How do I use the Beer-Lambert law?
Type any three of the absorbance, the molar absorptivity, the path length and the concentration. The Beer-Lambert law calculator works out the fourth from A = ε × l × c, and gives the transmittance that goes with the absorbance.
- Concentration (mol/L)
- 0.00000268116
A = 0.37, ε = 138,000 L/(mol·cm), l = 1 cm and c = 0.00000268116 mol/L.
- Transmittance (%)
- 42.658
Concentration (mol/L): 0.00000268116. A = 0.37, ε = 138,000 L/(mol·cm), l = 1 cm and c = 0.00000268116 mol/L.
How to calculate
Solves the Beer–Lambert law, A = ε × l × c, for the absorbance, molar absorptivity, path length or concentration from the other three, with the transmittance.
Example with the default inputs (Absorbance (A) 0.37, Molar absorptivity ε (L/(mol·cm)) 138,000, Path length (cm) 1): A = 0.37, ε = 138,000 L/(mol·cm), l = 1 cm and c = 0.00000268116 mol/L.
Formula: A = ε × l × c; ε = A ÷ (l × c); l = A ÷ (ε × c); c = A ÷ (ε × l); transmittance T = 100 × 10^−A %.
- The light is monochromatic, the solution is clear and dilute, and the substance does not react or scatter light, so A is proportional to c.
- The path length is in centimeters and the concentration in moles per liter, so ε is in L/(mol·cm).
- The absorbance is the decadic one, A = log₁₀(I₀ ÷ I), so T = I ÷ I₀ = 10^−A.
Worked examples
Each example is checked against the calculator on every build.
- Absorbance (A) 0.37, Molar absorptivity ε (L/(mol·cm)) 138,000, Path length (cm) 1 gives Concentration (mol/L) 0.000003, Transmittance (%) 42.657952.Source: OpenStax, Organic Chemistry, §14.7 Ultraviolet Spectroscopy (A = log(I₀ ÷ I); ε = A ÷ (c × l), c in mol/L and l in cm; β-carotene, ε = 138,000, absorbance 0.37 in a 1.0 cm cell: c = 2.7 × 10⁻⁶ mol/L), https://openstax.org/books/organic-chemistry/pages/14-7-ultraviolet-spectroscopy (retrieved 2026-10-18)
- Absorbance (A) 0.5, Path length (cm) 1, Concentration (mol/L) 0.0001 gives Molar absorptivity ε (L/(mol·cm)) 5,000.Source: IUPAC Compendium of Chemical Terminology (Gold Book), Beer–Lambert law: A = log₁₀(P₀ ÷ P) = εcl, ε the molar (decadic) absorption coefficient, https://doi.org/10.1351/goldbook.B00626
- Molar absorptivity ε (L/(mol·cm)) 138,000, Path length (cm) 1, Concentration (mol/L) 0.000004 gives Absorbance (A) 0.552, Transmittance (%) 28.054336.Source: OpenStax, Organic Chemistry, §14.7 Ultraviolet Spectroscopy (A = log(I₀ ÷ I); ε = A ÷ (c × l), c in mol/L and l in cm; β-carotene, ε = 138,000, absorbance 0.37 in a 1.0 cm cell: c = 2.7 × 10⁻⁶ mol/L), https://openstax.org/books/organic-chemistry/pages/14-7-ultraviolet-spectroscopy (retrieved 2026-10-18)
- Absorbance (A) 1, Molar absorptivity ε (L/(mol·cm)) 2,000, Concentration (mol/L) 0.005 gives Path length (cm) 0.1, Transmittance (%) 10.Source: IUPAC Compendium of Chemical Terminology (Gold Book), Beer–Lambert law: A = log₁₀(P₀ ÷ P) = εcl, ε the molar (decadic) absorption coefficient, https://doi.org/10.1351/goldbook.B00626
How it works
The Beer–Lambert law:
A = ε × l × c
- A is the absorbance, log₁₀(I₀ ÷ I), where I₀ is the light going into the sample and I the light coming out. It has no unit.
- ε is the molar absorptivity (molar absorption coefficient) in L/(mol·cm).
- l is the path length in cm.
- c is the concentration in mol/L.
Type any three; the page solves for the fourth:
- A = ε × l × c
- ε = A ÷ (l × c)
- l = A ÷ (ε × c)
- c = A ÷ (ε × l)
The transmittance is the share of the light that passes through: T = I ÷ I₀ = 10^−A, shown in percent as 100 × 10^−A. It is shown for an absorbance up to 300; above that it is below 10⁻²⁹⁸ % and is left out.
Rules. A is from 0 to 10¹², ε is above 0 and at most 10,000,000, l is above 0 and at most 1,000 cm, and c is from 0 to 1,000 mol/L. A solved value outside those limits has no answer, and so does solving for ε or l when c is 0. A result too large or too small to hold in a 64-bit float (beyond about 10³⁰⁸, or a nonzero value below about 10⁻³²³) has no answer rather than showing infinity or 0. Typed values are read as exact decimals, so A, ε, l and c are exact fractions until they are rounded for display; the transmittance is a power of ten worked in floating point.
Output format. Each value shows 6 significant figures; values below 0.000001 or from 10¹⁵ up show as d.ddddd × 10ⁿ.
Worked examples by hand
β-carotene (OpenStax §14.7). A = 0.37, ε = 138,000 L/(mol·cm), l = 1 cm. c = 0.37 ÷ (138,000 × 1) = 2.68116 × 10⁻⁶ mol/L, shown as 0.00000268116 mol/L (the book: 2.7 × 10⁻⁶). T = 100 × 10^−0.37 = 42.658%.
Molar absorptivity from a standard. A = 0.5, l = 1 cm, c = 0.0001 mol/L. ε = 0.5 ÷ (1 × 0.0001) = 5,000 L/(mol·cm).
Absorbance of a 4 µM solution. ε = 138,000, l = 1 cm, c = 0.000004 mol/L. A = 138,000 × 1 × 0.000004 = 0.552; T = 100 × 10^−0.552 = 28.0543%.
Path length. A = 1, ε = 2,000, c = 0.005 mol/L. l = 1 ÷ (2,000 × 0.005) = 0.1 cm; T = 100 × 10^−1 = 10%.
Other questions people ask
What is the Beer-Lambert law?
It says that the absorbance of a dilute solution is proportional to the path length of the light through it and to the concentration of the absorbing substance: A = ε × l × c. The constant ε, the molar absorptivity, depends on the substance and the wavelength.
How do I find the concentration from the absorbance?
Divide the absorbance by the molar absorptivity times the path length: c = A ÷ (ε × l). β-carotene has ε = 138,000 L/(mol·cm); an absorbance of 0.37 in a 1 cm cell gives c = 0.37 ÷ 138,000 = 2.68 × 10⁻⁶ mol/L.
How do I calculate molar absorptivity?
Measure the absorbance of a solution of known concentration and divide: ε = A ÷ (l × c). A 0.1 mM (0.0001 mol/L) solution that reads 0.5 in a 1 cm cuvette has ε = 0.5 ÷ (1 × 0.0001) = 5,000 L/(mol·cm). The units are often left off.
How are absorbance and transmittance related?
Transmittance is the fraction of the light that gets through, T = I ÷ I₀, and the absorbance is A = log₁₀(I₀ ÷ I) = −log₁₀ T. In percent, A = 2 − log₁₀(%T): 10% transmittance is an absorbance of 1, and 1% is an absorbance of 2.
What units does the calculator use?
Path length in centimeters and concentration in moles per liter, so ε is in liters per mole per centimeter, L/(mol·cm), also written M⁻¹ cm⁻¹. Absorbance has no unit.
When does the Beer-Lambert law not hold?
IUPAC states the law for a collimated beam of monochromatic light in a uniform (homogeneous) sample. It does not hold when the band of wavelengths is wide compared with the absorption band, or when the sample scatters light, so use a clear sample at a single wavelength.