OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2023: Question 8

1 mark · Medium difficulty · Multiple Choice

Identify the alkane that undergoes complete combustion given the volume of oxygen required at RTP and the moles of the alkane.

Practise this question

Question

Multiple choice question 8 asking to identify which alkane (pentane, hexane, heptane, or octane) undergoes complete combustion when 0.100 mol of the alkane requires 22.8 dm3 of O2 measured at RTP.
Question text

8 For complete combustion, 0.100 mol of an alkane requires 22.8 dm3 of O , measured at RTP.

Which alkane has undergone complete combustion?

A pentane

B hexane

C heptane

D octane

Your answer

[1]

Mark scheme

Show the mark scheme Mark scheme indicating the correct answer is option B for question number 8, worth 1 mark.

8 B 1 AO2.2

How to answer it

Identifying an Alkane via Combustion Stoichiometry

What this question tests

This question assesses your ability to combine gas molar volume concepts with stoichiometric reacting ratios for the complete combustion of alkanes (general formula CₙH₂ₙ₊₂ ). You must scale moles from given volumes at RTP and apply the general combustion equation balance.

Question 8 (Multiple Choice)

Exam Breakdown & Step-by-Step Solution

✅ Correct Answer

B (hexane)

Awarded 1 mark for correct identification (AO2.2).

💡 Key Knowledge

  • Molar volume of a gas at RTP = 24.0 dm³ mol⁻¹ .
  • General equation for complete combustion of an alkane: CₙH₂ₙ₊₂ + ((3n+1)/2) O₂ → n CO₂ + (n+1) H₂O
  • Relating mole ratios directly to reacting gas volumes under identical conditions.

📐 Calculation Steps

  1. Find moles of O₂ required:
    Moles = Volume / Molar Volume = 22.8 dm³ / 24.0 dm³ mol⁻¹ = 0.950 mol
  2. Determine the reacting mole ratio:
    Ratio of alkane to O₂ = 0.100 mol : 0.950 mol
    Simplifies to 1 : 9.5 (for every 1 mole of alkane, 9.5 moles of O₂ are needed).
  3. Use the balancing coefficient for O₂:
    From the general equation, coefficient for O₂ is (3n + 1) / 2 = 9.5
    3n + 1 = 19
    3n = 18 ⇒ n = 6
  4. Identify the alkane:
    Since n = 6 , the alkane is hexane ( C₆H₁₄ ).

🧠 Exam Technique

If you prefer not to use algebra, test the options by scaling up their balanced equations to match 0.100 mol of alkane:

  • Hexane: C₆H₁₄ + 9.5 O₂ → 6 CO₂ + 7 H₂O
  • For 0.100 mol of hexane, oxygen needed = 0.100 × 9.5 = 0.950 mol .
  • Volume of O₂ = 0.950 mol × 24.0 dm³ mol⁻¹ = 22.8 dm³ . Matches the question perfectly!

❌ Common Errors & Pitfalls

  • Forgetting RTP molar volume: Using 22.4 dm³ mol⁻¹ (which is standard temperature and pressure, STP, rather than room temperature and pressure, RTP). Always use 24.0 dm³ mol⁻¹ unless stated otherwise in OCR assessments.
  • Ratio inversion: Dividing moles of O₂ by moles of alkane incorrectly and messing up the (3n+1)/2 algebraic rearrangement. Always check your value of n against the carbon count in the name (e.g., hex = 6).

Topics

Module 2: Foundations in chemistry · Module 4: Core organic chemistry · 2.1 Atoms and reactions · 4.1 Basic concepts and hydrocarbons

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.