March 28, 2024, 09:33:11 PM
Forum Rules: Read This Before Posting


Topic: Theoretical/Percent Yield with Hydration?  (Read 15456 times)

0 Members and 1 Guest are viewing this topic.

Offline phoenixcelestius

  • New Member
  • **
  • Posts: 7
  • Mole Snacks: +0/-0
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  :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.  :)

Offline opti384

  • Full Member
  • ****
  • Posts: 434
  • Mole Snacks: +33/-25
  • Gender: Male
    • In the Search for the Laws of Nature
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.

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27633
  • Mole Snacks: +1799/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
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

Check your math.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline phoenixcelestius

  • New Member
  • **
  • Posts: 7
  • Mole Snacks: +0/-0
Re: Theoretical/Percent Yield with Hydration?
« Reply #3 on: April 07, 2011, 07:34:53 AM »
sorry, I meant [Cu(NH3)4]SO4•H2O = 245.79 g/mol
I mistyped the formula on that post...

Sponsored Links