Use the Beer–Lambert law in concentration calculations

A spectrophotometer compares the incident light intensity I₀ with the transmitted intensity I. The absorbance is A = lg(I₀/I), so greater light absorption gives a larger A.

GCE A-Level H3 Chemistry 9813-2027

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Learning objectives

Use the Beer–Lambert law in concentration calculations

Turn light loss into concentration

A spectrophotometer compares the incident light intensity I₀ with the transmitted intensity I. The absorbance is A = lg(I₀/I), so greater light absorption gives a larger A.

For a suitable solution measured at a fixed wavelength, the Beer–Lambert law connects absorbance with concentration and path length: A = εcl.

The Beer–Lambert quantities

A is absorbance and has no unit. ε is the molar absorption coefficient, usually in dm³ mol⁻¹ cm⁻¹ when c is mol dm⁻³ and l is cm.

c is the concentration of the absorbing species and l is the optical path length through the sample. With matched cuvettes, l is commonly 1.00 cm.

At fixed wavelength, temperature, solvent and chemical form, ε is constant for the absorbing species. Changing the wavelength can change ε substantially.

Use the law with its conditions

The relationship predicts A proportional to c when ε and l are fixed, and A proportional to l when ε and c are fixed. A graph of A against c should therefore be linear through the blank-corrected origin over the valid range.

Very concentrated solutions, stray light, chemical equilibria or instrumental limits can cause deviation from linearity. If absorbance is too high, dilute the sample and account for the dilution afterward.

Always check unit compatibility before substitution. If l is in cm and ε uses cm⁻¹, do not convert l to metres.

Worked example: find an unknown concentration

A solution has A = 0.576 at a wavelength where ε = 1.44 × 10⁴ dm³ mol⁻¹ cm⁻¹, using a 1.00 cm cuvette.

Rearrange before substituting: c = A/(εl) = 0.576/[(1.44 × 10⁴)(1.00)] = 4.00 × 10⁻⁵ mol dm⁻³.

Try this

Calculate c for A = 0.576, ε = 1.44 × 10⁴ dm³ mol⁻¹ cm⁻¹ and l = 1.00 cm.

Check your answer

c = A/(εl) = 4.00 × 10⁻⁵ mol dm⁻³.

Practise path-length reasoning

If concentration and ε remain fixed, doubling the path length doubles absorbance. This is a proportional comparison, so no logarithm calculation is needed.

The transmittance does not double in the same way because absorbance is logarithmic in I₀/I.

Try this

A solution gives A = 0.150 in a 1.00 cm cuvette. Predict A in a 2.00 cm cuvette under the same conditions.

Check your answer

A doubles to 0.300 because A = εcl and only l has doubled.

Link absorbance to transmitted intensity

For A = 1.00, lg(I₀/I) = 1.00, so I₀/I = 10 and I/I₀ = 0.10. Only 10% of the incident intensity is transmitted.

This logarithmic definition explains why absorbance is convenient for a linear concentration law even though transmittance itself is not linear in concentration.

Try this

A sample has A = 2.00. What fraction of the incident intensity is transmitted?

Check your answer

lg(I₀/I) = 2, so I₀/I = 100 and I/I₀ = 0.010: 1.0% is transmitted.

Common mistake: rearranging by eye

From A = εcl, concentration is A divided by εl. Multiplying A by ε gives an impossible large concentration and usually inconsistent units.

Do not attach units to absorbance. Its logarithmic intensity ratio is dimensionless.

Try this

Correct the rearrangement c = Aεl.

Check your answer

Divide both sides by εl: c = A/(εl).

Check your understanding

Show the equation, rearrangement, substitution, units and a quick sense check. A larger ε or l should require a smaller c to produce the same A.

Next, plan a complete quantitative analysis using a blank, standards or ε, a suitable wavelength and any necessary dilution.

Try this

State the two Beer–Lambert equations and the conditions that must remain fixed for A to be proportional to c.

Check your answer

A = lg(I₀/I) and A = εcl. Keep wavelength, path length, solvent, temperature and chemical form fixed within the linear range.