Answer to Question #197015 in General Chemistry for lea

Question #197015

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



1
Expert's answer
2021-05-26T07:13:18-0400

% hydration = "\\frac{mass of water molecules}{molar mass}\u00d7100"


KAl(SO4)2.12H2

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

Mass of water = 216g 

% hydration = = "\\frac{216g}{474g}\u00d7100" =46%


CuSO4.5H2

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

Mass of water = 90g 

% hydration = "\\frac{90g}{249.5g}\u00d7100" = 36%


MgSO4.7H2

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

Mass of water = 126g 

% hydration = "\\frac{126g}{230g}\u00d7100" = 54%


 FeSO4.7H2O

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

Mass of water = 126g 

% hydration = "\\frac{126g}{262g}\u00d7100" = 48%


CuSO4.2H2

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

Mass of water = 36g 

% hydration = "\\frac{36g}{179.5g}\u00d7100" = 20%


NaCO3.H2

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

Mass of water = 18g 

% hydration = "\\frac{18g}{101g}\u00d7100" = 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}\u00d7100" = 48%


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