The Beer–Lambert Law

Set a molar absorptivity, a concentration and a cuvette. The first picture follows the beam through the cell; the second says what the reading you got is worth.

Worked example — The Beer–Lambert Law

Absorbance A	0.6220
Light through the sample	23.88%
Light the sample stops	76.12%
What the instrument reports	0.6206
Highest absorbance this instrument can report	3.0004
Uncertainty in the concentration	1.46%
The best this instrument can do	1.36%
Cost of this reading against the best	1.08×

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

The first unit of absorbance costs 90% of the light. The second costs 9%. 🖖

Type 160.77 into the concentration, with ε at 6,220 and a 1 cm cell. Absorbance reads 1.0000 and 10.00% of the beam comes through. Type 321.54 and it reads 2.0000, with 1.00% through. Type 482.32 for 3.0000 and 0.10%.

Each of those steps adds the same amount to the number on the screen. The first took 90.00% of the incident beam; the second took a further 9.00%; the third took 0.90%. The Light the sample stops row prints the running total: 90.00%, then 99.00%, then 99.90%.

Look at where the dashed lines fall in the cuvette. They mark the depth at which each whole unit of absorbance has accumulated, and they are evenly spaced, because the exponential decays at the same rate everywhere in the cell. The light is what is not evenly spaced. By A = 3 the detector is being handed a thousandth of what went in and asked to measure it well.

The most precise reading available is A = 0.4343, and it is not the palest one 🖖

A spectrophotometer measures transmittance. Its photometric uncertainty — readout resolution, detector noise, lamp drift — is roughly a fixed number of transmittance units rather than a fixed fraction of the absorbance. Carry that through A = −log₁₀T and the relative uncertainty in concentration comes out as σ_T ÷ (T ln(1/T)).

Neither ε nor ℓ survives the division. Twelve different combinations of absorptivity, concentration and path length that land on A = 0.4343 all give the same 1.36% on an instrument good to 0.5 %T. The uncertainty depends on the reading and on nothing else about the sample.

That expression has a maximum in its denominator. T ln(1/T) is largest at T = 1/e, so the best reading sits at A = log₁₀e = 0.4343 with 36.79% of the light coming through, and the uncertainty there is e·σ_T.

The green band on the error panel is where a reading costs no more than twice that floor: A from 0.1007 to 1.1632. The curve climbs on both sides of the minimum, gently below and steeply above. At A = 1 the penalty is 1.60×, at A = 2 it is 7.99×, at A = 3 it is 53.26×. And a pale solution is no refuge: at A = 0.1 the penalty is already 2.01×.

Stray light puts a ceiling on absorbance at 3.0004, and no chemistry lifts it 🖖

Some of the source reaches the detector without passing through the sample. It scatters off the grating, leaks past a mirror, bounces around inside the monochromator. Call that fraction s. What the instrument reports is then −log₁₀[(10^−A + s) ÷ (1 + s)], and as the sample darkens that expression stops climbing: it approaches log₁₀((1 + s)/s) and goes no further.

Press Past the ceiling. A millimolar NADH solution has a true absorbance of 6.2200 and the instrument reports 3.0002. More than half the absorbance is missing. This page shows you both numbers because it computed them; a spectrophotometer has only the second one, and prints it without a warning.

The shorthand most books give is A_max = −log₁₀ s, which puts the ceiling at 3.0000. The extra 0.0004 is the reference beam seeing the same leak.

Trouble starts well below the ceiling. At 0.1% stray light the reported absorbance is 1% short by A = 1.5811 and 5% short by A = 2.5310, which is inside the range where a calibration curve still looks convincingly straight. Drag the stray-light slider and watch the red line move: at 1% the ceiling drops to 2.0043, and at 0.01% it rises to 4.0000.

Problem solved in full

A reading of 2.49, and a concentration worth ±107 9 steps
An enzyme assay is read at 340 nm in a 1 cm cuvette. The screen shows 2.49, and the analyst writes down 400 µmol/L of NADH. The arithmetic is correct. Work out what that number is worth, what should have been done instead, and what the instrument was actually reporting.

Open this problem in the tool
1. NADH absorbs at 340 nm with a molar absorptivity of 6,220 L mol⁻¹ cm⁻¹. The concentration is 400 µmol/L, which is 4.0000 × 10⁻⁴ mol/L, and the cell is 1 cm across. That is the whole of the law, and the tool prints the product under Absorbance A.
  
  $A = ε c ℓ = 6220 L mo l^{- 1} c m^{- 1} \cdot 4.0000 \times 1 0^{- 4} mol L^{- 1} \cdot 1 cm = 2.4880$
2. Absorbance is defined as −log₁₀T, so invert it. About three photons in a thousand get out of the cuvette. Everything the analyst is about to claim rests on the instrument telling that apart from two in a thousand, or from four.
  
  $T = 1 0^{- A} = 1 0^{- 2.4880} = 3.2509 \times 1 0^{- 3} = 0 . 33%$
3. Now put an uncertainty on it. The instrument measures T, so write c in terms of T and differentiate. The ε and the ℓ appear in both the derivative and the concentration itself, and they cancel: the relative uncertainty depends on the reading alone, not on what is in the cuvette or how wide the cuvette is.
  
  $c = \frac{- lo g _{10} T}{ε ℓ} ⟹ \frac{d c}{d T} = \frac{- 1}{ε ℓ T ln 10} ⟹ \frac{σ _{c}}{c} = \frac{σ _{T}}{ε ℓ T ln 10} \cdot \frac{ε ℓ ln 10}{ln ( 1/ T )} = \frac{σ _{T}}{T ln ( 1/ T )}$
4. A decent bench instrument is good to about 0.5% of the transmittance scale, so σ_T = 0.005. Rewriting the expression in terms of A rather than T lets you use the number the first step already produced. The tool prints 26.85% under Uncertainty in the concentration: the answer is 400 µmol/L give or take 107.
  
  $\frac{σ _{c}}{c} = \frac{σ _{T} 1 0 ^{A}}{A ln 10} = \frac{0 . 005 \cdot 307 . 61}{5 . 72883} = 0.26845400 \cdot 0.26845 = 107 μ mol L^{- 1}$
5. Where would that have been smallest? The denominator T ln(1/T) is what we want largest, and it has a single maximum: differentiating gives −ln T − 1, which is zero at T = 1/e. So the best available reading is A = log₁₀e = 0.4343, and the uncertainty there is e·σ_T = 1.36%. The analyst's reading cost 19.75 times that.
  
  $\frac{d}{d T} (- T ln T) = - ln T - 1 = 0 ⟹ T = \frac{1}{e} = 0.36788, A = lo g_{10} e = 0.4343, e σ_{T} = 0.013591 \frac{0 . 26845}{0 . 013591} = 19.75$
6. Reading too pale costs 2.01× at A = 0.1. Reading too dark costs 53.26× at A = 3. Those two readings are a comparable distance from the optimum and nowhere near a comparable distance in accuracy, which is why the error panel is drawn on a logarithmic axis and why the advice a lab gives is to dilute rather than to concentrate.
  
  $A 0.1000 0.2000 0.4343 0.7000 1.0000 1.5000 2.0000 2.4880 3.0000 T 79 . 43% 63 . 10% 36 . 79% 19 . 95% 10 . 00% 3 . 16% 1 . 00% 0 . 33% 0 . 10% σ_{c} / c 2 . 73% 1 . 72% 1 . 36% 1 . 55% 2 . 17% 4 . 58% 10 . 86% 26 . 85% 72 . 38% \times floor 2.01 1.27 1.00 1.14 1.60 3.37 7.99 19.75 53.26$
7. So dilute. 2.4880 ÷ 0.4343 = 5.73, which brings the sample to 69.82 µmol/L, the figure under Concentration at the optimum. One pipetting step, and the same instrument answers 19.75 times more precisely. The dilution carries an uncertainty of its own that adds in quadrature, and it would have to be worse than 26.8% to make the trade a bad one.
  
  $\frac{A}{lo g _{10} e} = \frac{2 . 4880}{0 . 4343} = 5.73, c_{opt} = \frac{lo g _{10} e}{ε ℓ} = \frac{0 . 4343}{6220} = 6.982 \times 1 0^{- 5} mol L^{- 1} = 69.82 μ mol L^{- 1} 0.2684 5^{2} - 0.01359 1^{2} = 0.268$
8. None of this is what the screen said. With 0.1% stray light the instrument reported 2.3720, which is 4.7% short of the absorbance the chemistry produced, and it could not have reported above 3.0004 whatever went into the cuvette. The analyst's 2.49 was already two errors deep before the uncertainty was counted.
  
  $A_{meas} = - lo g_{10} \frac{1 0 ^{- A} + s}{1 + s} = - lo g_{10} \frac{3 . 2509 \times 1 0 ^{- 3} + 0 . 001}{1 . 001} = 2.3720, 1 - \frac{2 . 3720}{2 . 4880} = 4 . 7%, A_{m a x} = lo g_{10} \frac{1 . 001}{0 . 001} = 3.0004$
9. A monochromator does not deliver a single wavelength. Suppose half the beam sits where ε is 20% lower, which is what a finite slit width does on the flank of an absorption peak. The detector adds the two intensities and the logarithm is taken once, afterwards, so the apparent absorbance falls below the intensity-weighted average of the two. At a nominal A of 1 the instrument reads 0.8886 against an average of 0.9000; at a nominal 2 it reads 1.7555 against 1.8000. The shortfall grows with concentration, so the calibration line bends toward the concentration axis instead of shifting. Anchor a single-point calibration at the optimum, use it at A = 3, and the concentration comes out 3.03% low. None of that is on the panel above. A second ε and a beam split are three more controls than this page has.
  
  $A_{app} = - lo g_{10} [\frac{1}{2} 1 0^{- A} + \frac{1}{2} 1 0^{- 0.8 A}] : A = 1 \to 0.8886 (mean 0.9000), A = 2 \to 1.7555 (mean 1.8000)$
Answer

400 µmol/L give or take 107. The arithmetic was right and the reading was in the wrong place: 0.33% of the beam through the cell, 26.85% on the concentration, 19.75 times the uncertainty the same instrument would have given at A = 0.4343.

Diluting 5.73× to 69.82 µmol/L gives 69.82 ± 0.95, which multiplies back to 400 ± 5.4. Same sample, same instrument, same afternoon.

And the 2.4880 was never on the screen. Stray light had taken 4.7% of it before the analyst read anything at all.

$\frac{σ _{c}}{c} = \frac{σ _{T}}{T ln ( 1/ T )} minimised at T = \frac{1}{e} ⟹ A = lo g_{10} e = 0.4343$

References (5)

The concentration half of the law, and the paper the name comes from: August Beer, "Bestimmung der Absorption des rothen Lichts in farbigen Flüssigkeiten", Annalen der Physik 162(5), 78–88, 1852. doi:10.1002/andp.18521620505
Who actually wrote what, which the usual attribution gets wrong: Heinz G. Pfeiffer and Herman A. Liebhafsky, "The origins of Beer's law", Journal of Chemical Education 28(3), 123, 1951. doi:10.1021/ed028p123
The photometric error curve that carries their names: F. Twyman and G. F. Lothian, "Conditions for securing accuracy in spectrophotometry", Proceedings of the Physical Society 45(5), 643–662, 1933. doi:10.1088/0959-5309/45/5/304
An openly readable derivation of the same result, in the same notation: "13.3: Effect of Noise on Transmittance and Absorbance Measurements", Instrumental Analysis, Chemistry LibreTexts (CC BY-NC-SA 4.0).
Where the 6,220 L mol⁻¹ cm⁻¹ for NADH at 340 nm comes from: B. L. Horecker and Arthur Kornberg, "The extinction coefficients of the reduced band of pyridine nucleotides", Journal of Biological Chemistry 175(1), 385–390, 1948. doi:10.1016/S0021-9258(18)57268-9

By the 42Math team · Reviewed against published sources. Corrections welcome.