If a mixture contains 75% of one compound and 25% of its enantiomer, what is the e. e. of the mixture

The enantiomeric excess ee is 50%

Explanation:

Given that;

compound A = 75%, B = 25%

ee = ?

so

ee = [( A - B) / ( A + B)] × 100

= [( 75 - 25) / ( 75 + 25)] × 100

= [50 / 100] × 100

= 0.5 × 100

= 50%

Therefore enantiomeric excess ee is 50%