The exercise presents a series of questions related to an organic compound. We are asked to determine the molar mass of the compound from its vaporisation data. Next, we are provided with the combustion products of a known mass of the compound with the formula $C_xH_yO_z$. We have to determine the combustion equation and deduce the molecular formula and mass composition.

1. Problem Description

The exercise presents a series of questions related to an organic compound. We are asked to determine the molar mass of the compound from its vaporisation data. Next, we are provided with the combustion products of a known mass of the compound with the formula $C_xH_yO_z$. We have to determine the combustion equation and deduce the molecular formula and mass composition.

2. Solution Steps

2. 1 Molar mass determination:

We have the mass ($m = 2.56 g$), temperature ($t = 100^\circ C$, so $T = 100 + 273 = 373 K$), pressure ($P = 750 mmHg$) and volume ($V = 1057 cm^3 = 1.057 L$). We want to calculate the molar mass $M$. We will use the ideal gas law to determine the number of moles $n$:

$PV = nRT$

where $R = 0.0821 L \cdot atm \cdot K^{-1} \cdot mol^{-1}$.

First, convert the pressure from $mmHg$ to $atm$.

$P = 750 mmHg = \frac{750}{760} atm \approx 0.9868 atm$.

Now, we can find the number of moles:

$n = \frac{PV}{RT} = \frac{0.9868 atm \cdot 1.057 L}{0.0821 L \cdot atm \cdot K^{-1} \cdot mol^{-1} \cdot 373 K} \approx \frac{1.043}{30.6233} mol \approx 0.03406 mol$.

The molar mass $M$ is given by:

$M = \frac{m}{n} = \frac{2.56 g}{0.03406 mol} \approx 75.16 g/mol$.

3. 2 Combustion analysis:

We are given that $1.491 g$ of $C_xH_yO_z$ produces $3.54 g$ of $CO_2$ and $1.81 g$ of $H_2O$.

Moles of $CO_2 = \frac{3.54 g}{44 g/mol} \approx 0.08045 mol$. Since each mole of $CO_2$ contains one mole of carbon, the moles of carbon in the original compound is $0.08045 mol$.

Mass of carbon $= 0.08045 mol \cdot 12 g/mol \approx 0.9654 g$.

Moles of $H_2O = \frac{1.81 g}{18 g/mol} \approx 0.10056 mol$. Since each mole of $H_2O$ contains two moles of hydrogen, the moles of hydrogen in the original compound is $2 \cdot 0.10056 = 0.20112 mol$.

Mass of hydrogen $= 0.20112 mol \cdot 1 g/mol \approx 0.20112 g$.

Mass of oxygen $= 1.491 g - 0.9654 g - 0.20112 g \approx 0.32448 g$.

Moles of oxygen $= \frac{0.32448 g}{16 g/mol} \approx 0.02028 mol$.

Now, we find the ratio of $C:H:O$.

$C:H:O = 0.08045:0.20112:0.02028$.

Divide by the smallest number (0.02028):

$C:H:O \approx 3.967 : 9.917 : 1 \approx 4:10:1$.

So the empirical formula is $C_4H_{10}O$.

Since the molar mass is approximately $75.16 g/mol$, and the molar mass of $C_4H_{10}O$ is $4(12) + 10(1) + 1(16) = 48 + 10 + 16 = 74 g/mol$. The empirical formula is the molecular formula. So $x=4, y=10, z=1$.

The combustion equation is:

$C_4H_{10}O + \frac{13}{2} O_2 \rightarrow 4CO_2 + 5H_2O$

Multiply by 2 to get whole number coefficients:

$2C_4H_{10}O + 13O_2 \rightarrow 8CO_2 + 10H_2O$

Mass composition:

Carbon: $\frac{0.9654 g}{1.491 g} \cdot 100\% \approx 64.75\%$

Hydrogen: $\frac{0.20112 g}{1.491 g} \cdot 100\% \approx 13.49\%$

Oxygen: $\frac{0.32448 g}{1.491 g} \cdot 100\% \approx 21.76\%$

Alternatively,

Carbon percentage: $\frac{4 \times 12}{74} \times 100\% = \frac{48}{74} \times 100\% \approx 64.86\%$

Hydrogen percentage: $\frac{10 \times 1}{74} \times 100\% = \frac{10}{74} \times 100\% \approx 13.51\%$

Oxygen percentage: $\frac{1 \times 16}{74} \times 100\% = \frac{16}{74} \times 100\% \approx 21.62\%$

3. Final Answer

Molar mass $\approx 75.16 g/mol$

Combustion equation: $2C_4H_{10}O + 13O_2 \rightarrow 8CO_2 + 10H_2O$

Molecular formula: $C_4H_{10}O$

Mass composition: C $\approx$ 64.86%, H $\approx$ 13.51%, O $\approx$ 21.62%