Given :
A certain compound contains 4.0 g of calcium and 7.1 g of chlorine.
Its relative molecular mass is 111.
To Find :
Its empirical and molecular formulas.
Solution :
Moles of calcium , [tex]n_1=\dfrac{4}{40}=0.1\ moles[/tex].
Moles of chlorine , [tex]n_2=\dfrac{7.1}{35.45}=0.2\ moles[/tex] .
The ratio calcium and chlorine is 1 : 2 .
So , the empirical formula is [tex]CaCl_2[/tex] .
Now , molecular mass of [tex]CaCl_2[/tex] is :
[tex]M=40+(2\times 35.45)\\\\M=110.9\ g[/tex]
So , [tex]n=\dfrac{111}{110.9} \approx1[/tex]
Therefore , the molecular formula is also [tex]CaCl_2[/tex] .
Hence , this is the required solution .
