OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2024: Question 6

1 mark · Medium difficulty · Multiple Choice

Calculate the volume of oxygen gas needed at RTP for the complete combustion of a given mass of an alcohol using its balanced equation and molar mass.

Practise this question

Question

Multiple choice question 6 asking to calculate the volume of oxygen gas in dm3 needed for the complete combustion of 4.30 g of C5H9OH (Mr = 86.0) at RTP, given the balanced equation C5H9OH(l) + 7O2(g) -> 5CO2(g) + 5H2O(l). Four options are provided: A 1.2, B 2.4, C 5.8, D 8.4, along with an answer box.
Question text

64.30 g of the alcohol C5H9OH, (Mr = 86.0), is burned in oxygen.

C5H9OH(l) + 7O2(g) 5CO2(g) + 5H2O(l)

Which volume of oxygen gas is needed, in dm3, for this complete combustion of C H OH,

at RTP?

A 1.2

B 2.4

C 5.8

D 8.4

Your answer

[1]

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer for question 6 is D.

6 D 1

How to answer it

Calculating Oxygen Volume in Alcohol Combustion

What this question tests

This question assesses your core quantitative chemistry skills: calculating moles from mass and molar mass, applying stoichiometric mole ratios from a balanced equation, and converting gas moles to volume at Room Temperature and Pressure (RTP) using the molar gas volume (24.0 dm³ mol⁻¹).

Question 6: Multiple Choice Solution

Correct Option: D (8.4 dm³)

✅ Correct Answer

Option D is correct because 4.30 g of C₅H₉OH corresponds to 0.050 mol. Using the 1 : 7 stoichiometric ratio, this requires 0.350 mol of O₂ , which occupies 8.4 dm³ at RTP ( 0.350 × 24.0 ).

💡 Key Knowledge

  • Moles from Mass: Moles = Mass ÷ Mᵣ
  • Reacting Ratios: Coefficients in balanced equations give exact molar ratios.
  • Gas Volume at RTP: 1 mole of any gas occupies 24.0 dm³ at RTP ( Volume = Moles × 24.0 ).

🧠 Exam Technique

Don't panic over unfamiliar alcohol formulas. Trust the balanced equation provided in the stem! Follow a logical 3-step calculation framework: find moles of the given substance, use the mole ratio to find moles of the target substance, then convert to the final requested unit.

❌ Common Errors

  • Forgetting to multiply the moles of alcohol by the 7 coefficient for oxygen (distractors A or B often result from ratio slips).
  • Using 22.4 dm³ instead of 24.0 dm³ for RTP calculations (a common GCSE holdover, but A-Level OCR strictly uses 24.0 dm³ unless stated otherwise).

📐 Step-by-Step Calculation Guide

  1. Step 1: Calculate moles of the alcohol ( C₅H₉OH )
    Moles = 4.30 g ÷ 86.0 g mol⁻¹ = 0.050 mol
  2. Step 2: Use the stoichiometric ratio to find moles of O₂
    From the equation, 1 mol C₅H₉OH : 7 mol O₂
    Moles of O₂ = 0.050 mol × 7 = 0.350 mol
  3. Step 3: Calculate the volume of O₂ at RTP
    Volume = Moles × 24.0 dm³ mol⁻¹ = 0.350 mol × 24.0 dm³ mol⁻¹ = 8.4 dm³
Mark Scheme Note: 1 mark awarded for selecting D. In multiple-choice questions, working is not assessed, but writing down steps clearly on your question paper prevents careless calculator slips.

Topics

Module 2: Foundations in chemistry · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.