Question

A crystalline hydrated salt on being rendered anhydrous loses 45.6% of its weight.
The percentage composition of the anhydrous salt is:
$Al=10.5\%,K=15.1\%,S=24.8\%$ and oxygen $=49.6\%$
[Molar mass of $Al=27,K=39,S=32$].

What is the empirical formula of the salt?

• A. $K_{2}Al{S}_2{O}_7$
• B. $K_{2}A{l}_2{S}_2{O}_7$
• C. $KAl{S}_2{O}_8$
• D. $K_{3}Al{S}_2{O}_{12}$

Answer 1

The correct option is C $KAl{S}_2{O}_8$

$\begin{array}{cccc}\text{Element}& \%& \frac{\%\text{mass}}{\text{Atomic mass}}& \text{Simplest ratio}\\ Al& 10.5& \frac{10.5}{27}=0.388& 1\\ K& 15.1& \frac{15.1}{39}=0.387& 1\\ S& 24.8& \frac{24.8}{32}=0.775& 2\\ O& 49.6& \frac{49.6}{16}=3.1& 8\end{array}$

$\therefore$ Empirical formula = $KAl{S}_2{O}_8$