Question

An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of \(31.57 \% \mathrm{C}\) and \(5.30 \% \mathrm{H}\). The molar mass is determined by measuring the freezing- point depression of an aqueous solution. A freezing point of \(-5.20^{\circ} \mathrm{C}\) is recorded for a solution made by dissolving \(10.56 \mathrm{~g}\) of the compound in \(25.0 \mathrm{~g}\) water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

Short answer

The empirical formula of the unknown compound is C_2H_4O, and its molecular formula is C_4H_8O_2 with a molar mass of approximately 152 g/mol.

Step-by-step solution

Determine the mass percentage of oxygen

We know that the mass percentages of carbon and hydrogen are 31.57% and 5.30%, respectively. Since the compound contains only carbon, hydrogen, and oxygen, the mass percentage of oxygen can be calculated by subtracting the other two mass percentages from 100%: Oxygen mass percentage = 100% - (Carbon mass percentage + Hydrogen mass percentage) Oxygen mass percentage = 100% - (31.57% + 5.30%) Oxygen mass percentage = 63.13%

Convert mass percentages to moles and find the empirical formula

Now, we need to convert mass percentages to moles for each element. Let's assume that we have 100g of the compound. Then, the masses of carbon, hydrogen, and oxygen in the compound would be 31.57g, 5.30g, and 63.13g, respectively. Divide these masses by their respective molar masses to find the moles: Carbon moles = 31.57g / 12.01g/mol = 2.63 mol Hydrogen moles = 5.30g / 1.01g/mol = 5.25 mol Oxygen moles = 63.13g / 16.00g/mol = 3.95 mol Now, divide each mole quantity by the smallest quantity to find the empirical formula. In this case, the smallest mole quantity is that of carbon (2.63 mol). Carbon ratio = 2.63 mol / 2.63 mol = 1 Hydrogen ratio = 5.25 mol / 2.63 mol ≈ 2 Oxygen ratio = 3.95 mol / 2.63 mol ≈ 1.5 Since the oxygen ratio is approximately 1.5, we can assume that the actual ratio is 2, which makes empirical formula C_2H_4O.

Use freezing point depression to calculate molar mass

The freezing point depression, ΔT, is given by the formula: ΔT = K_f * m where K_f is the cryoscopic constant for water (1.86 K·kg/mol) and m is the molality of the solution (moles of solute per kilogram of solvent). First, we need to find the molality of the solution, using the given freezing point depression (-5.20°C) and the K_f: molality = ΔT / K_f molality = -5.20 K / 1.86 K·kg/mol = -2.79 mol/kg Next, calculate the moles of solute (compound) in the solution: moles of solute = molality * mass of solvent (in kg) moles of solute = -2.79 mol/kg * 0.025 kg = -0.0697 mol Now, we can find the molar mass of the compound: molar mass = mass of solute / moles of solute molar mass = 10.56g / -0.0697 mol ≈ 152 g/mol

Determine the molecular formula

To find the molecular formula of the compound, we need to find the ratio between the molar mass and the molar mass of the empirical formula: n = molar mass / empirical formula mass n = 152 g/mol / (2 * 12.01g/mol + 4 * 1.01g/mol + 1 * 16.00g/mol) ≈ 2 Finally, we multiply the empirical formula by the calculated ratio to obtain the molecular formula: C_2H_4O * 2 → C_4H_8O_2 The molecular formula of the compound is C_4H_8O_2.