Ethyl benzoate (molar mass $150 g mo l^{- 1}$ ) weighing $12.00 g$ is saponified with excess NaOH and the mixture is acidified to give benzoic acid (molar mass $122 g mo l^{- 1}$ ). If $7.02 g$ of pure benzoic acid is isolated, what is the percentage yield? Give the answer to the nearest integer.

Numerical answer type, answer in %, tolerance 0.5.

Answer: 72 %

Moles of ester =

12.00/150 = 0.0800 mol

; theoretical benzoic acid =

0.0800 \times 122 = 9.76 g

. Yield =

7.02/9.76 \times 100 = 71.93%

, i.e. 72%.