Ethyl propanoate has a pineapple like odor and is used as a flavoring agent in fruit syrups. It is prepared as follows:

${\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{OH}}_{\text{(aq) }}+{\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{COOH}}_{\text{(aq) }}⟶{\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{COOC}}_2{\mathrm{H}}_{5\left(\mathrm{aq}\right)}+{\mathrm{H}}_2\mathrm{O}\left(\ell \right)$

In an experiment, 349 grams of ethyl propanoate was obtained from 250 grams of ethanol, with propanoic acid in excess:
(Molecular weight of ethyl propanoate: 102, molecular weight of ethanol : 46)

The percentage yield of the above reaction is :

From the given reaction, one mole of ethanol forms one mole of ethyl propanoate. Moles present in 250 g of ethanol is 5.43 moles.

$\mathrm{Moles}=\frac{\mathrm{Mass} \mathrm{of} \mathrm{ethanol}}{\mathrm{Molar} \mathrm{mass}}=\frac{250}{46}=5.43 \mathrm{mol}$

5.43 moles of ethanol forms 5.43 moles of ethyl propanoate. Mass present in 5.43 moles of ethyl propanoate is:

$$\begin{gathered} 5.43=\frac{\mathrm{Mass}}{102} \\ \mathrm{Mass}=553.86 \mathrm{g} \end{gathered}$$

Percentage yield is calculated as:

$$\begin{gathered} \% \mathrm{yield}=\frac{\mathrm{Actual} \mathrm{yield}}{\mathrm{Theoretical} \mathrm{yield}} \\ =\frac{349 \mathrm{g}}{553.86 \mathrm{g}}\times 100=63.01\% \end{gathered}$$