Question

In Duma’s method $0.52\, g$ of an organic compound on combustion gave $68.6$ mL $N_{2}$ at $27 ^{\circ}C$ and $756$ mm pressure. What is the percentage of nitrogen in the compound?

• A. $12.22\%$
• B. $14.93\%$
• C. $15.84\%$
• D. $16.23\%$

Answer 1

Answer: (B)

$V_{1}=68.6\, mL$, $P_{1}=756\, mm$, $T_{1}=300\, K$
$V_{2}= ?$, $p_{2}=760\, mm$, $T_{2}=273\, K$
$\frac{P_{1}V_{1}}{T_{1}}=\frac{P _{2}V_{2}}{T_{2}}$
At $NTP$, vol. of $N_{2} , V_{2} =\frac{P_{1}V_{1}}{T_{1}}\cdot\frac{T_{2}}{P_{2}}=\frac{756\times68.6}{300}\times\frac{273}{760}$
$=62.09\, mL$
Percentage of nitrogen in organic compound
$=\frac{28}{22400}\times\frac{V_{2}}{w}\times100$
$=\frac{28}{22400}\times\frac{62.09}{0.52}\times100$
$=14.93\%$