Question

Calculate the percentage composition of Copper sulphate CuSO4.5H2O

Answer 1

Explanation:

The molecular weight of water H2O=2(1)+16=18.

When 1 mole of CuSO4⋅5H2O is heated, 5 moles of water are lost.

The molecular weight of CuSO4⋅5H2O=64+32+4(16)+5(18)=250 g/mol

Mass of water lost =5(18)=90 g

Percentage loss of mass of CuSO4⋅5H2O=25090×100=36 %.

Answer 2

Answer:

The percentage composition of Copper sulphate  $CuSO_4.5H_2O$ is Cu is 25.45%, S is 12.82%, O is 57.77%, H is 4.01%.

Explanation:

Molecular formula of copper sulphate, $CuSO_4.5H_2O$.

Molar mass of $CuSO_4.5H_2O$ is,

$63.5+32+4\times16+5\times18=249.5$

Molar Mass of Cu is 63.5

Molar Mass percentage of Cu is,

$\frac{63.5}{249.5} \times 100=25.45$

Molar Mass of S is 32,

Molar Mass percentage of S is,

$\frac{32}{249.5} \times 100$=12.82%.

Molar Mass of O is $9\times 16=144$

Molar Mass percentage of O is $\frac{144}{249.5} \times 100$ =57.77%

Molar Mass of H is $10\times1=10$

Molar Mass percentage of H is

$\frac{10}{249.5} \times 100$ = 4.01%.

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