# How to relate the transmittance to the absorbance, concentration, and molar absorptivity?

## Apr 26, 2017

The absorbance $A$ can be expressed as function of the transmittance $T$:

$\boldsymbol{A = - \log T}$

a) The absorbance is therefore:

$\textcolor{b l u e}{A} = - \log \left(0.762\right) = \textcolor{b l u e}{0.118}$

What would the absorbance be if the percent transmittance was 100%? What does it physically represent to have a percent transmittance of 100%? (Highlight the sentence below to check.)

$\textcolor{w h i t e}{\text{The substance absorbs no incoming light at all.}}$

b) Given the concentration, we would write another relation of the absorbance, as a function of concentration, called Beer's Law:

$\boldsymbol{A = \epsilon b c}$,

where $\epsilon$ is the molar absorptivity in $\text{L/mol"cdot"cm}$, $b$ is the path length of the cuvette (usually $\text{1 cm}$), and $c$ is the concentration in $\text{mol/L}$. In this case, $b = \text{1.5 cm}$!

Therefore, the molar absorptivity is:

$\textcolor{b l u e}{\epsilon} = \frac{A}{b c}$

$= \frac{0.118}{\text{1.5 cm" cdot "0.0802 M}}$

$= \textcolor{b l u e}{\text{0.981 L/mol"cdot"cm}}$

This makes sense, because if the absorbance is low, the substance reflects most of the light that approaches it. Since absorbed light is not seen, the higher the absorbance, the darker the substance.

The darker the substance, the higher its $\epsilon$. Since this substance is light (not dark), its $\epsilon$ is low.

c) Now we're supposing that $\epsilon = \text{10000 L/mol"cdot"cm}$. Then the concentration with the same absorbance must be very low, since it is expected to absorb a lot of light, but it absorbs only a little since $A$ is low.

Let's see if that's true.

$\textcolor{b l u e}{c} = \frac{A}{\epsilon b}$

$= \frac{0.118}{\text{10000 L/mol"cdot"cm"cdot"1.5 cm}}$

$= \textcolor{b l u e}{7.87 \times {10}^{- 6} \text{M}}$

Indeed, the concentration is very low.