What is the percentage of nitrogen in an organic compound `0.14 g` of which gave by Dumas method `82.1 c.c`. Of nitrogen collected over water at `27^(@)C` and at a barometric pressure of `774.5 mm` ? `("aqueous tension of water at" 27^(@)C "is" 14.5 mm)`

Explanation

To find the percentage of nitrogen in the organic compound, we will follow these steps:

### Step 1: Calculate the effective pressure of nitrogen gas
The total barometric pressure is given as 774.5 mm and the aqueous tension at 27°C is 14.5 mm. The effective pressure of nitrogen (P) can be calculated using the formula:

\[ P = \text{Barometric Pressure} - \text{Aqueous Tension} = 774.5 \, \text{mm} - 14.5 \, \text{mm} = 760 \, \text{mm} \]

### Step 2: Convert the volume of nitrogen gas to standard temperature and pressure (STP)
The volume of nitrogen collected is given as 82.1 c.c. To convert this volume to STP (0°C or 273 K and 760 mm Hg), we can use the formula:

\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

Where:
- \(P_1 = 774.5 \, \text{mm}\)
- \(V_1 = 82.1 \, \text{c.c.}\)
- \(T_1 = 27°C = 300 \, \text{K}\)
- \(P_2 = 760 \, \text{mm}\)
- \(T_2 = 273 \, \text{K}\)

Rearranging for \(V_2\):

\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]

Substituting the values:

\[ V_2 = \frac{774.5 \times 82.1 \times 273}{760 \times 300} \]

Calculating \(V_2\):

\[ V_2 \approx 75.2 \, \text{c.c.} \]

### Step 3: Calculate the mass of nitrogen at STP
To find the mass of nitrogen gas at STP, we can use the molar volume of a gas at STP, which is 22,400 mL/mol. The molar mass of nitrogen (N₂) is approximately 28 g/mol.

Using the formula:

\[ \text{Mass of N}_2 = \frac{\text{Volume at STP} \times \text{Molar Mass}}{\text{Molar Volume}} \]

Substituting the values:

\[ \text{Mass of N}_2 = \frac{75.2 \, \text{c.c.} \times 28 \, \text{g/mol}}{22400 \, \text{mL/mol}} \approx 0.094 \, \text{g} \]

### Step 4: Calculate the percentage of nitrogen in the organic compound
Now, we can find the percentage of nitrogen in the organic compound using the formula:

\[ \text{Percentage of N} = \left( \frac{\text{Mass of N}_2}{\text{Mass of organic compound}} \right) \times 100 \]

Substituting the values:

\[ \text{Percentage of N} = \left( \frac{0.094 \, \text{g}}{0.14 \, \text{g}} \right) \times 100 \approx 66.7\% \]

### Final Answer:
The percentage of nitrogen in the organic compound is approximately **66.7%**.

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