AQA GCSE Chemistry Chemistry Paper 2 (Foundation), November 2021: Question 7

11 marks · Low Demand difficulty · Short Answer

Answer questions on organic compounds including uses of butane, the monomer of poly(propene), percentage yield of ethanol, balancing the combustion equation of ethanol, and comparing fermentation with hydration.

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Question

Question 7 consists of six parts: 07.1 asks candidates to select the use of butane from 'fertiliser', 'formulation', or 'fuel'; 07.2 asks to tick the monomer of poly(propene) among propane, propanoic acid, propanol, and propene; 07.3 gives an equation for the reversible production of ethanol from ethene and steam, provides values to calculate the percentage yield using the given formula; 07.4 asks to select two reasons from five options why percentage yield is less than 100%; 07.5 asks to balance the combustion equation C2H5OH + ___ O2 -> 3 H2O + 2 CO2; 07.6 presents Table 5 comparing fermentation and hydration on raw material, energy usage, rate of reaction, and purity of ethanol, then asks for two advantages and two disadvantages of fermentation.
Question text

07 This question is about organic compounds.

07.1 Butane is an alkane with small molecules.

Complete the sentence.

Choose the answer from the box.

[1 mark]

fertiliser formulation fuel

Butane can be used as a .

07.2 Poly(propene) is a polymer.

What is the name of the monomer used to produce poly(propene)?

[1 mark]

Tick ( ) one box.

Propane

Propanoic acid

Propanol

Propene 25

Ethene and steam react to produce ethanol.

The equation for the reversible reaction is:

ethene + steam ⇌ ethanol

07.3 The reaction produces a maximum theoretical mass of 400 kg of ethanol from 243 kg

of ethene and 157 kg of steam.

A company produces 380 kg of ethanol from 243 kg of ethene and 157 kg of steam.

The percentage yield of ethanol is less than 100%

Calculate the percentage yield of ethanol.

Use the equation:

mass of ethanol actually made

percentage yield of ethanol = × 100

maximum theoretical mass of ethanol

[2 marks]

Percentage yield = %

07.4 What are two possible reasons why the percentage yield of ethanol is less

than 100%?

[2 marks]

Tick ( ) two boxes.

Ethanol is the only product of the reaction.

Ethanol is very unreactive.

Some ethanol changes back into ethene and steam.

Some ethanol escapes from the apparatus.

Some ethanol reacts with steam. 26

07.5 Ethanol burns in oxygen.

Balance the equation for the reaction.

*25* [1 mark]

C2H5OH + ___ O2 → 3H2O + 2CO2

07.6 Two processes for producing ethanol are:

• fermentation

• hydration (reacting ethene with steam).

Table 5 shows information about the processes.

Table 5

Process

Feature

Fermentation Hydration

Raw material sugar crude oil

Energy usage low high

Rate of reaction slow fast

Purity of ethanol 15% 98%

Give two advantages and two disadvantages of using fermentation to

produce ethanol.

[4 marks]

Advantage of fermentation 1

Advantage of fermentation 2

Disadvantage of fermentation 1

Disadvantage of fermentation 2

Mark scheme

Show the mark scheme Mark scheme for Question 7 shows: 07.1 accepts 'fuel' (1 mark); 07.2 accepts 'propene' (1 mark); 07.3 awards 1 mark for working (380/400 x 100) and 1 mark for 95(%) (2 marks total); 07.4 awards 1 mark each for 'some ethanol changes back into ethene and steam' and 'some ethanol escapes from the apparatus' (2 marks total); 07.5 gives 1 mark for coefficient 3 in front of O2, allowing multiples; 07.6 awards 1 mark each for two advantages (low energy usage, uses renewable raw materials) and two disadvantages (produces impure ethanol, slow rate of reaction) (4 marks total). Total marks: 11.

Question 7

AO /

Question Answers Extra information Mark

Spec. Ref.

07.1 fuel 1 AO1

4.7.1.2

AO /

Spec. Ref.

07.2 propene 1 AO2

4.7.2.1

4.7.3.1

AO /

Spec. Ref.

07.3 (percentage yield =) AO2

380 1 4.3.3.1

× 100

= 95 (%) 1

AO /

Spec. Ref.

07.4 some ethanol changes back into 1 AO2

ethene and steam 4.3.3.1

some ethanol escapes from the 1

apparatus

AO /

Spec. Ref.

07.5 C2H5OH + 3 O2 → allow multiples 1 AO2

3 H2O + 2 CO2 4.1.1.1

4.3.1.1

4.7.2.3

Question 7 continued

AO /

Spec. Ref.

07.6 (advantages) AO3

4.7.2.2

(fermentation) low energy usage 1 4.7.2.3

4.10.1.1

18 (fermentation) uses renewable 1

raw materials

(disadvantages)

(fermentation) produces impure 1

ethanol

(fermentation) slow rate of 1

reaction

Total 11

How to answer it

Organic Chemistry: Fuels, Polymers & Ethanol

📌 What this question tests

This question assesses core understanding from Topic 7 (Organic Chemistry) and Topic 3 (Quantitative Chemistry):

  • Recall of the uses of short-chain alkanes and naming addition polymers.
  • Calculating percentage yield from given industrial masses.
  • Identifying real-world reasons why reactions do not achieve 100% yield.
  • Balancing symbol equations for the complete combustion of alcohols.
  • Evaluating competing industrial chemical processes (fermentation vs hydration).
Question 07.1

Uses of Small Alkanes

1 Mark • Assessment Objective: AO1

✅ Correct Answer

Butane can be used as a fuel.

1 Mark: Exactly identifies "fuel" from the options given.

💡 Key Knowledge

  • Butane (C₄H₁₀) is a four-carbon alkane.
  • Short-chain alkanes have low boiling points, ignite easily, and release large amounts of energy when burned, making them ideal fuels (e.g. in camping stoves and lighters).
Question 07.2

Identifying Monomers

1 Mark • Assessment Objective: AO2

✅ Correct Answer

Tick box 4: Propene

1 Mark: Only the box for "Propene" ticked.

🧠 Exam Technique

Addition polymers are named simply by placing "poly" in front of the monomer name in brackets:

poly(monomer) → monomer

To find the monomer, remove the "poly" prefix and brackets: poly(propene) comes from propene.

❌ Common Errors

Selecting propane. Remember that addition polymers require an unsaturated monomer containing a C=C double bond (an alkene, ending in -ene), not an alkane.

Question 07.3

Percentage Yield Calculation

2 Marks • Assessment Objective: AO2

📐 Step-by-Step Calculation

Formula:

Percentage Yield = (Actual Mass / Theoretical Mass) × 100

  1. Identify values from the text:
    Actual mass made = 380 kg
    Maximum theoretical mass = 400 kg
  2. Substitute into formula:
    (380 / 400) × 100 [1 mark]
  3. Calculate final percentage:
    = 95% [1 mark]

❌ Distractor Data Trap

The question provides reactant masses: 243 kg of ethene and 157 kg of steam .

Notice that 243 + 157 = 400 kg (Conservation of Mass). You do not need to divide by the reactant masses individually; use the theoretical product mass (400 kg) directly.

Question 07.4

Reasons for Incomplete Yield

2 Marks • Assessment Objective: AO2

✅ Correct Selections (Tick 2)

  • ☑ Some ethanol changes back into ethene and steam. [1 mark]
  • ☑ Some ethanol escapes from the apparatus. [1 mark]

💡 Understanding Yield Losses

  • Reversible Reaction: The equation displays a reversible symbol ( ⇌ ). As product forms, some decomposes back into reactants.
  • Separation/Transfer Loss: Ethanol is volatile (boils at 78 °C), so vapours easily escape during collection and purification.

❌ Incorrect Distractors

  • "Ethanol is the only product" — true, but having no side products would normally help yield, not reduce it.
  • "Ethanol is very unreactive" — ethanol is flammable and reactive.
  • "Some ethanol reacts with steam" — steam reacts with ethene, not ethanol.
Question 07.5

Balancing a Combustion Equation

1 Mark • Assessment Objective: AO2

✅ Correct Balanced Equation

C₂H₅OH + 3 O₂ → 3 H₂O + 2 CO₂

1 Mark: Correct balancing coefficient of 3 on the line.

🧠 Atom Balancing Step-by-Step

  1. Count atoms on Right-Hand Side (RHS):
    Carbons = 2 × 1 = 2
    Hydrogens = 3 × 2 = 6
    Oxygens = (3 × 1) + (2 × 2) = 3 + 4 = 7
  2. Balance Left-Hand Side (LHS):
    C₂H₅OH provides 2 C and (5 + 1) = 6 H (already balanced!).
    C₂H₅OH also provides 1 O.
  3. Find remaining oxygen:
    Total needed = 7 O atoms.
    7 - 1 = 6 O atoms must come from O₂ molecules.
    6 ÷ 2 = 3 O₂.

❌ Common Error

Forgetting the single oxygen atom already inside the ethanol molecule ( C₂H₅OH ). Students who forget this often try to balance with 3.5 or write 7 .

Question 07.6

Evaluating Industrial Methods: Fermentation vs Hydration

4 Marks • Assessment Objective: AO3

✅ Advantages of Fermentation (Any 2)

  • Low energy usage (operates at moderate temperatures, around 30–40 °C). [1 mark]
  • Uses renewable raw materials (sugar cane/plant crops vs non-renewable crude oil). [1 mark]

✅ Disadvantages of Fermentation (Any 2)

  • Low purity / produces impure ethanol (only ~15% yield before yeast is poisoned; requires fractional distillation). [1 mark]
  • Slow rate of reaction (takes days vs a continuous, fast industrial reaction). [1 mark]

🧠 How to Secure Full 4 Marks

  • Read the table carefully: The table lists raw material, energy usage, rate, and purity. Each row provides one comparison point.
  • Link raw materials to sustainability: When referring to "sugar", specify that it is a renewable resource compared to "crude oil", which is finite/non-renewable.
  • Match the question focus: Make sure you state advantages/disadvantages of fermentation, not hydration.

Topics

Chemistry · C3: Quantitative Chemistry · C7: Organic Chemistry · C10: Using Resources

Question and mark scheme from the AQA GCSE Chemistry examination, Chemistry Paper 2 (Foundation), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.