[phoenixcelestius]

Hi, I just need someone to verify the answer. I'm doing a pre-lab question that says:
If 4.52 g of [Cu(NH₃)₅Cl]SO₄•H₂O is prepared from 4.87 g of CuSO₄•5H₂O and an excess of ammonia, what is the percent yield?

The chemical equation is:
CuSO₄•5H₂O + 4NH₃  [Cu(NH₃)₅Cl]SO₄•H₂O + 4H₂O

I started by finding the molar masses of the compounds plus the hydration shells:
CuSO₄•5H₂O = 249.71 g/mol
[Cu(NH₃)₅Cl]SO₄•H₂O = 245.79 g/mol

Then I found the moles of CuSO₄•5H₂O:
4.87 g * (1 mol / 249.71 g) = 0.0195 mol CuSO₄•5H₂O

Since there are 0.0195 mol CuSO₄•5H₂O, the theoretical yield of the [Cu(NH₃)₅Cl]SO₄•H₂O is also 0.0195 mol based on the chemical equation. I converted 0.0195 mol [Cu(NH₃)₅Cl]SO₄•H₂O into grams:
0.0195 mol * (245.79 g /1 mol) = 4.79 g [Cu(NH₃)₅Cl]SO₄•H₂O

Since the given actual yield is 4.52 g, the percent yield is:
4.52 g / 4.79 g = 0.944 = 94.4%

Can someone tell me if this procedure and answer is correct? Thanks. 

[opti384]

Didn't check the numbers, but the process seems okay to me.

[Borek]

[Cu(NH₃)₅Cl]SO₄•H₂O = 245.79 g/mol

Check your math.

[phoenixcelestius]

sorry, I meant [Cu(NH3)₄]SO4•H2O = 245.79 g/mol
I mistyped the formula on that post...