Question

Combustion analysis of toluene, a common organic solvent, gives 7.03 mgmg of CO2 and 1.64 mgmg of H2O. If the compound contains only carbon and hydrogen, what is its empirical formula?

Answer 1

Answer:

The empirical formula is = $C_7H_8$

Explanation:

Mass of carbon dioxide obtained = 7.03 mg

1 mg = 10⁻³ g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of $CO_2$ = 7.03× 10⁻³ g  /44.01 g/mol = 0.1597×10⁻³ moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

Moles of C = 0.1597×10⁻³ moles

Mass of water obtained = 1.64 mg

Moles of $H_2O$ = 1.64× 10⁻³ g /18 g/mol = 0.0911×10⁻³ moles

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

Moles of H = 2 x 0.0911 ×10⁻³  = 0.1822 ×10⁻³ moles

Taking the simplest ratio for H and C as:

0.1822 ×10⁻³ : 0.1597×10⁻³

 = 8 : 7

The empirical formula is = $C_7H_8$