Question

A sample of copper sulphate pentahydrate contains 8.64 g of oxygen. How many gram of Cu is present in this sample ? (Atomic mass of Cu = 63.6, S = 32.06 , O= 16)​

Answer 1

Answer :-

3.816 grams of copper is present in the given sample.

Explanation :-

We have :-

→ Given mass of Oxygen = 8.64 g

→ Molar mass of Copper = 63.6 g/mol

→ Molar mass of Sulphur = 32.06 g/mol

→ Molar mass of Oxygen = 16 g/mol

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In the compound CuSO₄.5H₂O, (5 + 4) = 9 moles of Oxygen combine with 1 mole of Copper .

Mass of 9 moles of Oxygen :-

= No of moles × Molar mass

= 9 × 16

= 144 grams

And, mass of 1 mole of Copper is 63.6 grams.

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Now, we can calculate the required mass of Copper (Cu) by using unitary method .

∵ 144 g of 'O' combines with 63.6 g of 'Cu' .

∴ 8.64 g of 'O' will combine with :-

= (8.64 × 63.6/144) g of Cu

= (549.504/144) g of Cu

= 3.816 g of Cu

Answer 2

Given :-

A sample of copper sulphate pentahydrate contains 8.64 g of oxygen.

To Find :-

How many gram of Cu is present in this sample

Solution :-

Formula of copper sulphate pentahydrate = CuSO₄.5H₂O

Molecular mass of oxygen in it

(4 × 16) + (5 × 16)

64 + 80

144 g

Now

144 g of oxygen is present in 1 mole of CuSO₄

In 8.64 g

8.64/144

864/144 × 1/100

0.06 moles

Now,

Mole of Cu in 1 mole = 63.6 g

Mole of Cu in 0.06 mole = 0.06 × 63.6 = 3.816 g