How can absorbed light reveal concentration?
You will be able to: Use A = εbc and explain the effects of concentration, path length and wavelength.
How can absorbed light reveal concentration?
A colored solution absorbs some light passing through a cuvette. A longer path or more absorbing particles per volume can remove a larger fraction of the incident light.
A useful starting point: Why do atoms and molecules absorb selected photon energies? →
Words and symbols before equations
- Absorbance A
- A dimensionless logarithmic measure of attenuation, A = −log₁₀(I/I₀).
- Molar absorptivity ε
- How strongly a species absorbs at a specified wavelength, in L·mol⁻¹·cm⁻¹ here.
- Path length b
- Distance light travels through solution, in cm here.
- Concentration c
- Absorbing-species concentration in mol/L.
- Transmittance
- Fraction I/I₀ of incident intensity transmitted.
What this picture assumes
Clear sample in a linear Beer–Lambert regime, ε = 100 L·mol⁻¹·cm⁻¹ at one fixed wavelength. Concentration control is mmol/L: divide by 1000 for mol/L. A = εbc and transmission = 10⁻ᴬ. Beam thickness/color is schematic.
Read the picture in three steps
- Read the species and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- A = 0.5; transmission = 31.62%; c = 5 mmol/L; b = 1 cm.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the chemistry
In the Beer–Lambert linear regime, A = εbc. Doubling c doubles A when ε and b stay fixed; doubling b doubles A when ε and c stay fixed.
ε depends on the absorbing species and wavelength. Keep wavelength fixed during a concentration comparison. The wavelength of maximum absorbance often provides good sensitivity.
Absorbance is not the percentage of light absorbed. Because A is logarithmic, doubling A does not simply halve the transmitted fraction. At A = 1, transmittance is 0.10; at A = 2, it is 0.01.
The simple law assumes an appropriate dilute, clear sample with stable chemical form and suitable instrumentation. Scattering, stray light or chemical changes can break the simple relationship.
| Absorbance A | Transmitted fraction | Percent transmitted |
|---|---|---|
| 0 | 1 | 100% |
| 1 | 0.1 | 10% |
| 2 | 0.01 | 1% |
A worked example, step by step
Find A for ε = 100 L·mol⁻¹·cm⁻¹, b = 1.00 cm and c = 0.00500 mol/L.
- Check units: L·mol⁻¹·cm⁻¹ multiplied by cm and mol/L cancels completely.
- A = εbc = (100)(1.00)(0.00500) = 0.500.
- Transmittance = 10⁻⁰·⁵⁰⁰ ≈ 0.316, or 31.6%.
- Doubling concentration at unchanged ε and b gives A = 1.00 and 10.0% transmittance, not zero transmitted light.
A = 0.50 does not mean 50% of light was absorbed. Absorbance is logarithmic and dimensionless.
If the cuvette path doubles at fixed c and wavelength, what happens to absorbance?
Compare with an explanation
It doubles in the Beer–Lambert linear regime because A is proportional to b.
Predict. Change one thing. Explain.
Hold path length at 1 cm and increase concentration from 5 to 10 mmol/L. Compare the linear absorbance graph with the nonlinear transmittance readout.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
A = 0.5; transmission = 31.62%; c = 5 mmol/L; b = 1 cm.
Clear sample in a linear Beer–Lambert regime, ε = 100 L·mol⁻¹·cm⁻¹ at one fixed wavelength. Concentration control is mmol/L: divide by 1000 for mol/L. A = εbc and transmission = 10⁻ᴬ. Beam thickness/color is schematic.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using particle interactions, concentration, gas behavior or energy transfer. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA sample has A = 0.600 at b = 1.00 cm and ε = 150 L·mol⁻¹·cm⁻¹. Calculate c and predict A after diluting to twice the initial volume without losing solute.
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Compare with the answer and four-point rubric
- 1 point: Rearrange c = A/(εb).
- 1 point: c = 0.600/(150 × 1.00) = 0.00400 M.
- 1 point: Dilution to twice the volume halves concentration to 0.00200 M.
- 1 point: At unchanged wavelength and path, absorbance halves to 0.300 in the linear regime.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What three factors appear in Beer–Lambert law?
Molar absorptivity, path length and concentration.
RECALL 2Why hold wavelength fixed?
Molar absorptivity depends on wavelength.
RECALL 3Is absorbance measured in percent?
No; it is dimensionless and logarithmic.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How can absorbed light reveal concentration?
- A = εbc; ε in L·mol⁻¹·cm⁻¹, b in cm, c in mol/L.
- I/I₀ = 10⁻ᴬ; A has no units.
Remember: A = 0.50 does not mean 50% of light was absorbed. Absorbance is logarithmic and dimensionless.
Conditions: Clear sample in a linear Beer–Lambert regime, ε = 100 L·mol⁻¹·cm⁻¹ at one fixed wavelength. Concentration control is mmol/L: divide by 1000 for mol/L. A = εbc and transmission = 10⁻ᴬ. Beam thickness/color is schematic.
Refresh Kid · AP Chemistry Unit 3 · Objectives 3.13.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 3.13, objective 3.13.A. CED effective Fall 2024 and June 2026 clarifications checked September 16, 2026. Unit 3: Properties of Substances and Mixtures, Topics 3.1–3.13. Focused lesson names, examples, models and assessments are original Refresh Kid teaching materials, not additional official topics or official AP questions. Official corrections.
The model states its assumptions beside the diagram. Colligative-property calculations and solution molality/mass-percent/volume-percent calculations are not required here. The optional speed-density model illustrates distributions; it does not require memorizing its mathematical derivation.
Teaching resources: The Organic Chemistry Tutor video titles/descriptions and topic coverage were checked for optional links; no claim is made to have watched every video. No creator scripts, examples, worksheets or artwork were copied. GitHub’s 3D website collection and its Three.js camera-control example informed the idea of controllable spatial inspection. Scientific diagrams, geometry and interactions here are original. The self-hosted Three.js runtime retains its MIT license. Camera rotation changes the view, not the chemistry.
Independent teacher review and observation of students remain pending. Implementation checks do not certify scientific accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Chemistry questions and scoring guides. This archive contains questions across units; it is not an assignment of every question to this lesson.
The teaching sequence is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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