Question

When 1.00 g of a pheromone is dissolved in 8.50g of C6H6, a freezing point of 3.37 C is observed. What is the pheromone’s molecular weight? The freezing point depression constant of C6H6 are 5.5 C and 5.12 c/m.   

The percentage composition of this pheromone is 80.78% C, 13.56 H, 5.66% O. What is its empirical formula and molecular formula.

Answer 1

mass of pheromone = 1.00 g

delta Tf = Kf x m

5.5 - 3.37 = 5.12 x m

m = 0.416

molality = 0.416 m

molality = moles / mass of solvent

0.416 = moles / 8.50 x 10^-3

moles = 3.54 x 10^-3

moles = mass / molar mass

0.01387 = 1 / molar mass

molar mass = 282 g /mol

moles of C = 80.78 / 12 = 6.73

moles of H = 13.56 / 1 = 13.56

moles of O = 5.66 / 16 = 0.354

C            H            O

6.73      13.56       0.354

19          38             1

Emperical formula = C19H38O

emperical formula mass = 282 g/mol

n = 282 / 282 = 1

Molecular formula = C19H38O