The combustion of 135 mg of a hydrocarbon produces 440 mg of CO2 and 135 mg H2O. The molar mass of the hydrocarbon is 270 g/mol. Determine the molecular formula of this compound.

Molecular formula = C20H30

Explanation:

NB 440mg = 0.44g, 135mg = 0.135g

From the question, moles of CO2 = 0.44/44 = 0.01mol

Since 1 mol of CO2 contains 1mol of C, it implies mol of C = 0.01

Also from the question, moles of H2O = 0.135/18 = 0.0075mole

Since 1 mol of H2O contains 2mol of H, it implies mol of H = 0.0075*2 = 0.015 mol of H

To get the empirical formula, divide by smallest number of mole

Mol of C = 0.01/0.01=1

Mol of H = 0.015/0.01 = 1.5

Multiply both by 2 to obtain a whole number

Mol of C = 1*2 = 2

Mol of H = 1.5*2 = 3

Empirical formula = C2H3

[C2H3] not = 270

[ (2*12) + 3]n = 270

27n = 270

n=10

Molecular formula = [C2H3]10 = C20H30