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Title: How much R enantiomer is present in 10g of a mixture which has an enantiomeric excess of 60% of the R isomer? Post by: mib123 on Jun 1, 2013 Please show steps.
Title: How much R enantiomer is present in 10g of a mixture which has an enantiomeric excess of 60% of the R isomer? Post by: Juliana1984 on Jun 1, 2013 enantiomeric excess = (amount of R - amount of S) / (amount of R + amount of S).
We have 2 equations and 2 variables: r + s = 10 (grams) (r - s)/(r+s) = 0.6 so r - s = 0.6 * (r+s) = 0.6 * 10 = 6 (r+s) + (r-s) = 2r = 10 + 6 ---> r = 8 grams R enantiomer s = 10 - 8 = 2 grams S enantiomer |