question

A 0.1005g sample of menthol is combusted,producing 0.2829g of CO2 and 0.1159g of H2O.What is the empirical formula?If the compound has a molar mass of 156g/mol what is the molecular formula?

In 0.1005g of the compound :

mole of C = mole of C in CO2 = 0.2829g /44 = 0.006430

mole of H = mole of H in H2O = 2 x 0.1159g /18 = 0.01288

mole of O = (0.1005 - 0.006430 x 12 - 0.01288 x 1)/16 = 0.0006538

mole ratio C : H : O = 0.006430 : 0.01288 : 0.0006538 = 10 : 20 : 1

Hence empirical formula = C10H20O

Let molecular formula = (C10H20O)n

Relative molecular mass = (12x10 + 1x20 + 16x1)n = 156

n = 1

Thus, molecular formula = C10H20O