Infrared spectra are customarily recorded on a transmittance scale so that weak and strong bands can be displayed on the same scale. The region near $\mathbf{2000}{\mathbf{cm}}^{\mathbf{-}\mathbf{1}}$in the infrared spectra of compounds A and B is shown in the figure. Note that absorption corresponds to a downward peak on this scale. The spectra were recorded from a 0.0100M solution of each, in cells with 0.00500 - cm path lengths. A mixture of A and B in a 0.00500 - cm cell gave a transmittance 34.0 % of at2022cm and 383% at 1093 cm. Find the concentrations of A and B.

Beer's law states that through the sample and the concentration of the absorbing species, the absorbance is proportional to the path length.

$\mathrm{A}=\mathrm{\epsilon bC}$

is the absorbance,$\epsilon$ is the molar absorptivity, b is the length of light path, C is the concentration

Find the absorbance atfor,

$\begin{array}{r}{\mathrm{A}}_{\mathrm{A}}=-\log \left(\mathrm{T}\right)=-\log (0.310)=0.50\\ \mathrm{Use}\mathrm{Beer}’\mathrm{s}\mathrm{law}\mathrm{to}{\mathrm{calculate\epsilon }}_{\mathrm{A}},\\ {\dot{\mathrm{\partial }}}_{\mathrm{A}}=\frac{\mathrm{A}}{\mathrm{b}\cdot \mathrm{c}}=\frac{0.5086}{0.005\mathrm{cm}\cdot 0.01\mathrm{M}}\\ {\dot{\mathrm{\partial }}}_{\mathrm{A}}=10172{\mathrm{M}}^{-1}{\mathrm{cm}}^{-1}\end{array}$

Thus, absorbance for B,

${\mathrm{A}}_{\mathrm{B}}=-\log \left(T\right)=-\log \left(0.974\right)=0$

Calculate:${\mathrm{\epsilon }}_{\mathrm{B}}$

$\begin{array}{r}{\dot{o}}_{\mathrm{B}}=\frac{\mathrm{A}}{\mathrm{b}\cdot \mathrm{c}}=\frac{0.0114}{0.005\mathrm{cm}\cdot 0.01\mathrm{M}}\\ {\dot{o}}_{\mathrm{B}}=228{\mathrm{M}}^{-1}{\mathrm{cm}}^{-1}\end{array}$

Evaluate the absorbance at 1993${\mathrm{cm}}^{-1}$ for A

$A_A=-\log \left(T\right)=-\log \left(0.797\right)=0.0985$

Calculate:$\epsilon {\text{'}}_A$

$\begin{array}{r}{\dot{o}}_{\mathrm{A}}^{\mathrm{\prime }}=\frac{\mathrm{A}}{\mathrm{b}\cdot \mathrm{c}}=\frac{0.0985}{0.005\mathrm{cm}\cdot 0.01\mathrm{M}}\\ {\dot{o}}_{\mathrm{A}}^{\mathrm{\prime }}=197{\mathrm{OM}}^{-1}{\mathrm{cm}}^{-1}\end{array}$

Hence, absorbance for B

$A_B=-\log \left(T\right)=-\log \left(0.20\right)=0.69897$

Calculate:

$\begin{array}{r}{\dot{\mathrm{\partial }}}_{\mathrm{B}}^{\mathrm{\prime }}=\frac{\mathrm{A}}{\mathrm{b}\cdot \mathrm{c}}=\frac{0.69897}{0.005\mathrm{cm}\cdot 0.01}\\ {\dot{\mathrm{\partial }}}^{\mathrm{\prime }}=13979.4{\mathrm{M}}^{-1}{\mathrm{cm}}^{-1}\end{array}$

Hence, the mixture absorbance at 2022 ${\mathrm{cm}}^{-1}$is,

$\begin{array}{r}{\mathrm{A}}_1=-\log \left(\mathrm{T}\right)=-\log (0.34)\\ {\mathrm{A}}_1=0.4685\end{array}$

Thus, the mixture absorbance at $1993{\mathrm{cm}}^{-1}$is,

$\begin{array}{r}{\mathrm{A}}_2=-\log \left(\mathrm{T}\right)=-\log \left(0.383\right)\\ {\mathrm{A}}_2=0.4168\end{array}$

Use the formula to find the concentration of

$\begin{array}{r}\left[\mathrm{A}\right]=\frac{\left|\begin{array}{ll}{\mathrm{A}}_1& {\dot{o}}_{\mathrm{B}}\cdot \mathrm{b}\\ {\mathrm{A}}_2& {\dot{o}}_{\mathrm{B}}^{\mathrm{\prime }}\cdot \mathrm{b}\end{array}\right|}{\left|\begin{array}{cc}{\dot{o}}_{\mathrm{A}}\cdot \mathrm{b}& {\dot{o}}_{\mathrm{B}}\cdot \mathrm{b}\\ {\dot{o}}_{\mathrm{A}}^{\mathrm{\prime }}\cdot \mathrm{b}& {\dot{o}}_{\mathrm{B}}^{\mathrm{\prime }}\cdot \mathrm{b}\end{array}\right|}\\ \left[A\right]=\frac{\left|\begin{array}{cc}0.4685& 228\cdot 0.005\\ 0.4168& 13979.4\cdot 0.005\end{array}\right|}{\left|\begin{array}{cc}10172\cdot 0.005& 228\cdot 0.005\\ 1970\cdot 0.005& 13979.4\cdot 0.005\end{array}\right|}\end{array}$

$\begin{array}{r}\left[\mathrm{A}\right]=\frac{\left|\begin{array}{cc}0.4685& 1.14\\ 0.4168& 69.897\end{array}\right|}{\left|\begin{array}{cc}50.86& 1.14\\ 9.85& 69.897\end{array}\right|}\\ \left[\mathrm{A}\right]=\frac{(0.4685\cdot 69.897)-(1.14\cdot 0.4168)}{(50.86\cdot 69.897)-(1.14\cdot 9.85)}\\ \left[\mathrm{A}\right]=9.11\cdot {10}^{-3}\mathrm{M}\end{array}$

Evaluate the concentration for B,

$\begin{array}{r}\left[\mathrm{B}\right]=\frac{\left|\begin{array}{cc}{\dot{o}}_{\mathrm{A}}\cdot \mathrm{b}& {\mathrm{A}}_1\\ {\dot{o}}_{\mathrm{A}}^{\mathrm{\prime }}\cdot \mathrm{b}& {\mathrm{A}}_2\end{array}\right|}{\left|\begin{array}{ll}{\dot{o}}_{\mathrm{A}}\cdot \mathrm{b}& {\dot{o}}_{\mathrm{B}}\cdot \mathrm{b}\\ {\widehat{o}}_{\mathrm{A}}^{\mathrm{\prime }}\cdot \mathrm{b}& {\dot{o}}_{\mathrm{B}}^{\mathrm{\prime }}\cdot \mathrm{b}\end{array}\right|}\\ \left[B\right]=\frac{\left|\begin{array}{cc}10172\cdot 0.005& 0.4685\\ 1970\cdot 0.005& 0.4168\end{array}\right|}{\left|\begin{array}{cc}10172\cdot 0.005& 228\cdot 0.005\\ 1970\cdot 0.005& 13979.4\cdot 0.005\end{array}\right|}\\ \left[B\right]=\frac{\left|\begin{array}{cc}50.86& 0.4685\\ 9.85& 0.4168\end{array}\right|}{\left|\begin{array}{cc}50.86& 1.14\\ 9.85& 69.897\end{array}\right|}\\ \left[B\right]=\frac{(50.86\cdot 0.4168)-(9.85\cdot 0.4685)}{(50.86\cdot 69.897)-(1.14\cdot 9.85)}\\ \left[B\right]=4.68\cdot {10}^{-3}M\end{array}$

Therefore, the concentrations of A and B are $\left[\mathrm{A}\right]=9.11\cdot {10}^{-3}\mathrm{M}\text{and}\left[\mathrm{B}\right]=4.68\cdot {10}^{-3}\mathrm{M}$.