AQA A-Level Chemistry Paper 3, 2021: Question 29

1 mark · Medium difficulty · Multiple Choice

This question assesses your understanding of percentage yield calculations in organic synthesis. You’re given the masses and relative molecular masses (Mr) of the reactant and product in a nitration reaction, and you must apply stoichiometric ratios and percentage yield to determine the actual mass of product formed. This tests both your grasp of mole ratios and your skill with percentage yield formulas.

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Question

AQA A-Level Chemistry Paper 3, 2021: Question 29
Question text

29 Nitration of 1.70 g of methyl benzoate (Mr = 136.0) produces methyl 3-nitrobenzoate

(Mr = 181.0). The percentage yield is 65.0%

What mass, in g, of methyl 3-nitrobenzoate is produced?

[1 mark]

A 0.830

B 1.10

C 1.47

D 2.26

Mark scheme

Show the mark scheme Mark scheme for AQA A-Level Chemistry Paper 3, 2021: Question 29

29 C 1 1.47

How to answer it

Multiple Choice Explained: Yield of Methyl 3-nitrobenzoate

Correct Answer: C – 1.47 g

Step-by-step:

  • Moles of methyl benzoate = 1.70 / 136 = 0.0125 mol
  • Assuming 1:1 molar ratio, max moles of product = 0.0125 mol
  • Theoretical mass = 0.0125 × 181 = 2.26 g
  • Actual mass = 2.26 × 0.65 = 1.47 g

Why A is incorrect:

0.830 g would imply a yield far below 65%. This is too low based on the given percentage yield.

Why B is incorrect:

1.10 g suggests around 49% yield, which does not match the 65% stated. It underestimates the actual mass.

Why D is incorrect:

2.26 g is the maximum theoretical yield, not the actual mass after 65% yield is applied. This answer ignores the percentage yield completely.

Topics

Physical Chemistry · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.