Question

What is the percentage of enantiomeric excess of a mixture containing 12.8 mol (R)-2-bromobutane and 3.2 mol (S)-2-bromobutane?

Answer 1

$\text{Enantiomeric excess = moles (R) – moles(S)}$
= 12.8 – 3.2 = 9.6 mol

$\text{Enantiomeric excess}=\frac{\text{Excess of one enantiomer over other}}{\text{Entire mixture}}$

The percent enantiomeric excess can be calculated by dividing the excess (9.6 mol) of the R enantiomer by the total number of moles for both enantiomers.

$\text{Enantiomeric excess}=\frac{9.6}{12.8+3.2}=60\%$