How much of the R enantiomer is present in 10 g of a mixture that has an enantiomeric excess of 30% of the S isomer?
A)3)0 g
B)3)5 g
C)6)5 g
D)7)0 g

Name: How much of the R enantiomer is present in 10 g of a mixture that has an enantiomeric excess of 30% of the S isomer? A)3)0 g B)3)5 g C)6)5 g D)7)0 g
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Question

Quizplus · Accepted Answer

Enantiomeric excess (ee) = 30% of the S isomerTherefore, % of R isomer = 100% - 30% = 70%Mass of R isomer in 10 g of mixture = (70/100) x 10 g = 7 g

How Much of the R Enantiomer Is Present in 10