In Duma's method 0.206 g of an organic compound gave `18.8 cm^(3)` of moist `N_(2)` at `17^(@)C` and 760 mm Hg pressure. If aqueous at `17^(@)C` is 14.5 mm Hg, calculate the percentage of nitrogen in the given organic compound.

Explanation

Mass of organic compound =0.206 g
Calculation of the volume of `N_(2)` at N.T.P.
`{:("Experimental Conditions","N.T.P. conditions"),(V_(1)=18.8 cm^(3),V_(2)=?),(P_(1)=(760-14.5)mm Hg,P_(2)=760 mm Hg),(=745.5 mm Hg,),(T_(1)=(273+17)K=290 K,T_(2)=273 K):}`
By applying gas equation, `(P_(1)V_(1))/T_(1)=(P_(2)V_(2))/T_(2)`
`V_(2)=(P_(1)V_(1)T_(2))/(T_(1)P_(2))=((745.5mmHg)xx(18.8 cm^(3))xx(273 K))/((290 K)xx(760 mm Hg))=17.37 cm^(3)`
Step II. Calculation of percentage of nitrogen
`22400 cm^(3)` of `N_(2)` at N.T.P. has a mass =28 g
`17.37 cm^(3)` of `N_(2)` at N.T.P. has a mass `=((28g)xx(17.37 cm^(3)))/((22400 cm^(3)))=0.0217 g`
Percentage of nitrogen `=("Mass of "N_(2)" produced "xx100)/("Mass of organic compound")=((0.0217g)xx100)/((0.206 g))=10.54`