An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57% C and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of −5.20°C is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

Interpretation: The molar mass, empirical formula and molecular formula of the unknown compound has to be determined.

Concept Introduction:

The ratio of the elements present in a compound and not the arrangement of the atoms are called as Empirical formula

Molar mass of a compound can be given by mass of a substance to the amount of the substance. It can be given by expression,

Molar mass = Mass of substanceAmount of substance

Molecular formula is the representation of sum of number of atoms and molecules, not their arrangement in structure.

Record the given info

Mass percent of Carbon     = 31.57%

Mass percent of Hydrogen = 5.30%

Mass of the compound      = 10.56 g

Mass of Water                  = 25.0 g

Freezing point depression = -5.20°C

To calculate the moles of Carbon, Hydrogen and Oxygen.

Molecular Weight of Carbon = 12.01 g

Molecular Weight of Hydrogen = 1.008 g

Molecular weight of Oxygen = 16.00 g

Moles of Carbon = 31.57 g C×1 mol C12.01 g=2.629 moles

Moles of Hydrogen = 5.30 g×1 mol H1.008 g=5.26 moles

Moles of Oxygen = 63.13 g×1 mol O16.00 g=3.946 moles

Moles of Carbon = 2.629 moles

Moles of Hydrogen = 5.26 moles

Moles of Oxygen = 3.946 moles

To write the empirical formula of the Unknown compound

Moles of Carbon = 2.629 moles

Moles of Hydrogen = 5.26 moles

Moles of Oxygen = 3.946 moles

In order, to get empirical formula of the Unknown compound, the moles of individual elements is divided by the smallest number of moles

Empirical formula of the compound = C = 2.6292.629=1.000

                                                          H = 5.262.629=2.00

   O = 3.9462.629=1.501

The empirical formula of the compound = C2H4O3

Record the given info

Mass percent of Carbon     = 31.57%

Mass percent of Hydrogen = 5.30%

Mass of the compound      = 10.56 g

Mass of Water                  = 25