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2018 O Level Chem P1 Q12 — Stoichiometry / Percentage Yield Solution

Stoichiometry / Percentage Yield 2018 O Level Pure Chemistry · Paper 1 · Q12

Question

Bromobutane, C₄H₉Br, can be made from butanol using the reaction shown. C₄H₉OH + HBr → C₄H₉Br + H₂O In an experiment, 10 g of butanol produced 12 g of bromobutane. What is the percentage yield of bromobutane? [Mr: C₄H₉OH, 74; C₄H₉Br, 137]

A45%
B54%
C65%
D83%

Correct answer: C

Video Solution

Explanation

Answer: C (65%). The reaction \(\text{C}_4\text{H}_9\text{OH} + \text{HBr} \rightarrow \text{C}_4\text{H}_9\text{Br} + \text{H}_2\text{O}\) has a 1 : 1 mole ratio.

Moles of butanol used \(= 10 / 74 = 0.135\) mol, so the theoretical yield of bromobutane is also 0.135 mol. Theoretical mass \(= 0.135 \times 137 = 18.5\) g.

Percentage yield \(= (\text{actual} / \text{theoretical}) \times 100\% = (12 / 18.5) \times 100\% = 64.8\% \approx 65\%\). Answer C.

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