The problem describes an organic compound A composed of Carbon (C), Hydrogen (H), and Oxygen (O). 1) We need to determine the molar mass of A given that its vapor density relative to air is $d = 1.59$. 2) We are given that the complete combustion of $0.2$ moles of A produces $10.8$ g of liquid water and $9.6$ L of carbon dioxide gas at standard atmospheric pressure and $20^\circ$C. We need to determine the empirical formula of A, given that the molar volume at $20^\circ$C is $24$ L/mol.

1. Problem Description

The problem describes an organic compound A composed of Carbon (C), Hydrogen (H), and Oxygen (O).

1) We need to determine the molar mass of A given that its vapor density relative to air is $d = 1.59$.

2) We are given that the complete combustion of $0.2$ moles of A produces $10.8$ g of liquid water and $9.6$ L of carbon dioxide gas at standard atmospheric pressure and $20^\circ$C. We need to determine the empirical formula of A, given that the molar volume at $20^\circ$C is $24$ L/mol.

2. Solution Steps

1) Determining the molar mass of A:

The density of a gas relative to air is given by the ratio of their molar masses:

$d = \frac{M_A}{M_{air}}$, where $M_A$ is the molar mass of compound A and $M_{air}$ is the molar mass of air, which is approximately $29$ g/mol.

Therefore, $M_A = d \times M_{air} = 1.59 \times 29$ g/mol.

2) Determining the empirical formula of A:

We are given that $0.2$ moles of A are combusted. Let the empirical formula of A be $C_xH_yO_z$. The combustion reaction is:

$C_xH_yO_z + (x + \frac{y}{4} - \frac{z}{2})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O$

The number of moles of $CO_2$ produced is given by the volume divided by the molar volume:

$n_{CO_2} = \frac{V}{V_m} = \frac{9.6 \text{ L}}{24 \text{ L/mol}} = 0.4 \text{ mol}$

Since $0.2$ moles of A produces $0.4$ moles of $CO_2$, we have:

$0.2x = 0.4$, so $x = \frac{0.4}{0.2} = 2$

The number of moles of $H_2O$ produced is given by the mass divided by the molar mass:

$n_{H_2O} = \frac{m_{H_2O}}{M_{H_2O}} = \frac{10.8 \text{ g}}{18 \text{ g/mol}} = 0.6 \text{ mol}$

Since $0.2$ moles of A produces $0.6$ moles of $H_2O$, we have:

$0.2 \times \frac{y}{2} = 0.6$, so $\frac{y}{2} = \frac{0.6}{0.2} = 3$, and thus $y = 6$

Now, we know that the molar mass of A is $M_A = 1.59 \times 29 = 46.11$ g/mol.

The molar mass of $C_xH_yO_z$ is $12x + y + 16z = 12(2) + 6 + 16z = 24 + 6 + 16z = 30 + 16z$.

So, $30 + 16z = 46.11$, thus $16z = 46.11 - 30 = 16.11$, and $z = \frac{16.11}{16} \approx 1$.

Therefore, the empirical formula is $C_2H_6O$.

3. Final Answer

1) The molar mass of A is $46.11$ g/mol.

2) The empirical formula of A is $C_2H_6O$.