Calculate the concentration of the reactant from the absorbance of the product without molar absorptivity


> 
> Given the reaction $\ce{A->B}$, where the absorbance of the product $\ce{B}$ has been measured (see the table below), and the initial concentration of $\ce{A}$ is $\pu{10^{-4} M},$ calculate the concentration of the reactant at the time $t.$
> 
> 
> $$
> \begin{array}{rr}
> \hline
> t/\pu{min} & A \\
> \hline
> 0 & 0.000 \\
> 1 & 0.115 \\
> 2 & 0.188 \\
> 3 & 0.237 \\
> 4 & 0.273 \\
> 5 & 0.301 \\
> 6 & 0.321 \\
> 7 & 0.339 \\
> 8 & 0.353 \\
> 9 & 0.365 \\
> 10 & 0.375 \\
> \infty & 0.500 \\
> \hline
> \end{array}$$
> 
> 
> 


From the answer it is stated that one should use the following equation to calculate the concentration of $\ce{A}$:


$$[\ce{A}]\_t = [\ce{A}]\_0\left(1 – \frac{A\_t}{A\_\infty}\right)\label{eqn:1q}\tag{1}$$


I don't understand where this comes from since I have only ever converted absorbance to concentration using the equation:


$$A = \varepsilon lc\tag{2}$$


Could someone please explain the reasoning behind the equation \eqref{eqn:1q}?




Reactant $\ce{A}$ does not absorb the light, so the measured absorbance is only due to $\ce{B}$. Let $[\ce{B}]\_t$ and $A\_t$ be the measured values of $[\ce{B}]$ and $A$ at any time during the experiment:


$$A\_t = \varepsilon\_\ce{B} \cdot l \cdot [\ce{B}]\_t \label{eqn:1a} \tag{1}$$


At the end of the experiment $t = \infty$, $A\_\infty = 0.500$ and $[\ce{B}]\_\infty = [\ce{A}]\_0 = \pu{E-4 M}$. As a consequence


$$\varepsilon\_\ce{B}\cdot l = \frac{A\_\infty}{[\ce{B}]\_\infty} = \frac{A\_\infty}{[\ce{A}]\_0} \label{eqn:2a}\tag{2}$$


Introducing \eqref{eqn:2a} in \eqref{eqn:1a} yields


$$[\ce{B}]\_t = \frac{A\_t}{\varepsilon\_\ce{B}\cdot l} = [\ce{A}]\_0 \frac{A\_t}{A\_\infty} \tag{3}$$


$$
\begin{align}
[\ce{A}]\_t &= [\ce{A}]\_0 - [\ce{B}]\_t \\
 &= [\ce{A}]\_0 - [\ce{A}]\_0 \frac{A\_t}{A\_\infty} \\
 &= [\ce{A}]\_0\left(1 – \frac{A\_t}{A\_\infty}\right) \tag{4}
\end{align}
$$


And this is the wanted formula.


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