Question

11.6 g of an organic compound having formula $C_nH_{2n+2}$ is burnt in excess of $O_2(g)$ initially taken in a 22.41 litre steel vessel. Before reaction the gaseous mixture was at 273 K with pressure reading 2 atm. After complete combustion and loss of considerable amount of heat, the mixture of product and excess of $O_2$ had a temperature of 546 K and 4.6 atm pressure. The formula of organic compound is:

Answer 1

$C_4H_{10}$

Answer 2

The organic compound has the formula $C_nH_{2n+2}$. Its molar mass is $M = 12n + (2n+2) = 14n + 2$ g/mol.
The mass of the compound is 11.6 g, so the number of moles of the organic compound is $n_{compound} = \frac{11.6}{14n+2}$.

The combustion reaction is:
$C_nH_{2n+2} + (\frac{3n+1}{2}) O_2 \rightarrow n CO_2 + (n+1) H_2O$

The reaction is carried out in a steel vessel of constant volume $V = 22.41$ L.
Initial conditions: $T_1 = 273$ K, $P_1 = 2$ atm.
Final conditions: $T_2 = 546$ K, $P_2 = 4.6$ atm.

Let $n_{total, initial}$ be the total number of moles of gas initially present, and $n_{total, final}$ be the total number of moles of gas after combustion. Using the ideal gas law $PV=nRT$, and noting that $V$ is constant:
$\frac{P_1 V}{n_{total, initial} T_1} = \frac{P_2 V}{n_{total, final} T_2}$
$\frac{P_1}{n_{total, initial} T_1} = \frac{P_2}{n_{total, final} T_2}$
$\frac{2}{n_{total, initial} \times 273} = \frac{4.6}{n_{total, final} \times 546}$
$n_{total, final} = n_{total, initial} \times \frac{4.6}{2} \times \frac{273}{546} = n_{total, initial} \times 2.3 \times 0.5 = 1.15 \times n_{total, initial}$.

Alkanes $C_nH_{2n+2}$ with $n=1, 2, 3, 4$ (methane, ethane, propane, butane) are gases at 273 K and 1 atm. Alkanes with $n \ge 5$ are liquids or solids. Given the context of gas pressure, it is reasonable to assume the organic compound is also in gaseous state initially and contributes to the initial pressure.
Let $n_{compound, initial}$ and $n_{O_2, initial}$ be the initial moles of the organic compound and oxygen, respectively.
$n_{total, initial} = n_{compound, initial} + n_{O_2, initial}$.
Using the ideal gas law for the initial state:
$P_1 V = n_{total, initial} R T_1$
$2 \text{ atm} \times 22.41 \text{ L} = n_{total, initial} \times (0.0821 \text{ L atm/mol K}) \times 273 \text{ K}$
$n_{total, initial} = \frac{2 \times 22.41}{0.0821 \times 273} \approx \frac{44.82}{22.41} = 2$ moles.
So, $n_{compound, initial} + n_{O_2, initial} = 2$.

After complete combustion, $n_{compound, initial}$ moles of $C_nH_{2n+2}$ react.
From the stoichiometry:
Moles of $O_2$ consumed = $n_{compound, initial} \times (\frac{3n+1}{2})$
Moles of $CO_2$ produced = $n_{compound, initial} \times n$
Moles of $H_2O$ produced = $n_{compound, initial} \times (n+1)$
The problem states "excess of $O_2$", so $n_{O_2, initial} > n_{compound, initial} \times (\frac{3n+1}{2})$.
The final mixture contains $CO_2$, $H_2O$, and excess $O_2$. At $T_2 = 546$ K (273 °C), water is in gaseous state.
$n_{total, final} = n_{CO_2} + n_{H_2O} + n_{O_2, excess}$
$n_{O_2, excess} = n_{O_2, initial} - n_{compound, initial} \times (\frac{3n+1}{2})$
$n_{total, final} = n_{compound, initial} \times n + n_{compound, initial} \times (n+1) + n_{O_2, initial} - n_{compound, initial} \times (\frac{3n+1}{2})$
$n_{total, final} = n_{compound, initial} \times (n + n+1 - \frac{3n+1}{2}) + n_{O_2, initial}$
$n_{total, final} = n_{compound, initial} \times (\frac{2n + 2n+2 - 3n-1}{2}) + n_{O_2, initial}$
$n_{total, final} = n_{compound, initial} \times (\frac{n+1}{2}) + n_{O_2, initial}$

We found $n_{total, initial} = 2$. So $n_{O_2, initial} = 2 - n_{compound, initial}$.
$n_{total, final} = n_{compound, initial} \times (\frac{n+1}{2}) + (2 - n_{compound, initial})$
$n_{total, final} = n_{compound, initial} \times (\frac{n+1}{2} - 1) + 2$
$n_{total, final} = n_{compound, initial} \times (\frac{n-1}{2}) + 2$

Using $n_{total, final} = 1.15 \times n_{total, initial} = 1.15 \times 2 = 2.3$ moles.
$2.3 = n_{compound, initial} \times (\frac{n-1}{2}) + 2$
$0.3 = n_{compound, initial} \times (\frac{n-1}{2})$
$0.6 = n_{compound, initial} \times (n-1)$

Substitute $n_{compound, initial} = \frac{11.6}{14n+2}$:
$0.6 = \frac{11.6}{14n+2} \times (n-1)$
$0.6 (14n+2) = 11.6 (n-1)$
$8.4n + 1.2 = 11.6n - 11.6$
$1.2 + 11.6 = 11.6n - 8.4n$
$12.8 = 3.2n$
$n = \frac{12.8}{3.2} = 4$.

The formula of the organic compound is $C_nH_{2n+2} = C_4H_{2(4)+2} = C_4H_{10}$. This is butane, which is a gas at 273 K, validating our initial assumption.

Let's quickly check the alternative assumption that the compound is liquid initially, and only $O_2$ contributes to initial pressure. In this case, $n_{total, initial} = n_{O_2, initial} = 2$ moles. The organic compound is 11.6 g.
$n_{total, final} = n_{CO_2} + n_{H_2O} + n_{O_2, excess}$.
$n_{compound, initial} = \frac{11.6}{14n+2}$.
$n_{CO_2} = n \times n_{compound, initial}$
$n_{H_2O} = (n+1) \times n_{compound, initial}$
$n_{O_2, consumed} = (\frac{3n+1}{2}) \times n_{compound, initial}$
$n_{O_2, excess} = n_{O_2, initial} - n_{O_2, consumed} = 2 - (\frac{3n+1}{2}) \times n_{compound, initial}$.
$n_{total, final} = n \times n_{compound, initial} + (n+1) \times n_{compound, initial} + 2 - (\frac{3n+1}{2}) \times n_{compound, initial}$
$n_{total, final} = n_{compound, initial} \times (n + n+1 - \frac{3n+1}{2}) + 2 = n_{compound, initial} \times (\frac{n+1}{2}) + 2$.
This leads to the same equation as before: $2.3 = n_{compound, initial} \times (\frac{n-1}{2}) + 2$, which gave $n=4$.
Since $C_4H_{10}$ is a gas at 273 K, the initial assumption was correct. If $n$ had turned out to be 5 or more, we would have needed to reconsider the initial state.

The formula of the organic compound is $C_4H_{10}$.

Explanation of the solution:

1. Write the balanced chemical equation for the combustion of $C_nH_{2n+2}$.
2. Use the ideal gas law $PV=nRT$ and the given initial and final conditions ($P_1, T_1, P_2, T_2, V$ constant) to find the relationship between the total initial moles of gas ($n_{total, initial}$) and total final moles of gas ($n_{total, final}$).
3. Calculate $n_{total, initial}$ from the initial conditions using the ideal gas law. Assume the organic compound is gaseous and contributes to the initial pressure.
4. Express $n_{total, initial}$ in terms of initial moles of organic compound ($n_{compound, initial}$) and oxygen ($n_{O_2, initial}$).
5. Express $n_{total, final}$ in terms of $n_{compound, initial}$, $n_{O_2, initial}$, and $n$ using the stoichiometry of the reaction and the fact that water is gaseous at the final temperature.
6. Substitute the expression for $n_{O_2, initial}$ from step 4 into the expression for $n_{total, final}$ from step 5.
7. Substitute the calculated value of $n_{total, final}$ from step 2 and the expression for $n_{compound, initial}$ in terms of $n$ (using the given mass and molar mass) into the equation from step 6.
8. Solve the resulting algebraic equation for $n$.
9. Determine the formula of the organic compound using the value of $n$.
10. Verify the assumption about the initial state of the organic compound based on the determined value of $n$.