The molecular formula : Al₂Cl₆
Given
20.2% by mass of Aluminum and 79.8% by mass of Chlorine.
Required
the molecular formula
Solution
Aluminium (Ar=27 g/mol)
[tex]\tt \dfrac{20.2}{27}=0.748[/tex]
Chlorine(Ar=35.5 g/mol)
[tex]\tt \dfrac{79.8}{35.5}=2.248[/tex]
mol ratio Al : Cl =
[tex]\tt \dfrac{0.748}{0.748}\div \dfrac{2.248}{0.748}=1\div 3[/tex]
So the empirical formula = AlCl₃
(Empirical formula)n=Molecular formula
(27+3.35.5)n=267
(133.5)n=267⇒n=2
The molecular formula : Al₂Cl₆