2B - % Yield calculations

    1)         3.924g of 2-chloropropane reacted with an excess of aqueous sodium hydroxide.

                2.955g of propan-2-ol was formed.  calculate the % yield of propan-2-ol.

                        CH₃CH₂(Cl)CH₃    +    NaOH    à    CH₃CH₂(OH)CH₃    +    NaCl   

                                 3.925g                                           2.995g

                    moles = 3.925                                moles = 2.995

                                  78.5                                                61

                    moles = 0.0500                              moles = 0.0490

                    moles = 0.0500  -------------------->   moles = 0.0494

                                             % yield = 0.0490   x  100

                                                            0.0500

                                             % yield = 98%

    2)         Ethanol reacts with ethanoic acid to produce an ester and water.

                5.5g of the ester is produced from 4.0g of ethanol and 4.5g of ethanoic acid.

                calculate the % yield of the ester.

                        CH₃CH₂OH    +    CH₃COOH    à    CH₃COOCH₂CH₃    +    H₂O   

                           4.0g                            4.5g                            5.50g

            moles = 4.0                   moles = 4.5                  moles = 5.50

                         46                                  60                                 88

            moles = 0.087               moles = 0.075               moles = 0.0625

                             (in excess)                            (limiting reagent)

                                                 moles = 0.075  ------>   moles = 0.075

                                             % yield = 0.0625   x  100

                                                             0.075

                                             % yield = 83.3%