OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2019: Question 11

1 mark · Medium difficulty · Multiple Choice

Calculate the percentage yield of an organic product formed from the reaction of phenylamine with ethanoyl chloride.

Practise this question

Question

Multiple choice question 11. A student reacts 4.50 g of C6H5NH2 with excess CH3COCl according to the reaction equation C6H5NH2 + CH3COCl -> C6H5NHCOCH3 + HCl, where Mr of C6H5NH2 is 93.0 and Mr of C6H5NHCOCH3 is 135.0. Given that the reaction produces 3.25 g of C6H5NHCOCH3, four options A (49.8), B (68.9), C (72.2), and D (95.4) are provided to determine the percentage yield.
Question text

11 A student reacts 4.50 g of C6H5NH2 with excess CH3COCl in the reaction below.

C6H5NH2 + CH3COCl → C6H5NHCOCH3 + HCl

Mr = 93.0 Mr = 135.0

The reaction produces 3.25 g of C6H5NHCOCH3.

What is the percentage yield of C6H5NHCOCH3?

A 49.8

B 68.9

C 72.2

D 95.4

Your answer

[1]

Mark scheme

Show the mark scheme The correct answer for question 11 is A, worth 1 mark.

11 A 1 AO2.4

How to answer it

Percentage Yield Calculation (Aromatic Amine Acylation)

What this question tests

This question assesses quantitative chemistry skills applied to organic synthesis (specifically the acylation of phenylamine). You must demonstrate proficiency in calculating moles from mass, using stoichiometric reacting ratios from balanced chemical equations, and determining percentage yield to appropriate significant figures.

Question 11 (Multiple Choice)

Calculating Percentage Yield

✅ Correct Answer: A (49.8)

Option A is the correct percentage yield derived from stepping through moles, theoretical yield, and comparing it to the actual mass produced.

💡 Key Knowledge

  • Moles formula: Moles = Mass / M_r
  • Reacting Ratio: The balanced equation shows a 1 : 1 stoichiometric ratio between C₆H₅NH₂ and C₆H₅NHCOCH₃.
  • Percentage Yield formula: (Actual Yield / Theoretical Yield) × 100

🧠 Exam Technique

For multiple-choice calculations, never guess. Write out your steps clearly on rough paper. Watch out for common distractor values created by inverted ratios or incorrect M_r usage.

❌ Common Errors

  • Using the mass of the reactant directly in place of the theoretical mass.
  • Failing to account for the 1 : 1 mole ratio properly (though 1:1 here, students often panic-multiply by coefficients blindly).
  • Rounding intermediate values too early, leading to off-target final numbers.

📐 Step-by-Step Calculation Breakdown

  1. Step 1: Calculate moles of reactant (C₆H₅NH₂)
    Moles = Mass / M_r = 4.50 / 93.0 = 0.048387 mol
  2. Step 2: Determine theoretical moles of product (C₆H₅NHCOCH₃)
    Using the 1 : 1 stoichiometric ratio from the balanced equation:
    Theoretical moles of product = 0.048387 mol
  3. Step 3: Calculate theoretical mass of product
    Mass = Moles × M_r = 0.048387 × 135.0 = 6.5323 g
  4. Step 4: Calculate percentage yield
    Percentage Yield = (Actual Mass / Theoretical Mass) × 100
    Percentage Yield = ( 3.25 / 6.5323 ) × 100 = 49.7527% = 49.8% (to 3 sig fig)
Mark Scheme Allocation: 1 mark awarded for selecting A (AO2.4: Applying chemical knowledge and quantitative skills to unfamiliar contexts).

Topics

Module 2: Foundations in chemistry · Module 6: Organic chemistry and analysis · 6.2 Nitrogen compounds, polymers and synthesis · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.