© All Rights reserved @ LearnWithDash
Step-by-Step Solution Using Duma’s Method
Step 1: Identify the given data
• Mass of the organic compound, $m_{\text{organic}} = 0.25\text{ g}$
• Volume of nitrogen collected, $V_{1} = 40\text{ mL}$
• Temperature at which $V_{1}$ is collected, $T_{1} = 300\text{ K}$
• Total pressure, $P_{\text{total}} = 725\text{ mmHg}$
• Aqueous tension (water vapor pressure), $P_{\text{aq}} = 25\text{ mmHg}$
• Standard pressure, $P_{2} = 760\text{ mmHg}$
• Standard temperature, $T_{2} = 273\text{ K}$
Step 2: Determine the partial pressure of nitrogen
The actual pressure of nitrogen is the total pressure minus the aqueous tension:
$$P_{1} = P_{\text{total}} - P_{\text{aq}} = 725 - 25 = 700 \text{ mmHg}.$$
Step 3: Use the combined gas law to find the volume at STP
The combined gas law states:
$$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}.$$
Rearranging to find $V_{2}$ (the volume of $N_{2}$ at STP):
$$
V_{2} = \frac{P_{1} \times V_{1} \times T_{2}}{T_{1} \times P_{2}}
= \frac{700 \times 40 \times 273}{300 \times 760}.
$$
Calculating this gives:
$$
V_{2} \approx 33.52 \text{ mL}.
$$
Step 4: Relate volume of $N_{2}$ at STP to its mass
• At STP, 1 mole of any gas occupies $22400\text{ mL}$.
• Molar mass of $N_{2} = 28\text{ g/mol}$.
So, the mass of $N_{2}$ obtained, $m_{N_{2}}$, can be found by:
$$
m_{N_{2}} = \frac{28 \times V_{2}}{22400} = \frac{28 \times 33.52}{22400} \text{ g}.
$$
Step 5: Calculate the percentage of nitrogen in the organic compound
Finally, the percentage of nitrogen is:
$$
\%N = \frac{m_{N_{2}}}{m_{\text{organic}}} \times 100\%.
$$
Substituting the values directly in a combined expression often used in Duma’s method:
$$
\%N = \frac{28 \times V_{2} \times 100}{22400 \times m_{\text{organic}}}.
$$
Plugging in $V_{2} = 33.52\text{ mL}$ and $m_{\text{organic}} = 0.25\text{ g}$:
$$
\%N
= \frac{28 \times 33.52 \times 100}{22400 \times 0.25}
\approx 16.75\%.
$$
Final Answer
The percentage of nitrogen in the given organic compound is approximately 16.75%.
