Question

A salt has the following percentage composition : $Al=$ 10.50%, $K=$ 15.1 %, $S=$ 24.8 % and the remaining oxygen. The empirical formula of the salt is:

($Al=$ 27 g/mol, $K=$ 39 g/mol, $S=$ 32 g/mol, $O=$16 g/mol)

• A. $A{l}_{2}K(S{O}_4{)}_2$
• B. $Al{K}_2(S{O}_4{)}_2$
• C. $AlK\left(S{O}_4\right)$
• D. $AlK(S{O}_4{)}_2$

Answer 1

The correct option is D $AlK(S{O}_4{)}_2$

A salt has the following % composition : $Al$ $=$ 10.50%, $K$ $=$ 15.1 %, $S$ $=$ 24.8 % and the remaining oxygen.
Thus, $100g$ of salt will have $10.50gAl,15.1gK,24.8gS$ and $100-10.50-15.1-24.8=49.6g$ oxygen.
The atomic weights are $Al=27,K=39,S=32,O=16$.
When mass is divided with atomic weight, we get number of moles.
Number of moles of $Al$ $=\frac{10.50}{27}=0.389$ moles.

Number of moles of $K$ $=\frac{15.1}{39}=0.387$ moles.

Number of moles of $S$ $=\frac{24.80}{32}=0.775$ moles.

Number of moles of $O$ $=\frac{49.6}{16}=3.1$ moles.
The mole ratio $Al:K:S:O=0.389:0.387:0.775:3.1=1:1:2:8$ .
Hence, the empirical formula is $AlK(S{O}_4{)}_2$