A sample of 0 . 125   g of an organic compound when analysed by Duma's method yields 22 . 78  …

Let the mass of organic compound = m g

Volume of nitrogen collected = V₁ mL

Room temperature = T₁K

$\mathrm{V}=22.78\mathrm{ml},\mathrm{T}=280 \mathrm{K}$

${\mathrm{P}}_{\mathrm{total}}=759\mathrm{mmHg}$

${\mathrm{P}}_{{\mathrm{N}}_2}$= Atmospheric pressure – Aqueous tension

${\mathrm{P}}_{{\mathrm{N}}_2}=759-14.2=744.8\mathrm{mmHg}$

$22400 \mathrm{mL} {\mathrm{N}}_2 \mathrm{at} \mathrm{STP} \mathrm{weighs} 28 \mathrm{g}.$

$\mathrm{V} \mathrm{mL} {\mathrm{N}}_2 \mathrm{gas} \mathrm{at} \mathrm{STP} \mathrm{weighs} =\frac{ 28\times \mathrm{V}}{22400}\mathrm{g}$

${\mathrm{n}}_{{\mathrm{N}}_z}=\frac{744.8\times 22.78}{760\times 1000\times 0.082\times 280}=0.00097$

${\mathrm{W}}_{\mathrm{Nitrogen}}=0.02716$g

$\% \mathrm{N}=\frac{0.02716}{0.125}\times 1000=21.728$