Calculate the percentage compostion of various elements in the following compounds:
Blue vitriol `(CuSO_(4). 5H_(2) O)`
b. Green vitriol `(Fe SO_(4) . 7H_(2) O)`
c. White vitriol `(ZnSO_(4). 5H_(2) O)`
d. Ethanol `(C_(2) H_(5) OH)`
e. Mohr's salt `[(NH_(4))_(2) SO_(4). FeSO_(4). 6H_(2) O]`

Explanation

a. Molar mass of `CuSO_(4). 5H_(2)O = 63.5 + 32 4 xx 16 + 5 xx 18`
`249.5g`
Mass % of `Cu = (63.5 xx 100)/(249.5) = 25.45%`
Mass % of `s = (32 xx 100)/(249.5) = 12.82%`
Mass % of `O = (16 xx 9 xx 100)/(249.5) = 57.71%`
Mass % of `H = (10 xx 1.008 xx 100)/(249.5) = 4.040%`
b. Molar mass of `FeSO_(4). 7H_(2)O = 56 + 32 + 64 + 7 xx 18`
`= 278 g`
Mass % of `Fe = (56 xx 100)/(278) = 20.1%`
Mass % of `S = (32 xxx 100)/(278) = 11.5 %`
Mass % of `O = (11 xx 16 xx 100)/(278) = 63.3%`
Mass % of `H = (14 xx 1.008 xx 100)/(278) = 5.0%`
c. Molar mass of `ZnSO_(4). 5H_(2)O = 65.37 + 32 + 64 + 5 xx 18`
`= 251.37 g`
Mass % of `Zn = (65 xx 37 xx 100)/(251.37) = 26.0%`
Mass % of `S = (32 xx 100)/(251.37) = 12.7%`
Mass % of `O = (16 xx 9 xx 100)/(251.37) = 57.2%`
Mass % of `H = (10 xx 1.008 xx 100)/(251.37) = 4.01%`
d. Molar mass of `C_(2) H_(5) OH = 2 xx 12.01 + 6 xx 1.008 + 16.00 g`
`= 46.068 g`
Mass % of `C = (24.02 g)/(46.068 g) xx 100 = 52.14^`
Mass % of `H = (6.048)/(46.068g) xx 100 = 13.13%`
Mass % of `O = (16.00 g)/(46.068 g) = 100 = 34.73 %`
e. Molar mass of `(NH_(4))_(2) SO_(4). FeSO_(4) 6 H_(2) O`
`2 (14 + 4) + 32 + 16 xx 4 + 56 + 32 + 64 + 108 = 392 g`
Mass % of `N = (28 g)/(392 g) xx 100 = 7.14 %`
Mass % of `H = (20 xx 1.0 g)/(392 g) xx 100 = 5.10%`
Mass % of `S = (2 xx 32 g)/(392 g) xx 16.32 %`
Mass of `O = (14 xx 16.00 g)/(392 g) xx 100 = 57.14 %`
Mass % of `Fe = (56 g)/(382 g) xx 100 = 14.28 %`