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Calculate optical purity and enantiomeric excess

Key idea: For a matched two-enantiomer mixture, rotation relative to a pure reference gives enantiomeric excess. An excess of e% means (100 + e)/2% major enantiomer, not e% major enantiomer.

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  • GCE A-Level H3 Chemistry 2027
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A chiral drug is often made as a mixture of its two enantiomers, and only one of them may work. Measuring how far a sample rotates plane-polarised light shows how much one enantiomer is in excess. This lesson shows how to calculate optical purity and the composition of a mixture of enantiomers.

Check your understanding 1

A sample contains 70% of one enantiomer and 30% of the other. What is its enantiomeric excess?

Unit: %
Show answer

40%. The 30% of the minor enantiomer cancels the rotation of 30% of the major one, leaving a 40% excess that rotates the light.

Optical rotation measures a sample

R/S labels describe molecular configuration. Optical rotation describes the behaviour of a sample in plane-polarised light. The two enantiomers of one compound give rotations of equal magnitude and opposite sign under the same conditions.

A racemic 50:50 mixture has zero net rotation. An unequal mixture can rotate the light because the contributions do not cancel completely. A positive sign does not itself mean R.

Compare like with like

Specific rotation, written [α], accounts for the sample concentration and light path length. Comparing specific rotations still requires the same temperature, light wavelength and solvent, and a known pure-enantiomer reference.

For the examples below, the solute contains only the stated R and S enantiomers, and rotation contributions are assumed to add in proportion to their fractions. Other optically active solutes, incorrect concentrations or unmatched reference conditions would invalidate this simple calculation.

Calculate optical purity

For a mixture of one enantiomeric pair under these assumptions:

optical purity (%) = |[α]ₘᵢₓₜᵤᵣₑ/[α]ₚᵤᵣₑ| × 100%.

The magnitude is the enantiomeric excess (ee): the percentage of the major enantiomer minus the percentage of the minor one. Use the sign relative to the identified pure reference to decide which enantiomer is in excess.

If the excess is e%, the composition is:

major enantiomer = (100 + e)/2%, minor enantiomer = (100-e)/2%.

The percentages add to 100 and differ by e. Optical purity is not simply the percentage of the major enantiomer.

Find ee when the signs match

Worked example 1

Reference and mixture with the same sign

Problem

Illustrative specific rotations, in ° cm³ g⁻¹ dm⁻¹, are +30.0 for pure R and +12.0 for a mixture of R and S under identical conditions. Find the enantiomeric excess and the percentage of each enantiomer.

Show full solution

  1. Pure S

    Method

    Reverse the sign of pure R.

    Reason

    Enantiomers rotate light by equal amounts in opposite directions.

    Working

    Answer

    Pure S: −30.0

  2. Enantiomeric excess

    Method

    Divide the mixture’s magnitude by the pure magnitude.

    Reason

    Optical rotation is proportional to the excess of one enantiomer.

    Working

    Answer

    12.0/30.0 = 0.400, so 40.0% ee

  3. Convert excess to composition

    Method

    Use the sign to identify the major enantiomer, then solve for percentages that sum to 100 and differ by 40.0.

    Reason

    The excess is the difference between the two percentages, not the major enantiomer’s percentage.

    Working

    Answer

    The mixture has the same sign as pure R, so R is in excess:

    %R = (100 + 40.0)/2 = 70.0%, %S = 30.0%.

  4. Report and check the mixture

    Working

    Answer

    40.0% ee in favour of R: 70.0% R and 30.0% S. The percentages sum to 100.0% and differ by 40.0 percentage points.

Check your understanding 2

A mixture contains 80.0% R and 20.0% S. The pure-enantiomer specific rotations are +30.0 for R and −30.0 for S, in ° cm³ g⁻¹ dm⁻¹ under matched conditions. Calculate the mixture’s specific rotation and its enantiomeric excess.

Show the steps Optional working
Unit: ° cm³ g⁻¹ dm⁻¹
Unit: ° cm³ g⁻¹ dm⁻¹
Unit: ° cm³ g⁻¹ dm⁻¹
Unit: %
Show answer

Weight by the fractions: 0.800(+30.0) + 0.200(-30.0) = 24.0-6.0 = +18.0 ° cm³ g⁻¹ dm⁻¹. The 60.0 percentage-point R excess is distinct from the 80.0% R content.

Interpret the opposite sign

Check your understanding 3

Pure R has specific rotation +24.0, while a mixture of R and S gives −18.0 in the same units under identical conditions. Find its ee and composition.

Unit: %
Enantiomer in excess
Show the steps Optional working
Unit: %
Unit: %
Show answer

The magnitude ratio gives 18.0/24.0 × 100% = 75.0% ee. The mixture is negative while pure R is positive, so S is in excess. %S = (100 + 75.0)/2 = 87.5% and %R = 12.5%.

A negative pure reference

Check your understanding 4

Pure S has specific rotation −40.0. A mixture of R and S gives −10.0 in the same units under matched conditions. Find its composition and explain why “negative means R” fails here.

Unit: %
Show the steps Optional working
Unit: %
Unit: %
Show answer

The ee is 10.0/40.0 × 100% = 25.0%. The negative sign matches pure S, so the mixture is 62.5% S and 37.5% R. Rotation sign identifies an enantiomer only through a stated reference; it has no universal relationship with R/S.

Check the assumptions before trusting a result

A ratio above 100% cannot describe the stated two-enantiomer mixture in this simple model. Check units, concentrations, conditions, reference identity and other solutes rather than accepting an impossible composition.

Comparing raw observed angles from cells of different lengths or solutions of different concentrations is also invalid unless those differences are accounted for. Zero rotation can also come from an achiral compound or from a sample containing other solutes; check what the sample contains before calling it racemic.

Check your understanding 5

Pure R has specific rotation +30.0. A student’s mixture of R and S gives +36.0 under what they say are the same conditions. What should they conclude?

Conclusion
Show answer

The ratio 36.0/30.0 = 1.2 would mean more than 100% excess, which no mixture of the two enantiomers can give. Check the concentration and path length, the temperature, wavelength and solvent, the identity of the reference, and whether another optically active substance is present.

Common mistakes

  • Taking the optical purity as the percentage of the major enantiomer. 40% ee means 70% of one enantiomer and 30% of the other.
  • Assuming a positive rotation means R. The sign identifies an enantiomer only by comparison with a stated pure reference.
  • Comparing rotations measured at different temperatures, wavelengths or in different solvents.
  • Accepting a ratio above 100%, which cannot describe a mixture of the two enantiomers alone.
  • Taking zero rotation as proof of a racemic mixture. The compound may be achiral.

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