Question

Combustion of 47.60 mg of cumene CxHy produces 156.85 mg C02 (g) and 42.80 mg H20 (1). The molar mass of H2O is 18.01 g / mol. The molar mass of C02 is 44.01 g / mol.

What is the mass of hydrogen H in 47.60 mg of cumene?

What is mass of carbon C in 47.60 mg of cumene? What is the empirical formula for cumene CxHy?

The molar mass of cumene is between 90 and 150 g / mol. What is the molecular formula for cumene?

Answer 1

Use the mas of CO2 and H2O formed to calculate the moles and mass of C anf H respectively in the cumene sample.

All the C that was in the cumene ends up in the CO2
moles CO2 formed = mass / molar mass
= 156.85 mg / 44.01 g/mol
= 0.003564 moles CO2

Each mole CO2 has 1 mole C
Thus moles C = moles CO2 = 0.003564 moles C

mass C = moles x molar mass
= 0.003564 moles x 12.01 g/mol
= 0.042804 g C
= 42.803 mg C

All the H from the sample ends up in the H2O
moles H2O = mass / molar mass
= 42.80 mg / 18.01 g/mol
= 0.0023764 moles H2O

There are 2 H in each H2O
moles H = 2 x moles H2O
= 0.004753 moles H in the sample of cumene

mass H in cumene = moles H x molar mass
= 0.004753 mol x 1.008 g/mol
= 0.004791 g H in the cumene
= 4.7909 mg H

Now that we know moles of each element in the sample you can work out the empirical formula.

ratio of moles C : moles H

42.803 : 4.7909

Divide each number in the ratio by the smallest number to get the smallest whole number ratio

42.803/4.7909 : 4.7909 /4.7909 == 9:1

so the empirical formula comes to be C9H1

To get the molecular formula from the molar mass and empirical formula divide the molar mass by the mass of the empirical formula
This tells you how many times the empirical formula fits into the molar mass

empirical formula mass =109g/mol

lets take molecular weight be 109g /mol

molar mass / empirical formula mass =109 g/mol / 109 g/mol
= 1

so here it comes ,

molecular formula = empirical formula

i.e, C9H1