If a sample contains 76.0% of the R enantiomer and 24.0% of the S enantiomer, what...

Question

Question

If a sample contains 76.0% of the R enantiomer and 24.0% of the S enantiomer, what is the enantiomeric excess of the mixture?

Answer 1

Answer #1

enantiomeric excess = [% R - %S / % R + %S] x 100

= [76-24 / 76+24] x 100

= 52 %

Therefore, enantiomeric excess of the mixture = 52 %

Answer 2

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