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Eduardwww [97]
2 years ago
15

A sample of the chiral molecule limonene is 79% enantiopure. what percentage of each enantiomer is present? what is the percent

enantiomeric excess?
Chemistry
2 answers:
Degger [83]2 years ago
7 0

Answer :  The % of (+) limonene isomer = 79%


                The % of (-) limonene isomer = 0%


                The % of enantiomeric excess = 58%


Explanation :   Enantiomeric excess (ee) is the measurement of purity used for chiral substances.


Given,


% of pure limonene enantiomer = The % of (+) limonene isomer = 79%


Therefore, The % of (-) limonene isomer = 0%


Formula used :  

\%(+)\text{ isomer}=\frac{ee}{2}+50\%


Where,         ee → enantiomeric excess


Now, put all the values in above formula, we get the value of enantiomeric excess (ee).


     {ee}=\frac{\%(+)-50\%}{2}


            =\frac{79\%-50\%}{2}


              = 58%



DaniilM [7]2 years ago
7 0

Answer:

The major enatiomer makes up 89.5% of the mixture

The minor enatiomer makes up 10.5% of the mixture

The percent enantiometric excess is 79%

Explanation:

% (+) = ee/2

79 % / 2 + 50%

= 39.5% + 50%

= 89.5%

and

% (-) = 100 % - 89.5 %

= 10.5%

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