WJEC A-Level Chemistry AS Unit 1, June 2025: Question 8

2 marks · Medium difficulty · Short Answer

Calculate the number of nitrogen atoms present in 66.0 g of dinitrogen monoxide, N2O.

Practise this question

Question

Question 8 asks: 'Calculate the number of nitrogen atoms in 66.0 g of N2O.' with 2 marks allocated. An answer line is provided reading 'Number of nitrogen atoms = '.

Mark scheme

Show the mark scheme Mark scheme for Question 8 shows two marking points: first mark for calculating the amount in moles of N2O as n(N2O) = 66.0 / 44.0 = 1.5 mol; second mark for calculating the number of nitrogen atoms as 1.5 × 2 × 6.02 × 10^23 = 1.81 × 10^24, noting error carried forward (ecf) is possible.

How to answer it

Calculating the Number of Atoms in a Given Mass of N₂O

📋 What this question tests

This question assesses fundamental mole concept calculations from AS Chemistry Unit 1:

  • Calculating relative formula mass ( Mᵣ ) from chemical formulas using the Periodic Table.
  • Converting mass into moles using the formula: moles = mass / Mᵣ .
  • Applying the Avogadro constant ( Nₐ = 6.02 × 10²³ mol⁻¹ ) to find the total count of molecules.
  • Using molecular stoichiometry to determine the count of a specific constituent atom within a compound (recognising the 2 : 1 ratio of N to N₂O).

Question 8

Calculate the number of nitrogen atoms in 66.0 g of N₂O. [2 Marks]

📐 Step-by-Step Calculation

  1. Calculate Mᵣ of N₂O:
    Mᵣ(N₂O) = (2 × 14.0) + 16.0 = 44.0 g mol⁻¹
  2. Calculate moles of N₂O:
    n(N₂O) = mass / Mᵣ = 66.0 / 44.0 = 1.50 mol
    [1 mark awarded here]
  3. Account for the number of nitrogen atoms per molecule:
    Each molecule of N₂O contains 2 nitrogen atoms.
    Moles of N atoms = 1.50 × 2 = 3.00 mol
  4. Multiply by the Avogadro constant:
    Number of N atoms = 3.00 × 6.02 × 10²³ = 1.81 × 10²⁴ (or 1.806 × 10²⁴)
    [1 mark awarded here]

✅ Correct Answer & Mark Scheme

Final Answer: 1.81 × 10²⁴ (or 1.806 × 10²⁴ )

Mark Breakdown:
  • Mark 1: Moles of N₂O = 66.0 / 44.0 = 1.5 mol
  • Mark 2: 1.5 × 2 × 6.02 × 10²³ = 1.81 × 10²⁴

Note: Error Carried Forward (ecf) is permitted if an incorrect number of moles is calculated in step 1, provided the remaining operations (multiplying by 2 and 6.02 × 10²³) are executed correctly.

💡 Key Knowledge

  • The Avogadro Constant: L or Nₐ = 6.02 × 10²³ mol⁻¹ is provided in the WJEC data booklet.
  • Atoms vs Molecules: 1 mole of N₂O molecules contains 1 mole of oxygen atoms, but 2 moles of nitrogen atoms. Always check the subscript!
  • Significant Figures: The input data has 3 significant figures ( 66.0 g ), so standard practice is to state the final answer to 3 significant figures ( 1.81 × 10²⁴ ).

🧠 Exam Technique

  • Write out the formula first: Writing n = m / Mᵣ clearly shows your method and secures method marks if arithmetic slips happen.
  • Highlight the target particle: Read carefully whether the question asks for molecules of N₂O or atoms of N. Underline "nitrogen atoms" in the question stem.
  • Scientific Notation: Always write standard form properly using powers of 10. Avoid typing calculator shorthand like 1.81E24 .

❌ Common Errors to Avoid

  • Forgetting to multiply by 2: Calculating 1.5 × 6.02 × 10²³ = 9.03 × 10²³ . This is the number of N₂O molecules, not nitrogen atoms.
  • Incorrect Mᵣ: Forgetting that nitrogen is diatomic in N₂O, e.g., using 14.0 + 16.0 = 30.0 instead of 44.0 .
  • Multiplying mass by Avogadro constant directly: Forgetting to convert mass into moles first.

Topics

Physical Chemistry · 1.3 Chemical calculations

Question and mark scheme from the WJEC A-Level Chemistry examination, AS Unit 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.