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

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%