Question

A solution contains 10.75g of an unknown compound dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL for water.) The freezing point of the solution is -3.38 C. The mass percent composition of the compound is 60.98% C, 11.94% H, and the rest is O.

-What is the molecular formula of the compound?

Express your answer as a molecular formula..

Answer 1

depression in freezing point, delta Tf = -3.38 oC
Freezing point depression constant for watre: Kf = 1.86 Kg-oC/mol {from google}
Mass of water = V*density = 50 *1 = 50 gm = 0.05 Kg
molality, m = number of moles of solute / mass of solvent inKg
m = (mass/molar mass) / mass of solvent inKg
m = (mass/molar mass * mass of solvent inKg )
   =10.75/(M*0.05)

Now use:
delta Tf = Kf *m
3.38 = 1.86 * 10.75/(M*0.05)
M = 118. 3 g

Now Let us find the empirical formula for compound:
C = 60.98%
Mass of C = 60.98*10.75/100 = 6.55 gm
number of moles of C = 6.55/12 = 0.55 mol
H = 11.94%
Mass of H = 11.94*10.75/100 = 1.28 gm
number of moles of H = 1.28/1 = 1.28 mol
O = 100 -11.94 - 60.98 = 27.08 %
Mass of O = 27.08*10.75/100 = 2.91 gm
number of moles of O = 2.91/16 = 0.18 mol
Simplest ratio of C = 0.55/0.18 = 3
Simplest ratio of H= 1.28/0.18 = 7
Simplest ratio of O = 0.18/0.18 = 1
Empirical formula : C3H7O
Empirical formula mass = 3*12+1*7+16 = 59 gm

But molar mass calculated above = 118.3 gm
so, multiplying factor = 118.3/59 = 2
So Formula of the compound is 2* (C3H7O) = C6H14O2

Answer: C6H14O2