Question

Verdigris, copper (II) acetate monohydrate, was prepared using the three reactions shown

below. In a 125 mL Erlenmeyer flask were combined 4.20 g copper (II) sulfate pentahydrate, 35 mL of a 3.0 M aqueous solution of ammonia, and 25 mL of water (density 0.997 g/mL). The light blue solid CuSO4·3Cu(OH)2 that formed was collected by vacuum filtration and combined with 40. mL of water and 8.5 mL of a 2.0 M aqueous solution of sodium hydroxide in a second 125 mL Erlenmeyer flask to give the blue solid copper (II) hydroxide. The blue solid was collected by vacuum filtration, transferred to a third 125 mL Erlenmeyer flask, and combined with 40. mL of a 0.82 M aqueous acetic acid solution. The resultant solution was evaporated to one-half of its original volume; the dark green crystals of verdigris that formed were isolated by vacuum filtration. Collected was 1.25 g of solid verdigris. Calculate the percent yield of verdigris.

CuSO4·5H2O(aq) + 6NH3(aq) → CuSO4·3Cu(OH)2(s) +3(NH4)2SO4(aq) +14H2O(l)   (1)

CuSO4·3Cu(OH)2(s) + 2NaOH(aq) → 4Cu(OH)2(s) + Na2SO4(aq)           (2)

Cu(OH)2(s) + 2HC2H3O2(aq) → Cu(C2H3O2)2·H2O(aq) + H2O(l)               (3)

Answer 1

Ans. Reaction 1: Moles of CuSO₄.5H₂O taken = Mass/ Molar mass

                                                = 4.20 g / (249.686 g / mol)

                                                = 0.01682113 mol

Moles of NH₃ taken = Molarity x Volume of solution in liters

                                    = 3.0 M x 0.035 L

                                    = 0.105 mol

Experimental molar ration of reactants = Moles of CuSO₄.5H₂O : Moles of NH₃

                                    = 0.01682113 mol : 0.105 mol

                                    = 1 : 6.24

In balanced reaction, 1 mol CuSO₄.5H₂O reacts with 6 mol NH₃.

Since experimental ratio of NH₃ is greater than 6 when that of CuSO₄.5H₂O is kept constant at 1 mol, NH3 is the reagent in excess.

So, CuSO₄.5H₂O is the limiting reactant.

The formation of product follows stoichiometry of limiting reactant.

Thus, Moles of product CuSO₄.3Cu(OH)₂ formed = moles of CuSO₄.5H₂O taken

So, Moles of product CuSO₄.3Cu(OH)₂ formed = 0.01682113 mol

# Reaction 2:

We have (from #1), Moles of product CuSO₄.3Cu(OH)₂ formed = 0.01682113 mol

Moles of NaOH taken = 2.0 M x 0.0085 L = 0.017 mol

Experimental molar ration of reactants = Moles of CuSO₄.3Cu(OH)₂: Moles of NaOH

                                    = 0.01682113 mol : 0.017 mol

                                    = 1 : 1.01

Theoretical molar ration of reactants = Moles of CuSO₄.3Cu(OH)₂: Moles of NaOH

                                    = 1 : 2 (in balanced reaction)

Comparing the experimental and theoretical molar ration, the moles of NaOH is less than 2 when that of CuSO₄.3Cu(OH)₂ is kept constant at 1 mol.

Thus, NaOH is the limiting reactant.

The formation of product follows the stoichiometry of limiting reactant.

So,

Moles of product Cu(OH)₂ formed = (2 / 4) x moles of NaOH

                                                = (2 / 4) x x 0.017 mol

                                                = 0.0085 mol                      

# Reaction 3:

We have (from #2), moles of Cu(OH)₂ formed = 0.0085 mol

Moles of acetic acid taken = 0.82 M x 0.040 mol = 0.0328 mol

Experimental molar ration of reactants = Moles of Cu(OH)₂: Moles of CH3COOH

                                    = 0.0085 mol : 0.0328 mol

                                    = 1 : 3.86

Theoretical molar ration of reactants = Moles of Cu(OH)₂: Moles of CH3COOH = 1 : 2

# Comparing the experimental and theoretical molar ration, the moles of CH3COOH is greater than 2 when that of Cu(OH)₂ is kept constant at 1 mol.

Thus, CH3COOH is the reagent in excess and Cu(OH)₂ is the limiting reactant.

The formation of product follows the stoichiometry of limiting reactant.

So,

Moles of product Cu(C₂H₃O₂)₂.2H₂O formed = Moles of Cu(OH)₂ = 0.0085 mol

Therefore,

Theoretical moles of Cu(C₂H₃O₂)₂.2H₂O formed = 0.0085 mol

Theoretical mass of Cu(C₂H₃O₂)₂.2H₂O formed = Theoretical moles x Molar mass

                                                            = 0.0085 mol x (217.6658 g/ mol)

                                                            = 1.850 g

Now,

% yield = (Mass of product obtained / Theoretical mass of product produced) x 100

                        = (1.25 g / 1.85 g) x 100

                        = 67.568 %