Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: phoenixcelestius on April 06, 2011, 10:43:48 PM
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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 :rarrow: [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. :)
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Didn't check the numbers, but the process seems okay to me.
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[Cu(NH3)5Cl]SO4•H2O = 245.79 g/mol
Check your math.
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sorry, I meant [Cu(NH3)4]SO4•H2O = 245.79 g/mol
I mistyped the formula on that post...