Question

An organic compound was analysed by Dumas method. $0.45gm$ of the compound on combustion gave $48.6ml$ nitrogen at $27\circ C$ and $756mm$ pressure. Calculate the percentage of nitrogen.

Answer 1

$V_1=48.6ml$ $V_2=?$
$P_1=756mm$ $P_2=760mm$
$T_1=27+273=300K$; $T_2=273K$
Applying the general gas equation $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$, we get
The volume of nitrogen at $STP$,
$V_2=\frac{P_1{V}_1}{T_1}\frac{T_2}{P_2}$
$=\frac{756\times 48.6}{300}\times \frac{273}{760}$
$=43.99ml$
Mass of organic compound = $0.45gm$
Percentage of nitrogen in the compound = $\frac{28}{22400}\times V_{2}W\times 100$
$=\frac{28}{22400}\times \frac{43.99}{0.45}\times 100$
$=12.22\%$ [$ANSWER$]