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MariettaO [177]

3 years 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?

2 answers:

ikadub [295]3 years ago

6 0

The mixture contains 62 % one isomer and 38 % the enantiomer.

Let’s say that the mixture contains 62 % of the (<em>R</em>)-isomer.

Then % (<em>S</em>) = 100 % -62 % = 38 %

ee = % (<em>R</em>) - % (<em>S</em>) = 62 % -38 % = 24 %

Yakvenalex [24]3 years ago

3 0

<u>Answer:</u> The percentage of R-enantiomer is 62 %, the percentage of S-enantiomer is 38 % and the percent enantiomer excess is 24 %

<u>Explanation:</u>

We are given:

A chiral molecule limonene which is 62 % enantiopure.

This molecule has two enantiomers, which are R-limonene and S-limonene.

Let us consider that 62% of the given molecule is R-limonene.

So, the remaining S-limonene will be = (100 - 62) = 38 %

Percent enantiomeric excess is defined as the difference between the percentage major enantiomer and the percentage minor enantiomer.

Mathematically,

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

% major enantiomer = 62 %

% minor enantiomer = 38 %

Putting values in above equation, we get:

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

Hence, the percentage of R-enantiomer is 62 %, the percentage of S-enantiomer is 38 % and the percent enantiomer excess is 24 %

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