When a 0.860 g sample of an organic compound containing C, H, and O was burned completely in oxygen, 1.64 g of CO2 and 1.01 g of H2O were produced. What is the empirical formula of the compound?

Empirical formula is the simplest ratio of atoms present in a molecule. To solve this question we must find the moles of C, H and O in order to find the simplest ratio of this atoms.

To solve this question we must find the moles of C from CO2, moles of H from H2O and moles of O from the difference of masses of C and H and the sample:

Moles C:

1.64g CO2 * (1mol / 44g) = 0.03727 moles CO2 * (1mol C / 1mol CO2) = 0.03727 moles C

0.03727 moles C * (12g / mol) = 0.4473g C

Moles H:

1.01g H2O * (1mol / 18g) = 0.05611 moles H2O * (2mol H / 1mol H2O) =

0.1122 moles H

0.1122 moles H * (1g / mol) = 0.1122g H

Moles O:

0.860g - 0.1122g H - 0.4473g C = 0.3005g O * (1mol / 16g) =

0.01878 moles O

The ratio of the moles of atoms regard to the moles of O (Lower number of moles) is:

C = 0.03727 moles C / 0.01878 moles O = 1.98 ≈ 2

H = 0.1122 moles H / 0.01878 moles O = 5.97 ≈ 6

O = 0.01878 moles O / 0.01878 moles O = 1

Empirical formula is: