AQA GCSE Chemistry Chemistry Paper 1 (Foundation), June 2022: Question 6

8 marks · Low Demand difficulty · Short Answer

Identify structures, properties, and formulae of carbon allotropes, simple molecules, and calculate the relative formula mass of carbonic acid.

Practise this question

Question

Question 6 consists of multiple parts about carbon and carbon compounds. Figure 9 shows four carbon-based structures: A (buckminsterfullerene), B (carbon nanotube), C (layers of graphite), and D (polymer chain). Students identify graphite and poly(ethene) from these options. Figure 10 shows the tetrahedral giant covalent lattice of diamond, asking how many covalent bonds each carbon forms and which property it possesses. Figure 11 displays a ball-and-stick model of an ethane molecule with two carbon atoms and six hydrogen atoms, requiring the molecular formula. The final parts involve identifying the ion produced by carbonic acid (H2CO3) in aqueous solution and calculating its relative formula mass (Mr) given Ar values.
Question text

06 This question is about carbon and compounds of carbon.

Figure 9 shows diagrams that represent different structures.

Figure 9

Use Figure 9 to answer questions 06.1 and 06.2.

06.1 Which diagram represents graphite?

[1 mark]

Tick ( ) one box.

A B C D

06.2 Which diagram represents poly(ethene)?

[1 mark]

Tick ( ) one box.

A B 23 C D

Figure 10 represents the structure of diamond.

Figure 10

06.3 How many covalent bonds does each carbon atom form in diamond?

*22* [1 mark]

06.4 Which is a property of diamond?

[1 mark]

Tick ( ) one box.

Conducts electricity

Low melting point

Very hard 24

06.5 Figure 11 shows a model of a molecule.

Figure 11

Complete the molecular formula of the molecule.

[1 mark]

Molecular formula = C__ H__

Carbonic acid is a compound of carbon.

The formula of carbonic acid is H2CO3

06.6 Which ion is produced by carbonic acid in aqueous solution?

[1 mark]

Tick ( ) one box.

H+ OH− O2−

06.7 Calculate the relative formula mass (Mr) of carbonic acid (H2CO3).

Relative atomic masses (Ar): H = 1 C = 12 O = 16

[2 marks]

Relative formula mass (Mr) =

Mark scheme

Show the mark scheme Mark scheme for Question 6: 06.1 awards 1 mark for C; 06.2 awards 1 mark for D; 06.3 awards 1 mark for 4 or four; 06.4 awards 1 mark for 'very hard'; 06.5 awards 1 mark for C2H6; 06.6 awards 1 mark for H+; 06.7 awards 1 mark for working (1 x 2) + 12 + (16 x 3) [or 2 + 12 + 48] and 1 mark for the final answer 62. Total marks: 8.

Question 6

AO /

Question Answers Extra information Mark

Spec. Ref.

06.1 C 1 AO1

4.2.3.2

AO /

Spec. Ref.

06.2 D 1 AO1

4.2.1.4

4.2.2.5

AO /

Spec. Ref.

06.3 4 / four 1 AO1

4.2.3.1

AO /

Spec. Ref.

06.4 very hard 1 AO1

4.2.2.6

4.2.3.1

AO /

Spec. Ref.

06.5 C2H6 1 AO2

4.2.1.4

AO /

18 Spec. Ref.

H+ 1 AO1

06.6

4.4.2.4

AO /

Spec. Ref.

(Mr =)

06.7 (1 × 2) + 12 + (16 × 3) allow (Mr =) 2 + 12 + 48 1 AO2

4.3.1.2

= 62 1

Total Question 6 8

How to answer it

Allotropes and Compounds of Carbon

📋 Revision Overview

What this question tests

  • Structural recognition: Identifying giant covalent allotropes (graphite, diamond), fullerenes (nanotubes, buckminsterfullerene), and polymers (poly(ethene)) from visual models.
  • Bonding and properties: Linking covalent bonding in diamond to its physical hardness and understanding its lack of electrical conductivity.
  • Molecular formulae: Translating 3D ball-and-stick representations into molecular formulae ( C₂H₆ ).
  • Acids in aqueous solution: Recalling that all acids release hydrogen ions ( H⁺ ) in solution.
  • Quantitative chemistry: Calculating relative formula mass ( Mᵣ ) accurately using relative atomic masses ( Aᵣ ).
Questions 06.1 & 06.2

Identifying Carbon Structures from Diagrams

Figure 9: Structures A, B, C, and D

✅ Correct Selections

06.1 Graphite: C (1 mark)

06.2 Poly(ethene): D (1 mark)

💡 Key Knowledge: Identifying Allotropes

  • A = Buckminsterfullerene (C₆₀): Spherical cage of 60 carbon atoms with 5- and 6-membered rings.
  • B = Carbon nanotube: Cylindrical fullerene; a rolled sheet of graphene.
  • C = Graphite: Giant covalent structure made of parallel, hexagonal layers with weak intermolecular forces between layers.
  • D = Poly(ethene): Long chain of covalently bonded carbon atoms, each attached to two hydrogen atoms.

🧠 Exam Technique: Elimination Strategy

Look at the features: graphite always shows separate flat layers of hexagonal rings. Polymers are represented as long repeating chains of -C-C- backbones with single bonds to side atoms.

❌ Common Errors

Confusing B (nanotube) with graphite because both have hexagonal rings. Remember, nanotubes are cylindrical tubes, whereas graphite is stacked flat planes.

Question 06.3

Bonding in Diamond

Number of covalent bonds per carbon atom

✅ Correct Answer

4 (or four ) (1 mark)

Award 1 mark for the correct number.

💡 Key Knowledge: Diamond vs. Graphite

  • In diamond, each carbon atom forms 4 strong covalent bonds in a rigid tetrahedral giant covalent lattice.
  • In graphite, each carbon atom forms only 3 covalent bonds, leaving one delocalised electron per atom that can move and carry charge.
Question 06.4

Physical Properties of Diamond

Selecting the correct property

✅ Correct Answer

Tick the box: Very hard (1 mark)

❌ Incorrect Options & Why

  • Conducts electricity: False. Diamond has no free delocalised electrons or ions. Graphite conducts, but diamond does not.
  • Low melting point: False. Diamond has a very high melting point because enormous energy is needed to break many strong covalent bonds.
Question 06.5

Molecular Formula of a Hydrocarbon

Figure 11: Ball-and-stick model of ethane

✅ Correct Answer

Molecular formula = C₂H₆ (1 mark)

Gives 2 after C and 6 after H.

🧠 Exam Technique: Systematic Counting

Always count the atoms systematically from the diagram:

  • Dark spheres = Carbon atoms = 2
  • Light spheres = Hydrogen atoms = 6
  • Check using alkane general formula: CnH2n+2 → for n = 2 , 2(2) + 2 = 6 .
Question 06.6

Aqueous Solutions of Acids

Ions produced by carbonic acid (H₂CO₃)

✅ Correct Answer

Tick the box: H⁺ (1 mark)

💡 Key Knowledge: Definition of Acids & Alkalis

  • Acids: Form H⁺ (hydrogen) ions when dissolved in water.
  • Alkalis: Form OH⁻ (hydroxide) ions in aqueous solution.
  • Even though carbonic acid is a weak acid, it still produces H⁺ ions by partially ionising in water.
Question 06.7

Calculating Relative Formula Mass (Mᵣ)

Carbonic acid: H₂CO₃

📐 Step-by-Step Calculation

Step 1: Identify atomic masses and counts:
  • Hydrogen (H): 2 atoms × 1 = 2
  • Carbon (C): 1 atom × 12 = 12
  • Oxygen (O): 3 atoms × 16 = 48
Step 2: Add the values together:

Mᵣ = (1 × 2) + 12 + (16 × 3)

Mᵣ = 2 + 12 + 48 = 62

[1 mark] for correct substitution: (1 × 2) + 12 + (16 × 3) or 2 + 12 + 48
[1 mark] for final answer: 62

❌ Common Errors to Avoid

  • Multiplying by subscripts incorrectly: Forgetting that oxygen has a subscript of 3, resulting in adding only 16 instead of 48.
  • Ignoring hydrogen subscript: Using 1 instead of 2 for H₂ .
  • Adding units: Relative formula mass ( Mᵣ ) is a ratio and has no units. Do not add 'g' or 'g/mol'.

Topics

Chemistry · C2: Bonding, Structure and the Properties of Matter · C3: Quantitative Chemistry · C4: Chemical Changes

Question and mark scheme from the AQA GCSE Chemistry examination, Chemistry Paper 1 (Foundation), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.