Question

Consider the below reaction where $6.1g$ of benzoic acid is used to get $7.8g$ of m-bromo benzoic acid.
The percentage yield of the product is_____. (Round off to the nearest integer)
[Given : Atomic masses : C : 12.0 u, H : 1.0 u, O : 16.0 u, Br : 80.0 u]

Answer 1

Answer: 78
Molecular mass of $PhCOOH$ is $122g/mol$
Molecular mass of $C_6{H}_{4}COOHBr$ is $201g/mol$

$\frac{\text{Moles of}PhCOOH}{1}=\frac{\text{Moles of}{C}_6{H}_{4}COOHBr}{1}$
$\text{Moles of}{C}_6{H}_{4}COOBr=\frac{6.1}{122}=\frac{1}{20}\text{mol}$
$\text{Mass of}{C}_6{H}_{4}COOBr=201\times \frac{1}{20}g$
$\text{\% yeild}=\frac{7.8}{201/20}\times 100$
$=77.612$%
$\simeq 78\text{(Nearest Integer)}$