Question:
When 0.15 g of an organic compound was analyzed using Carius method for estimation of bromine, 0.2397 g of AgBr was obtained. The percentage of bromine in the organic compound is _______. (Nearest integer)
[Atomic mass : Silver = 108, Bromine = 80]
Solution:
Moles of $\mathrm{Br}=$ Moles of $\mathrm{AgBr}$ obtained
$\Rightarrow$ Mass of $\mathrm{Br}=\frac{0.2397}{188} \times 80 \mathrm{~g}$
therefore $\% \mathrm{Br}$ in the organic compound
$=\frac{\mathrm{W}_{\mathrm{Br}}}{\mathrm{W}_{\mathrm{T}}} \times 100$
$=\frac{0.2397 \times 80}{188 \times 0.15} \times 100=0.85 \times 80$
= 68
$\Rightarrow$ Nearest integer is '68'