Question

Suppose that an HPLC column produces Gaussian peaks. The detector measures absorbance at 254 nm. A sample containing equal moles of compounds A and B was injected into the column. Compound A ${\mathbf{E}}_{\mathbf{254}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{26}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}{\mathbf{M}}^{\mathbf{-}\mathbf{1}}{\mathbf{cm}}^{\mathbf{-}\mathbf{1}}$has a height h=128mm and a half-width ${\mathit{w}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{1}$mm. Compound B ${\mathbf{E}}_{\mathbf{254}}\mathbf{=}\mathbf{168}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}{\mathbf{M}}^{\mathbf{-}\mathbf{1}}{\mathbf{cm}}^{\mathbf{-}\mathbf{1}}$has ${\mathit{w}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{6}$mm. What is the height of peak B in millimeters?

Short answer

The height of peak B in millimeters${\mathrm{h}}_{\mathrm{B}}=126.45\mathrm{mm}$

Step-by-step solution

Step 1:

Since the sample containing equal moles of compounds A and B, the peak Ares are proportional to molar absorptivity:

$$\begin{gathered} \frac{\mathrm{Peak}\mathrm{A}}{\mathrm{peak}\mathrm{B}} \\ =\frac{2.26.{10}^4}{1.68.{10}^4} \\ =\frac{1.064.h_A.w_{1/2}}{1.064.h_B.w_{1/2}} \end{gathered}$$

Step 2:

The height of peak B is:

$$\begin{gathered} h_B=\frac{1.064.h_A.w_{1/2}.1.68.{10}^4}{1.064.w_{1/2}.2.26.{10}^4} \\ {h}_B=\frac{1.064.128\mathrm{mm}.10.1\mathrm{mm}.1.68.{10}^4}{1.064.76\mathrm{mm}.2.26.{10}^4} \\ {h}_B=126.45\mathrm{mm} \\ \mathrm{Here},\mathrm{the}\mathrm{final}\mathrm{result}\mathrm{is} \\ {\mathrm{h}}_{\mathrm{B}}=126.45\mathrm{mm} \end{gathered}$$