Question

If a sample contains 78.0 % of the R enantiomer and 22.0 % of the S enantiomer, what is the enantiomeric excess of the mixture?

Answer 1

Answer: The enantiomeric excess of the mixture =56%

Explanation:

As per given : R isomer = 78%

S isomer = 22%

In chiral substances, Enantiomeric excess is the measure of purity.

So, Enantiomeric excess of the mixture = $|\frac{R-S}{R+S}|\times100$

$=|\frac{78-22}{78+22}|\times100$

$=|\frac{56}{100}|\times100\% \\\\=56\%$

Hence,  the enantiomeric excess of the mixture =56%