AQA GCSE Chemistry Chemistry Paper 1 (Foundation), June 2023: Question 3

7 marks · Low Demand difficulty · Short Answer

Identify properties, isotopic composition, mass calculations, and allotropes of carbon including graphite and Buckminsterfullerene.

Practise this question

Question

Question 3 contains five sub-questions about carbon. 03.1 asks to classify carbon as a compound, element, or mixture via checkboxes. 03.2 provides a word box (electrons, ions, molecules, neutrons, protons) to complete sentences about carbon isotopes having the same number of one subatomic particle and different numbers of another. 03.3 gives the statement '12 g of carbon contains 6.02 × 10²³ atoms' and asks to choose the expression for the mass of one atom. 03.4 shows three diagrams of carbon allotropes labeled A (diamond), B (Buckminsterfullerene sphere), and C (carbon nanotube), asking to identify Buckminsterfullerene. 03.5 shows a layer structure of graphite and asks to draw lines matching two properties ('Graphite conducts electricity' and 'Graphite is soft') to four structural features.
Question text

03 This question is about carbon.

03.1 Which type of substance is carbon?

[1 mark]

Tick ( ) one box.

Compound

Element

Mixture

03.2 Carbon has isotopes with mass numbers 12, 13 and 14.

Complete the sentences.

Choose answers from the box.

[2 marks]

electrons ions molecules neutrons protons

The isotopes of carbon have the same number of .

The isotopes of carbon have a different number of .

03.3 12 g of carbon contains 6.02 ×1023 atoms.

Which expression is used to calculate the mass of one atom of carbon?

[1 mark]

Tick ( ) one box.

6.02 × 10

6.02 × 10

12 × 6.02 × 1023

03.4 Figure 2 shows diagrams that represent different forms of carbon.

Figure 2

Which diagram in Figure 2 represents Buckminsterfullerene?

[1 mark]

Tick ( ) one box.

A B10 C

03.5 Figure 3 represents part of the structure of graphite.

Figure 3

Draw one line from each property of graphite to the structural feature that is the

reason for that property.

[2 marks]

Property Structural feature

Graphite has hexagonal rings

of carbon atoms.

Graphite conducts electricity.

The bonds between carbon

atoms in the layers are strong.

Graphite is soft. There are no covalent bonds

between layers of atoms.

There are delocalised

electrons in graphite.

Mark scheme

Show the mark scheme Mark scheme for Question 3: 03.1 awards 1 mark for 'element'. 03.2 awards 1 mark for 'protons' (allow electrons) and 1 mark for 'neutrons' in that order. 03.3 awards 1 mark for the expression '12 / (6.02 × 10²³)'. 03.4 awards 1 mark for 'B'. 03.5 awards 1 mark for connecting 'Graphite conducts electricity' to 'There are delocalised electrons in graphite', and 1 mark for connecting 'Graphite is soft' to 'There are no covalent bonds between layers of atoms'.

AO /

Question Answers Extra information Mark

Spec. Ref.

03.1 element 1 AO1

4.1.1.1

AO /

Spec. Ref.

03.2 must be in this order AO1

4.1.1.5

protons allow electrons 1

neutrons 1

AO /

Spec. Ref.

03.3 12 1 AO2

6.02 × 1023 4.1.1.1

AO /

Spec. Ref.

03.4 B 1 AO1

4.2.3.3

– HEMISTRY – –

AO /

Question Answers Mark

Spec. Ref.

03.5 AO1

4.2.3.2

do not accept more than one line from a box on the left

– HEMISTRY – –

Total Question 3 7

Question 4

How to answer it

Carbon: Atomic Structure, Isotopes, and Allotropes

What this question tests

This question assesses foundational knowledge across Atomic Structure and the Periodic Table and Structure and Bonding:

  • Classifying substances as elements, compounds, or mixtures.
  • Defining isotopes in terms of subatomic particles (protons and neutrons).
  • Setting up mathematical expressions involving Avogadro's constant.
  • Identifying carbon allotropes (diamond, Buckminsterfullerene, nanotubes).
  • Linking microscopic structures of graphite to macroscopic properties (conductivity and softness).
Question 03.1 • 1 Mark

Classifying Carbon as a Substance

Type of substance: Compound, Element, or Mixture

✅ Correct Answer

Element

💡 Key Knowledge

An element consists of only one type of atom and cannot be broken down chemically into simpler substances. Carbon has the chemical symbol C and is found directly on the Periodic Table.

❌ Common Errors

Students sometimes tick compound or mixture because they confuse elemental forms (such as diamond, coal, or charcoal) with impure mixtures. Pure carbon in any allotrope is strictly an element.

Mark Scheme: 1 mark for ticking "Element". Any other selection scores 0.
Question 03.2 • 2 Marks

Isotopes of Carbon

Completing sentences using: electrons, ions, molecules, neutrons, protons

✅ Correct Answer

The isotopes of carbon have the same number of protons.

The isotopes of carbon have a different number of neutrons.

(Note: "electrons" is accepted for the first blank).

🧠 Exam Technique

  • The words must be in this exact order: protons first, neutrons second.
  • Atomic number (bottom number) = number of protons. Since all carbon atoms have atomic number 6, they always have 6 protons.
  • Mass number (top number) = protons + neutrons. Different mass numbers (12, 13, 14) mean different numbers of neutrons (6, 7, and 8).

❌ Common Errors

Mixing up the order (putting neutrons first and protons second) gives 0 marks. Selecting irrelevant distractor words like ions or molecules shows a failure to recall subatomic particles.

Mark Scheme: 1 mark for "protons" (allow "electrons"); 1 mark for "neutrons". Must be in this order.
Question 03.3 • 1 Mark

Calculating the Mass of a Single Atom

Working with Avogadro's constant and sample mass

✅ Correct Answer

Tick the first box:

12 6.02 × 10²³

📐 Step-by-Step Logic

  1. Total mass of sample = 12 g.
  2. Total number of atoms in sample = 6.02 × 10²³ atoms.
  3. To find the mass of one single atom, divide the total mass by the total number of atoms:
    Total Mass (g) Number of Atoms = 12 6.02 × 10²³

❌ Common Errors

Choosing the inverted option 6.02 × 10²³ 12 . That would calculate how many atoms are in 1 gram, not the mass of 1 atom.

Mark Scheme: 1 mark for selecting 12 / (6.02 × 10²³).
Question 03.4 • 1 Mark

Identifying Allotropes of Carbon

Recognising Buckminsterfullerene from molecular diagrams

✅ Correct Answer

Tick box B

💡 Knowing All Carbon Allotropes in Figure 2

  • A: Diamond — Rigid 3D giant tetrahedral network where each carbon forms 4 covalent bonds.
  • B: Buckminsterfullerene (C₆₀) — Spherical cage molecule composed of hexagons and pentagons (like a football).
  • C: Carbon Nanotube — Cylindrical fullerene tube with high tensile strength.
Mark Scheme: 1 mark for B.
Question 03.5 • 2 Marks

Relating Graphite's Structure to Its Properties

Matching properties to structural features

✅ Correct Matches

  • Graphite conducts electricity ➔
    "There are delocalised electrons in graphite."
  • Graphite is soft ➔
    "There are no covalent bonds between layers of atoms."

💡 Why These Match

  • Electrical conductivity: In graphite, each carbon atom bonds to only 3 others. The 4th outer electron is delocalised and free to move throughout the structure carrying charge.
  • Soft / Slippery: The hexagonal layers are held together only by weak intermolecular forces (no covalent bonds between layers), allowing the layers to easily slide over one another.

❌ Critical Exam Warning

Rule: The question states: "Draw one line from each property..." If you draw more than one line emerging from a box on the left, the mark scheme instructs the examiner to award 0 marks for that property.

Distractor Alert: "The bonds between carbon atoms in the layers are strong" explains graphite's high melting point, NOT why it is soft!

Mark Scheme: 1 mark per correct linking line. Do NOT accept more than one line from a box on the left. Maximum 2 marks.

Topics

Chemistry · C1: Atomic Structure and the Periodic Table · C2: Bonding, Structure and the Properties of Matter · C3: Quantitative Chemistry

Question and mark scheme from the AQA GCSE Chemistry examination, Chemistry Paper 1 (Foundation), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.