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) ${(N{H_4})_2}S{O_4}$

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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.

-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:
${\left( {\dfrac{{PV}}{T}} \right)_1} = {\left( {\dfrac{{PV}}{T}} \right)_2}$
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.

$\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P^ \circ }{V^ \circ }}}{{{T^ \circ }}}$
$\dfrac{{\left( {P - {P^ \circ }} \right){V_1}}}{{{T_1}}} = \dfrac{{760 \times {V^ \circ }}}{{273}}$
${V^ \circ } = \dfrac{{(P - {P^ \circ }){V_1} \times 273}}{{760 \times {T_1}}}$
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 = $\dfrac{{28 \times {V^ \circ }}}{{22400}}$ g
So, now the percentage of nitrogen will be calculated by:
% = (weight of ${N_2}$ / weight of organic compound) × 100
% = $\dfrac{{28 \times {V^ \circ }}}{{22400 \times m}} \times 100$
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.

$Nitrogen oxides + Cu \to {N_2} + CuO$

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.