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Beer–Lambert Law A = εlc — Absorbance and Concentration

By Aniket Bhardwaj · 5 September 2026 · Calculator/Formula Guide

When you shine light through a coloured solution, some of it is absorbed. The Beer–Lambert law is the simple equation that connects how much light is absorbed to how much substance is dissolved. It is the working principle of every colorimeter and UV–visible spectrophotometer, and it appears in Class 12 practicals, IIT-JAM, GATE and CSIR-NET analytical chemistry. This guide explains each symbol, works four numericals fully, and lists the errors students actually make.

The formula

A = ε l c   and   A = log₁₀(I₀ / I) = −log₁₀ T

The two halves of that box are the whole law. The first says absorbance is proportional to concentration. The second defines what absorbance is: a logarithm of the ratio of light going in to light coming out.

What each symbol means

SymbolNameUsual unitWhat it really is
AAbsorbancenone (a pure number)How strongly the sample absorbs at one chosen wavelength
εMolar absorptivity (molar extinction coefficient)L mol⁻¹ cm⁻¹A property of the substance at that wavelength — big ε means intensely coloured
lPath lengthcmThickness of solution the light crosses; a standard cuvette is 1.00 cm
cConcentrationmol L⁻¹ (molarity)How much absorbing species is dissolved
I₀, IIncident and transmitted intensitysame units, they cancelLight before and after the sample
TTransmittancefraction, or × 100 for %TT = I / I₀ — the fraction of light that gets through

Notice that the units of ε, l and c are chosen so that their product has no unit at all: (L mol⁻¹ cm⁻¹) × (cm) × (mol L⁻¹) cancels completely. If your answer for A comes out with a unit attached, you have used the wrong units somewhere.

Absorbance and transmittance are not the same thing

Many instruments display %T, but the law is linear in A, not in T. The relationship is logarithmic, so the two behave very differently:

Transmittance T%TAbsorbance A = −log₁₀ T
1.000100%0.000
0.50050%0.301
0.25025%0.602
0.10010%1.000
0.0101%2.000

Read that table once more: halving the transmitted light does not halve the absorbance, it adds 0.301 to it. Every time A rises by exactly 1, ten times less light gets through.

Worked example 1 — absorbance from concentration

Question: A dye has ε = 5500 L mol⁻¹ cm⁻¹ at 520 nm. What is the absorbance of a 2.4 × 10⁻⁵ mol L⁻¹ solution in a 1.00 cm cuvette?

A = ε l c
A = 5500 × 1.00 × 2.4 × 10⁻⁵
A = 5500 × 2.4 × 10⁻⁵ = 13 200 × 10⁻⁵
A = 0.132

Check: A is small, so most light passes through — T = 10⁻⁰·¹³² ≈ 0.74, i.e. about 74% of the light gets through. A pale solution. That is consistent.

Worked example 2 — concentration from absorbance

Question: A complex with ε = 15 600 L mol⁻¹ cm⁻¹ gives A = 0.62 in a 1.00 cm cell. Find its concentration.

Rearrange: c = A / (ε l)
c = 0.62 / (15 600 × 1.00)
c = 0.62 / 15 600 = 3.974 × 10⁻⁵
c = 3.97 × 10⁻⁵ mol L⁻¹

Check by substituting back: 15 600 × 1.00 × 3.974 × 10⁻⁵ = 0.620 ✓

Worked example 3 — starting from %T

Question: A sample transmits 25.0% of the incident light in a 1.00 cm cell. If ε = 12 040 L mol⁻¹ cm⁻¹, find the concentration.

Step 1 — convert %T to T: T = 25.0 / 100 = 0.250
Step 2 — convert T to A: A = −log₁₀(0.250) = 0.602
Step 3 — apply the law: c = A / (ε l) = 0.602 / (12 040 × 1.00)
c = 0.602 / 12 040 = 5.00 × 10⁻⁵
c = 5.00 × 10⁻⁵ mol L⁻¹

Step 1 and Step 2 are where marks are lost. Never put 25.0 or 0.250 straight into A = εlc.

Worked example 4 — the calibration (comparison) method

In a school or college laboratory you usually do not know ε. You do not need it. Measure a standard of known concentration and an unknown under identical conditions, and ε and l cancel.

A₁ / A₂ = c₁ / c₂  ⟹  c₂ = c₁ × (A₂ / A₁)

Question: A standard of 4.0 × 10⁻⁵ mol L⁻¹ reads A = 0.480. An unknown of the same substance in the same cuvette reads A = 0.300. Find the unknown concentration.

c₂ = 4.0 × 10⁻⁵ × (0.300 / 0.480)
0.300 / 0.480 = 0.625
c₂ = 4.0 × 10⁻⁵ × 0.625
c₂ = 2.5 × 10⁻⁵ mol L⁻¹

Sanity check: the unknown absorbs less than the standard, so it must be more dilute — and 2.5 × 10⁻⁵ is indeed smaller than 4.0 × 10⁻⁵ ✓

When the straight line bends

The law is a limiting law, not a universal truth. A plot of A against c is straight only while the assumptions hold. It curves when:

Common mistakes that cost marks

  • Using %T as if it were T. 25.0% must become 0.250 before taking the log.
  • Using natural log instead of log₁₀. The Beer–Lambert absorbance in chemistry is defined with base-10 logarithms. Using ln gives an answer 2.303 times too big.
  • Path length in millimetres. ε is quoted per centimetre. A 5 mm cell is l = 0.5 cm, not 5.
  • Concentration in g/L. Molar absorptivity needs mol L⁻¹. Convert with c = (mass concentration) ÷ (molar mass) first.
  • Forgetting the blank. The instrument must be zeroed with the solvent and reagents but no analyte, otherwise you are measuring the cuvette and the solvent too.
  • Assuming A adds up for a mixture without checking wavelength. Absorbances of different species do add at the same wavelength, but only if each obeys the law there.

Where this appears in exams

ExamTypical use
CBSE/ICSE Class 12Colorimetry in practicals; the idea that colour intensity measures concentration
BSc analytical chemistryCalibration curves, determination of ε, deviations from linearity
IIT-JAM / CUET-PGA ⇄ T conversions, path-length and dilution scaling
GATE / CSIR-NETTwo-component mixtures, back-calculating ε, spectrophotometric titrations

Do the conversion without slipping a decimal. The Beer–Lambert calculator takes any three of A, ε, l and c and returns the fourth, and handles the transmittance step for you so a %T value never gets used as an absorbance by accident.

Open the Beer–Lambert Law Calculator →

If logarithms and concentration units are where your numericals keep breaking down, that is a fixable gap. ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre and online classes across India — details at abcchemistry.in.