A crystalline salt on being rendered anhydrous loses 45.6% of its weight. The percentage composition of the anhydrous salt is Aluminium = 10.50%, Potassium = 15.1% Sulphur = 24.96%, Oxygen = 49.92%. Find the simplest formula of the anhydrous and crystalline salt.
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 2710.5=0.39 moles 15.1 g K corresponds to 3915.1=0.39 moles 24.8 g S corresponds to 3224.8=0.775 moles 49.6 g I corresponds to 12749.6=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 KAlS2I. The empirical formula mass is 39+27+32+32+127=257g/mol. The empirical formula weight of hydrated salt is 100−45.6100×257=472. The mass of water present in 1 formula unit of crystalline salt is 472−257=215g. This corresponds to 18215=12 water molecules. Hence, the empirical formula of hydrated salt is KAlS2I.12H2O.
