Problem 3-105

Cumene: C_xH_y (unknown values of x and y)

Complete combustion (reaction with O₂) of any molecule containing only "C", "H" and "O" ALWAYS leads to the formation of CO₂(g) and H₂O.

Here: C_xH_y + nO₂ → xCO₂ + ^y/₂H₂O

The knowledge of the stoichiometry of O₂ reactant is useless here.
Additionally, you can work in grams instead of mg, the results will remain the same and the numbers will be easier to manage.

Mass of cumene: 47.6 g
Mass of H₂O: 42.8 g
Mass of CO₂: unknown

All the hydrogen present in the H₂O collected is coming from cumene:
42.8 g H₂O / 18.01 g/mol = 2.377 mol H₂O = 4.753 mol H (since there is 2 H in each H₂O)
Mass of H in cumene = 4.753 mol H x 1.008 g/mol = 4.791 g H

The mass of carbon in cumene can be known by difference

mass of C in cumene = 47.6 g (total) - 4.791 g (Hydrogen) = 42.81 g C
mole of C in cumene: 42.81 g / 12.01 g/mol = 3.564 mol.

At this point, cumene = C_3.564H_4.753

By dividing this formula by 3.564 (the smallest number) cumene = C₁H_1.333

Since the empirical formula has to be made of integer only: 3 x (C₁H_1.333) = C₃H₄ (40 g / empirical unit)

Knowing that cumene molar mass is 115-125 g/mol, then the molecular formula = C₉H₁₂ (120 g / mol)