If a sample contains 84.0 % of the R enantiomer and 16.0 % of the S enantiomer, what is the enantiomeric excess of the mixture

enantiomeric excess = 68%

Explanation:

Enantiomeric excess is a value used to determine the purity of chiral molecules. It is possible to determine enantiomeric excess (ee) using:

ee = R - S / R + S * 100

Where R is the mass (In this case percentage) of the R enantiomer and S of the S enantiomer.

Replacing with values of the problem:

ee = 84% - 16% / 84% + 16% * 100

i think   the answer is c hope i'm right