In Dumas' method for estimation of nitrogen,$0.3 \ g$ of an organic compound gave $50 \ mL$ of nitrogen collected at $300 \ K$ temperature and $715 \ mm$ pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at $300 \ K = 15 \ mm$)

(N/A) The pressure of dry nitrogen is $P_{N_2} = P_{total} - P_{aqueous} = 715 \ mm - 15 \ mm = 700 \ mm$.
Using the combined gas law to convert the volume to $STP$ ($273 \ K$ and $760 \ mm$):
$V_{STP} = \frac{P_{N_2} \times V \times 273}{P_{STP} \times T} = \frac{700 \times 50 \times 273}{760 \times 300} \approx 41.91 \ mL$.
Since $22,400 \ mL$ of $N_2$ at $STP$ weighs $28 \ g$,the mass of $N_2$ is:
$Mass_{N_2} = \frac{28 \times 41.91}{22400} \approx 0.05239 \ g$.
Percentage of nitrogen $= \frac{Mass_{N_2}}{Mass_{compound}} \times 100 = \frac{0.05239}{0.3} \times 100 \approx 17.46 \%$.