In the Claisen-Schmidt reaction to prepare dibenzalacetone from 5.3 g of benzaldehyde, a total of 3.51 g of product was obtained. The percentage yield in this reaction was _____%. [2025]

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Solution

Q.
In the Claisen-Schmidt reaction to prepare dibenzalacetone from 5.3 g of benzaldehyde, a total of 3.51 g of product was obtained. The percentage yield in this reaction was _____%. [2025]

Ans.
(60)

Moles of benzaldehyde initially taken $(n_{C_{6} H_{5} C H O})$

$= \frac{Mass of C_{6} H_{5} C H O}{Molar mass of C_{6} H_{5} C H O} = \frac{5.3}{106} = 0.05 mol$

For 100% yield of the reaction, 2 mol of benzaldehyde give 1 mol of dibenzalacetone, so 0.05 mol of benzaldehyde give 0.025 mol of dibenzalacetone.

Actual moles of dibenzalacetone obtained

$= \frac{Mass of dibenzalacetone obtained}{Molar mass of dibenzalacetone} = \frac{3.51}{234} = 0.015 mol$

Percentage yield

$= \frac{Actual moles of dibenzalacetone obtained}{Moles of dibenzalacetone for 100% yield} \times 100$

$= \frac{0.015}{0.025} \times 100 = 60 %$

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