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

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mutzy188
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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
 
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Discussion

Find how many moles of CuSO4*5H2O are in 5g of the hydrate and then multiply by the molar mass of the anhydrate.
 
mutzy188 said:
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.
mutzy188 said:
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...