Chemistry and laboratory
Beer–Lambert law calculator
Solve A = εlc for concentration, absorbance, molar absorption coefficient or path length, from absorbance or percent transmittance.
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Result
- Concentration c
- 50.0 µM
- Absorbance
- 0.311
- Transmittance
- 48.9 %
- Concentration
- 50.0 µM
Working
- A = −log₁₀(T) = ε l c
- c = A ÷ (ε l) = 0.311 ÷ (6220 M⁻¹ cm⁻¹ × 1 cm) = 50.0 µM
- Valid for dilute, non-scattering solutions at a single wavelength. Subtract the blank first.
- Rounded to 3 significant figures.
Method and limits
Choose the unknown, then enter the other quantities. Absorbance can be typed directly or as percent transmittance.
A = log₁₀(I₀ ÷ I) = −log₁₀ T = ε l c
ε is the molar (decadic) absorption coefficient in M⁻¹ cm⁻¹ (L mol⁻¹ cm⁻¹), l the path length in cm and c the concentration in mol/L. With a mass coefficient (the absorbance of a 1 mg/mL solution, in mL mg⁻¹ cm⁻¹) the concentration is in mg/mL. A molar mass converts between the two.
Limits
- The law holds for dilute, clear solutions at one wavelength. Most instruments become non-linear above an absorbance of about 1 to 1.5 (under 3–10 % of the light gets through): dilute and measure again.
- Subtract a blank of the solvent and cuvette, and use the ε measured under the same conditions (wavelength, pH, solvent, temperature).
- Scattering samples (cells, precipitates) and fluorescent samples break the law.
Worked example
NADH at 340 nm has ε = 6220 M⁻¹ cm⁻¹. An absorbance of 0.311 in a 1.00 cm cuvette gives c = 0.311 ÷ (6220 × 1.00) = 5.00 × 10⁻⁵ M = 50.0 µM; that absorbance transmits 48.9 % of the light.
References
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