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%

Step-by-step 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 =
`|(R-S)/(R+S)|*100`

`=|(78-22)/(78+22)|*100`

`=|(56)/(100)|*100\% \\\\=56\%`

Hence, the enantiomeric excess of the mixture =56%