Question

A known mixture of compounds A and B gave the following HPLC results:

A solution was prepared by mixing $\mathbf{12}\mathbf{.}\mathbf{49}\mathit{m}\mathit{g}$of $\mathit{B}$plus $\mathbf{10}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{mL}$of unknown containing just and diluting to $\mathbf{25}\mathbf{.}\mathbf{00}\mathbf{mL}$. Peak areas of $\mathbf{5}\mathbf{.}\mathbf{97}$and $\mathbf{6}\mathbf{.}\mathbf{38}$were observed for $\mathit{A}$and$\mathit{B}$, respectively. Find the concentration of $\mathbf{A}\mathbf{(}\mathbf{mg}\mathbf{/}\mathbf{mL}\mathbf{)}$in the unknown.

Short answer

The concentration of $\mathrm{A}$is $0.418\mathrm{mg}/\mathrm{mL}.$.

Step-by-step solution

Definition of compound.

A compound is a substance made up of two or more chemical components that have been chemically bonded together. Covalent bonds and ionic bonds are two typical forms of bonds that hold components in a compound together. In any compound, the components are always present in set ratios.

The concentration of A(mg/mL) in the unknown.

$$\begin{gathered} \mathrm{Peak}\mathrm{area}\mathrm{of}A\mathrm{is}5.97. \\ \mathrm{Peak}\mathrm{area}\mathrm{of}B\mathrm{is}6.38. \\ \frac{\mathrm{Area}{}_{\mathrm{A}}}{\left[\mathrm{A}\right]}=\mathrm{F}\left(\frac{{\mathrm{Area}}_{\mathrm{B}}}{\left|\mathrm{B}\right|}\right) \\ \frac{10.86}{\left|1.03\right|}=\mathrm{F}\left(\frac{4.37}{\left[1,16\right]}\right) \\ F\mathit{=}\frac{\mathit{10}\mathit{.}\mathit{86}}{\left|1.03\right|}\mathit{\times }\mathit{}\frac{\mathit{|}\mathit{1}\mathit{.}\mathit{16}\mathit{]}}{\mathit{4}\mathit{.}\mathit{37}} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\frac{12.5976}{4.5011} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\mathit{2}\mathit{.}\mathit{789} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\mathit{2}\mathit{.}\mathit{79} \\ \mathit{25}\mathit{.}\mathit{00}\mathit{}mL\mathit{}i\mathrm{s}\mathrm{the}\mathrm{total}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{unknown}\mathrm{solution}. \\ \mathrm{The}\mathrm{additional}\mathrm{weight}\mathrm{of}B\mathrm{is}12.49\mathit{}g\mathit{.} \\ \mathrm{The}\mathrm{total}\mathrm{volume}\mathrm{of}A\mathrm{that}\mathrm{has}\mathrm{been}\mathrm{added}\mathrm{is}10\mathit{}mL\mathit{.} \\ As\mathit{}demonstrated\mathit{}below\mathit{,}\mathit{}the\mathit{}concentration\mathit{}of\mathit{}B\mathit{}in\mathit{}the\mathit{}unknown\mathit{}solution\mathit{}may\mathit{}be\mathit{}determined\mathit{.} \end{gathered}$$

$12.49\mathrm{g}/25.00\mathrm{ml}=0.4996\mathrm{g}/\mathrm{mL}$

$$\begin{gathered} \mathrm{The}\mathrm{concentration}\mathrm{of}\mathrm{is}\mathrm{represented}\mathrm{below}. \\ \frac{{\mathrm{Arca}}_{\mathrm{A}}}{\left|\mathrm{A}\right|}=\mathrm{F}\left(\frac{{\mathrm{Arca}}_{\mathrm{B}}}{\left|\mathrm{B}\right|}\right) \\ \frac{5.97}{\left[\mathrm{A}\right]}=2.79\left(\frac{6.38}{\left|0.4996\right|}\right) \\ \left[\mathrm{A}\right]=\frac{5.97}{2.79}\times \frac{10.4996]}{6.38} \\ =\frac{2.982612}{17.8002} \\ =0.1675\mathrm{g}/\mathrm{mL} \\ \mathrm{The}\mathrm{concentration}\mathrm{of}A\mathrm{in}\mathrm{the}\mathrm{initial}\mathrm{unknown}\mathrm{solution}\mathrm{is}\mathrm{determined}\mathrm{to}\mathrm{be}\mathrm{as}\mathrm{follows}: \\ A\mathit{}\mathrm{in}\mathrm{original}\mathrm{unknown} \\ =\frac{25.00}{10.00}(0.167\mathrm{mg}/\mathrm{mL}) \\ =0.418\mathrm{mg}/\mathrm{mL} \\ \mathrm{Therefore},\mathrm{the}\mathrm{concentration}\mathrm{of}A\mathrm{in}\mathrm{the}\mathrm{unknown}\mathrm{is}\mathrm{found}\mathrm{as}0.418\mathrm{mg}/\mathrm{mL}. \end{gathered}$$