OCR A-Level Chemistry AS Breadth in chemistry (01), November 2020: Question 18

1 mark · Medium difficulty · Multiple Choice

Calculate the percentage yield of CH3CH2CHOHCH3 given the mass of 2-bromobutane reacted and the mass of product formed.

Practise this question

Question

A multiple choice question numbered 18. It presents a chemical reaction equation for the reaction of 2-bromobutane with sodium hydroxide to form butan-2-ol and sodium bromide, along with relative molecular masses for the reactant and product. Below the equation, it states the mass of 2-bromobutane and the actual mass of butan-2-ol produced, asking for the percentage yield with options A (10.7%), B (19.8%), C (36.7%), and D (54.1%).
Question text

18 A student reacts 24.24 g of 2-bromobutane in the reaction below.

CH3CH2CHBrCH3 + NaOH → CH3CH2CHOHCH3 + NaBr

Mr = 136.9 Mr = 74.0

The reaction produces 4.81 g of CH3CH2CHOHCH3.

What is the percentage yield of CH3CH2CHOHCH3?

A 10.7%

B 19.8%

C 36.7%

D 54.1%

Your answer [1]

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer for question 18 is option C.

18 C 1 2.8 ALLOW 36.7

How to answer it

Percentage Yield Calculation

📌 What this question tests

This question assesses your ability to calculate the percentage yield of a chemical reaction using reactant mass, molar masses (Mᵣ), stoichiometric ratios, and actual vs. theoretical yields. It tests core quantitative chemistry skills required across AS Level OCR Chemistry.

Question 18: Multiple Choice Calculation

Determining the percentage yield ofbutan-2-ol from 2-bromobutane

✅ Correct Answer: C (36.7%)

Following the step-by-step stoichiometric calculation yields a theoretical mass of 13.12 g, leading to a percentage yield of 36.7%.

💡 Key Knowledge

  • Moles formula: Moles = Mass / Mᵣ
  • Stoichiometry: The balanced equation shows a 1:1 molar ratio between CH₃CH₂CHBrCH₃ and CH₃CH₂CHOHCH₃ .
  • Percentage Yield formula: (Actual Yield / Theoretical Yield) × 100

🧠 Exam Technique

Always calculate the maximum theoretical yield of the product first by finding the moles of the limiting reagent. Avoid mixing up reactant and product Mᵣ values when scaling masses.

❌ Common Errors

  • Dividing the actual mass by the reactant mass directly instead of converting both to moles.
  • Using the wrong Mᵣ value for the theoretical mass calculation.

📐 Step-by-Step Calculation Guide

  1. Step 1: Calculate moles of 2-bromobutane (reactant)
    Moles = 24.24 g / 136.9 g mol⁻¹ = 0.17706 mol
  2. Step 2: Determine moles of the product (butan-2-ol)
    Due to the 1:1 molar ratio in the equation, maximum moles of product = 0.17706 mol .
  3. Step 3: Calculate the maximum theoretical mass of the product
    Mass = Moles × Mᵣ = 0.17706 mol × 74.0 g mol⁻¹ = 13.102 g
  4. Step 4: Calculate the percentage yield
    Percentage Yield = (4.81 g / 13.102 g) × 100 = 36.7%
Examiner Note: Top-level responses structure their working clearly by stating formula substitutions at each step. Option C is directly supported by the mark scheme.

Topics

Module 2: Foundations in chemistry · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.