OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2022: Question 9

1 mark · Medium difficulty · Multiple Choice

Calculate the volume of hydrogen gas required to react with 0.0500 mol of an unsaturated nitrile containing two double bonds to form a saturated compound.

Practise this question

Question

Multiple choice question 9 showing a skeletal formula of an unsaturated organic compound containing a nitrile group (-CN) and two carbon-carbon double bonds. The text asks for the volume of hydrogen gas, measured at room temperature and pressure, that reacts with 0.0500 mol of the compound to form a saturated compound. Four options are provided: A (2.40 dm3), B (3.60 dm3), C (4.80 dm3), and D (6.00 dm3), alongside an answer box.
Question text

9 The compound below reacts with hydrogen gas to form a saturated compound.

CN

What is the volume of hydrogen, measured at room temperature and pressure, that reacts with

0.0500 mol of the compound?

A 2.40 dm3

B 3.60 dm3

C 4.80 dm3

D 6.00 dm3

Your answer

[1]

Mark scheme

Show the mark scheme Mark scheme indicating the correct answer is C, with 1 mark awarded, showing the correct volume calculation resulting in 4.80 dm3.

9 C 1 AO2.6 ALLOW 4.8 (This is the correct volume)

How to answer it

Hydrogen Addition & Gas Volume Calculation

📌 What this question tests

This question assesses your ability to interpret skeletal organic structures, identify functional groups capable of reacting with hydrogen (alkenes/saturation), determine stoichiometric reacting ratios, and apply molar gas volume conversions at room temperature and pressure (rtp).

Question 9 Overview

Multiple Choice Strategy & Breakdown

✅ Correct Answer: C

The correct volume of hydrogen gas required is 4.80 dm³ (Option C).

💡 Key Knowledge

  • Skeletal structures: Every vertex and line terminus represents a carbon atom.
  • Functional groups: The molecule contains two double bonds ( C=C ) and one nitrile group ( -C≡N ). Only the carbon-carbon double bonds undergo catalytic hydrogenation under standard conditions to form a fully saturated compound.
  • Molar volume at rtp: 1 mole of any gas occupies 24.0 dm³ (or 24,000 cm³) at room temperature and pressure.

🧠 Exam Technique

Always count functional groups systematically before jumping into calculations. Watch out for hidden reactive sites or groups (like the nitrile triple bond -C≡N ) that do not react under the specified gentle conditions, preventing stoichiometric over-calculation.

❌ Common Errors

  • Over-reduction: Mistakenly assuming the nitrile ( -C≡N ) group reacts with H₂ to form an amine, leading to an incorrect 3:1 ratio instead of 2:1.
  • Ratio confusion: Forgetting that each double bond requires two moles of H atoms, meaning one molecule of H₂ per C=C bond.
  • Gas volume multiplier slip: Multiplying by 22.4 dm³ ( STP value) instead of the standard A-Level rtp value of 24.0 dm³ .

📐 Step-by-Step Calculation

  1. Identify the reacting ratio: The skeletal formula shows two C=C double bonds. To achieve a fully saturated compound, 2 moles of H₂ molecules are required per 1 mole of the organic compound. (Ratio = 1 : 2).
  2. Calculate moles of H₂ needed:
    Moles of organic compound = 0.0500 mol
    Moles of H₂ = 0.0500 × 2 = 0.100 mol
  3. Convert moles to volume at rtp:
    Volume = Moles × Molar Volume (24.0 dm³)
    Volume = 0.100 mol × 24.0 dm³ mol⁻¹ = 4.80 dm³
Mark Allocation: 1 mark for identifying option C (AO2.6: Applying quantitative chemistry and structural analysis to unfamiliar contexts).

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 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.