Question

Consider the hydrate FeSO4 • 7H2O.

Answer 1

The molar mass of the hydrate is 278.06
The molar mass of the anhydrous salt is 151.92
The molar mass of water in the hydrate is 126.14

Answer 2

Answer: The molar mass of the hydrate $FeSO_4$.$7H_2O$ is 278g. The molar mass of the anhydrous salt  $FeSO_4$ is 152 g. The molar mass of the water in the hydrate is $H_2O$ is 126g.

Explanation: Molar mass of Fe = 56 g

Molar mass of of S = 32 g

Molar mass of O = 16 g

Molar mass of H = 1 g

Molar mass of  $FeSO_4$•$7H_2O$=$1\times \text{molar mass of Fe}+1\times \text{molar mass of S}+11\times \text{molar mass of O}+14\times \text{molar mass of H}$

Molar mass of  $FeSO_4$•$7H_2O$=$1\times 56+1\times 32+11\times 16+14\times 1=278g$

Molar mass of  $FeSO_4=1\times \text{molar mass of Fe}+1\times \text{molar mass of S}+4\times \text{molar mass of O}$

Molar mass of  $FeSO_4=1\times 56+1\times 32+4\times 16=152g$

Molar mass of  $7H_2O=14\times \text{molar mass of H}+7\times \text{molar mass of O}$

Molar mass of  $7H_2O=14\times 1+7\times 16=126g$