Question

$0.5 \text{ gm}$ of unknown organic compound undergo Duma's method for estimation of nitrogen. Percentage of nitrogen gas collected over water at $P = 715 \text{ mm}$ and $27^\circ \text{C}$ has volume $= 70 \text{ ml}$. Calculate $% \text{ N}$ in the unknown organic compound. ($\text{aq}$. $\text{Tension} = 15 \text{ mm}$)

• A. $14.65%$
• B. $15.50%$
• C. $16.80%$
• D. $13.20%$

Hint:

For gas volumetric analysis (Duma's method), carefully convert wet volume pressure to dry volume pressure ($\text{P}_{\text{dry}} = \text{P}_{\text{total}} - \text{P}_{\text{water}}$) and use the ideal gas law to find moles of $\text{N}_2$.

Answer 1

The Correct Option is A

Mass of organic compound ($W$) $= 0.5 \text{ gm}$.
Pressure of dry $\text{N}_2$ ($P$) $= P_{\text{total}} - P_{\text{water}} = 715 - 15 = 700 \text{ mm Hg}$.
Volume of $\text{N}_2$ ($V$) $= 70 \text{ ml} = 0.070 \text{ L}$. $T = 300 \text{ K}$.
Calculate moles of $\text{N}_2$ ($n_{\text{N}_2}$): $\text{PV} = n\text{RT}$. ($R = 0.0821 \text{ L atm/mol K}$).
$P = 700/760 \text{ atm}$.
$n_{\text{N}_2} = \frac{(700/760) \times 0.070}{0.0821 \times 300} \approx 0.002631 \text{ mol}$.
Mass of Nitrogen ($W_{\text{N}}$): $W_{\text{N}} = 2 \times n_{\text{N}_2} \times 14 \text{ g/mol} \approx 0.07367 \text{ g}$.
Percentage of Nitrogen: $% \text{N} = \frac{W_{\text{N}}}{W} \times 100$.
$% \text{N} = \frac{0.07367}{0.5} \times 100 \approx 14.73%$.
Using the simplified calculation provided in the source leading to $14.65%$:
$% \text{N} = \frac{28}{22400} \times \frac{700}{760} \times \frac{273}{300} \times \frac{100}{0.5} \approx 14.65%$.