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

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

pure substance: iron,copper

impure substance: water,charcoal

samuel de champlain. in 1608, king henry iv sponsored pierre dugua, sieur de mons and samuel de champlain as founders of the city of quebec with 28 men. this was the second permanent french settlement in the colony of canada.