If a sample contains 74.0% of the R enantiomer and 26.0% of the S enantiomer, what is the enantiomeric excess of the mixture?

The enantiomeric excess of the mixture is 48%

Explanation:

Enantiomeric excess refers to the excess of one enantiomer over the other in a mixture of enantiomers.

Enantiomeric excess = % of major enantiomer (R) - % of minor enantiomer (S)

= 74% - 26% = 48%.

it takes on an electronical energy

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