Question

Cumene is a compound containing only carbon and hydrogen that is used in the production of acetone and phenol in the chemical industry. Combustion of 47.6 mg cumene produces some CO2 and 42.8 mg water. The molar mass of cumene is between 115 and 125 g/mol. Determine the empirical and molecular formulas.

Answer 1

Answer:

$C_9H_{12}$

Explanation:

$Moles =Given\ mass \times {Molar\ mass}$

Mass of water obtained = 42.8 mg

1 mg = 10⁻³ g

Moles of $H_2O$ = 42.8/18 = 2.3778×10⁻³ moles

2 moles of hydrogen atoms are present in 1 mole of water. So,

Moles of H = 2 x 2.4 = 4.7556×10⁻³ moles

Molar mass of H atom = 1.008 g/mol

Mass of H in molecule = 4.7556×10⁻³ x 1.008 = 4.7936×10⁻³ g

Given that the cumene only contains hydrogen and carbon. So,

Mass of C in the sample = Total mass - Mass of H

Mass of the sample = 47.6 mg = 47.6×10⁻³ g

Mass of C in sample = 47.6×10⁻³ - 4.80×10⁻³ = 42.8064×10⁻³ g

Moles of C  = 42.8064×10⁻³  / 12 = 3.5672×10⁻³ moles

Taking the simplest ratio for H and C as:

4.7556×10⁻³ : 3.5672×10⁻³  = 4 : 3

The empirical formula is = $C_3H_4$

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 3×12 + 4×1 = 40 g/mol

115 g/mol < Molar mass > 125 g/mol

So,  

Molecular mass = n × Empirical mass

115 g/mol < 40 n > 125 g/mol

⇒ n ≅ 3

The formula of cumene = $C_9H_{12}$