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evablogger
1 month ago
11

A sample of the chiral molecule limonene is 62% enantiopure. what percentage of each enantiomer is present? what is the percent

enantiomeric excess?
Chemistry
2 answers:
alisha [2.7K]1 month ago
6 0

The composition consists of 62 % one isomer and 38 % its enantiomer.

Assuming that the mixture comprises 62 % of the (R)-isomer.

Then the percentage of the (S) is calculated as 100 % - 62 % = 38 %.

Enantiomeric excess = %  (R) - % (S) = 62 % - 38 % = 24 %.


lions [2.6K]1 month ago
3 0

Answer: The proportion of the R-enantiomer is 62 %, the S-enantiomer is at 38 %, and the enantiomeric excess is 24 %

Explanation:

We know that:

A chiral molecule limonene is recorded as 62 % enantiopure.

This compound has two enantiomeric forms, R-limonene, and S-limonene.

Let's state that 62 % of the sample is R-limonene.

Consequently, the remaining S-limonene will amount to = (100 - 62) = 38 %

Enantiomeric excess is determined by the difference between the percentage of the major and the minor enantiomer.

In mathematical terms,

\%\text{ Enantiomer excess}=\%\text{ Major enantiomer}-\%\text{ Minor enantiomer}

% major enantiomer = 62 %

% minor enantiomer = 38 %

Substituting the values into the equation, we have:

\%\text{ Enantiomer excess}=62\%-38\%=24\%

Thus, the R-enantiomer is 62 %, the S-enantiomer is 38 %, and the enantiomeric excess is 24 %.

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