Chemistry empirical and molecular formula

Question

A 7.893 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 20.41 grams of CO₂ and 4.179 grams of H₂O are produced.

In a separate experiment, the molecular weight is found to be 136.2 amu. Determine the empirical formula and the molecular formula of the organic compound.

Answer 1

FInd moles of C, H and O:

20.41 g CO2 x 1 mol CO2 / 44 g = 0.4629 moles x 1 mol C/mol CO2 = 0.4629 mols C

4.179 g H2O x 1 mol H2O/18 g = 0.2322 moles x 2 mol H/mol H2O = 0.4644 mols H

To find mols O, we convert back to mass and subtract from original mass of 7.893 g

0.4639 mol C x 12 g C/ mol = 5.567 g C

0.4644 mol H x 1 g H/mol = 0.4644 g H

g of O = 7.893 g - 5.567 g - 0.4644 g = 1.862 g O

1.863 g O x 1 mol O / 16 g = 0.1164 mol O

Divide all moles by the smallest values (0.1164) to try to get whole numbers:

moles C = 0.4629 / 0.1164 = ~4

moles H = 0.4644 / 0.1164 = ~4

moles O = 0.1164/0.1164 = 1

Empirical formula = C₄H₄O

Molar mass of C4H4O = 48 + 4 + 16 = 68 g/mol

How many of the empirical form. fit in the molecular form.? 136.2 / 68 = 2 empiricals in 1 molecular

Molecular formula = C₈H₈O₂