combustion of CHO

Question

A 3.920 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 9.889 grams of CO₂ and 1.735 grams of H₂O are produced.

In a separate experiment, the molar mass is found to be 122.1 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

Answer 1

Notice that the unknown compound contains C, H and O.

We can determine moles of C from the grams of CO2

We can determine moles of H from the grams of H2O

To find moles of O, we'll need to subtract grams C and grams of H from total grams to get grams O and then convert that to moles of O (see below)

moles C = 9.889 g CO2 x 1 mol CO2 / 44.01 g x 1 mol C / mol CO2 = 0.2247 moles C

moles H = 1.735 g H2O x 1 mol H2O / 18.02 g H2O x 2 mol H / mol H2O = 0.193 moles H

grams C = 0.2247 mols C x 12 g / mol = 2.696 g C

grams H = 0.193 mols H x 1 g / mol = 0.193 g H

grams O = 3.920 g - 2.696 g - 0.193 g = 1.031 g O

moles O = 1.031 g O x 1 mol / 16 g = 0.0644 moles O

Dividing all moles by the smallest value (0.0644), we get...

moles C = 0.2247 / 0.0524 = 3.49 mols C

moles H = 0.193 / 0.0524 = 3.00 moles H

moles O = 0.0644 / 0.0644 = 1.00 moles O

Trying to get all values to whole numbers, we can multiply all by 2 to get ...

moles C = 3.49 x 2 = 7

moles H = 3.00 x 2 = 6

moles O = 1 x 2 = 2

Empirical formula = C₇H₆O₂

molar mass of empirical formula = 7x12 + 6x1 + 2x16 = 84 + 6 + 32 = 122 g / mol

Since this is the same as the molar mass of the molecular formula, the two are the same.

Molecular formula = C₇H₆O₂