Answer to Question #129065 in Chemistry for CLASEY

Question

Benzene is a liquid hydrocarbon. When a 342 mg sample of benzene is burned in an excess of
oxygen 1156 mg of carbon dioxide is produced. Determine the empirical formula of benzene. If
the molar mass of benzene is 78.1134 g/mol, what is the molecular formula of benzene?

Answer 1

The reaction of benzene combustion can be described as following:

C_xH_z + O₂ = CO₂ + H₂O

From here:

mass of carbon = (1156 mg / 1000 mg of CO₂) × (1 mol CO2 / 44.01 g CO₂) × (1 mol C / 1 mol CO₂) × (12.011 g / 1 mol C)= 0.315 g of carbon is present in the sample.

As mass of the sample equals 342 mg and the mass of carbon in the sample is 315 mg, than the mass of the hydrogen in the sample equals:

mass of hydrogen = 342 mg - 315 mg = 27 mg = 0.027 g.

The number of moles of carbon and hydrogen in the sample are:

moles of carbon = 0.315 g × 1 mol / 12.011 g = 0.026 mol

moles of hydrogen = 0.027 g × 1 mol / 1.0079 g = 0.027 mol

From here, benzene contains equal ratio of moles of carbon to moles of hydrogen: C₁H₁ or CH.

The molecular formula of benzene can be found as a ratio between the molar mass of the compound and the empirical formula mass (Mr(CH) = 13.0189 g/mol):

78.1134 g/mol / 13.0189 g/mol = 6

From here, the molecular formular of benzene is C₆H₆.

Answer: CH and C₆H₆.