Question

A 12.55 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 18.40 grams of CO2 and 7.532 grams of H2O are produced. In a separate experiment, the molar mass is found to be 60.05 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

Answer 1

1.

1 mol of CO2 has 1 mol of C

44.00 g. of CO2 has 12.00 g. of C

then, 18.40 g. of CO2 has 12.00 * 18.40 / 44.00 = 5.018 g. of C

This amount of C is present in 12.55 g. of sample

then 100g. of sample has 5.018*100 / 12.55 = 39.98 % of C

2.

1 mol H2O has 2 mol H

18 g. of H2O has 2 g. of H

Then, 7.532 g. of H2O has 2.000 * 7.532 / 18.00 = 0.8369 g. of H

This much amount of H is in 12.55 g. of sample.

So, 100 g. of sample has 0.8369 * 100 / 12.55 = 6.668 % of H

Now, % O in the sample = 100 - 39.98 - 6.668 = 53.35 %

3. Emperical formula,

Element   Mass Number of moles        Division by small number       simple whole number

   C        39.98 39.98 / 12 = 3.332                3.332/3.332 = 1                         1

   H        6.668    6.668 / 1 = 6.668                 6.668 / 3.332 = 2                       2

   O       53.35    53.35 / 16 = 3.334                3.334/3.332 = 1                        1

The emperical formula = CH2O

Emperical formula mass = 12 + 2 + 16 = 30

Molecular mass = 60.05 g/mol

n = Molecular mass / emperical formula mass

n = 60.05 / 30

n = 2

Therefore,

Molecular formula = [Emperical formula]n = (CH₂O)_n = C₂H₄O₂