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Beer–Lambert Law Calculator

Find absorbance with the Beer–Lambert law.

How it is calculated

The Beer-Lambert law gives absorbance as A = ε · b · c, where ε is the molar absorptivity in L/(mol·cm), b is the path length in cm (1 cm for a standard cuvette) and c is the concentration in mol/L. Transmittance follows from T = 10−A × 100, in percent. All inputs must be zero or positive. Absorbance has no unit and is shown with four decimals. The law holds for dilute solutions and a single absorbing species; at high concentration the relationship stops being linear. Absorbance of 1 means 10% of the light gets through, 2 means 1%.

Worked example

A dye has ε = 1,500 L/(mol·cm), is measured in a 1 cm cuvette, and has concentration 0.0004 mol/L. Absorbance: 1,500 × 1 × 0.0004 = 0.6000. Transmittance: 10−0.6 = 0.2512, which is 25.12 % of the light passing through. The absorbance is 0.6000 and the transmittance 25.12 %.

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Common mistakes

  • Entering concentration in mmol/L or µmol/L instead of mol/L. A value off by 1,000 gives an absorbance off by 1,000 too.
  • Entering the path length in mm or metres. It must be in cm, and a typical cuvette is 1.
  • Treating absorbance and transmittance as the same scale. Absorbance is logarithmic, so doubling it does not halve the transmittance; it squares the fraction.
  • Trusting a high result. Real instruments lose linearity at large absorbance, so dilute the sample if the value is far above about 1 to 2.

About this calculator

This calculator applies the Beer–Lambert law to find absorbance: A = ε·b·c, where ε is the molar absorptivity (extinction coefficient, L·mol⁻¹·cm⁻¹), b is the path length through the sample (cm), and c is the molar concentration. It also converts the result to percent transmittance: %T = 10^(2−A).

Analytical chemists, biochemists, and lab technicians use this relationship constantly with UV-Vis spectrophotometers to determine an unknown sample's concentration from its measured absorbance, or to predict the absorbance expected from a known concentration and cuvette path length. Any two of ε, b, and c are enough to solve for absorbance and transmittance.

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Comments (1)

  • Editorial team

    Treat the result as trustworthy only in a moderate range. On typical spectrophotometers, readings above roughly 1 to 2 absorbance units become unreliable, because very little light reaches the detector. If your calculated A is far higher, dilute the sample and multiply back, rather than trusting the transmittance figure. The extinction coefficient also depends on wavelength and solvent, so use the value measured for your own conditions. Which compound and wavelength are you working with?

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