In Dumas' method for estimation of nitrogen, `0.3 g` of an organic compound released `50 mL` of nitrgen gas collected under `715 mm Hg` and at `300 K`. If aqueous tension at `300 K` is `15 mm Hg`, calculate the percentage composition of nitrogen in the organic compound.
Strategy:
Step `1`. Calculate the volume of `N_(2)` gas at `STP//NTP` using combined gas equation.
Step `2`. Find the munber of moles of `N_(2)` gas:
`n_(N_(2)) = ("Volume of "N_(2) "gas at" NTP)/(22400 mL mol^(-1))`
Step `3`. Find the mass of `N_(2)` gas:
`m_(N_(2)) = n_(N_(2)) xx` Molar mass of `N_(2)`
Step `4`. Find the `%` of `N_(2)` gas
`% N = ("Mass of nitrogen")/("Mass of organic compound") xx 100%`
or use Eq. `(13.7)` where steps `(2)` to `(4)` are combined.

Explanation

Step 1. `P_(1) = ("Atmospheric pressure") - ("Aqueous tension")`
`= (715 mm Hg) - (15 mm Hg)`
`= 700 mm Hg`
`V_(1) = 50 mL, T_(1) = 300 K`
Thus, Volume of `N_(2)` gas at `STP//NTP = (P_(1) V_(1))/(T_(1)) xx (273)/(760)`
`- (700 xx 50 xx 273)/(300 xx 760)`
`= 41.9 mL`
Steps `(2)` to `(4)`. Applying Eq. `(13.7)`,
`% N = (28)/(22400) xx ("Volume of" N_(2) gas at NTP)/("Mass of organic compound") xx 100%`
`= (28)/(22400) xx (41.9)/(0.3) xx 100%`
`= 17.46 %`