A chemist is investigating an unknown ionic compound. The compound contains only magnesium and oxygen. The chemist determines that the compound contains 60.3% magnesium and 39.7% oxygen by mass. Use this information to determine the empirical formula of the compound, showing your reasoning clearly.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Magnesium has a relative atomic mass of 24.3. Oxygen has a relative atomic mass of 16.0.
Model answer (5 marks)
1. 60.3 % Mg ÷ 24.3 g mol⁻¹ = 2.48 mol Mg
2. 39.7 % O ÷ 16.0 g mol⁻¹ = 2.48 mol O
3. 2.48 ÷ 2.48 = 1 : 1 (simplest ratio)
4. Empirical formula = MgO
5. An empirical formula is the simplest whole‑number ratio of atoms of each element in a compound.
2. 39.7 % O ÷ 16.0 g mol⁻¹ = 2.48 mol O
3. 2.48 ÷ 2.48 = 1 : 1 (simplest ratio)
4. Empirical formula = MgO
5. An empirical formula is the simplest whole‑number ratio of atoms of each element in a compound.
Examiner tips
- Show the two division steps and the ratio calculation clearly; use the word ‘simplest’ when stating the formula.
- Explain that the empirical formula is the whole‑number ratio of atoms, not the molar mass.
Mark scheme (5 marks)
- Divides percentage of magnesium by its relative atomic mass: 60.3 ÷ 24.3
- Divides percentage of oxygen by its relative atomic mass: 39.7 ÷ 16.0
- Divides both values by the smallest to find the simplest ratio
- States the correct empirical formula as MgO
- Correctly defines empirical formula as the simplest whole number ratio of atoms of each element in the compound
Key terms in this question
Related
- All OCR A-Level Chemistry A (H432) revision notes →
- How to answer a "Justify" question →
- Decode the mark scheme abbreviations →
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