A compound containing carbon, hydrogen, and oxygen is analyzed. 1) 1.491g of the substance yields 3.540g of carbon dioxide ($CO_2$) and 1.810g of water ($H_2O$) upon complete combustion. Determine the percentage composition of the substance. 2) 2.52g of the substance, when vaporized at $100^\circ C$ and under a pressure of 750 mmHg, produces a vapor of 1057 $cm^3$. Determine: a) The density and the approximate molar mass of the substance. b) The empirical formula. Given: $C = 12$, $H = 1$, $O = 16$ (g/mol), Density of air $\rho_{air} = 1.29 g/L$.

1. Problem Description

A compound containing carbon, hydrogen, and oxygen is analyzed.

1) 1.491g of the substance yields 3.540g of carbon dioxide ($CO_2$) and 1.810g of water ($H_2O$) upon complete combustion. Determine the percentage composition of the substance.

2) 2.52g of the substance, when vaporized at $100^\circ C$ and under a pressure of 750 mmHg, produces a vapor of 1057 $cm^3$. Determine:

a) The density and the approximate molar mass of the substance.

b) The empirical formula.

Given: $C = 12$, $H = 1$, $O = 16$ (g/mol), Density of air $\rho_{air} = 1.29 g/L$.

2. Solution Steps

1) Determining the percentage composition:

First, we calculate the mass of carbon in $CO_2$ and the mass of hydrogen in $H_2O$.

Mass of C in $CO_2$:

$m_C = \frac{12}{44} \times 3.540g = 0.965 g$

Mass of H in $H_2O$:

$m_H = \frac{2}{18} \times 1.810g = 0.201 g$

Now, we can find the mass of oxygen by subtracting the masses of carbon and hydrogen from the total mass of the substance:

$m_O = 1.491g - 0.965g - 0.201g = 0.325g$

Next, we calculate the percentage composition:

Percentage of C:

$\%C = \frac{0.965}{1.491} \times 100 = 64.72 \%$

Percentage of H:

$\%H = \frac{0.201}{1.491} \times 100 = 13.48 \%$

Percentage of O:

$\%O = \frac{0.325}{1.491} \times 100 = 21.79 \%$

2) Determining the density and molar mass:

a) The density of the vapor is given by:

$\rho = \frac{mass}{volume} = \frac{2.52g}{1057 cm^3} = \frac{2.52g}{1057 mL} = 0.002384 g/mL = 2.384 g/L$

To find the molar mass, we use the ideal gas law: $PV = nRT$, where $n = \frac{m}{M}$, so $PV = \frac{m}{M}RT$ and $M = \frac{mRT}{PV}$.

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

$V = 1057 cm^3 = 1.057 L$

$T = 100^\circ C = 373.15 K$

$R = 0.0821 \frac{L \cdot atm}{mol \cdot K}$

$m = 2.52g$

$M = \frac{2.52g \times 0.0821 \frac{L \cdot atm}{mol \cdot K} \times 373.15 K}{0.9868 atm \times 1.057 L} = \frac{77.18}{1.043} = 73.99 g/mol$

So, approximate molar mass is 74 g/mol

b) Determining the empirical formula:

Divide the percentage composition by the respective atomic masses:

$C: \frac{64.72}{12} = 5.39$

$H: \frac{13.48}{1} = 13.48$

$O: \frac{21.79}{16} = 1.36$

Divide each value by the smallest value (1.36):

$C: \frac{5.39}{1.36} \approx 4$

$H: \frac{13.48}{1.36} \approx 10$

$O: \frac{1.36}{1.36} = 1$

Therefore, the empirical formula is $C_4H_{10}O$

The molar mass of $C_4H_{10}O$ is $4(12) + 10(1) + 1(16) = 48 + 10 + 16 = 74 g/mol$, which matches the molar mass we calculated above. Hence the molecular formula is also $C_4H_{10}O$

3. Final Answer

1) Composition centesimale:

C: 64.72%

H: 13.48%

O: 21.79%

2) a) Density: 2.384 g/L, Molar mass: 74 g/mol

b) Molecular Formula: $C_4H_{10}O$