Topic 3.1: Empirical Formula
Past Paper Examination Practice
EXAMINER'S TRAP WARNING
Empirical vs Molecular: Students frequently confuse these! Empirical formula = simplest whole number ratio. Molecular formula = actual number of atoms in one molecule. Example: Glucose is C₆H₁₂O₆ (molecular) but CH₂O (empirical).
ASection A: Definitions & Theory (5 Questions)
Define the term empirical formula.
[2 marks]
Define the term molecular formula.
[2 marks]
State the relationship between empirical formula and molecular formula.
[1 mark]
Glucose has the molecular formula C₆H₁₂O₆.
Explain why the empirical formula is different from the molecular formula.
[2 marks]
A compound has an empirical formula of CH₂ and a relative molecular mass (Mᵣ) of 84.
State what additional information is needed to determine the molecular formula.
[1 mark]
BSection B: Formula Conversion (5 Questions)
Deduce the empirical formula of propene, C₃H₆.
[1 mark]
Deduce the empirical formula of benzene, C₆H₆.
[1 mark]
Which of the following compounds has the empirical formula CH₂O?
A. Methanoic acid (HCOOH)
B. Ethanoic acid (CH₃COOH)
C. Glucose (C₆H₁₂O₆)
D. Ethanol (C₂H₅OH)
[1 mark]
A hydrocarbon has the empirical formula CH₂ and a relative molecular mass of 42.
Determine the molecular formula of the hydrocarbon.
[2 marks]
Octane has the molecular formula C₈H₁₈.
State the empirical formula of octane.
[1 mark]
CSection C: Calculations from Data (5 Questions)
A compound contains 72.0% magnesium and 28.0% nitrogen by mass.
Determine the empirical formula of this compound.
[Aᵣ: Mg = 24, N = 14]
[3 marks]
A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.
Calculate the empirical formula of this compound.
[Aᵣ: C = 12, H = 1, O = 16]
[3 marks]
When 2.70 g of aluminium is completely burned in oxygen, 5.10 g of aluminium oxide is formed.
Determine the empirical formula of aluminium oxide.
[Aᵣ: Al = 27, O = 16]
[4 marks]
A compound of iron and sulfur contains 63.6% iron by mass.
Determine the empirical formula of this compound.
[Aᵣ: Fe = 56, S = 32]
[3 marks]
A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its relative molecular mass is 56.
(i) Calculate the empirical formula.
(ii) Determine the molecular formula.
[Aᵣ: C = 12, H = 1]
[5 marks]
🎯EXAM SUCCESS CHECKLIST
✓ Always show your working — method marks awarded even if answer wrong
✓ Use the correct significant figures (usually 2 or 3 s.f.)
✓ Remember: Empirical ≠ Molecular (most common mistake!)
✓ When given %, the other element = 100% - given%
✓ Standard method: % → ÷Aᵣ → ÷smallest → simplify ratio
✓ For molecular formula: Find multiplier = Mᵣ(molecule) ÷ Mᵣ(empirical)
✓ Units matter: Always show g, mol, g/mol in calculations
💡THE EMPIRICAL FORMULA ALGORITHM
STEP 1: Get % by mass (if mass given, treat as %) STEP 2: Divide each % by Aᵣ → gives mole ratio STEP 3: Divide all by SMALLEST number → simplest ratio STEP 4: If decimals (like 1.33, 1.5), multiply ALL to get whole numbers STEP 5: Write formula using whole number subscripts EXAMPLE: 40% Ca, 12% C, 48% O (Aᵣ: Ca=40, C=12, O=16) Ca: 40÷40 = 1.0 C: 12÷12 = 1.0 O: 48÷16 = 3.0 Ratio = 1:1:3 → CaCO₃ ✓
📊 EXAM STATS:
15 Questions | 35 Total Marks | Time: ~30 minutes
🎓 CAMBRIDGE TIP:
Paper 4 empirical formula questions = FREE MARKS if you follow the method!
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