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\(\%\) 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.

The problem already gave us mass percents of carbon (31.57%) and hydrogen (5.30%). To determine the mass percentage of oxygen, we know that the total mass percentage is 100%. Therefore, \( \%~O = 100\% - \%~C - \%~H \) Calculating the mass percent of oxygen, we get: \( \%~O = 100\% - 31.57\% - 5.30\% = 63.13\% \)

Assume there is 100 g of the compound. So, we have: Carbon: \( 31.57\% \times 100~g = 31.57~g~C \) \\ Hydrogen: \( 5.30\% \times 100~g = 5.30~g~H \) \\ Oxygen: \( 63.13\% \times 100~g = 63.13~g~O \)

Divide each mass of the element by its molar mass to find the moles: Carbon: \( \frac{31.57~g}{12.01~g/mol} = 2.63~mol \) \\ Hydrogen: \( \frac{5.30~g}{1.008~g/mol} = 5.26~mol \) \\ Oxygen: \( \frac{63.13~g}{16.00~g/mol} = 3.95~mol \)

Divide the moles of each element by the smallest moles value to get the ratio of atoms: \( \frac{2.63~mol}{2.63} : \frac{5.26~mol}{2.63} : \frac{3.95~mol}{2.63} = 1 : 2 : 1.5 \) To get whole numbers, multiply each ratio by 2: \( 1 \times 2 : 2 \times 2 : 1.5 \times 2 = 2 : 4 : 3 \) So, the empirical formula of the compound is \(C_{2}H_{4}O_{3}\).

We are given the freezing point depression (\( ΔT_{f} \)) of the solution, which is \( -5.20^{\circ}C \), the mass of the compound (10.56 g), and the mass of the solvent (25.0 g water). The molality of the solution can be determined using the formula: \( ΔT_{f} = iK_{f}m \) Here, i is the van't Hoff factor (assumed to be 1 since the compound is a nonelectrolyte), \( K_{f} \) is the molal freezing point depression constant for water (\( 1.86~K~kg/mol \)), and m is the molality of the solution. Solving for m, we get: \( m = \frac{ΔT_{f}}{iK_{f}} = \frac{-5.20^{\circ}C}{1 \times 1.86~K~kg/mol} = -2.79~mol/kg \) To find moles of solute: \( moles = molality \times mass~of~solvent~(in~kg) = -2.79~mol/kg \times 0.025~kg = -0.0699~mol \) Now, the molar mass of the compound can be calculated: \( Molar~mass = \frac{mass~of~solute}{moles} = \frac{10.56~g}{-0.0699~mol} = 151.1~g/mol \)

Divide the molar mass by the empirical formula mass to find the ratio between the molecular and empirical formulas: \( Ratio = \frac{Molar~mass}{Empirical~formula~mass} = \frac{151.1~g/mol}{(2 \times 12.01 + 4 \times 1.008 + 3 \times 16.00)~g/mol} = \frac{151.1}{92.102} \approx 1 \) With the ratio being approximately 1, the molecular formula is the same as the empirical formula, which is \( C_{2}H_{4}O_{3} \).