A 3.0 g sample of a compound that is 80% (R) and 20% (S) is mixed with 2.0 g of its racemate. The enantiomeric excess of the combined 5.0 g sample is

A60%
B48%
C24%
D36%

Answer: D

(R) = 3.0 x 0.80 + 1.0 = 3.4 g; (S) = 3.0 x 0.20 + 1.0 = 1.6 g. ee = (3.4 - 1.6)/5.0 = 36%. The racemate adds equal amounts of both, diluting the excess.

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