Question

A hydrocarbon Z with molecular mass 78 on combustion gave 3.385g of CO2 and 0.692g of H2O. Determine the molecular formula of Z [H=1, C=12, O=16]

Answer 1

Z + O₂ = CO₂ + H₂O

$moles \space of \space CO_2 = \frac {3.385 \space g}{44 \space g/mol} = 0.0769 mol$

$mass \space of \space carbon \space in \space Z= 12 g/mol \cdot 0.0769 mol= 0.923 g$

$moles \space of H_2O=\frac {0.692 g}{18g/mol}=0.0384 mol$

$mass \space of \space hydrogen \space in \space Z= 2 g/mol \cdot 0.0384 mol = 0.077 g$

Empirical formula of Z is (CH)_n = 78 g/mol

13n =78

n=78/13

n= 6

Molecular formula of Z is C₆H₆