Question:
Consider the above reaction where $6.1 \mathrm{~g}$ of benzoic acid is used to get $7.8 \mathrm{~g}$ of $\mathrm{m}$-bromo
benzoic acid. The percentage yield of the product is _______________ . (Round off to the Nearest integer)
[Given : Atomic masses : $\mathrm{C}=12.0 \mathrm{u}, \mathrm{H}: 1.0 \mathrm{u}$,$\mathrm{O}: 16.0 \mathrm{u}, \mathrm{Br}=80.0 \mathrm{u}]$
Solution:
(78)
Moles of Benzoic acid $=\frac{6.1}{122}=$ moles of $\mathrm{m}$-bromobenzoic acid So, weight thel mobenzoic acid
$=\frac{6.1}{122} \times 201 \mathrm{gm}$
$=10.05 \mathrm{gm}$
$\%$ yield $=\frac{\text { Actual weight }}{\text { Theoretical weight }} \times 100$
$=\frac{7.8}{10.05} \times 100$
$=77.61 \%$