Question

1.615g of an anhydrous salt was placed in moist air.After a few days it was found to be 2.875g.Assuming that complete hydration has taken place ,calculate the simplest formula of hydrated salt.The composition of anhydrous salt is(Zn=40.6%,S=19.8%,O=39.6%)

Answer 1

The percentage of elements in anhydrous salt is as follows:

Zn – 40%

S- 19.8 %

O – 39.6 %

Therefore, in hundred grams of salt,

Mass of Zn = 40 g

Mass of S = 19.8 g

Mass of O = 39.6 g

Moles of Zn = 40/ 65.3 = 0.6

Moles of S = 19.8 / 32 = 0.6

Moles of O = 39.6/16 = 2.4

Ratio of moles of Zn, S and O = 0.6 : 0.6 : 2.4

Lowest ratio = 1 : 1: 4

Therefore, the formula of the anhydrous salt = ZnSO₄

Molar mass of ZnSO₄ = 161.5 g/mol

Number of moles in 1.615 g of ZnSO₄ = 1.615 / 161.5 = 0.01

Now,

Mass of water absorbed = 2.875 g - 1.615 g = 1.26 g

Moles of H₂O absorbed = 1.26 g / 18.0 g/mol = 0.07 moles

Moles of ZnSO₄ present = 0.01 mol

Therefore, moles H₂O /mol ZnSO₄ = 0.07 / 0.01 = 7

Hence, formula of hydrates salt is: ZnSO₄.7 H₂O