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A 8.919 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 12.65 grams of CO2 and 7.768 grams of H2O are produced. In a separate experiment, the molar mass is found to be 62.07 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

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Answer:

Empirical formula = CH₃O

Molecular formula = C₂H₆O₂

Explanation:

The empirical formula is the simplest whole-number ratio of atoms present in a molecule. To find the ratio of atoms we must find the moles of C, H and O. With the molar mass we can find the molecular formula:

Moles C = Moles CO₂:

12.65g * (1mol / 44.01g) = 0.2874 moles of C * 12.01g/mol = 3.452g C

Moles H = 1/2 moles H₂O:

7.768g * (1mol / 18.015g) = 0.4312 moles of H₂O * 2 = 0.8624 moles H * 1.008g/mol = 0.869g H

Moles O:

Mass O = 8.919g - 0.869g H - 3.452g C = 4.598g O * (1mol / 16g) = 0.2874 moles of O

Ratio of atoms -Dividing in the lower number of moles-:

C = 0.2874 mol C / 0.2874 mol = 1

H = 0.8624 mol H / 0.2874 mol = 3

O = 0.2874 mol O / 0.2874 mol = 1

Empirical formula = CH₃O

The molar mass of this formula is:

1C = 12.01g/mol

3H = 3*1.008g/mol = 3.024g/mol

1O = 16g/mol

12.01+3.024+16 = 31.034g/mol

Twice this formula gives the molecular formula = 31.034g/mol*2 = 62.068g/mol ≅ the molar mass of the compound. That means molecular formula of the compound is:

C₂H₆O₂

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