# An organic compound is subjected to chlorination

Question:

An organic compound is subjected to chlorination to get compound A using $5.0 \mathrm{~g}$ of chlorine. When $0.5 \mathrm{~g}$ of compound $\mathrm{A}$ is reacted with $\mathrm{AgNO}_{3}$ [Carius Method], the percentage of chlorine in compound $\mathrm{A}$ is___________ when it forms $0.3849 \mathrm{~g}$ of $\mathrm{AgCl}$. (Round off to the Nearest Integer)

$\mathrm{AgCl}$. (Round off to the Nearest Integer)

(Atomic masses of $\mathrm{Ag}$ and $\mathrm{Cl}$ are $107.87$ and $35.5$ respectively)

Solution:

$\mathrm{n}_{\mathrm{c} \ell}$ in compound $=\mathrm{n}_{\mathrm{AgCl}}=\frac{0.3849 \mathrm{~g}}{(107.87+35.5)} \mathrm{g} / \mathrm{mol}$

$\Rightarrow$ mass of chlorine $=\mathrm{n}_{\mathrm{Cl}} \times 35.5=0.0953 \mathrm{gm}$

$\Rightarrow \%$ wt of chlorine $=\frac{0.0953}{0.5} \times 100$

$=19.06 \%$

OR

Mass of organic compound $=0.5 \mathrm{gm}$.

mass of formed $\mathrm{AgCl}=0.3849 \mathrm{gm}$

$\%$ of $\mathrm{Cl}=\frac{\text { atomic mass of } \mathrm{Cl} \times \text { mass formed } \mathrm{AgCl}}{\text { molecular mass of } \mathrm{AgCl} \times \text { mass of organic compound }} \times 100$

$=\frac{35.5 \times 0.3849}{143.37 \times 0.5} \times 100$

$=19.06$

$\approx 19$