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Spectrophotometry: Beer–Lambert law

Convert absorbance, transmittance and percent transmittance; solve any variable of A = εbc (absorbance, molar absorptivity, path length or concentration) and work the typical exam problem: a sample with two, three or more species measured at several wavelengths, where the concentrations are found from the molar absorptivities of each species.

What you can do

The law

A = −log₁₀ T = ε b c. In a mixture of species that do not react with each other the absorbances add up: A(λ) = b Σ εᵢ(λ) cᵢ, one linear equation per wavelength.

Guide: the Beer–Lambert law and spectrophotometry

When light passes through a colored solution, part of it is absorbed, and the more substance there is, the more light is absorbed. The Beer–Lambert law turns that observation into a linear relationship that lets you measure concentrations with a spectrophotometer. This guide explains absorbance and transmittance, the law, its limits and how a mixture of two species is analyzed.

Transmittance and absorbance

If I0 is the intensity of the light entering the sample and I the intensity leaving it, the transmittance is T = I/I0 (or %T = 100·T) and the absorbance is

A = −log10 T

For example, %T = 50 corresponds to A ≈ 0.301, and A = 1 means only 10 % of the light gets through. Absorbance has no units.

The law

A = ε · b · c

b is the path length (the width of the cuvette, usually 1 cm), c the concentration and ε the molar absorptivity (in L·mol⁻¹·cm⁻¹), specific to each substance at each wavelength. Since A is proportional to c, measuring A gives c, provided ε is known or a calibration curve has been prepared with standards.

Example 1: one species

A sample gives A = 0.45 in a 1 cm cuvette, at a wavelength where the species has ε = 2000 L·mol⁻¹·cm⁻¹ (given in the exercise).

  1. c = A / (ε·b) = 0.45 / (2000 · 1) = 2.25 × 10⁻⁴ mol/L.
  2. The transmittance is T = 10−0.45 ≈ 0.355, that is, %T ≈ 35.5 %: the sample absorbs 64.5 % of the light.

Example 2: a mixture of two species

The absorbances of species that do not react with each other add up. With two species X and Y, measurements at two wavelengths are needed, giving a system of two equations. With b = 1 cm and these absorptivities (given in the exercise):

1000·cX + 200·cY = 0.50 100·cX + 800·cY = 0.42

Solving the first for cX and substituting in the second: 780·cY = 0.37, so cY ≈ 4.74 × 10⁻⁴ M and cX ≈ 4.05 × 10⁻⁴ M. The calculator solves this system for any number of species; if there are more wavelengths than species, it uses least squares. Both examples give the same results in it.

Limits of the law

That is why real analysis prefers a calibration curve with standards of known concentration, measured on the same instrument, to using an ε from the literature.

How to use the calculator

  1. Convert between A, T and %T.
  2. In A = εbc, type three of the values and leave blank the one you want.
  3. For mixtures, follow the steps: the instrument (path length and units), the species (the ones you are looking for and the ones you know), the measurements (one row per wavelength, with each species' ε or the reading of a standard) and the result.

Common mistakes

Further reading

D. C. Harris, Quantitative Chemical Analysis (W. H. Freeman); D. A. Skoog, F. J. Holler and S. R. Crouch, Principles of Instrumental Analysis (Cengage). Simuveliq is not affiliated with these authors or publishers.

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