Beer–Lambert Law Calculator
Find absorbance with the Beer–Lambert law.
Result
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 %.
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. Enter any two of ε, b, and c to solve for absorbance and transmittance.
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