Properties of hydrocarbons - AQA GCSE Chemistry Exam Questions

Practise 24 exam-style questions on Properties of hydrocarbons for AQA GCSE Chemistry. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

24 questions
1–6 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

Describe how the boiling point, viscosity and flammability of hydrocarbons change as the size of their molecules increases.

Model answer

As the molecules get bigger, the boiling point increases and the viscosity increases, but the flammability decreases.

Explanation

Larger hydrocarbon molecules have stronger forces between them, so they boil at higher temperatures, flow less easily and are harder to set alight. These properties decide which fractions make good fuels.

Mark scheme

1 mark
boiling point increases and viscosity increases
both needed for this mark
1 mark
flammability decreases
allow they become harder to ignite

Question 2

3 marks

Butane (C₄H₁₀) is used as a fuel in camping stoves.

Write a balanced symbol equation, including state symbols, for the complete combustion of butane at room temperature.

Model answer

2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(l)

Explanation

Each butane molecule gives 4 CO₂ and 5 H₂O, needing 8 + 5 = 13 oxygen atoms, which is 6½ O₂. Doubling everything gives whole numbers. Butane, oxygen and carbon dioxide are gases; water is a liquid at room temperature (but a gas in the hot flame).

Mark scheme

1 mark
correct formulae: C₄H₁₀ + O₂ → CO₂ + H₂O
1 mark
balanced: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
allow C₄H₁₀ + 6½O₂ → 4CO₂ + 5H₂O; allow multiples
1 mark
state symbols: C₄H₁₀(g), O₂(g), CO₂(g), H₂O(l)
allow H₂O(g)

Question 3

3 marks

When 1 mole of methane (CH₄) burns completely, 890 kJ of energy is released.
Relative atomic masses (Ar): H = 1, C = 12

Calculate the energy released when 4.0 g of methane burns completely.

Model answer

Mr of CH₄ = 12 + (4 × 1) = 16
moles of methane = 4.0 ÷ 16 = 0.25 mol
energy released = 0.25 × 890 = 222.5 kJ

Explanation

The combustion of hydrocarbon fuels releases energy, and the amount released is proportional to the amount of fuel burned. Find the moles of methane first, then scale the energy per mole.

Mark scheme

1 mark
Mr of CH₄ = 16
1 mark
moles = 4.0 ÷ 16 = 0.25 (mol)
allow ecf from an incorrect Mr
1 mark
energy = 0.25 × 890 = 222.5 kJ
allow 220-223 kJ; allow ecf; correct answer with no working scores 3 marks

Question 4

5 marks

A sample of a hydrocarbon was burned completely in oxygen. It produced 26.4 g of carbon dioxide and 12.6 g of water. The relative formula mass (Mr) of the hydrocarbon is 86.
Relative formula masses (Mr): CO₂ = 44, H₂O = 18
Relative atomic masses (Ar): H = 1, C = 12

Determine the empirical formula and the molecular formula of the hydrocarbon.

Model answer

All the carbon ends up in CO₂: moles of CO₂ = 26.4 ÷ 44 = 0.60 mol, so there are 0.60 mol of C atoms.
All the hydrogen ends up in H₂O: moles of H₂O = 12.6 ÷ 18 = 0.70 mol, so there are 2 × 0.70 = 1.40 mol of H atoms.
ratio C : H = 0.60 : 1.40 = 3 : 7, so the empirical formula is C₃H₇
Mr of C₃H₇ = (3 × 12) + (7 × 1) = 43
86 ÷ 43 = 2, so the molecular formula is C₆H₁₄

Explanation

In complete combustion, every carbon atom in the fuel becomes CO₂ and every hydrogen atom becomes H₂O. Remember that each water molecule contains 2 hydrogen atoms, so double the moles of water to get the moles of hydrogen.

Mark scheme

1 mark
moles of CO₂ = 26.4 ÷ 44 = 0.60, so moles of C = 0.60
correct answer with no working scores full marks
1 mark
moles of H₂O = 12.6 ÷ 18 = 0.70, so moles of H = 1.40
1 mark
ratio C : H = 0.60 : 1.40 = 3 : 7, so empirical formula = C₃H₇
allow ecf from incorrect moles
1 mark
Mr of C₃H₇ = 43, and 86 ÷ 43 = 2
1 mark
molecular formula = C₆H₁₄
allow ecf from an incorrect empirical formula

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