An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives the mass percent of 31.57% and 5.30%. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of –5.30°C is recorded for a solution made by dissolving 10.8 g of the compound in 25.0 g water. Determine the molar mass and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

C4H8O6 is the molecular formula

The molar mass is 152 g/mol

Explanation:

Step 1: Data given

Mass % C = 31.57%

Mass % H = 5.30 %

Mass % O = 100 - 31.57 - 5.30 = 63.13 %

The freezing point = –5.30°C

Mass of the compound = 10.8 grams

Mass of water = 25.0 grams

compound is a nonelectrolyte.

Step 2: Calculate moles of elements

moles C = 31.57 grams/ 12.01 g/mol = 2.629 moles

moles H = 5.30 grams/ 1.01 g/mol = 5.25 moles

moles O = 63.13 grams / 16.0 g/mol = 3.95  moles

Step 3: Calculate the mol ratio

We divide by the smallest number

C: 2.629 moles / 2.629 moles = 1

H: 5.25 moles / 2.629 moles = 2

O: 3.95 moles / 2.629 moles = 1.5

The empirical formula is C2H4O3

The molecular mass of this is 76.06 g/mol

Step 4: Calculate moles solute

5.30 = m * 1.86

m = 2.85 = moles solute / 0.025 Kg

moles solute = 2.85 * 0.025 = 0.07125 moles

Step 5: Calculate molar mass

molar mass = 10.8 grams / 0.07125 moles = 152 g/mol

Step 6: Calculate molecular formula

152/76.06 ≈ 2

We have to multiply thr empirical formula by 2

2*(C2H4O3) = C4H8O6

C4H8O6 is the molecular formula