Question

A hydrocarbon has a weight of 78 gives carbon dioxide and water on combustion with 3.385 of carbon and 0.682 of water find the formula

Answer 1

Calculate the moles of C and of H formed in the combustion:

$\dfrac{3.385}{44}=0.0769 moles$ CO₂ * ($\dfrac{1 mole}{1moles_{CO_{2}}})$ $=0.0769moles$ of C

$\dfrac{0.682}{18}=0.0379moles$ H₂O * ($\dfrac{2 mole}{1moles_{H_{2}O}})$ = $0.0758moles$ of H

Divide each by the smaller:

C = $\dfrac{0.0769}{0.0758}=1.0$

H = $\dfrac{0.0758}{0.0758} = 1.0$

The empirical formula is CH

Since the empirical formula mass is 13 $( \tfrac{g}{mole})$ dividing the molar mass of the compound by the formula mass gives $\tfrac{78}{16} =6$ .

The molecular formula is C₆H₆