Question

1. Combustion of 6.38g of an organic compound containing C, H and O gives 9.06g of CO2 and 5.58g of H2O. What is the empirical formula of the compound?

Answer 1

moles of CO2 = 9.06 / 44 = 0.2059

moles of C= 0.2059

mass of Carbon = 12 x 0.2059 = 2.471 g

moles of H2O = 5.58 / 18 = 0.31

moles of H = 2 x 0.31 = 0.62

mass of hydroegen = 0.62 g

Oxygen mass = 6.38 - (3.091 ) = 3.289 g

moles of O = 3.289 /16 = 0.2056

C                O               H

0.2059       0.2056        0.62

1               1                    3

CH3O -----------------------> empirical formula