In Duma's method of estimation of nitrogen, 0.1840 g of an organic compound gave 30 mL of nitrogen c

Q: Why can't we just use the 758 mm Hg pressure directly in the ideal gas equation?

Because the gas was collected over water! The 758 mm Hg is the *total* pressure, which includes the pressure of the water vapor (aqueous tension) mixed with the nitrogen. We only want the pressure of the dry nitrogen.

Q: Do I always have to convert pressure to atm and volume to Liters?

It depends on the Universal Gas Constant ($R$) you choose. If you use $R = 0.0821 \text{ L atm K}^{-1} \text{mol}^{-1}$, then yes, pressure must be in atm and volume in Liters. It's the safest habit to avoid unit mismatch errors.

Q: Why is the molar mass of nitrogen taken as 28 and not 14?

Nitrogen in its elemental gaseous state exists as a diatomic molecule, $N_2$. Since each nitrogen atom has an atomic mass of 14, the molecule has a molar mass of $14 \times 2 = 28 \text{ g/mol}$.

Q: What if the aqueous tension wasn't given in the problem?

If the problem explicitly states the gas is collected over water but doesn't give aqueous tension, it's usually a poorly framed question, or you might be expected to assume it's negligible (though rare in JEE). Always look for it!

Q: Why did we round off to 19% at the very end instead of rounding intermediate steps?

Rounding off intermediate values (like moles or mass) can introduce compounding errors, leading to a final answer that is slightly off. Always keep a few extra decimal places during calculations and round off only the final result.

Question

In Duma's method of estimation of nitrogen, $0.1840	ext{ g}$ of an organic compound gave $30	ext{ mL}$ of nitrogen collected at $287	ext{ K}$ and $758	ext{ mm}$ of Hg pressure. The percentage composition of nitrogen in the compound is ......... . (Round off to the nearest integer).
[Given: Aqueous tension at $287	ext{ K} = 14	ext{ mm}$ of Hg]

Sigma Percentile · Accepted Answer

The quest to determine exactly how much of a specific element is hidden inside an organic molecule is one of the most fascinating puzzles in chemistry. When it comes to nitrogen, chemists have developed ingenious traps. One such trap is **Duma's Method**, a brilliant technique that forces the organic compound to surrender all its nitrogen in the form of pure $N_2$ gas.

But catching the gas is only half the battle; the real magic happens when we apply the laws of physical chemistry to weigh something we can barely see. Let's embark on this thrilling journey to solve the problem step-by-step.

### The Setup: Catching Gas Over Water

Imagine the experimental setup. We take $0.1840 	ext{ g}$ of our mysterious organic compound and subject it to intense heat in the presence of copper oxide. The compound breaks down, and the nitrogen is liberated as a gas. We collect this gas in an inverted tube over a trough of water.

We observe that the gas occupies a volume of $30 	ext{ mL}$ at a temperature of $287 	ext{ K}$. The pressure gauge reads $758 	ext{ mm Hg}$. But here is the catch—the trap we used to catch the gas (the water) has contaminated our sample! Because the gas is collected over water, some water evaporates and mixes with the nitrogen.

This means the total pressure of $758 	ext{ mm Hg}$ is not just from the nitrogen; it is a combined pressure of nitrogen gas and water vapour. This water vapour pressure is known as **aqueous tension**, and at $287 	ext{ K}$, it is given as $14 	ext{ mm Hg}$.

### Dalton's Law: Separating the Vapour

To find out how much nitrogen we actually have, we need to mathematically dry the gas. We call upon **Dalton's Law of Partial Pressures**, which states that the total pressure of a gas mixture is the sum of the partial pressures of its individual components.

$$P_{	ext{total}} = P_{N_2} + P_{	ext{aqueous}}$$

By rearranging this, we can isolate the pressure of the dry nitrogen:

$$P_{N_2} = P_{	ext{total}} - P_{	ext{aqueous}}$$

Substituting our values:

$$P_{N_2} = 758 	ext{ mm Hg} - 14 	ext{ mm Hg} = 744 	ext{ mm Hg}$$

Now we have the true pressure exerted solely by the nitrogen molecules.

### The Ideal Gas Law: Counting the Molecules

We know the pressure, volume, and temperature of the nitrogen gas. How do we convert these physical properties into a tangible amount, like moles? We use the master key of gas chemistry: the **Ideal Gas Equation**.

$$PV = nRT$$

Before we plug in the numbers, we must respect the units. The universal gas constant $R$ is typically $0.0821 	ext{ L atm K}^{-1} 	ext{mol}^{-1}$. This dictates that our pressure must be in atmospheres and our volume in liters.

Let's convert the pressure:

$$P = \frac{744}{760} 	ext{ atm}$$

And the volume:

$$V = 30 	ext{ mL} = 0.03 	ext{ L}$$

Now, we rearrange the ideal gas equation to solve for $n$ (moles):

$$n = \frac{PV}{RT}$$

Substituting the pristine values:

$$n = \frac{\left(\frac{744}{760}ight) 	imes 0.03}{0.0821 	imes 287}$$

Calculating this carefully, we find:

$$n \approx 1.246 	imes 10^{-3} 	ext{ moles}$$

### Stoichiometry: From Moles to Mass

We have successfully counted the moles of nitrogen gas. But a percentage composition requires mass. We know that nitrogen gas exists as diatomic molecules ($N_2$). Since the atomic mass of a single nitrogen atom is $14 	ext{ u}$, the molar mass of $N_2$ gas is $28 	ext{ g/mol}$.

To find the mass, we simply multiply the moles by the molar mass:

$$	ext{Mass of } N_2 = n 	imes 	ext{Molar Mass}$$

$$	ext{Mass of } N_2 = 1.246 	imes 10^{-3} 	ext{ moles} 	imes 28 	ext{ g/mol}$$

$$	ext{Mass of } N_2 \approx 0.0349 	ext{ g}$$

This tiny number, $0.0349 	ext{ g}$, is the exact weight of the nitrogen that was originally trapped inside the $0.1840 	ext{ g}$ of the organic compound.

### The Final Verdict: Percentage Composition

The final step is a simple percentage calculation. We want to know what fraction of the total compound's mass is made up of nitrogen.

$$\% N = \frac{	ext{Mass of } N_2}{	ext{Total Mass of Compound}} 	imes 100$$

Substituting our hard-earned values:

$$\% N = \frac{0.0349 	ext{ g}}{0.1840 	ext{ g}} 	imes 100$$

$$\% N = 18.96\%$$

The question asks us to round off to the nearest integer. Therefore, the final percentage composition of nitrogen in the compound is **19%**.

Through a beautiful symphony of Dalton's Law, the Ideal Gas Equation, and basic stoichiometry, we have successfully unmasked the nitrogen content. This is the true power of quantitative analysis!

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Setup}

\text{Dalton's Law of Partial Pressures}

\text{Substituting Pressure Values}

\text{Calculating Dry } N_2 \text{ Pressure}

\text{The Ideal Gas Equation}

\text{Substituting Values for Moles}

\text{Calculating Moles of } N_2

\text{Mass of Nitrogen}

\text{Calculating Mass of } N_2

\text{Percentage Composition Formula}

\text{Final Percentage Calculation}

\text{The Way Forward}

The Sigma Insight: Nomenclature and Characterisation

The Setup

Dalton's Law

The Ideal Gas Law

Stoichiometry

The Final Verdict

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