The Beer–Lambert Law with Worked Examples

Hold up a glass of weak blackcurrant squash and a glass of strong squash to the light. The strong one looks darker because it absorbs more light. The Beer–Lambert law turns that everyday observation into a precise equation, and it’s the basis of colorimetry and UV–visible spectroscopy, techniques used to measure concentrations in hospitals, water labs, breweries and research labs every day.

The law

A = ε × c × l

where:

• A is the absorbance (no units)
• ε (epsilon) is the molar absorptivity (or molar extinction coefficient), in dm³ mol⁻¹ cm⁻¹. It’s a property of the substance at a particular wavelength.
• c is the concentration, in mol/dm³
• l is the path length of light through the sample, in cm (usually 1.00 cm for a standard cuvette)

In words: absorbance is directly proportional to concentration (and to path length). Double the concentration and you double the absorbance.

The law combines work by Johann Lambert (1760, on path length) and August Beer (1852, on concentration), building on earlier observations by Pierre Bouguer.

Absorbance and transmittance

A spectrophotometer shines light of intensity I₀ into the sample and measures the intensity I that comes out.

• Transmittance, T = I ÷ I₀ (often given as a percentage)
• Absorbance, A = log₁₀(I₀ ÷ I) = −log₁₀ T

% transmittance	Absorbance
100%	0
50%	0.30
10%	1.0
1%	2.0
0.1%	3.0

Each increase of 1 in absorbance means ten times less light gets through. Absorbance is used rather than transmittance because it’s the quantity that’s proportional to concentration.

Worked example 1: finding concentration

A solution in a 1.00 cm cuvette has an absorbance of 0.600 at 630 nm. The dye’s molar absorptivity at that wavelength is 1.20 × 10⁴ dm³ mol⁻¹ cm⁻¹. Find its concentration.

c = A ÷ (ε × l) = 0.600 ÷ (1.20 × 10⁴ × 1.00) = 5.00 × 10⁻⁵ mol/dm³

Worked example 2: finding absorbance

Potassium permanganate has ε = 2.4 × 10³ dm³ mol⁻¹ cm⁻¹ at 525 nm. What absorbance does a 2.0 × 10⁻⁴ mol/dm³ solution give in a 1.00 cm cell?

A = 2.4 × 10³ × 2.0 × 10⁻⁴ × 1.00 = 0.48

Worked example 3: finding ε

A 3.5 × 10⁻⁵ mol/dm³ solution of a compound gives A = 0.420 in a 1.00 cm cell. Find ε.

ε = A ÷ (c × l) = 0.420 ÷ (3.5 × 10⁻⁵ × 1.00) = 1.2 × 10⁴ dm³ mol⁻¹ cm⁻¹

Worked example 4: changing the path length

The solution in example 1 is placed in a 0.500 cm cuvette. What’s the new absorbance?

Absorbance is proportional to path length, so halving the path halves the absorbance: A = 0.300.

This is a practical trick: if a solution is too concentrated to measure, use a shorter cuvette (or dilute it).

Worked example 5: from transmittance

A sample transmits 25.0% of the incident light. What is its absorbance?

A = −log₁₀(0.250) = 0.602

Worked example 6: dilution and a real sample

A 5.00 cm³ sample of pond water is treated with a reagent that forms a coloured complex with phosphate, then made up to 50.0 cm³. The absorbance is 0.345. A calibration shows A = 0.0230 × [phosphate in μmol/dm³]. Find the phosphate concentration in the pond water.

• In the diluted solution: 0.345 ÷ 0.0230 = 15.0 μmol/dm³
• The sample was diluted tenfold (5.00 → 50.0 cm³)
• Pond water: 150 μmol/dm³

This combines Beer–Lambert with a calibration curve, which is how the law is usually applied in practice.

Why ε matters

Molar absorptivity tells you how strongly a substance absorbs light at a given wavelength.

• Large ε (10⁴–10⁵ dm³ mol⁻¹ cm⁻¹): strongly absorbing, such as many dyes and highly conjugated molecules. Tiny concentrations can be measured.
• Small ε (1–100 dm³ mol⁻¹ cm⁻¹): weakly absorbing, such as hydrated transition metal ions like [Cu(H₂O)₆]²⁺. Higher concentrations are needed, or a reagent is added to form a more intensely coloured complex.

That’s why analytical methods often add a colour-forming reagent: for example, ammonia turns pale blue copper(II) solutions into a much deeper blue complex, and thiocyanate turns iron(III) into an intense blood-red complex, making them easier to measure at low concentrations.

Choosing the wavelength

Measurements are made at the wavelength of maximum absorbance (λmax) for the substance. At λmax:

• sensitivity is highest (largest absorbance for a given concentration)
• small errors in the wavelength setting cause the smallest changes in absorbance

In a simple colorimeter, you choose a filter of the complementary colour to the solution. A blue solution absorbs orange-red light, so a red or orange filter is used. See colorimetry.

When the law breaks down

The Beer–Lambert law works best for dilute solutions, typically at absorbances between about 0.1 and 1.0. It can fail when:

• Concentration is high. Molecules interact with each other, changing how they absorb, and the refractive index of the solution changes.
• Absorbance is very high. So little light reaches the detector (A = 2 means only 1% transmitted) that stray light and detector noise dominate, and the graph curves.
• The chemical form changes with concentration. For example, an indicator whose ionisation depends on dilution, or molecules that pair up (dimerise) at higher concentrations.
• Light isn’t monochromatic. Wide filters let through a range of wavelengths where ε varies.
• The sample scatters light. Cloudy suspensions scatter light, which looks like absorbance but isn’t.
• The sample fluoresces, re-emitting some absorbed light.

That’s why calibration curves are always checked for linearity, and samples that read too high are diluted.

Where the Beer–Lambert law is used

• Medicine: measuring haemoglobin, bilirubin and glucose (after colour-forming reactions) in blood; pulse oximeters use a related principle to estimate oxygen saturation.
• Biochemistry: measuring DNA and protein concentrations by UV absorbance at 260 nm and 280 nm.
• Environmental testing: nitrate, phosphate and metal ions in water.
• Food and drink: colour and bitterness in beer, sugar and dye content.
• Kinetics: following how the concentration of a coloured reactant or product changes over time during a reaction.

Practice questions

1. A solution has A = 0.750 in a 1.00 cm cell and ε = 1.50 × 10⁴ dm³ mol⁻¹ cm⁻¹. Find c.
2. What percentage of light is transmitted by a solution with A = 0.500?

Answers: (1) c = 0.750 ÷ 1.50 × 10⁴ = 5.00 × 10⁻⁵ mol/dm³. (2) T = 10⁻⁰·⁵⁰⁰ = 0.316, so 31.6%.

Common mistakes

• Using transmittance in A = εcl. Convert to absorbance first.
• Wrong units. If ε is in dm³ mol⁻¹ cm⁻¹, path length must be in cm and concentration in mol/dm³.
• Forgetting the blank. The instrument must be zeroed with a cuvette of solvent so that only the analyte’s absorbance is measured.
• Using the law outside its linear range.
• Handling cuvettes on the clear faces. Fingerprints absorb and scatter light.

Key takeaways

• A = εcl: absorbance is proportional to concentration and path length.
• A = −log₁₀ T, so an absorbance of 1 means 10% of the light gets through.
• ε measures how strongly a substance absorbs at a given wavelength; measure at λmax.
• The law works best for dilute solutions with absorbances below about 1.
• In practice, it’s applied through calibration curves made with standard solutions.