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Author: Subject: heating CuSO4.xH2O
ionic bond
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[*] posted on 22-7-2004 at 06:07
heating CuSO4.xH2O


Hi

when we heat CuSO4.xH2O

CuSO4.xH2O --------> CuSO4+ xH2O

If 36 g of CuSO4.xH2O gives 15.88 g of H2O.

Calculate x

I tried to calcualte it like this:

36/(195.5+18x) = 15.88/18x

x = 7

is that right?

thanks
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vulture
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[*] posted on 22-7-2004 at 06:26


Don't think that is right...

Try it this way:

36g - 15.88g = 20.12g CuSO4 --> calculate how many moles CuSO4 this is.

15.88g/18g = 0.88 mole H2O

divide 0.88 mole H2O through # moles of CuSO4 et voila...




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ionic bond
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[*] posted on 22-7-2004 at 09:59


Hi

but we don't know if all of the CuSO4.xH2O will react!!

[Edited on 22-7-2004 by ionic bond]
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[*] posted on 22-7-2004 at 10:08


Oh, if you heat it up enough, surely all the CuSO4x xH2O will decompose.
I would also do it vulture's way- you want the ratio of the moles H2O versus the # of moles CuSO4 .




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[*] posted on 22-7-2004 at 17:26


That all the water leaves the salt is implied in your statement of the problem:


Quote:

CuSO4.xH2O --------> CuSO4+ xH2O




The single most important condition for a successful synthesis is good mixing - Nicodem
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vulture
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[*] posted on 23-7-2004 at 06:39


If you don't heat the reaction untill it's completed, it's impossible to calculate it.

Furthermore, completion of the reaction can be easily observed visually.




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[*] posted on 23-7-2004 at 16:43


Don't you see that the results are the same!
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[*] posted on 24-7-2004 at 01:30


Indeed, you are right, 7 is the result.

I didn't calculate it through because this is in the beginnings section.

I'm not aware of the existance of a heptahydrate, but it could be possible.

Another possibility is that your pentahydrate was wet, eg not properly dried/crystallized.

It would be a fluke though to get a nice seven if it was residual water.




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