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1 year ago

If a sample contains 70.0 % of the R enantiomer and 30.0 % of the S enantiomer, what is the enantiomeric excess of the mixture?

Chemistry

1 answer:

NeTakaya1 year ago

3 0

Answer:

the enantiomeric excess of the mixture is 40%

Explanation:

The computation of the enantiomeric excess of the mixture is shown below:

As we know that

$= |\frac{R - S}{R + S} |\times 100\\\\= |\frac{70 - 30}{70 + 30}| \times 100\\\\= 40\%$

Hence, the enantiomeric excess of the mixture is 40%

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