KEY2CHEM

The Beer-Lambert Law: Determination of Concentration

Absorption of a photon by a molecule causes a change in energy of the molecule equal to the energy of the photon  ($$\Delta{E}_{system} = E_{photon} = h\nu$$). Due to the large number of molecules in a sample, the bands/peaks observed by UV-vis spectroscopy are broad. The higher the concentration of molecules in solution, the more photons of light that can be absorbed. The concentration can be related to the absorbed light by the Beer-Lambert law ($$A=\epsilon b c$$), where $$A$$ = absorbance (related to the amount of light absorbed, usually in arbitrary units), $$\epsilon$$ = molar extinction coefficient (also called molar absorptivity), which is specific for a specific molecule/species, $$b$$ = cell path length of the sample, and $$c$$ = concentration of the sample.

Example 1.

A protein shows an absorbance of $$0.725$$ at $$280\text{ nm}$$. If the concentration of the sample is $$1.50 \times 10^{-5}\text{ M}$$ and the cell path length is $$1\text{ cm}$$, what is the molar extinction coefficient of the protein at this wavelength?

A. $$4.83\times 10 ^{4}\text{ M}^{-1}\text{ cm}^{-1}$$

B. $$1.09\times 10 ^{-5}\text{ M}^{-1}\text{ cm}^{-1}$$

C. $$2.87\times 10 ^{3}\text{ M}^{-1}\text{ cm}^{-1}$$

Solution

A. $$4.83\times 10 ^{4}\text{ M}^{-1}\text{ cm}^{-1}$$

The Beer-Lambert law can be used to evaluate molar extinction coefficients.

\begin{align}A&=\epsilon b c \\ 0.725&=\epsilon (1\text{ cm}) (1.50\times 10^{-5}\text{M})\\4.83\times 10^{4}\text{ M}^{-1}\text{ cm}^{-1} &= \epsilon\end{align}

Example 2.

What is the concentration of a dye that has a molar absorptivity of $$2.13\times 10^{4}\text{ M}^{-1}\text{cm}^{-1}$$ and gives an absorbance reading of $$0.445$$? The spectrometer has a $$1\text{ cm}$$ path length.

A. $$9.48\times 10^{-3}\text{ M}$$

B. $$2.09\times 10^{-5}\text{ M}$$

C. $$5.42\times 10^{-2}\text{ M}$$

Solution

B. $$2.09\times 10^{-5}\text{ M}$$

\begin{align}A&=\epsilon b c \\ 0.445&=(2.13\times 10^{4}\text{M}^{-1}\text{cm}^{-1}) (1\text{ cm}) c\\2.09\times 10^{-5}\text{ M} &= c\end{align}

Example 3.

Compounds with large molar absorptivity values can be detected at ______ concentrations than compounds with small molar absorptivity values.

A. higher

B. lower

C. only the same

Solution

B. lower

Compounds with large molar absorptivity values absorb visible light strongly, so the give large absorbance values at low concentrations. As such, they can be detected at lower concentrations than compounds that do not absorb strongly.