Question

An unknown compound contains only C,H and O. Combustion of 5.80 g of this compound produced 14.2 g of CO2 and 5.80 g of H2O. What is the empirical formula of the unknown compound?

Answer 1

Atomic mass of H = 1 g/mol

Atomic mass of C = 12g/mol

Atomic mass of O = 16 g/mol

Atomic mass of H2O = 18g/mol

Atomic mass of CO2 = 44 g/mol

Mass of C in 14.2 g

CO2 = 14.2g / 44 g/mol * 12 g/mol = 3.87 g

Mass of H in 5.80g H2O

H2O = 5.80g / 18g/mol. * 2g/mol = 0.64 g

These two amounts of C and H are obtained from 5.80g compound containing C,H and O

So the remaining ( 5.80-3.87-0.64)g = 1.29 g is the mass of oxygen present

The ratio of number of atoms of C,H and O in the molecule of the compound

C : H: O = (3.87/12) : (0.64/1) : (1.29/16)

= 0.322 : 0.64 : 0.08

= (0.322/0.08) : (0.64/0.08) :(0.08/0.08)

= 4: 8 :1

The empirical formula is C4H8O