Question

A research chemist isolates a new compound with an empirical formula C_3H_ O_2. Dissolving 2.51 g of the compound in 100.0 g of water produces a solution with a freezing point of -0.325 degree C. What is the molecular formula of the compound? For water K_ = 1.86 degree C/m.

Answer 1

3.

Depresion of freezing point is expressed as

delta T = K_f x molality

molality = delta T / K_f

Here delta T = 0.325 ^oC

K_f = 1.86 ^oC/m

So, molality = delta T / K_f

                    = 0.325 ^oC / (1.86 ^oC/m)

                    = 0.1747 m

                    = 0.1747 moles/Kg

we have 2.51 g of our new compound in 100 g of water.

So, 25.1 g / 1000 g of water = 25.1 g/Kg of water.

Hence,

0.1747 moles = 25.1 g

1 mole = 25.1 / 0.1747 = 143.7 g

Empirical Formula = C₃H₃O₂.

So, empirical mass = (3 x 12) + (3 x 1) + (2 x 16)

                               = 71g

Which indicates there are 2 empirical units per molecular formula.

Hence the molecular formula is 2(C₃H₃O₂) = C₆H₆O₄