Calculate the percentage loss of mass of hydrated copper ll sulphate when completely dehydrated

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Solution

Hydrated copper (II) sulphate - ${C u S O}_{4} . x H_{2} O$
Considering that the Salt is completely hydrated... x=5

Hydrated copper (II) sulphate - ${C u S O}_{4} .5 H_{2} O$

Dehydrated copper (II) sulphate - ${C u S O}_{4}$

Molecular weights of -

1. ${C u S O}_{4} .5 H_{2} O$
64 + 32 + (16x4) + 5 [2 + 16]
64 + 32 + 64 + 90
250 grams

2. ${C u S O}_{4}$
64 + 32 + (16x4)
64 + 32 + 64
160 grams

Therefore, Difference in the molecular weights: 250 - 160 = 90 grams.
$P e r c e n t a g e l o s s = \frac{D i f f e r e n c e i n w e i g h t}{M o l e c u l a r w e i g h t o f i n i t i a l c o m p o u n d} \times 100$
$P e r c e n t a g e l o s s = \frac{90}{250} \times 100$
$P e r c e n t a g e l o s s = 36$

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

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Similar questions

[C u S O_{4} .5 H_{2} O]

C u S O_{4} .5 H_{2} O \to C u S O_{4} + 5 H_{2} O

C u = 64, S = 32, O = 16, H = 1

2

(C u S O_{4} .5 H_{2} O)

C u = 63.5 g/mol

[C u S O_{4} .5 H_{2} O]

H_{2} S O_{4}

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