Beer–Lambert law calculator (absorbance to concentration)

The formula

$$A = \varepsilon\, l\, c$$

$A$ is absorbance (unitless), $\varepsilon$ the molar absorptivity in L/(mol·cm), $l$ the path length in cm and $c$ the concentration in mol/L.

Where it comes from

Picture light crossing a thin slice of solution of thickness $dx$. The chance that a photon is absorbed in that slice is proportional to how many absorbing molecules it meets, which depends on the concentration and the thickness. So the intensity drops by a fixed fraction per slice:

$$\frac{dI}{dx} = -k\,c\,I$$

This says intensity decays exponentially. Separate the variables and integrate across the whole path, from intensity $I_0$ at the front to $I$ at depth $l$:

$$\int_{I_0}^{I} \frac{dI}{I} = -k c \int_0^l dx \quad\Rightarrow\quad \ln\frac{I}{I_0} = -k\,c\,l$$

Spectrophotometers report absorbance in base 10, defined as $A = -\log_{10}(I/I_0)$. Converting the natural log, $\ln x = \ln 10 \cdot \log_{10} x$:

$$A = \frac{k}{\ln 10}\, c\, l = \varepsilon\, l\, c$$

All the constants fold into $\varepsilon$, a property of the molecule at one wavelength.

Worked example

NADH absorbs at 340 nm with $\varepsilon = 6{,}220$ L/(mol·cm). A reading of $A = 0.50$ in a 1 cm cuvette means

$$c = \frac{A}{\varepsilon\, l} = \frac{0.50}{6{,}220 \times 1} = 8.04 \times 10^{-5}\ \text{M} = 80.4\ \mu\text{M}$$

Watch out for

• Stay in the linear range. Above an absorbance of about 1 to 1.5, stray light and molecular interactions bend the line. Dilute and multiply back.
• Blank first. Absorbance assumes $I_0$ is light through the solvent alone, so zero the instrument on a blank.

Last checked November 11, 2026