Question

calculate the theoretical yield for this experiment of the synthesis of Copper (II) Aspirinate:

1 g potassium bicarbonate dissolved in 20 mL H2O.

0.9 g of asprin is dissolved in the bicarbonate solution.

0.65 g of CuSO4•5H2O is dissolved in 10 mL H2O and added to asprin solution.

caluclate the theoretical yield of copper (II) aspirinate from this experiment.

Answer 1

Molar mass of Aspirin = 180.158 g/mol

0.9 g of asprin = 0.9g /180.158 g/mol = 0.0049956 Moles

1 g potassium bicarbonate = 1/  100.115 g/mol = 0.009988 Moles

0.65 g of CuSO4•5H2O = 0.65g / 159.609 g/mol = 0.00407 Moles

4 Molecules of aspirin can form complex with one copper atom. Hence

theoretical yield of copper (II) aspirinate in Moles =  0.0049956 Moles /4 = 0.0012489 Moles

Molar mass of copper (II) aspirinate =  843.69g/mol

theoretical yield of copper (II) aspirinate in gm = 0.0012489 Moles x 843.69g/mol = 1.053 gm