Consider the above reaction where 6.1 g of Benzoic acid is used to get 7.8 g of m-bromo benzoic acid. The percentage yield to the product is ______

(Round off to the Nearest Integer)  [Given: Atomic masses: C:$12.0u,H:1.0u,O:16.0u,\text{}Br:80.0u]$

Molecular mass of Ph-COOH is 122 g/mol and C₆H₄COOHBr is 201 g/mol.

Moles of Ph-COOH is:

$$\begin{gathered} 1 \mathrm{mole} \mathrm{of} \mathrm{benzoic} \mathrm{acid}=1 \mathrm{mole} \mathrm{of} \mathrm{bromo} \mathrm{benzoic} \mathrm{acid} \\ 123 \mathrm{g} \mathrm{benzoic} \mathrm{acid}=203 \mathrm{g} \mathrm{bromo} \mathrm{benzoic} \mathrm{acid} \\ 6.1 \mathrm{g} \mathrm{benzoic} \mathrm{acid}=\frac{203 \mathrm{g} \mathrm{bromo} \mathrm{benzoic} \mathrm{acid}}{123 \mathrm{g} \mathrm{benzoic} \mathrm{acid}}\times 6.1 \mathrm{g} \mathrm{benzoci} \mathrm{acid}=10.06 \mathrm{g} \end{gathered}$$

Percentage yield:

$\frac{\mathrm{theoretical} \mathrm{mass}}{\mathrm{Observed} \mathrm{mass}}\times 100=\frac{7.8 \mathrm{g}}{10.08 \mathrm{g}}\times 100\cong 78\%$