### Topic: Theoretical/Percent Yield with Hydration?  (Read 11561 times)

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#### phoenixcelestius ##### Theoretical/Percent Yield with Hydration?
« on: April 06, 2011, 10:43:48 PM »
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

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• Gender:  ##### Re: Theoretical/Percent Yield with Hydration?
« Reply #1 on: April 07, 2011, 01:13:45 AM »
Didn't check the numbers, but the process seems okay to me.

#### Borek ##### Re: Theoretical/Percent Yield with Hydration?
« Reply #2 on: April 07, 2011, 04:21:04 AM »
[Cu(NH3)5Cl]SO4•H2O = 245.79 g/mol

#### phoenixcelestius 