H2O + CuSO4 · 5H2O + 4NH3+ -> [Cu(NH3)4]SO4 · H2O + 5H2O 1) Determine the mass o

Question

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

1) Determine the mass of CuSO4 · 5H2O you need to prepare 0.25g of [Cu(NH3)4]SO4 · H2O. Assume a 90% yield.

2) Calculate the volume of 6 M ammonia necessary to convert all of your copper(II) sulfate to the tetraamminecopper(II) sulfate.

3) If 2.75 g of [Cu(NH3)4]SO4 · H2O is obtained from 3.25g of CuSO4 · 5H2O, what is the percent yield?

4) Which of the three ligands investigated do you think will be the strongest field ligand? Rank the three ligands from strongest to weakest and explain your answer based on the molecular-level evidence.

Explanation / Answer

-----Calculations………..

M.W. of [Cu(NH3)4]SO4 · H2O = 245.73 g/mol

M.W of CuSO4 · 5H2O = 249.677 g/mol

0.25 g of [Cu(NH3)4]SO4 · H2O = 0.25 / 245.73 = 0.00101 moles

The reaction is 1:1, but the percentage yield is 90 %, amount of CuSO4 · 5H2O required = 0.00101 /90*100 = 0.00112 moles

0.00112 moles of CuSO4 · 5H2O = 0.00112 x 249.677 = 0.279 g.

Volume of 6 M ammonia required to convert 0.00112 moles of CuSO4 · 5H2O to [Cu(NH3)4]SO4 · H2O

For each Cu2+ ions it need 4 NH3 molecules to form the product.

i.e. moles of NH3 = 0.00112 x 4 = 0.00448 moles

Find volume, 0.00448 mole / 6M = 0.000746 L = 0.746 mL

If 2.75 g of [Cu(NH3)4]SO4 · H2O is obtained from 3.25g of CuSO4 · 5H2O, what is the percent yield

2.75 g of [Cu(NH3)4]SO4 · H2O = 0.01111 moles

3.25g of CuSO4 · 5H2O = 0.013 moles

% yield = 0.0111/0.013 x 100 = 85 %

Ligand filed strength (Magnitude of Crystal Field Splitting), according to Spectrochemical series is NH3> H2O > SO42-

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