Answer to Question #141962 in General Chemistry for gia

Question

Analysis of a compound gave 39.50% C, 2.21% H, and 58.30% Cl. When 0.855 g of this solid was dissolved in 7.50 g of naphthalene, the solution had a freezing point of 78.0°C. The pure solvent freezes at 80.0°C; its molal freezing point constant is 6.8°C/m. What is the molecular formula of the compound?

Answer 1

Use the freezing-point-depression equation to find the molality of the solution. The freezing point of naphthalene is 80.0 °C and Kf = 6.8°C/m. Thus

The moles of the compound are

Moles of compound = 0.294 mol / 1 kg solvent x 0.00750 kg = 2.21 x 10−3 mol

The molar mass of the compound is

Molar mass = 0.855 g / 2.21 x 10-3 mol = 387 g/mol

The moles of the elements in 100 g of the compound are

Moles of C = 39.50 g C x 1 mol / 12.01 g C = 3.2889 mol

Moles of H = 2.21 g H x 1 mol / 1.008 g H = 2.192 mol

Moles of Cl = 58.30 g Cl x 1 mol / 35.45 g Cl = 1.6446 mol

This gives mole ratios of 2 mol C to 1.33 mol H to 1 mol Cl. Multiplying by 3 gives ratios of 6 mol C to 4 mol H to 3 mol Cl. Therefore, the empirical formula of the compound is C6H4Cl3. The mass of this formula unit is approximately 182.5 amu. Since the molar mass of the compound is 387. g/mol, the value of n is

n = 387 g/mol / 182.5 g/unit = 2.12, or 2

Therefore, the formula of the compound is C12H8Cl6.

b.    The molar mass of the compound is (to the nearest tenth of a gram)

12(12.01) + 8(1.008) + 6(35.45) = 364.884 = 364.9 g/mol