What is the percentage of enantiomeric excess of a mixture containing 12.8 mol (R)-2-bromobutane and 3.2 mol (S)-2-bromobutane?

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Solution

$Enantiomeric excess = moles (R) - moles(S)$
= 12.8 – 3.2 = 9.6 mol

$Enantiomeric excess = \frac{Excess of one enantiomer over other}{Entire mixture}$

The percent enantiomeric excess can be calculated by dividing the excess (9.6 mol) of the R enantiomer by the total number of moles for both enantiomers.

$Enantiomeric excess = \frac{9.6}{12.8 + 3.2} = 60 %$

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