Question

5. Fill in the following table to show amounts to use and theoretical yield. Product Epoxide product Moles MW Density Amount Amounts Reactant Reactant MCPBA Product from last experiment 22 mmol 22 mmol 172.57 g/mol 82.14 g/mol NA 0.83 g/ml Calculate and Calculate and write write in pre-lab in pre-lab notes notes 22 mmol 98.14 g/mol NA This amount will be your theoretical yield.

Answer 1

1. Reactant: MCPBA Molecular weight (MW) = 172.57 g/mol. (This means Imol of MCPBA weights 172.579) we need 22 mmol. = 22X10-3 mol. Weight of 1 molo = 172.57g. weight of 22x10-3 mol = 172.57X22X103 = 3,796 g (ans) So we need 3.796 g of MCPBA. 2. Reactant! from Product from last experiment, similarly, Amount = moles X MW = 22 mmol x 82.14 g/mol 8 = 22x10-3 mol X 82.14 g/mol = 1807.08X10-3g = 1.807 Now Mass density = volume 0.83 g/mL = or 1.807 g rolumea

a 1.807 g Volume = 0.83 gone = 2.18 mb.(ans) 3. Product : Amount = moles y Molecular weight = 22 mmol x 98.14 g/mol = 22x10-3 mol X 98.14 8/mol = 2.159 g (ans)