# Explain Dumas method of estimation of nitrogen.

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Explain Dumas method of estimation of nitrogen.

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Dumas method:

Principle:

This method is based on the fact that nitrogeneous compound when heated with cupric oxide in an atmosphere of CO2 yields free nitrogen.

Traces of nitrogen are reduced to elemental nitrogen by passing over heated copper spiral.

Description of the apparatus: CO2 Generator:

CO2 needed in this process’is prepared by heating magnetite or sodium bicarbonate contained in a hard glass tube (or) by the action of dil. HCl on marble in a kipp's apparatus. The gas is passed through the combustion tube after dried by bubbling through cone. H2SO4 .

Combustion tube:

The combustion tube is heated in a furnace is charged with:-

(a) A roll of oxidised copper gauze to prevent the back diffusion of products of combustion and to heat the organic substance mixed with CuO by radiation

(b) a weighed amount of organic substance mixed with excess of CuO

(c) a layer of CitO packed in about $\frac{2}{3}$ length of the tube and kept in position by loose asbestos plug on either side and

(d) a reduced copper piral which reduces any oxides of nitrogen formed during combustion of nitrogen.

Schiff’s nitromctc:

The nitrogen gas obtained by the decomposition of the substance in the combustion tube is mixed with considerable excess of CO2 . It is estimated by passing nitro meter when CO2 is absorbed by KOH and the nitrogen gas gets collected in the upper part of the graduated tube.

$C_xH_yN_z+\Big(2x+\frac{1}{2}\Big)CuO$$\longrightarrow$$xCO_2+\frac{y}{2}H_2O+\frac{z}{2}N_2\Big(2x+\frac{y}{2}\Big)Cu$

Calculation:

Weight of the substance taken Wg

Volume of nitrogen = $V_1L$

Room temperature = $T_1K$

Atmospheric pressure = p mm Hg

Aqueous tension at room temperature P’ nun of Hg

Pressure of dry nitrogen = P—P’ = P’ mm Fig

$P_0$ , $V_0$ and $T_0$ be the pressure, volume and temperature respectively of dry nitrogen at S.T.P.

Then, $\frac{P_0V_0}{T_0}=\frac{P_1V_1}{T_1}$

V0 = $\frac{P_1V_1}{T_1}$ x $\frac{T_0}{P_0}$

V0$\frac{P_1V_1}{T_1}$ x $\frac{273K}{760mm\,Hg}$

22.4 L of N2 at STP weigh 28 g of N2

V0L of N2 at STP weigh $\frac{28}{22.4}\times V_0$

W g of organic compound contain $\frac{28}{22.4}\times V_0$ g of N2

100 g of organic contain $\frac{28}{22.4}\times \frac{V_0}{w}\times100$ = % of Nitrogen