Combustion analysis of an unknown compound containing only carbon and hydrogen produced 2.277 g of CO2 and 1.161 g of H2O. What is the empirical formula of the compound?

Answer 1

moles of C = 2.277 g CO₂ x 1 mol CO₂ / 44 g x 1 mol C / mol CO₂ = 0.0518 mols C

moles of H = 1.161 g H₂O x 1 mol H₂O / 18 g x 2 mol H / mol H₂O = 0.129 mols H

Divide both by 0.0518 to try to get whole numbers:

C = 0.0518 / 0.0518 = 1

H = 0.129 / 0.0518 = 2.5

Multiply both by 2 to get whole numbers:

C = 2

H = 5

Empirical formula = C₂H₅

Answer 2

First you convert both amounts from g to mol.

2.277 g CO₂ × (1 mol)/(44 g) = 2.277/44 = 0.05175 mol CO₂

1.161 g H₂O × (1 mol)/(18 g) = 1.161/18 = 0.0645 mol H₂O

Next you convert mol of CO₂ to mol of C atoms and mol of H₂O to mol of H atoms

0.05715 mol CO₂ × (1 mol C atoms)/(1 mol CO₂) = 0.05175 mol C atoms

0.0645 mol H₂O × (2 mol H atoms)/(1 mol H₂O) = 2(0.0645) = 0.129 mol H aroms

Next you take the ratio of the larger number (H aroms) to the smaller (C aroms)

(0.129 mol H)/(0.05175 mol C) = 2.49 which we'll round off to 2.5

This is the ratio of H to C in the empirical formula

Now you could write a formula C₁H_2.5

But all numbers in the formula need to be whole numbers, so you multiply both numbers by 2 to get

C₂H₅

This is your empirical formula.

Notes

Often in the process of determining an empirical formula you'll get numbers that a close to whole numbers and you round them off.

If you do come up with a number 1 for one of the atoms in your empirical formula, you don't include the 1 in your formula.