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What is the ee for the following mixture of enantiomers A and B? 80% A and...

What is the ee for the following mixture of enantiomers A and B?

80% A and 20% B

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Answer #1

enantiomeric excess (ee) = [ (A- B) / (A + B) ] x 100

                                         = (80-20 / 80 + 20) x 100

                                        = 60 %

enantiomeric excess (ee) = 60 %

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