Question

A crystalline hydrated salt on being rendered anhydrous, loses 45.6% of its weight. The percentage composition of anhydrous salt is : Al = 10.5%, K = 15.1%, S = 24.8% and I = 49.6%. Find the empirical formula of the salt.

• A. $KAl{S}_{2}I$
• B. $K_{2}Al{S}_{2}I$
• C. $KAl{S}_{2}I.12H_{2}O$
• D. $K_{2}Al{S}_{2}I.12H_{2}O$

Answer 1

The correct option is C $KAl{S}_{2}I.12H_{2}O$

The molar masses of Al, K, S and I are 27 g/mol, 39 g/mol, 32 g/mol and 127 g/mol respectively.
The percentage composition of anhydrous salt is : Al = 10.5%, K = 15.1%, S = 24.8% and I = 49.6%.

Thus, 100 g of anhydrous salt contains 10.5 g Al, 15.1 g K, 24.8 g S and 49.6 g I.

10.5 g Al corresponds to $\frac{10.5}{27}=0.39$ moles

15.1 g K corresponds to $\frac{15.1}{39}=0.39$ moles

24.8 g S corresponds to $\frac{24.8}{32}=0.775$ moles

49.6 g I corresponds to $\frac{49.6}{127}=0.39$ moles

Thus, the ratio of the number of moles of $Al:K:S:I$ is $0.39:0.39:0.775:0.39$

To get the whole number ratio, divide by 0.39

Thus, the ratio of the number of moles of $Al:K:S:I$ is $1:1:2:1$.

Hence, the empirical formula of the anhydrous salt is $KAl{S}_{2}I$.

The empirical formula mass is $39+27+32+32+127=257g/mol$.

The empirical formula weight of hydrated salt is $\frac{100}{100-45.6}\times 257=472$.

The mass of water present in 1 formula unit of crystalline salt is $472-257=215g$.

This corresponds to $\frac{215}{18}=12$ water molecules.

Hence, the empirical formula of hydrated salt is $KAl{S}_{2}I.12H_{2}O$.