# The transmittance for a sample of a solution, measured at 590nm in a 1.5-cm cuvette, was 76.2%. What is the corresponding absorbance?

Sep 9, 2017

Absorbance as it relates to transmittance is given by a logarithmic relationship, just as loudness is related to sound intensity in that manner.

$A = \log \left(\frac{1}{T}\right)$

where $T$ is the transmittance. Why must this be $\frac{1}{T}$ since $0 < T < 1$?

So we know the absorbance to be:

$\textcolor{b l u e}{A} = \log \left(\frac{1}{0.762}\right) = \textcolor{b l u e}{0.118}$

Knowing the concentration, we can use Beer's law at low concentrations to see a linear relationship:

$A = \epsilon b c$

where:

• $\epsilon$ is the molar absorptivity in "L"/("mol"cdot"cm").
• $b$ is the path length, standardized to be $\text{1 cm}$ in many labs.
• $c$ is the concentration in $\text{mol/L}$.

Thus, the molar absorptivity would be:

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

$= \frac{0.118}{\text{1 cm" cdot "0.0802 mol/L}}$

$= \textcolor{b l u e}{\text{1.472 L/mol"cdot"cm}}$

which is fairly low.

It makes physical sense because the absorbance was quite low, even though the concentration was also low.

If the absorptivity was already known to be $\text{1000 L/mol"cdot"cm}$, one could find the concentration to be:

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

$= \frac{0.118}{\text{1000 L/mol"cdot"cm" cdot "1 cm}}$

$= \textcolor{b l u e}{1.181 \times {10}^{- 4} \text{M}}$

If the molar absorptivity was this high (which is quite large), and the absorbance was still so low, it would only be because there was a low concentration of particles to absorb the incoming excitation light source.