Determining a Chemical Formula

Question

A 9.749 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 21.01 grams of CO₂ and 8.601 grams of H₂O are produced.

In a separate experiment, the molar mass is found to be 102.1 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

1. Enter the elements in the order C, H, O

Empirical formula=

molecular formula =

Answer 1

The combustion reaction would look like this:'

CHO + O₂ ==> CO2 + H2O ... unbalanced

From the mass of CO2 and H2O collected, we can determine the number of moles of C and H in the original sample:

21.01 g CO2 x 2 mol CO2 / 44 g x 1 mol C / mol CO2 = 0.4775 moles C in sample

8.601 g H2O x 1 mol H2O / 18 g x 2 mol H / mol H2O = 0.9557 moles H in sample

To find moles O in sample, we first need to find grams of O and we do this by subtracting g C and g H from total grams, as follows:

grams C = 0.4775 mols C x 12 g /mol = 5.73 g C

grams H = 0.9557 mols H x 1 g/mol = 0.9557 g H

grams O = 9.749 g - 5.73 g - 0.9557 g = 3.063 g O

moles O = 3.063 g O x 1 mol/16 g = 0.1915 moles O in sample

Summary of moles in sample:

C = 0.4775 mols

H = 0.9557 mols

O = 0.1915 mols

Divide all by 0.1915 to try to get whole numbers:

C = 2.5 mols

H = 4.99 mols

O = 1.00 mols

Now multiply all by 2 to get whole numbers:

C = 5 mols

H = 10 mols

O = 2 mols

Empirical formula = C₅H₁₀O₂ molar mass = 60 + 10 + 32 = 102 g /mol

Since molar mass of empirical formula is the same as that given for the molecular formula, then ...

Molecular formula = C₅H₁₀O₂