DP Chemistry (first assessment 2025)

Question 22M.2.SL.TZ1.1b(iii)

Date	May 2022	Marks available	[Maximum mark: 2]	Reference code	22M.2.SL.TZ1.1b(iii)
Level	SL	Paper	2	Time zone	TZ1
Command term	Deduce	Question number	b(iii)	Adapted from	N/A

b(iii).

[Maximum mark: 2]

22M.2.SL.TZ1.1b(iii)

(b(iii))

Assume the reaction in (a)(i) is the only one occurring and it goes to completion, but some product has been lost from the crucible. Deduce the percentage yield of magnesium oxide in the crucible.

[2]

Markscheme

$⟨ ⟨ 0.2614 mol \times (24.31 g {mol}^{- 1} + 16.00 g {mol}^{- 1}) = 0.2614 mol \times 40.31 g {mol}^{- 1} ⟩ ⟩ = 10.536 « g »$ ✔

$⟨ ⟨ 100 \times \frac{9.569 g}{10.536 g} = 90.822 ⟩ ⟩ = 91 « % »$ ✔

Award «0.2614 mol x 40.31 g mol^–1»

Accept alternative methods to arrive at the correct answer.

Accept final answers in the range 91-92%

[2] for correct final answer.

Examiners report

This was not well answered, but definitely better than the previous question with quite a few gaining some credit for correctly determining the theoretical yield.

Syllabus sections

Reactivity 2. How much, how fast and how far? » Reactivity 2.1—How much? The amount of chemical change » Reactivity 2.1.4—The percentage yield is calculated from the ratio of experimental yield to theoretical yield. Solve problems involving reacting quantities, limiting and excess reactants, theoretical, experimental and percentage yields.

Question 22M.2.SL.TZ1.1b(iii)

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