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14 November, 07:53

If the enantiomeric excess of a mixture is 75%, what are the % compositions of the major and minor enantiomer?

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  1. 14 November, 10:35
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    Let us say that R is the major enantiomer, while S is the minor enantiomer, therefore the formula for enantiomeric excess (ee) is:

    ee = (R - S) * 100%

    Let us further say that the fraction of R is x (R = x), and therefore fraction of S is 1 - x (S = 1 - x), therefore:

    75 = (x - (1 - x)) * 100

    75 = 100 x - 100 + 100 x

    200 x = 175

    x = 0.875

    Summary of answers:

    R = major enantiomer = 0.875 or 87.5%

    S = minor enantiomer = (1 - 0.875) = 0.125 or 12.5%
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