A pharmaceutical company is checking the purity of a newly synthesised compound. The compound contains only nitrogen and oxygen. A sample of the compound is found to contain 1.4 g of nitrogen and 3.2 g of oxygen. Use this information to determine the empirical formula of the compound. Show your reasoning clearly.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Nitrogen has a relative atomic mass of 14. Oxygen has a relative atomic mass of 16.
Model answer (5 marks)
Divide the mass of N by its relative atomic mass: 1.4 g ÷ 14 g mol⁻¹ = 0.10 mol N.
Divide the mass of O by its relative atomic mass: 3.2 g ÷ 16 g mol⁻¹ = 0.20 mol O.
Divide both mole values by the smaller value (0.10 mol) to obtain the simplest whole‑number ratio:
N: 0.10 ÷ 0.10 = 1
O: 0.20 ÷ 0.10 = 2
Thus the empirical formula is NO₂, the simplest whole‑number ratio of atoms of each element in the compound.
Divide the mass of O by its relative atomic mass: 3.2 g ÷ 16 g mol⁻¹ = 0.20 mol O.
Divide both mole values by the smaller value (0.10 mol) to obtain the simplest whole‑number ratio:
N: 0.10 ÷ 0.10 = 1
O: 0.20 ÷ 0.10 = 2
Thus the empirical formula is NO₂, the simplest whole‑number ratio of atoms of each element in the compound.
Examiner tips
- Show all calculations with units; use the correct relative atomic masses. State the final empirical formula clearly. Explain that the ratio represents the simplest whole‑number ratio of atoms.
Common mistakes
- Using the wrong relative atomic mass (e.g. 15 for O). Failing to divide by the smallest mole value. Writing the formula as N₁O₂ instead of the standard NO₂.
Mark scheme (5 marks)
- Divide mass of nitrogen by its relative atomic mass: 1.4 ÷ 14 = 0.1 (moles of nitrogen)
- Divide mass of oxygen by its relative atomic mass: 3.2 ÷ 16 = 0.2 (moles of oxygen)
- Divide both values by the smallest (0.1) to find the simplest ratio: N = 0.1 ÷ 0.1 = 1, O = 0.2 ÷ 0.1 = 2
- States the empirical formula as NO₂
- Correctly describes the 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 "Explain" question →
- Decode the mark scheme abbreviations →
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