Question

A 1.802 gram sample of unknown compound contains 0.721 grams of carbon, 0.121 grams of hydrogen and 0.960 grams of oxygen. The compound has a molar mass of 180.2 g/mol. Determine the empirical formula of the compound and the molecular formula. Show all work.

Answer 1

Fraction of Carbon present in the compound = Mass of Carbon / Mass of Compound

Fraction of Carbon present in the compound = (0.721 g / 1.802 g) = 0.4

Fraction of Hydrogen present in the compound = Mass of Hydrogen / Mass of Compound

Fraction of Hydrogen present in the compound = 0.121 g / 1.802 g = 0.067

Fraction of Oxygen present in the compound = Mass of Oxygen / Mass of Compound

Fraction of Oxygen present in the compound = 0.960 g / 1.802 g = 0.533

1 mol of compound weighs 180.2 g

In 1 mol of compound:

Mass of Carbon present = Fraction of Carbon x Mass of Compound

Mass of Carbon present = 0.4 x 180.2 g = 72.08 g

Moles of Carbon present = Mass of Carbon / Molar Mass of Carbon = 72.08 g / 12 g/mol = 6.01 mol ~ 6 mol

Mass of Hydrogen present = Fraction of Hydrogen x Mass of Compound

Mass of Hydrogen present = 0.067 x 180.2 g = 12.07 g

Moles of Hydrogen present = Mass of Hydrogen / Molar Mass of Hydrogen = 12.07g / 1 g/mol = 12.07 mol ~ 12 mol

Mass of Oxygen present = Fraction of Oxygen x Mass of Compound

Mass of Oxygen present = 0.533 x 180.2 g = 96.05 g

Moles of Oxygen present = Mass of Oxygen / Molar Mass of Oxygen = 96.05 g / 16g/mol = 6 mol

Molecular Formula = C₆H₁₂O₆ Since 1 mol of the compound contains 6 mol of Carbon, 12 mol of Hydrogen, 6 mol of Oxygen

Empirical Formula = CH₂O Since the highest common factor was 6

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