Beer–Lambert Law Calculator
Absorbance from the molar absorption coefficient, path length and concentration by A = εlc, or any one of them from the others, with absorbance and transmittance converted.
Result
Absorbance, A: 0.622
Holds for dilute solutions and narrow-band light. At high absorbance, readings stop being proportional to concentration.
- Transmittance
- 23.8781%
- Concentration, in the unit chosen
- 0.1
- Molar absorption coefficient, L/(mol·cm)
- 6,220
- Path length
- 1 cm (10 mm)
https://calcgallery.com/chemistry/beer-lambert-law-calculator/
Results come from the published equations applied to the values you enter. They are not electrical design, conductor sizing, laboratory safety or medical guidance. Weather indices describe a standard person in standard conditions and are not a forecast. Where codes apply, local rules vary and must be checked. Read the full disclaimer
How to use the Beer–Lambert Law Calculator
- Choose the unknown, absorbance, concentration, the molar absorption coefficient ε or the path length.
- Enter the other values. A reading can be typed as absorbance or as percent transmittance, and the concentration in M, mM or µM.
- Read the unknown, with the absorbance and transmittance of the sample beside it.
The formulas
A = ε × l × cc = A ÷ (ε × l) ε = A ÷ (l × c) l = A ÷ (ε × c)A = −log₁₀ T T = 10^(−A)ε is in litres per mole per centimetre, l in centimetres and c in moles per litre, so A has no unit. Path lengths entered in millimetres and concentrations in mM or µM are converted before the calculation.
What the diagram shows
A beam of light entering a cuvette of path length l and leaving it fainter. The label on the outgoing beam is the transmittance: the share of the light that gets through.
With your numbers
- A = ε × l × c = 6,220 L/(mol·cm) × 1 cm × 0.0001 mol/L = 0.622
- T = 10^−A = 23.8781%
Quick reference
Holds for dilute solutions and narrow-band light. At high absorbance, readings stop being proportional to concentration.
| Example | Result |
|---|---|
| ε 6,220, 1 cm, 0.1 mM | Absorbance, A: 0.622 |
| Concentration at A 0.5, ε 6,220, 1 cm | Concentration, in the unit chosen: 80.385852 |
| Concentration at 25% transmittance | Concentration, in the unit chosen: 0.096794 |
| ε from A 0.8, 0.5 cm, 50 µM | Molar absorption coefficient, L/(mol·cm): 32,000 |
| Path length for A 1.2, ε 15,000, 40 µM | Path length: 2 cm |
| Concentration in µM at 50% transmittance, ε 6,220, 1 cm | Concentration, in the unit chosen: 48.397105 |
Common questions about the Beer–Lambert Law Calculator
- What does the Beer–Lambert law say?
- That absorbance is proportional to how far the light travels through the sample and to the concentration of the absorbing substance: A = ε × l × c. With ε of 6,220 L/(mol·cm), a 1 cm cuvette and a 0.1 mM solution, the absorbance is 0.622.
- How do I find a concentration from an absorbance?
- Divide the absorbance by ε times the path length: c = A ÷ (ε × l). An absorbance of 0.5 with ε of 6,220 L/(mol·cm) in a 1 cm cuvette is about 80.4 µM. The ε must be for your substance at the wavelength you measured.
- How are absorbance and transmittance related?
- Transmittance is the share of the light that gets through, T = I ÷ I₀, and absorbance is its negative base-10 logarithm, A = −log₁₀ T. So 50% transmittance is an absorbance of about 0.301, 10% is 1 and 1% is 2. Each extra unit of absorbance lets through ten times less light.
- When does the law stop working?
- It holds for dilute solutions and light that is narrow compared with the substance's absorption band. At high concentrations molecules interact, and at high absorbance so little light reaches the detector that stray light dominates, so the reading flattens off. A calibration curve from standards shows where your own instrument stays linear.