Question

1. Calculate the theoretical yields of the compounds to be prepared in this experiment. The metal ion in both cases is the limiting reagent. Hint: Find the number of moles of Cu(II) in the sample of CuSO4 5H20 that you used. That will equal the number of moles of [Cu(NH3)4] SO4 H20 that could theoretically be prepare. Proceed in a similar way for the synthesis involving Co(II).

a.

b.

2.How could you establish that the metal ion is the limiting reagent?

3.How would you formally name the compound [Cu(NH3)4] SO4 H20 prepared in part A?

4. Give the formulas of the following compounds:

a. Dichlorotetrmminechromium (III) nitrate.

b. Dichlorodiaquoplatinum (II)

Answer 1

1. The chemical equation being,

                        CuSO4.5H2O + 4NH3 ------> [Cu(NH3)4]SO4.H2O

molar mass      249.68 g/mol                               245.74 g/mol  

So, the mass of complex formed from 249.68 g of CuSO4.5H2O would be 245.74 g

This is the theoretical yield.

Now for x g it would be (x/molar mass) CuSO4.5H2O = moles CuSO4.5H2O

This would yield = moles of CuSO4.5H2O x molar mass of complex = g theoretical yield.

c. The metal ion is the limiting reagent, we can confirm it by adding excess moles of NH3, we still would get the same amount of complex formed in the reaction.

d. Name of [Cu(NH3)4]SO4.H2O = tetramminecopper(II)sulfate

4. Formulas of the compounds

a. Dichlorotetramminechromium(III)nitrate = [Cr(NH3)4Cl2]NO3

b. Dichloroaquoplatinum(II) = [Pt(Cl)2(H2O)]