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Chapter 11: Problem 103

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

Expert verified
The empirical formula of the unknown compound is \(C_2H_4O_3\), its molar mass is \(150.86g/mol\), and its molecular formula is \(C_4H_8O_6\).

Step by step solution

01

Calculate the mass of carbon, hydrogen, and oxygen

The mass percentage of each element is given, so we can assume that we have a 100g sample of the unknown compound, and calculate the masses of each element. - Mass of carbon = \(31.57\%\) = \(31.57g\) - Mass of hydrogen = \(5.30\%\) = \(5.30g\) - The sum of mass percentages of carbon and hydrogen is \(31.57\% + 5.30\% = 36.87\%\). Since the compound contains only carbon, hydrogen, and oxygen, mass percentage of oxygen is \(100\% - 36.87\% = 63.13\%\). Thus, the mass of oxygen = \(63.13g\).
02

Determine the moles of carbon, hydrogen, and oxygen

Divide the mass of each element by their respective molar masses to find the moles: - Moles of carbon = \(\frac{31.57g}{12.01g/mol} = 2.63 mol\) - Moles of hydrogen = \(\frac{5.30g}{1.008g/mol} = 5.26 mol\) - Moles of oxygen = \(\frac{63.13g}{16.00g/mol} = 3.95 mol\)
03

Calculate the mole ratios and find the empirical formula

Divide the moles of each element by the smallest number of moles to get the mole ratios: - Carbon: \(\frac{2.63}{2.63} = 1\) - Hydrogen: \(\frac{5.26}{2.63} = 2\) - Oxygen: \(\frac{3.95}{2.63} ≈ 1.5\) Since we got a ratio close to 1.5 for oxygen, we will multiply all ratios by 2 to get the whole number ratios: - Carbon = 1 × 2 = 2 - Hydrogen = 2 × 2 = 4 - Oxygen = 1.5 × 2 = 3 So, the empirical formula is \(C_2H_4O_3\).
04

Find the molar mass using the freezing-point depression data

We are given that the freezing point depression of the solution is \(-5.20^{\circ} C\), and molality is defined as the change in freezing point divided by the cryoscopic constant of the solvent (\(K_f = 1.86 ^{\circ}C/mol\cdot kg\)). Molality = \(\frac{-5.20^{\circ}C}{1.86^\circ C/mol\cdot kg} = -2.80 mol/kg\) Now we can find the moles of the compound present in the solution: \(\text{Moles of compound} = \text{molality} \times \text{mass of solvent (in kg)} = -2.80 mol/kg \times \frac{25.0g}{1000g/kg} = -0.0700 mol\) We can then find the molar mass: \[Molar\ mass = \frac{mass}{moles} = \frac{10.56g}{-0.0700 mol} = 150.86g/mol\]
05

Determine the molecular formula using the molar mass and empirical formula

First, find the molar mass of the empirical formula: \(C_2H_4O_3\). - Molar mass of \(C_2H_4O_3 = 2(12.01g/mol) + 4(1.008g/mol) + 3(16.00 g/mol) = 74.06 g/mol\) Next, determine the ratio of the molar mass of the unknown compound to the molar mass of the empirical formula: \(\frac{150.86g/mol}{74.06g/mol} \approx 2\) Since the ratio is approximately 2, the molecular formula is twice the empirical formula: \(C_4H_8O_6\) Thus, the empirical formula is \(C_2H_4O_3\), the molar mass is \(150.86g/mol\), and the molecular formula is \(C_4H_8O_6\).

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Most popular questions from this chapter

Which of the following statements is(are) true? Correct the false statements. a. The vapor pressure of a solution is directly related to the mole fraction of solute. b. When a solute is added to water, the water in solution has a lower vapor pressure than that of pure ice at \(0^{\circ} \mathrm{C}\). c. Colligative properties depend only on the identity of the solute and not on the number of solute particles present. d. When sugar is added to water, the boiling point of the solution increases above \(100^{\circ} \mathrm{C}\) because sugar has a higher boiling point than water.

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