Given : 

i. Atomic mass : 

Fe = 56; 

S = 32; 

O = 16 

ii. Mass of crystal = 4.54 kg 

To find :

i. Mass percentage of Fe, S, H and O 

ii. Mass of iron and water of crystallisation in 4.54 kg of crystal

Formula :

Percentage (by weight) = \(\frac{Mass\,of\,the\,element\,in\,1\,mole\,of\,compound}{Molar\,mass\,of\,the\,compound}\) \(\times 100\)

i. Molar mass of FeSO₄.7H₂O 

= 1 × (56) + 1 × (32) + 14 × (1) + 11 × (16) 

= 56 + 32 + 14+ 176 

= 278 g mol^-1

Percentage of Fe = \(\frac{56}{278}\) \(\times 100\)

= 20.14%

Percentage of S = \(\frac{32}{278}\)\(\times 100\)

= 11.51%

Percentage of H = \(\frac{14}{278}\)\(\times 100\)

= 5.04%

Percentage of O = \(\frac{176}{278}\)\(\times 100\)

= 63.31%

ii. 278 kg green vitriol 

= 56 kg iron 

∴ 4.54 kg green vitriol = x 

∴ x = \(\frac{56\times 4.54}{278}\)

Mass of 7H₂O in 278 kg green vitriol = 7 × 18

= 126 kg

∴ 4.54 kg green vitriol = y

∴ y = \(\frac{126\times 4.54}{278}\)

∴ i. Mass percentage of Fe, S, H and O in FeSO₄ .7H₂O are 20.14%, 11.51%, 5.04% and 63.31% respectively.

ii. Mass of iron in 4.54 kg green vitriol = 0.915 kg

Mass of water of crystallisation in 4.54 kg 

green vitriol = 2.058 kg