Answer to Question #197015 in General Chemistry for lea

Question

Calculate the %hydration for each common crystalline hydrate and complete the table.

Formula %hydration

KAl(SO4)2·12H2O

CuSO4·5H2O

MgSO4·7H2O

FeSO4·7H2O

CaSO4·2H2O

NaCO3·H2O

Please show calculation to FeSO4·7H2O

Answer 1

% hydration = $\frac{mass of water molecules}{molar mass}×100$

KAl(SO₄)₂.12H₂O 

Molar mass = ( 39 + 27 + 32×2 + 16×8 + 12×18) = 474g 

Mass of water = 216g 

% hydration = = $\frac{216g}{474g}×100$ =46%

CuSO₄.5H₂O 

Molar mass = ( 63.5 + 32 + 16×4 + 5×18)=249.5g

Mass of water = 90g 

% hydration = $\frac{90g}{249.5g}×100$ = 36%

MgSO₄.7H₂O 

Molar mass = ( 24 + 32 + 16×4 + 7×18) = 230g 

Mass of water = 126g 

% hydration = $\frac{126g}{230g}×100$ = 54%

 FeSO₄.7H₂O

Molar mass = ( 56 + 32 + 16×4 + 7×18) = 262g 

Mass of water = 126g 

% hydration = $\frac{126g}{262g}×100$ = 48%

CuSO₄.2H₂O 

Molar mass = ( 63.5 + 32 + 16×4 + 2×18) = 179.5g 

Mass of water = 36g 

% hydration = $\frac{36g}{179.5g}×100$ = 20%

NaCO₃.H₂O 

Molar mass = ( 23 + 12 + 16×3 + 1×18) = 101g 

Mass of water = 18g 

% hydration = $\frac{18g}{101g}×100$ = 17%

Hence the calculations for

FeSO₄.7H₂O is as follows

Molar mass = ( 56 + 32 + 16×4 + 7×18) = 262g 

Mass of water = 126g 

% hydration = $\frac{126g}{262g}×100$ = 48%