Question

Calculate the percentage loss of mass of hydrated copper [II] Sulphate when it is completely dehydrated.

Atomic wts are Cu = 64, S = 32, O =16 and H = 1

Answer 1

Hey Friend,

Hydrated copper (II) sulphate - CuSO4.xH2O
Considering that the Salt is completely hydrated... x=5

Hydrated copper (II) sulphate - CuSO4.5H2O 

Dehydrated copper (II) sulphate - CuSO4

Molecular weights of -

1. CuSO4.5H2O 
    64 + 32 + (16x4) + 5 [2 + 16]
    64 + 32 + 64 + 90
    250 grams

2. CuSO4
    64 + 32 + (16x4)
    64 + 32 + 64
    160 grams

Therefore,

Difference in the molecular weights... 250 - 160 = 90 grams

Percentage loss =      Difference in weights
                              --------------------------------------    x 100
                         Molecular weight of intial compound

Percentage loss =  90 x 100 / 250
                           = 36 %

Therefore, the percentage loss in the mass when hydrated copper sulphate is completely dehydrated is 36 %.

Hope it helps!

Answer 2

Answer:

CuSO4.5H2O --------> CuSO4 + 5 H2O

64+32+4(16)+5(2+16)

=> 250 grams

i.e 250 grams of hydrated copper sulphate contains 90 grams of water.

=> 90×100/250

= 36%

This is your Ans.....

Hope it helps you