Question

Dumas’ method involved the determination of nitrogen content in the organic compound in the form of:
(A) $N{H}_3$
(B) $N_2$
(C) NaCN
(D) $(N{H}_4{)}_{2}S{O}_4$

Answer 1

Hint: Dumas’ method is a method used for quantitative determination of nitrogen content. The product of this method is a mixture of gases from which a diatomic gas is isolated which has atomic number 7 and is present in abundant amounts in the earth’s atmosphere.


Complete answer:
-Dumas method is a method for quantitative determination of nitrogen in the chemical substances. In this method the given amount of substance is heated with copper oxide in the presence of $C{O}_2$ atmosphere, at a temperature of 800 – 900°C. The combustion causes release of carbon dioxide ($C{O}_2$), water ($H_{2}O$) and nitrogen ($N_2$). These gases are then passed over special columns which absorb the carbon dioxide and water. The nitrogen content is then measured by a thermal conductivity detector at the end.
-This reaction for Dumas’ method can be written as:
   $C_x{H}_y{N}_z+(2x+y/2)CuO\to xC{O}_2+(y/2)H_{2}O+(z/2)N_2+(2x+y/2)Cu$
This mixture of gases is placed over KOH solution which absorbs $C{O}_2$.
-We will take the mass of organic compound containing N to be = m g
Volume of $N_2$ collected = $V_1$ L
Room temperature = T K
Atmospheric pressure at which this nitrogen is collected= P mm of Hg
Aqueous tension at room temperature or pressure of water vapour at room temperature = $P^{\circ }$ mm of Hg
Pressure of dry nitrogen will be = $P-P^{\circ }$ = $P_1$ mm of Hg
First we will find the volume of nitrogen at standard conditions (STP).

According to the gas law:
            ${\left({\displaystyle \frac{PV}{T}}\right)}_1={\left({\displaystyle \frac{PV}{T}}\right)}_2$
     where (1) is for the given room conditions and (2) is for STP conditions.
Let $P^{\circ }$(760 mm of Hg), $V^{\circ }$ and $T^{\circ }$(273 K) will represent the respective conditions at STP.

          ${\displaystyle \frac{P_1{V}_1}{T_1}}=\frac{P^{\circ }{V}^{\circ }}{T^{\circ }}$
     ${\displaystyle \frac{\left(P-P^{\circ }\right)V_1}{T_1}}={\displaystyle \frac{760\times V^{\circ }}{273}}$
            $V^{\circ }={\displaystyle \frac{(P-P^{\circ })V_1\times 273}{760\times T_1}}$
This is the volume of nitrogen collected at STP conditions.

-Now we will calculate the percentage of nitrogen.
 We all know that 22.4 L or 22400$c{m}^3$$N_2$ at STP weighs = 28 g
So, $V^{\circ }c{m}^3$ of $N_2$ at STP weights = ${\displaystyle \frac{28\times V^{\circ }}{22400}}$ g
So, now the percentage of nitrogen will be calculated by:
                         % = (weight of $N_2$ / weight of organic compound) × 100
                        % = ${\displaystyle \frac{28\times V^{\circ }}{22400\times m}}\times 100$
So, the correct option is: $N_2$

So, the correct answer is “Option B”.

Note: Sometimes some oxides of nitrogen are also released in the final products. These nitrogen oxides are reduced back to $N_2$ by passing through heated copper gauze.

                                      $Nitrogenoxides+Cu\to N_2+CuO$

All the nitrogen from the mixed products is isolated by passing it through a solution of KOH which absorbs all the $C{O}_2$ and $H_{2}O$.
Also remember that the volume term in the final formula ($V^{\circ }$) is always volume at STP conditions.