Upon combustion, a 0.8009 g sample of a compound containing only carbon, hydrogen, and oxygen produces 1.6004 g co2 and 0.6551 g h2o. determine the empirical formula of the compound.

The compound contains Carbon, Hydroxide and Oxide

1 mole of carbon iv oxide contains 44 g, out of which 12 g are carbon.

Therefore, 1.6004 g of CO2 will contain;

1.6004 * 12/44 = 0.4365 g of carbon

1 mole of water contains 18 g of which 2 g is hydrogen,

Therefore, 0.6551 g of H2O will hace;

0.6551 * 2/18 = 0.0728 g of hydrogen.

The total mass of the compound is 0.8009 g,

Thus the mass of oxygen = 0.8009 - (0.4365 + 0.0728)

= 0.2916 g

To get the empirical formula we first get the number of moles of each element; /

Carbon = 0.4365/12 = 0.036375 moles

Hydrogen = 0.0728/1 = 0.0728 moles

Oxygen = 0.2916/16 = 0.018225 moles

Then, to get the smallest ratio we divide each with the smallest value;

Carbon : Hydrogen : Oxygen

= (0.036375/0.018225) : (0.0728/0.018225) : (0.018225/0.018225)

= 1.996 : 3.995 : 1

≈ 2 : 4 : 1

Therefore, the empirical formula is C2H4O