Problem: Hydrated sulphate of a divalent metal of atomic weight 65.4g loses 43.85% of its weight on dehydration then find the number of molecules of water of crystallization and formula weight of hydrated salt.

Solution:

Calculating the number of water molecules of crystallization:

Let the formula of hydrated salt be MSO4.xH2O.

Let the weight of hydrated salt be w grams.

Let the weight of anhydrous salt be a grams.

Therefore, weight of water of crystallization = w - a grams.

Given that on dehydration, the weight of hydrated salt reduces by 43.85%.

Hence, the weight of water lost is 43.85% of weight of hydrated salt = 0.4385w grams.

Therefore, weight of anhydrous salt = w - 0.4385w = 0.5615w grams.

Also, the weight of anhydrous salt (a) is equal to the atomic weight of the metal (M) + the weight of the sulphate group (SO4).

Atomic weight of M = 65.4 g/mol

Weight of SO4 = 32 + 4(16) = 96 g/mol

Therefore, weight of anhydrous salt (a) = 65.4 + 96 = 161.4 g/mol

Now, the weight of water of crystallization = w - a grams.

Therefore, weight of water of crystallization = w - 0.5615w = 0.4385w grams.

The molecular weight of water is 18 g/mol.

Hence, the number of water molecules of crystallization = weight of water of crystallization / molecular weight of water.

Therefore, number of water molecules of crystallization = (0.4385w / 18) = 0.0243w.

Calculating the formula weight of hydrated salt:

The formula weight of hydrated salt is the sum of atomic weights of all atoms in the formula unit.

Formula unit of hydrated salt = MSO4.xH2O

Atomic weight of M = 65.4 g/mol

Atomic weight of S = 32 g/mol

Atomic weight of O = 16 g/mol

Molecular weight of H2O = 18 g/mol

Therefore, formula weight of hydrated salt = 65.4 + 32 + 4(16) + (18x) = 65.4 + 32 + 64 + 18x = 161.4 + 18x g/mol

Hence, the formula weight of hydrated salt is 161.4 + 18x g/mol.

Conclusion: The number of water molecules of crystallization is 0.0243w and the formula weight of hydrated salt is 161.4 + 18x g/mol.