for the synthesis to be performed in this experiment, based on 1.00 g of copper (II) sulfate pentahydrate (CuSO4) taken initially, what is the theoretical yield of product tetraaminecopper (II) sulfate hydrate? [Cu(NH3)4]SO4 - H2O
the amounts of reactants i'm using in this experiment are:
1 g of CuSO4
10 mL of distilled water
5 mL of concentrated NH3 (what does concentrated mean? 1.0 M i'm assuming?)
and approximately 10 mL of ethyl alcohol
what i have set up as the equation so far:
CuSO4 + H2O + 4NH3 --> [Cu(NH3)4]SO4 - H2O
does anyone have any idea on how i can calculate theoretical yield of the product? any help/suggestion would be greatly appreciated!
2 replies to this topic
#2
Primarygun
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Posted 20 July 2005 - 05:02 AM
The lack of practice in laboratory reactions frustrates me from helping you.
Therefore, I could only suggest some ideas that I am certain of them.
To have tetraaminecopper(II) ions, you have to have excess ammonia, so that the precipitate copper (II) hydroxide produced from the precipitation reaction can dissolve by forming complex ion(tetraaminecopper(II) ions).
If I am right, with my correct calculations, I don't think you are having excess ammonia.
Assuming all the precipitates dissolved and a deep blue solution comes out,
you may get
Cu^2+(aq) + 4NH3 (aq)---> [Cu(NH3)4]^2+ (aq)
This time, ammonia is a limiting agent, so finding the total amount of NH3 has reacted, so that you can find the amount of tetraaminecopper(II) ions formed.
Then, one tetraaminecopper(II) ions will bind with one sulphate ion and one water molecule, then you can get it.
But, in this case, I'm afraid the experiment cannot succeed as there are still some excess copper(II) ions, you need to precipitate them first.
Or maybe my calculation is wrong!
Therefore, I could only suggest some ideas that I am certain of them.
To have tetraaminecopper(II) ions, you have to have excess ammonia, so that the precipitate copper (II) hydroxide produced from the precipitation reaction can dissolve by forming complex ion(tetraaminecopper(II) ions).
If I am right, with my correct calculations, I don't think you are having excess ammonia.
Assuming all the precipitates dissolved and a deep blue solution comes out,
you may get
Cu^2+(aq) + 4NH3 (aq)---> [Cu(NH3)4]^2+ (aq)
This time, ammonia is a limiting agent, so finding the total amount of NH3 has reacted, so that you can find the amount of tetraaminecopper(II) ions formed.
Then, one tetraaminecopper(II) ions will bind with one sulphate ion and one water molecule, then you can get it.
But, in this case, I'm afraid the experiment cannot succeed as there are still some excess copper(II) ions, you need to precipitate them first.
Or maybe my calculation is wrong!
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#3
BeYeu05
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Quark
Posted 20 July 2005 - 02:53 PM
Cu^2+(aq) + 4NH3 (aq)---> [Cu(NH3)4]^2+ (aq)
This time, ammonia is a limiting agent, so finding the total amount of NH3 has reacted, so that you can find the amount of tetraaminecopper(II) ions formed.
Then, one tetraaminecopper(II) ions will bind with one sulphate ion and one water molecule, then you can get it.
Can you explain that a little more?
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