Pheromones are compounds secreted by the females of many insect species to attract males. One of these compounds contains 80.78 percent \(\mathrm{C}, 13.56\) percent \(\mathrm{H},\) and 5.66 percent \(\mathrm{O} .\) A solution of \(1.00 \mathrm{~g}\) of this pheromone in \(8.50 \mathrm{~g}\) of benzene freezes at \(3.37^{\circ} \mathrm{C}\) What are the molecular formula and molar mass of the compound? (The normal freezing point of pure benzene is \(5.50^{\circ} \mathrm{C} .\)

To find the empirical formula, it is assumed that the total mass of the compound is 100g. Therefore the mass of Carbon (C) is 80.78g, Hydrogen (H) is 13.56g and that of Oxygen (O) is 5.66g. Next, convert these masses to moles by using their atomic masses (C=12.01; H=1.008; O=16.00). Hence, moles of carbon = 80.78g ÷ 12.01g/mol =6.72mol; moles of hydrogen = 13.56g ÷ 1.008g/mol = 13.44 mol, and for oxygen = 5.66g ÷ 16.00g/mol = 0.354 mol. The ratio of C:H:O is then found by dividing these values by the smallest value: mole ratio C:H:O = 6.72 : 13.44 : 0.354 = 19:38:1. The empirical formula is therefore C19H38O.

Calculate the molar mass of the empirical formula C19H38O by adding the molar masses of all the atoms in the formula. The molar mass of C19H38O is thus (19x12.01) + (38x1.008) + 16.00 = 298.57g/mol.

The depression of the freezing point \(\Delta T_f\) of benzene can be calculated using the formula \(\Delta T = T_{f, pure} - T_{f, solution}\). Substituting the given values gives \(\Delta T = 5.50℃ - 3.37℃ = 2.13℃\). The molar mass can then be calculated using the formula for freezing point depression \(\Delta T = K_f*m\), where \(K_f\) is the molal freezing point depression constant for benzene (5.12 °C/m) and \(m\) is the molality of the solution = moles of solute/kg of solvent. Solving for molar mass gives: \(Molar mass = (1.00g / Molar mass (unknown)) / (8.50g/1000g) = \Delta T / K_f = 2.13℃ / 5.12 °C/m = 0.416mol/kg\). Solving the equation for Molar mass (unknown) gives Molar mass (unknown) = 1.00g / (0.416 mol/kg * 0.00850 kg) = 297.9 g/mol.

The molar mass from the freezing point depression (297.9 g/mol) is very close to the molar mass of the empirical formula (298.57 g/mol). Hence, the empirical formula and molecular formula are the same: C19H38O.