Absorbance · free worksheet
From light to concentration
Four exercises with worked answers. All numbers are synthetic teaching data; the coefficient is illustrative and is not assigned to a real substance.
Content checked: 3 October 2026. These calculations do not establish experimental suitability or laboratory method validity. Measurement uncertainty is not calculated.
Try the calculations
1. A logarithmic measurement
Calculate percentage transmittance for A = 0, 1 and 2. Then calculate absorbance for 50% transmittance. Use base-10 logarithms.
%T = 100 × 10−A A = −log10(%T / 100)
2. Match the units
Use ε = 20,000 L·mol⁻¹·cm⁻¹, l = 10 mm and c = 25 µM. Convert to centimetres and mol/L, calculate A, then calculate %T.
A = ε × l × c
3. Keep the two solutions separate
A measured solution has A = 0.5, with the same ε and path length as exercise 2. It was made by diluting 1.00 mL of an original solution to a final volume of 10.00 mL. Find the measured and original concentrations.
4. Fit a calibration with an intercept
Fit absorbance y against concentration x using ordinary least squares. Keep all four rows. Find the slope, intercept, residuals and R². Estimate x for unknown A = 0.55 and A = 0.85, and identify any extrapolation.
| Standard | Concentration (µM) | Absorbance |
|---|---|---|
| 1 | 0 | 0.05 |
| 2 | 1 | 0.26 |
| 3 | 2 | 0.44 |
| 4 | 3 | 0.65 |
Absorbance · worked answers
Check your reasoning
1. Absorbance and transmission
A = 0, 1 and 2 give 100%, 10% and 1% transmittance respectively. For 50% transmission, T = 0.50 and A = −log₁₀(0.50) ≈ 0.301029995664. Exactly 0% transmittance has no finite absorbance.
2. Beer–Lambert calculation
10 mm = 1 cm and 25 µM = 25 × 10⁻⁶ mol/L. Therefore A = 20,000 × 1 × 0.000025 = 0.5. %T = 100 × 10⁻⁰·⁵ ≈ 31.62277660%.
3. Dilution
cmeasured = 0.5 / (20,000 × 1) = 0.000025 mol/L = 25 µM. D = 10.00 / 1.00 = 10. The original concentration is 25 × 10 = 250 µM. The concentration of the solution in the cell remains 25 µM.
4. Ordinary least squares and inversion
The means are x̄ = 1.5 µM and ȳ = 0.35. Σ(x − x̄)² = 5 µM² and Σ(x − x̄)(y − ȳ) = 0.99 µM. Thus m = 0.99 / 5 = 0.198 µM⁻¹, and b = 0.35 − 0.198 × 1.5 = 0.053.
ŷ = (0.198 µM⁻¹) × x + 0.053
| x (µM) | Observed A | Fitted A | Residual |
|---|---|---|---|
| 0 | 0.05 | 0.053 | −0.003 |
| 1 | 0.26 | 0.251 | 0.009 |
| 2 | 0.44 | 0.449 | −0.009 |
| 3 | 0.65 | 0.647 | 0.003 |
SSE = 0.00018; SST = 0.1962. R² = 1 − SSE / SST ≈ 0.999082568807. For A = 0.55, x = (0.55 − 0.053) / 0.198 ≈ 2.5101010101 µM. For A = 0.85, x ≈ 4.0252525253 µM, which is extrapolation beyond the standards’ 0–3 µM range.
This fit treats standards’ concentrations as fixed/exact and assumes a suitable linear response with comparable measurement variability. Inspect residuals; high R² does not validate a method. The intercept is fitted, with no automatic blank correction. Uncertainty is not calculated.