Chemistry · General chemistry II · Worked example

Find a concentration from transmittance

A dye solution in a 1.00 cm cell transmits 35.5% of the light at the dye’s measuring wavelength, where ε = 1.50 × 10⁴ L mol⁻¹ cm⁻¹. Find the dye’s concentration.

A = ε b c

Convert transmittance to absorbance

35.5% is T = 0.355, so A = −log 0.355 = 0.450.

A = - log (0.355) = 0.450

Solve the Beer–Lambert law for c

Divide both sides of A = εbc by εb.

c = \frac{A}{ε b}

Substitute

The units of ε and b cancel with the path length, leaving mol/L.

c = \frac{0.450}{(1.50 \times 10^{4} L {mol}^{- 1} {cm}^{- 1}) (1.00 cm)} = 3.00 \times 10^{- 5} mol/L

\begin{array}{c} c = \frac{0.450}{(1.50 \times 10^{4}) (1.00)} mol/L \\ = 3.00 \times 10^{- 5} mol/L \end{array}

Interpret

3.00 × 10⁻⁵ mol/L is 30.0 µmol/L. An absorbance near 0.45 is comfortably in the range where the law usually holds.

Result

c = 3.00 × 10⁻⁵ mol/L, which is 30.0 µmol/L.

Your turn

The same dye in a 0.500 cm cell has A = 0.300. What is its concentration?

Show the answer and explanation

4.00 × 10⁻⁵ mol/L.

c = A/(εb) = 0.300 ÷ (1.50 × 10⁴ L mol⁻¹ cm⁻¹ × 0.500 cm) = 4.00 × 10⁻⁵ mol/L.

c = \frac{0.300}{(1.50 \times 10^{4}) (0.500)} = 4.00 \times 10^{- 5} mol/L

\begin{array}{c} c = \frac{0.300}{(1.50 \times 10^{4}) (0.500)} \\ = 4.00 \times 10^{- 5} mol/L \end{array}

Keep exploring

In Photons & spectrophotometry, solve for absorbance instead and enter twice the concentration, 6.00 × 10⁻⁵ mol/L. The absorbance doubles to 0.900, but the transmittance falls from 35.5% to 12.6%, not to half.

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Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

David Harvey, Analytical Chemistry 2.1 — Spectroscopy based on absorption

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Solve it in Photons & spectrophotometry Convert 35.5% transmittance to absorbance Check the calculation in a Chemistry box Open worked example on a board Chemistry formulas: organic (Beer–Lambert law)

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