Question

In Duma’s method of estimation of nitrogen, 0.1840 g of an organic compound gave 30 mL of nitrogen collected at 287 K and 758 mm of Hg pressure. The percentage composition of nitrogen in the compound is __________. (Round off to the nearest integer). [Given: Aqueous tension at 287 K = 14 mm of Hg].

Answer 1

Aqueous tension at $287\mathrm{K}=14\mathrm{mm}\mathrm{of}\mathrm{Hg}.$

Hence actual pressure $=(758–14)=744\mathrm{mm}\mathrm{of}\mathrm{Hg}.$

$\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{N}}_2=\left[\frac{(758-14)\times 30\times {10}^{-2}}{760\times 0.0821\times 287}\right]=1.246\times {10}^{-3}\mathrm{mol}$

Molar mass of N₂ = Moles × atomic mass

Molar mass of N₂ = $1.246\times {10}^{-3}\times 28=34.888\times {10}^{-3}g$

Mass percentage $=\frac{\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{substance}}{\mathrm{Total}\mathrm{mass}}\mathrm{x}100$

$Mass\%of{N}_2=\frac{massofN\times 100}{totalmass}$

$=\frac{34.888\times {10}^{-3}\times 100}{0.184}$

= 19%

So, in Duma’s method of estimation of nitrogen, 0.1840 g of an organic compound gave 30 mL of nitrogen collected at 287 K and 758 mm of Hg pressure. The percentage composition of nitrogen in the compound is 19%.