Beer–Lambert Law Calculator (Absorbance, ε, Concentration)
Solve the Beer–Lambert law A = ε·l·c for concentration, absorbance, molar extinction coefficient or path length, with unit menus (nM–M, mm–m, M⁻¹ or mM⁻¹ cm⁻¹). Enter absorbance, percent transmittance or I and I₀, subtract a blank, correct for dilution, get mass concentration and a linear-range check.
How it works
A = ε × l × c · c = A ÷ (ε l) · ε = A ÷ (l c) · l = A ÷ (ε c) · A = −log₁₀(%T ÷ 100) = log₁₀(I₀ ÷ I) · %T = 100 × 10^(−A) · sample c = cuvette c × dilution · g/L = M × molar mass
When light passes through a solution, each thin layer absorbs the same fraction of what reaches it, so intensity falls exponentially with distance and with the number of absorbing molecules in the way. Taking the logarithm turns that into absorbance, which is simply proportional to path length and concentration; the extinction coefficient ε captures how strongly a particular molecule absorbs at a particular wavelength. Pick the unknown — concentration, absorbance, ε or path length — and enter the other three in whatever units you have: concentrations from nM to M, path lengths in µm, mm, cm or m, and ε per M or per mM. The reading can be an absorbance, a percent transmittance or the raw intensities I and I₀, and a blank and a dilution factor can be applied. Add a molar mass to get mg/mL and the specific absorbance (ε 0.1% and E1%). The relationship holds only while the solution is dilute and the light nearly monochromatic. At high absorbance very little light reaches the detector and stray light bends the curve, so readings above about 1.5 should be diluted rather than trusted. The table shows what the same sample would read in other cells.
Worked example
An NADH solution reads A340 = 0.450 in a 1 cm cuvette. With ε = 6,220 M⁻¹cm⁻¹ the concentration is 0.450 ÷ 6,220 = 7.235 × 10⁻⁵ M = 72.35 µM, and 100 × 10^(−0.45) = 35.5% of the light is transmitted. With NADH’s molar mass of 709.4 g/mol that is 72.35 µM × 709.4 = 51.3 µg/mL. Had the sample been diluted 1:10 before reading, the original would be 723.5 µM. Going the other way, 43.3 µM of a dye with ε = 8,400 M⁻¹cm⁻¹ in a 1 cm cell should read A = 8,400 × 1 × 4.33 × 10⁻⁵ = 0.364.
Sources: IUPAC Gold Book — Beer–Lambert law · IUPAC Gold Book — absorbance
Frequently asked questions
How do I convert transmittance to absorbance?
A = −log₁₀(%T ÷ 100), or log₁₀(I₀ ÷ I) from raw intensities. 10% T is A = 1, 1% T is A = 2, and 50% T is A = 0.301. Choose “A ↔ %T only” for a quick conversion and a reference table.
Where do I find ε for my compound?
Supplier data sheets, the literature, or measure it from a dilution series of known concentration (the slope of A vs c at 1 cm). Choose “ε” in Solve for to get it from one standard. For proteins, compute it from the tryptophan, tyrosine and cystine content (ExPASy ProtParam).
Are M⁻¹cm⁻¹ and L·mol⁻¹·cm⁻¹ the same?
Yes — M is mol/L, so M⁻¹cm⁻¹ = L·mol⁻¹·cm⁻¹. Values per mM (mM⁻¹cm⁻¹, as in 6.22 for NADH) are 1,000 times smaller; pick that unit and the calculator converts.
Can I use a plate reader?
Yes, but the path length is the liquid height, not 1 cm — typically 0.3–0.6 cm for 100–200 µL in a 96-well plate. Many readers offer path-length correction using the water peak at 977 nm. You can also solve for the path length from a standard of known concentration.
What absorbance range should I aim for?
Roughly 0.1 to 1.0 on most instruments. Below 0.1 noise and the blank dominate; above 1.5 stray light flattens the response. Dilute strong samples and enter the dilution factor so the calculator reports the original concentration.
Related calculators
Standard laboratory formulas and textbook recipes (Sambrook, Bio-Rad, Thermo Fisher protocol conventions). Follow your own kit or instrument manual where it differs, and validate with controls — this tool does not replace lab training or safety procedures.