Question

An organic acid was analyzed and the following percent composition was obtained: 68.85% carbon, 4.9% hydrogen, and 26.2% oxygen. A solution of 1.02 g of this compound dissolved in 10.0 g of benzene freezes at 3.37 °C. The normal freezing point of benzene is 5.50 °C and it has a Kf of 5.12 °C/m. Determine the molecular formula for this compound.C3H3OC7H6O2C10H12O7C14H12O4C15H16O3

Answer 1

Answer:

The molecular formula of the compound is: C14H12O4

Explanation:

Step 1: find molality

ΔT = Kf x m

ΔT = Tfinal - Tinitial = 3.37 - 5.50 = -2.13

molality (m) = ΔT/Kf = -2.13/-5.12 = 0.416 m

step 2: find number of moles

m = number of moles (n)/Kg solvent which is the benzene

n = molality x Kg solvent = 0.416 molality X (10 x 10^-3 Kg) = 0.00416 mole

Step 3: find molecular mass of the organic compound

n = mass/Molecular mass = m/Mm

Mm = m/n = 1.02/0.00416 = 245.19 g/mole of organic compound molecular formula

Step 4: find number of mole of each compound

let consider it being in 100g where each % will correspond to the mass of the element in 100g of the compound.

C = mass/molar mass = 68.85/12.01 = 5.7327 mole

H = mass/molar mass = 4.9/1.01 = 4.8515 mole

O = mass/molar mass = 26.2/16 = 1.6375 mole

Step 5: divide by smallest number of mole

C = 5.7327/1.6375 = 3.5 = 7/2

H = 4.8515/1.6375 = 3

O = 1.6375/1.6375 = 1

C7/2H3O1

multiply by two to remove the fraction: C7H6O2 in which the empirical molecular mass = 122.13 g/mole

to find the factor for the molecular formula x = Mm of molecular/Mm of empirical = 245.19/122.13 = 2

Multiply the empirical by 2 = 2 x (C7H6O2) = C14H12O4