Question

Calculate the theoretical percentage of water for the following hydrates. a

(a) manganese(II) monohydrate, MnSO4 H2O

(b) manganese(II) tetrahydrate, MnSO4 4H2O  

Answer 1

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

a)

Molar mass of MnSO4.H2O,

MM = 1*MM(Mn) + 1*MM(S) + 5*MM(O) + 2*MM(H)

= 1*54.94 + 1*32.07 + 5*16.0 + 2*1.008

= 169.026 g/mol

It has only 1 H2O

So,

mass % of H2O = 1*Molar mass of H2O * 100 / Molar mass of MnSO4.H2O

= 1* 18.016 * 100 / 169.026

= 10.66 %

Answer: 10.66 %

B)

Molar mass of MnSO4.4H2O,

MM = 1*MM(Mn) + 1*MM(S) + 8*MM(O) + 8*MM(H)

= 1*54.94 + 1*32.07 + 8*16.0 + 8*1.008

= 223.074 g/mol

It has only 4 H2O

So,

mass % of H2O = 4*Molar mass of H2O * 100 / Molar mass of MnSO4.4H2O

= 4* 18.016 * 100 / 223.074

= 32.30 %

Answer: 32.30 %