[BeYeu05]

hi everyone,

ok i am doing a lab outline and unfortunately LOST on how to do this:

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

With the answer I have, assuming that CuSO4 is the limiting reactant, the number of moles of [Cu(NH3)4]SO4 - H2O will be the same as that of CuSO4. 1.00 g of CuSo4 is 0.00626 moles. 0.00626 moles of CuSO4 = 0.00626 moles of [Cu(NH3)4]SO4 - H2O. The mass of 0.00626 moles of [Cu(NH3)4]SO4 - H2O is 1.22 grams.

Does it seem unreasonable to not take into account the # of moles of NH3 or H2O?
any help/suggestion would be greatly appreciated!

[AWK]

You use copper sulfate pentahydrate - it is mean you use 1/250 moles of salt and also 1/250 mole of pure CuSO4 - this should produce 1/250 moles of  complex, and a final result, assuming 100 % yield should be less than 1 g.

[Borek]

5 mL of concentrated NH3 (what does concentrated mean? 1.0 M i'm assuming?)

Stock ammonia is 30% w/w.

[BeYeu05]

You use copper sulfate pentahydrate - it is mean you use 1/250 moles of salt and also 1/250 mole of pure CuSO4 - this should produce 1/250 moles of  complex, and a final result, assuming 100 % yield should be less than 1 g.

How do you know it's only 1/250 moles of salt and 1/250 mole of pure CuSO4? Is there a way to calculate that fraction?

[Borek]

Check molar mass of copper sulfate pentahydrate.