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5.7.1.3Properties of hydrocarbons

AQA GCSE Combined Science (8464), Higher tier · Chemistry › Organic chemistry › Carbon compounds as fuels and feedstock

Practise Properties of hydrocarbons. 12 exam-style questions on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.

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Revision notes

How the size of a hydrocarbon molecule affects its boiling point, viscosity and flammability, why this decides how it is used as a fuel, and how to write balanced equations for complete combustion. Expect data-table questions, 'compare' questions and 'write a balanced equation' questions.

Grade by grade

What you need to be able to do, from the first marks up to the top grade.

  1. 3
    Name the products of complete combustionA hydrocarbon burns completely in oxygen to form carbon dioxide and water.
  2. 4
    Recall how properties change with molecule sizeBigger molecules: higher boiling point, higher viscosity and lower flammability.
  3. 5
    Choose a suitable fuel using property dataFor example, a fuel for a gas stove must have a boiling point below room temperature and be very flammable.
  4. 5
    State what happens to the fuel in combustionThe carbon and hydrogen in the fuel are oxidised, and energy is released.
  5. 6
    Balance complete combustion equationsBalance C, then H, then O; if you need half an O2, double every number.
  6. 7
    Explain the boiling point trend using intermolecular forcesLarger molecules have stronger forces between the molecules, so more energy is needed to overcome them.

Notes

Properties depend on molecule size

  • Some properties of hydrocarbons depend on the size of their molecules (the number of carbon atoms).
  • As the molecules get bigger: the boiling point increases, the viscosity increases (the liquid is thicker and flows less easily) and the flammability decreases (they are harder to set alight).
  • So small hydrocarbons are gases or runny liquids that ignite easily, and large ones are thick liquids or solids that are hard to ignite.
  • Why the boiling point rises: larger molecules have stronger intermolecular forces (forces between molecules), so more energy is needed to overcome them. The covalent bonds inside the molecules do not break when a hydrocarbon boils. grade 7+

Using the properties

  • These properties decide how a hydrocarbon is used as a fuel.
  • Hydrocarbons with small molecules make good fuels: they are very flammable and flow easily. Petrol is an example.
  • Hydrocarbons with large molecules are less useful as fuels, because they are viscous and hard to ignite.

Complete combustion

  • Burning (combustion) of hydrocarbon fuels releases energy.
  • During combustion, the carbon and hydrogen in the fuel are oxidised: they combine with oxygen.
  • Complete combustion (in plenty of oxygen): hydrocarbon + oxygen → carbon dioxide + water.
  • Example: CH4 + 2O2 → CO2 + 2H2O
  • To balance: (1) put the number of C atoms in front of CO2; (2) put half the number of H atoms in front of H2O; (3) count the O atoms on the right and halve that number for O2; (4) if you get a half, double everything.
  • Example: C2H6 gives 2CO2 + 3H2O, which has 7 O atoms, so 3½O2. Doubling: 2C2H6 + 7O2 → 4CO2 + 6H2O.

Cheatsheet

  • Bigger molecules → higher boiling point
  • Bigger molecules → more viscous (flow less easily)
  • Bigger molecules → less flammable
  • Viscosity: how thick a liquid is (high viscosity = flows slowly)
  • Complete combustion: hydrocarbon + oxygen → carbon dioxide + water
  • Combustion releases energy; the C and H in the fuel are oxidised
  • CH4 + 2O2 → CO2 + 2H2O
  • C3H8 + 5O2 → 3CO2 + 4H2O
  • Higher boiling point = stronger intermolecular forces, not stronger covalent bonds grade 7+

How to answer each type of question

Compare the properties of two hydrocarbons

1 to 3 marks4
  1. Decide which hydrocarbon has the bigger molecules.
  2. State how each property compares, using 'higher', 'more viscous' and 'less flammable'.

Example. Hydrocarbon A has 6 carbon atoms in each molecule. Hydrocarbon B has 20 carbon atoms in each molecule.
Compare the boiling point, viscosity and flammability of A and B.

Show the model answer
B has a higher boiling point than A (1). B is more viscous than A (1). B is less flammable than A (1).

Write a balanced combustion equation

1 to 2 marks6
  1. Write the formulae: hydrocarbon + O2 → CO2 + H2O.
  2. Balance carbon, then hydrogen.
  3. Count the O atoms on the right, then balance O2. Double everything if you get a half.

Example. Heptane, C7H16, is one of the hydrocarbons in petrol.
Write a balanced equation for the complete combustion of heptane.

Show the model answer
Correct formulae: C7H16 + O2 → CO2 + H2O (1)
Balanced: C7H16 + 11O2 → 7CO2 + 8H2O (1)
(Check: (7 × 2) + 8 = 22 O atoms on the right, which is 11O2.)

Use data to choose a fuel

2 to 3 marks5
  1. Compare each boiling point with room temperature (about 20 °C) to find the state.
  2. Use molecule size to judge flammability and viscosity.
  3. Quote the data in your answer.

Example. Data for three hydrocarbons:
P: C3H8, boiling point −42 °C
Q: C10H22, boiling point 174 °C
R: C18H38, boiling point 317 °C
Which hydrocarbon would be best as the fuel for a gas barbecue? Explain your answer.

Show the model answer
P (1). Its boiling point (−42 °C) is below room temperature, so it is a gas (1). It has the smallest molecules, so it is the most flammable and ignites easily (1).

Explain the trend in boiling point

2 to 3 marks7
  1. Say that the molecules are larger.
  2. Say that the intermolecular forces (forces between the molecules) are stronger.
  3. Say that more energy is needed to overcome these forces.

Example. Explain why dodecane, C12H26, has a higher boiling point than hexane, C6H14.

Show the model answer
Dodecane molecules are larger (1). So the intermolecular forces between dodecane molecules are stronger (1). So more energy is needed to overcome these forces (1). (Writing about breaking covalent bonds loses the last two marks.)

Shortcuts and memory tricks

  • Think honey and petrol: long chains are like honey (thick, hard to light); short chains are like petrol (runny, catch fire easily).
  • Two up, one down: boiling point and viscosity go up with size; flammability goes down.
  • Balancing order: C, then H, then O. Oxygen goes last because O2 is on its own.
  • Odd number of O atoms on the right? Double every number in the equation.
  • Alkane check: n carbons need (3n + 1) ÷ 2 molecules of O2. Propane: (9 + 1) ÷ 2 = 5.

Where marks are lost

  • Saying larger molecules are more flammable 'because there is more fuel'. They are less flammable.
  • Explaining boiling points by breaking covalent bonds. Boiling overcomes the weak forces between molecules.
  • Giving carbon monoxide or soot as products of complete combustion. They come from incomplete combustion.
  • Changing a formula to balance an equation, e.g. writing CO3. Only change the numbers in front.
  • Doubling only some terms when clearing a half, e.g. 2C2H6 + 7O2 → 2CO2 + 3H2O.

Exam technique

  • 'Compare' means say how the two are different, using comparative words: higher, more viscous, less flammable.
  • When you use data, quote the numbers with units, e.g. 'its boiling point, −42 °C, is below room temperature'.
  • After balancing an equation, count every atom on both sides. It takes 20 seconds and catches most slips.
  • Use the words 'viscosity' and 'flammability' rather than 'thickness' and 'burns better'.

Quick recall

Cover the answers and test yourself. The app has these as flashcards that come back just before you'd forget them.

Name the two products of the complete combustion of a hydrocarbon.
Carbon dioxide and water
What is meant by the viscosity of a liquid?
How easily it flows (how thick it is).
Write a balanced symbol equation for the complete combustion of hexane.
2C6H14 + 19O2 → 12CO2 + 14H2O

Sample questions

Written for this site in the style of AQA exam questions. They are not taken from real past papers.

Question 1Easy2 marks
(a) Which statement about alkanes is correct?
Tick (✓) one box.[1]
  • As the chain length increases, the boiling point increases
  • As the chain length increases, the flammability increases
  • As the chain length increases, the viscosity decreases
  • Short-chain alkanes are more viscous than long-chain alkanes
(b) What is meant by the viscosity of a liquid?[1]
Show the answer and mark scheme
(a) Answer: As the chain length increases, the boiling point increases
(b) Answer: How easily it flows (how thick it is).
  • how easily the liquid flows / its resistance to flowing / how thick it is
Question 2Medium5 marks
Alkanes are used as fuels.
(a) Balance the equation for the complete combustion of propane.
C3H8 + .....O2 → .....CO2 + .....H2O[1]
(b) Write a balanced equation for the complete combustion of butane, C4H10.[2]
(c) During combustion, the carbon and hydrogen in the fuel are oxidised.
What does 'oxidised' mean here?[1]
(d) Explain why the combustion of hydrocarbons is useful.[1]
Show the answer and mark scheme
(a) Answer: C3H8 + 5O2 → 3CO2 + 4H2O
  • C3H8 + 5O2 → 3CO2 + 4H2O
(b) Answer: 2C4H10 + 13O2 → 8CO2 + 10H2O
  • correct formulae: C4H10 + O2 → CO2 + H2O
  • balanced: 2C4H10 + 13O2 → 8CO2 + 10H2O
(c) Answer: They gain oxygen.
  • they gain oxygen / combine with oxygen
(d) Answer: It releases a lot of energy, so hydrocarbons can be used as fuels.
  • it releases (a lot of) energy / is exothermic, so hydrocarbons are useful fuels
Question 3Hard7 marks
(a) Explain why the boiling points of the alkanes increase as the number of carbon atoms increases.[3]
(b) Short-chain hydrocarbons are in high demand as fuels.
Give two reasons why short-chain hydrocarbons are better fuels than long-chain hydrocarbons.[2]
(c) Write a balanced equation for the complete combustion of dodecane, C12H26.[2]
Show the answer and mark scheme
(a) Answer: The molecules get larger, so the intermolecular forces are stronger and more energy is needed to overcome them.
  • the molecules are larger
  • so the intermolecular forces (forces between the molecules) are stronger
  • so more energy is needed to overcome these forces
(b) Answer: They are more flammable (easier to ignite) and less viscous (flow and evaporate more easily).
  • they are more flammable / ignite more easily
  • they are less viscous / flow or evaporate more easily
(c) Answer: 2C12H26 + 37O2 → 24CO2 + 26H2O
  • correct formulae: C12H26 + O2 → CO2 + H2O
  • balanced: 2C12H26 + 37O2 → 24CO2 + 26H2O

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