Question

Data:

8.0 grams of CuSO₄ • 5H₂O in a 125 mL Erlenmeyer flask: 8.130 grams

1) Calculate the theoretical yield of Cu(NH₃)₄SO₄ x H₂O (Molecular Weight= 245.7 g/mol) from the mass of the starting material, CuSO₄ x 5H₂O (Molecular Weight= 249.7 g/mol)

2) Calculate the percent yield of the product.

( experimental yield/theoretical yield ) x 100

3) Calculate the theoretical percents of copper, ammonia, sulfate and water from the formula, Cu(NH₃)₄SO₄ x H₂O

Answer 1

8.0 grams of CuSO4 • 5H2O in a 125 mL Erlenmeyer flask: 8.130 grams
For 1 mole of CuSO4 • 5H2O, moles of Cu(NH3)4SO4 x H2O forms is= 1 mole
(Considering no information on limiting reagents is given, it may vary depending on the moles of NH3 used in the reaction.)

1) Theoretical yield of Cu(NH3)4SO4 x H2O (Molecular Weight= 245.7 g/mol) from the mass of the starting material, CuSO4 x 5H2O (Molecular Weight= 249.7 g/mol)
moles of CuSO4.5H2O= 8 g/ 249.6850 g/mol = 0.032
So, theoretical moles of Cu(NH3)4SO4 x H2O to be formed = 0.032
Theoretical yield= 0.032 mol*245.7 g/mol= 7.86 g

2) Percent yield of the product( experimental yield/theoretical yield ) x 100
moles of Cu(NH3)4SO4 x H2O formed = 8.13 g/ 245.7 g/mol = 0.033
Grams of Cu(NH3)4SO4 x H2O formed = 0.033*245.7 = 8.13 g
So, theoretical yield = (experimental yield/theoretical yield)*100= (8.13/7.86)*100 = 103.43 %

3) Theoretical percents of copper, ammonia, sulfate and water from the formula, Cu(NH3)4SO4 x H2O
Theoretical % of Copper= (63.54/245.7) *103.43= 26.75%
theoretical % of Ammonia= (4*17/245.7)*103.43 = 28.62 %
theoretical % of sulfate= (80/245.7)*103.43 = 33.67 %
theoretical % of water= (18/245.7)*103.43 = 7.57%