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I’ve been given two formulas to use for enantiomeric excess or optical purity: $$\frac{\text{observed rotation}}{\text{specific rotation of pure enantiomer}}\:\:\:\:\:\times\:100\%\tag{1}$$ $$\frac{|d-l|}{d+l}\:\times\:100\%\tag{2}$$ However, I cannot figure out how they are equal to each other. You could rewrite equation (1) as $$\frac{\alpha_d-\alpha_l}{[\alpha]}\:\times\:100\%$$ and because $\alpha=[\alpha]cL$, $$(c_dL-c_lL)\:\times\:100\%$$ However, that’s as far as I got. I can’t see how you would eliminate L, and even doing so would not bring it to resemble the second equation. I would appreciate it if someone could offer a mathematical proof or—if this isn’t even a matter of mathematics at all but rather a conceptual misunderstanding—clarify things for me.

EDIT

$d$ is the mass of the dextrorotary enantiomer and $l$ is the mass of the levorotary enantiomer. $c$ refers to concentration, $L$ the pathlength travelled by light, [$\alpha$] specific rotatio, and $\alpha$ (without the brackets) means observed rotation.

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  • $\begingroup$ The better formulation of equation (1) is $$\dfrac{[\alpha]_{\mathrm{sample}}}{[\alpha]_\mathrm{pure\ enantiomer}}$$ Units need to cancel, and with this version they do. $\endgroup$ – Ben Norris Jul 3 '16 at 12:55
  • $\begingroup$ @BenNorris Equation (1) as I’ve written it also has units that cancel. Both the numerator and denominator are measured in degrees. $\endgroup$ – lightweaver Jul 4 '16 at 2:50
  • $\begingroup$ Observed rotation $\alpha$ has units of $^\circ$, but specific rotation $[\alpha]$ must have units of $^\circ \mathrm{mL\cdot dm \cdot g^{-1}}$ in order for the equation $\alpha =[\alpha]cL$ to be valid. For convenience, we shorten specific rotation to degrees. $\endgroup$ – Ben Norris Jul 4 '16 at 16:12
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The equations are trivially equal since each is equal to the enantiomeric excess. To prove this, we need only show each equal to the enantiomeric excess.


Start with the second equation. Because both enantiomers have the same molar mass, we can divide both numerator and denominator by the molar mass of the compound and rewrite the equation as $$\text{EE} = \frac{|\text{difference in moles of enantiomers}|}{\text{total moles of enantiomers}} = |\text{difference in mole fractions of enantiomers}|,$$ which is the definition of enantiomeric excess. This equation is not very useful, since most of the time the masses of each of the enantiomers in solution are not known.

The first equation, as you have it, is incorrect. Following Ben Norris's suggestion, we write $$\text{EE} = \frac{|[\alpha]_\text{obs}|}{[\alpha]_\text{pure}} = \frac{|[\alpha]_\text{obs}|}{[\alpha]_\text{pure}} = \frac{|\chi[\alpha]_\text{pure}-(1-\chi)[\alpha]_\text{pure}|}{[\alpha]_\text{pure}} = |\chi-(1-\chi)|,$$ where $\chi$ represents the mole fraction of one enantiomer, and we once again obtain the same result.


Had your original equation been right, $L$ would have cancelled in your derivation and you would have obtained $(c_d-c_l)/(c_d+c_l)$, which is readily brought to equation $(2)$.

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