 [1]
Examiner note: CH2 is the empirical formula for all alkenes (CnH2n). The molecular formula C4H8 tells us this is butene. If the Mr had been 42, the molecular formula would be C3H6 (propene).
Question 6 (4 marks, Supplement)
Complete combustion of 2.30 g of a compound containing carbon, hydrogen, and oxygen produces 4.40 g of CO2 and 2.70 g of H2O.
Calculate the empirical formula of the compound.
(Ar: H = 1, C = 12, O = 16)
Mark scheme
- Moles of CO2 = 4.40/44 = 0.10; moles of C = 0.10; mass of C = 0.10 x 12 = 1.20 g [1]
- Moles of H2O = 2.70/18 = 0.15; moles of H = 0.30; mass of H = 0.30 x 1 = 0.30 g [1]
- Mass of O = 2.30 - 1.20 - 0.30 = 0.80 g; moles of O = 0.80/16 = 0.05 [1]
- Ratio C : H : O = 0.10 : 0.30 : 0.05 = 2 : 6 : 1; empirical formula = C2H6O [1]
Examiner note: In combustion analysis, all C goes to CO2 and all H goes to H2O. One mole of CO2 contains one mole of C; one mole of H2O contains two moles of H. The mass of O is found by subtraction. This is C2H6O, which is the molecular formula of ethanol.
Question 7 (3 marks, Supplement)
An iron oxide contains 72.4% iron and 27.6% oxygen. Determine its empirical formula and name the compound.
(Ar: O = 16, Fe = 56)
Mark scheme
| Element | Fe | O |
|---|---|---|
| Mass (%) | 72.4 | 27.6 |
| Moles | 72.4/56 = 1.293 | 27.6/16 = 1.725 |
| Ratio | 1.293/1.293 = 1 | 1.725/1.293 = 1.33 |
| Multiply by 3 | 3 | 4 |
Empirical formula = Fe3O4 [1] [1] Name: iron(II,III) oxide / tri-iron tetraoxide [1]
Examiner note: When the ratio gives a non-integer (1.33 = 4/3), multiply all values by the denominator (3) to get whole numbers. Common fractions: 0.33 → multiply by 3; 0.5 → multiply by 2; 0.25 → multiply by 4.
What to revise if you scored below 5
If the table method was unclear, practise it for at least five different compounds. If combustion analysis (Question 6) was difficult, memorise: C comes from CO2, H comes from H2O, and O comes by subtraction. Revisit the empirical and molecular formulae notes.
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