AQA A-Level Chemistry Paper 3, 2017: Question 33

1 mark · Medium difficulty · Multiple Choice

Calculate the mass of phenylethanone produced given the mass of ethanoyl chloride reacting with excess benzene and a 62% yield.

Practise this question

Question

Question 33 shows a chemical reaction: ethanoyl chloride (CH3COCl) reacts with benzene to produce phenylethanone (C6H5COCH3) and HCl. It states: 'In a preparation, with an excess of benzene, the mass of ethanoyl chloride (Mr = 78.5) used was 5.7 x 10^-2 kg. The percentage yield of phenylethanone was 62%. What mass, in grams, of phenylethanone was produced?' Four multiple-choice options are given: A 35 g, B 54 g, C 87 g, D 102 g.
Question text

33 Phenylethanone can be prepared by the reaction between ethanoyl chloride and

benzene.

In a preparation, with an excess of benzene, the mass of ethanoyl chloride (Mr = 78.5)

used was 5.7 × 10−2 kg.

The percentage yield of phenylethanone was 62%.

What mass, in grams, of phenylethanone was produced?

[1 mark]

A 35 g

B 54 g

C 87 g

D 102 g

Mark scheme

Show the mark scheme Mark scheme table showing question number 33 with correct answer option B.

33 B

How to answer it

Preparation of Phenylethanone: Yield Calculation

📌 What this question tests

This question assesses your ability to perform a multi-step reacting mass and percentage yield calculation. Key competencies include: converting units from kg to g , calculating the relative formula mass (Mr) of an aromatic compound from its skeletal/structural formula, using 1:1 reaction stoichiometry, and determining actual mass produced from a percentage yield.

Question 33 (1 Mark)

Calculation Breakdown & Analysis

Reaction: CH₃COCl + C₆H₆ → C₆H₅COCH₃ + HCl

✅ Correct Answer

B — 54 g

Ethylethanone (Mr = 120.0) produced at 62% yield gives an actual product mass of 54.0 g.

💡 Key Knowledge

  • Stoichiometry: 1 mole of ethanoyl chloride forms 1 mole of phenylethanone.
  • Molecular Formula: Phenylethanone has a benzene ring with an acetyl group ( C₆H₅COCH₃ ), giving the molecular formula C₈H₈O .
  • Percentage Yield Formula:
    Actual Mass = (Theoretical Mass × % Yield) / 100

📐 Step-by-Step Calculation

  1. Convert the given mass to grams:
    Mass of CH₃COCl = 5.7 × 10⁻² kg = 5.7 × 10⁻² × 1000 g = 57.0 g
  2. Calculate moles of reactant (ethanoyl chloride):
    Moles = mass / Mr = 57.0 / 78.5 = 0.7261 mol
  3. Determine Mr of product (phenylethanone, C₈H₈O):
    Mr = (8 × 12.0) + (8 × 1.0) + (1 × 16.0) = 96.0 + 8.0 + 16.0 = 120.0
  4. Calculate maximum theoretical mass of phenylethanone:
    1:1 molar ratio → Theoretical moles of phenylethanone = 0.7261 mol
    Theoretical mass = 0.7261 mol × 120.0 g mol⁻¹ = 87.13 g
  5. Apply the 62% yield to find actual mass:
    Actual mass = 87.13 g × (62 / 100) = 54.02 g ≈ 54 g

❌ Common Distractor Traps

  • Distractor C (87 g): The theoretical maximum mass (100% yield). Students selected this if they forgot to multiply by the 62% yield.
  • Distractor A (35 g): Directly taking 62% of the starting reactant mass (57 g × 0.62 = 35.3 g), completely ignoring the difference in Mr between reactant and product.
  • Distractor D (102 g): Calculation error, often caused by incorrectly calculating the Mr of phenylethanone (e.g. including Cl or missing ring carbons).

🧠 Exam Technique & Speed Tips

  • Single line calculator entry: In multiple-choice papers, save time by chaining the operations:
    (57.0 / 78.5) × 120.0 × 0.62 = 54.02 g
  • Sanity Check: Because phenylethanone (Mr = 120) is significantly heavier than ethanoyl chloride (Mr = 78.5), 100% yield must be larger than 57 g (~87 g). 62% of ~87 g must be around 54 g.
  • Watch the prefix: Don't overlook × 10⁻² kg ; converting to standard lab units ( g ) first prevents power-of-ten errors.
Mark Scheme Reference: Multiple Choice Question 33 — Correct answer is B [1 mark].

Topics

Physical Chemistry · Organic Chemistry · 3.1.2 Amount of Substance · 3.3.10 Aromatic Chemistry

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.