Calculate Weight Lost by CuSO4 5H2O at 120°C for 12hrs

[mutzy188]

Homework Statement

How much weight would 5g of CuSO4 5H2O lose if ti were dried at 120 degrees C for 12 hours (i.e. if it were heated to drive off all the water of hydration)?

The Attempt at a Solution

CuSO4 = 159.61 g/mol
H2O = 18 g/mol = 90 g for 5 mols

159.61/(159.61 + 90) = 6.25%

5 g * 6.25% = amount of CuSO4 = .313 g

5 - .313 = how much H2O lost = 4.687 g

but this isn't the correct answer.

Any help would be greatly appreciated

Thanks

 

[bomba923]

Find how many moles of CuSO₄*5H₂O are in 5g of the hydrate and then multiply by the molar mass of the anhydrate.

 

[chemisttree]

CuSO4 = 159.61 g/mol
H2O = 18 g/mol = 90 g for 5 mols

159.61/(159.61 + 90) = 6.25%

I get 0.6394 for this expression. What does this value represent?
Everything from here on is wrong because both the arithmetic and the logic is faulty.

5 g * 6.25% = amount of CuSO4 = .313 g

5 - .313 = how much H2O lost = 4.687 g

but this isn't the correct answer.

Find the formula weight of the pentahydrate. From that information, calculate the number of moles of the pentahydrate in 5 grams of the pentahydrate. This will equal the number of moles of the anhydride. You can do it from there...