Question

$0.2g$ of an organic compound was subjected to estimation of nitrogen by Duma’s method in which volume of $N_2$ evolved (at STP) was found to be $22.400mL.$ The percentage of nitrogen in the compound is _____. [nearest integer]
(Given : Molar mass of $N_2$ is $28gmo{l}^{-1}$,
Molar volume of $N_2$ at STP : $22.4L$)

• A. 14
• B. 14.00
• C. 14.0

Answer 1

Answer: (A), (B), (C)

Dumas method is a method for quantitative determination of nitrogen in the chemical substances.
Nitrogen containing organic compound, when heated with copper oxide in an atmosphre of carbon dioxide, yields free nitrogen in addition to carbon dioxide and water.
$C_x{H}_y{N}_z+(2x+\frac{y}{2})CuO\to$
$xC{O}_2+\frac{y}{2}{H}_{2}O+\frac{z}{2}{N}_2+(2x+\frac{y}{2})Cu$
Traces of nitrogen oxides formed, if any, are reduced to nitrogen by passing the gasous mixture over a heated copper gauze. The mixture of gases so prdoced is collected over an aqueous soution of potassium hydroxide which absorbs carbon dioxide. Nitrogen is collected in the upper part of the graduated tube.
Let the mass of organic compounnd $=mg$
Volume of nitrogen collected $=V_{1}mL$
Room temperature $=T_{1}K$
Volumme of nitrogen at $STP=\frac{P_1{V}_1\times 273}{\underset{\text{Let it be}vmL}{760\times T_1}}$
Where $p_1$ and $V_1$ are the pressure and volume of nitrogen, $p_1$ is obtained by the rellation;
$p_1=$ atmosphre pressure - Aqueous tension
$22400mL{N}_2\text{at}STP\text{weighs}28g.$

$VmL{N}_2\text{at}STP\text{weighs}=\frac{28\times V}{22400}g$
Percentage of nitrogen $=\frac{28\times V\times 100}{22400\times m}$

Here,
$\text{Mass of organic compound}=0.2g$
$\text{Volume of}{N}_2\text{gas evolved at STP}=22.4mL$
$\text{Mass of}{N}_2\text{gas evolved}=\frac{22.4\times {10}^{-3}\times 28}{22.4}$
$=0.028g$
$\text{Percentage of nitrogen in the compound}=\frac{0.028\times 100}{0.2}=14\%$