Question

In Duma's method of 0.52 g of an organic compound on combustion gave 68.6 mL $N_2$ at ${27}^{o}C$ and 76 mm pressure. What is the percentage of nitrogen in the compound?

• A. 12.22%
• B. 14.93%
• C. 15.84%
• D. 16.23%

Answer 1

The correct option is B 14.93%

Organic compound $\underset{DumasMethod}{\to }{N}_2+$ other product

volume of $N_2.V=68.6ml$

pressure $P=76mmHg=1atm$

Temprature $T=273+27=300k$

$n=\frac{Pv}{RT}=\frac{1\times 68.6\times {10}^{-3}}{0.0821\times 300}=2.785\times {10}^{-2}$

amount of nitrogen $2.785\times {10}^{-2}\times 28=0.078$

% of nitrogen $=\frac{0.078}{0.52}=14.93\%$