An organic compound containing only C, H, and O is analyzed. 1.491 g of the substance yields 3.540 g of $CO_2$ and 1.810 g of $H_2O$ upon complete combustion. Vaporization of a sample of the substance at 100°C and 75 cm Hg pressure produces 2.560 g of vapor with a volume of 1072 $cm^3$. The tasks are: 1. Calculate the molar mass of the compound and give its molecular formula.

1. Problem Description

An organic compound containing only C, H, and O is analyzed. 1.491 g of the substance yields 3.540 g of $CO_2$ and 1.810 g of $H_2O$ upon complete combustion. Vaporization of a sample of the substance at 100°C and 75 cm Hg pressure produces 2.560 g of vapor with a volume of 1072 $cm^3$. The tasks are:

1. Calculate the molar mass of the compound and give its molecular formula.

2. Determine the centesimal composition (percentage composition) of the compound.

2. Solution Steps

Part 1: Determine the molar mass and molecular formula.

Step 1: Calculate the mass of carbon in the compound.

The molar mass of $CO_2$ is 44.01 g/mol and the molar mass of C is 12.01 g/mol.

$mass \, of \, C = 3.540 \, g \, CO_2 * (12.01 \, g \, C / 44.01 \, g \, CO_2) = 0.9666 \, g \, C$

Step 2: Calculate the mass of hydrogen in the compound.

The molar mass of $H_2O$ is 18.02 g/mol and the molar mass of $H$ is 1.008 g/mol.

$mass \, of \, H = 1.810 \, g \, H_2O * (2 * 1.008 \, g \, H / 18.02 \, g \, H_2O) = 0.2025 \, g \, H$

Step 3: Calculate the mass of oxygen in the compound.

$mass \, of \, O = 1.491 \, g \, compound - 0.9666 \, g \, C - 0.2025 \, g \, H = 0.3219 \, g \, O$

Step 4: Determine the empirical formula.

Convert masses to moles:

$moles \, of \, C = 0.9666 \, g / 12.01 \, g/mol = 0.0805 \, mol$

$moles \, of \, H = 0.2025 \, g / 1.008 \, g/mol = 0.2009 \, mol$

$moles \, of \, O = 0.3219 \, g / 16.00 \, g/mol = 0.0201 \, mol$

Divide by the smallest number of moles (0.0201):

$C: 0.0805 / 0.0201 \approx 4$

$H: 0.2009 / 0.0201 \approx 10$

$O: 0.0201 / 0.0201 = 1$

The empirical formula is $C_4H_{10}O$. The empirical formula mass is (4 * 12.01) + (10 * 1.008) + 16.00 = 48.04 + 10.08 + 16.00 = 74.12 g/mol.

Step 5: Determine the molar mass using the ideal gas law.

Convert pressure to atm: $75 \, cm \, Hg = 750 \, mm \, Hg * (1 \, atm / 760 \, mm \, Hg) = 0.9868 \, atm$

Convert volume to liters: $1072 \, cm^3 = 1.072 \, L$

Convert temperature to Kelvin: $100°C = 373.15 \, K$

$PV = nRT$

$n = PV / RT = (0.9868 \, atm * 1.072 \, L) / (0.0821 \, L*atm/mol*K * 373.15 \, K) = 0.0345 \, mol$

$Molar \, Mass = mass / moles = 2.560 \, g / 0.0345 \, mol = 74.20 \, g/mol$

Since the empirical formula mass is approximately equal to the molar mass, the molecular formula is the same as the empirical formula.

Part 2: Determine the centesimal composition.

$Percent \, C = (0.9666 \, g \, C / 1.491 \, g \, compound) * 100\% = 64.83\%$

$Percent \, H = (0.2025 \, g \, H / 1.491 \, g \, compound) * 100\% = 13.58\%$

$Percent \, O = (0.3219 \, g \, O / 1.491 \, g \, compound) * 100\% = 21.59\%$

3. Final Answer

1. The molar mass of the compound is 74.20 g/mol, and its molecular formula is $C_4H_{10}O$.

2. The centesimal composition is:

C: 64.83%

H: 13.58%

O: 21.59%