Calculate the enantiomeric excess of a mixtur | vedclass

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(B) The enantiomeric excess $(e.e.)$ is defined as the ratio of the excess of one enantiomer to the total mixture,expressed as a percentage.$e.e. = \f

Vedclass · Accepted Answer

$20$

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(B) The enantiomeric excess $(e.e.)$ is defined as the ratio of the excess of one enantiomer to the total mixture,expressed as a percentage.$e.e. = \frac{|	ext{mass of } (+)-2	ext{-butanol} - 	ext{mass of } (-)-2	ext{-butanol}|}{	ext{total mass of mixture}} 	imes 100$Given:Mass of $(+)-2$-butanol $= 6 \ g$Mass of $(-)-2$-butanol $= 4 \ g$Total mass $= 6 \ g + 4 \ g = 10 \ g$$e.e. = \frac{6 \ g - 4 \ g}{10 \ g} 	imes 100 = \frac{2 \ g}{10 \ g} 	imes 100 = 20 \ \%$

Calculate the enantiomeric excess of a mixture containing $6 \ g$ of $(+)-2$-butanol and $4 \ g$ of $(-)-2$-butanol.

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