(a) When a solution containing$\mathbf{234}\mathit{m}\mathit{g}$ of pentanol (FM 88.15) and$\mathbf{237}\mathbf{mg}$ of 2,3 -dimethyl-2-butanol (FM 102.17) in$\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{ml}$ was separated, relative peak areas were pentanol: 2,3 -dimethyl-2-butanol = 0.913 : 1.00. Considering pentanol to be the internal standard, find the response factor for 2,3 -dimethyl-2-butanol.

(b) Use Equation 24-8 to find the areas for pentanol and 2,3 -dimethyl-2-butanol in Figure 24-8.

(c) The concentration of pentanol internal standard in the unknown solution was$\mathbf{93}\mathbf{.}\mathbf{7}\mathbf{mM}$ . What was the concentration of 2,3 -dimethyl2-butanol?

• A molecule with two methyl groups often used in combination.
• In C4H9OH is one of two combustible isomers generated from straight-chain butane.

(a)

It is necessary to compute the response factor for 2,3-dimethyl-2-butanol.

The formula for quantitative analysis with internal standard can be found here,

$\frac{{\mathrm{A}}_{\mathrm{x}}}{\left[\mathrm{X}\right]}=\mathrm{F}\frac{{\mathrm{A}}_{\mathrm{s}}}{\left[\mathrm{S}\right]}$

Here localid="1654846542109" ${\mathrm{A}}_{\mathrm{X}}=$ signal region of analyte

$$\begin{gathered} {\mathrm{A}}_{\mathrm{s}}=\mathrm{area}\mathrm{of}\mathrm{internal}\mathrm{standard} \\ \left[\mathrm{X}\right]=\mathrm{analyte}\text{'}\mathrm{s}\mathrm{concentration} \\ \left[\mathrm{S}\right]=\mathrm{concentration}\mathrm{of}\mathrm{internal}\mathrm{standard} \\ \mathrm{F}=\mathrm{response}\mathrm{factor} \\ \mathrm{The}\mathrm{response}\mathrm{factor}\mathrm{for}2,3-\mathrm{dimethly}-2\mathrm{butanol}\mathrm{was}\mathrm{calculated}. \\ \mathrm{Given},\mathrm{Mass}\mathrm{of}\mathrm{Pentanol}=234\mathrm{mg} \\ \mathrm{Mass}\mathrm{of}2,3-\mathrm{dimethyl}-2-\mathrm{butanol}=237\mathrm{mg} \\ \mathrm{Volume}\mathrm{of}\mathrm{solution}=10.0\mathrm{ml} \\ \mathrm{Internal}\mathrm{standard}\mathrm{concentration}\left(\mathrm{pentanol}\right)\left(\mathrm{S}\right)=\frac{234\mathrm{mg}/88.15{\mathrm{gmol}}^{-1}}{10.0\mathrm{ml}} \end{gathered}$$

Internal standard concentration (S)=0.2655m

An alyte's concentration

$$\begin{gathered} \left(2,3-\mathrm{dimethyl}-2-\mathrm{butanol}\right)\left[X\right]=\frac{237\mathrm{MG}/102.17{\mathrm{gmol}}^{-1}}{10.0\mathrm{ml}} \\ \mathrm{Analyte}\mathrm{concentration}\left[\mathrm{X}\right]=0.2320\mathrm{M} \\ \mathrm{The}\mathrm{response}\mathrm{factor}\mathrm{is}\mathrm{determined}\mathrm{as}\mathrm{follows}: \\ \frac{{\mathrm{A}}_{\mathrm{x}}}{\left[\mathrm{X}\right]}=\mathrm{F}\frac{{\mathrm{A}}_{\mathrm{s}}}{\left[\mathrm{S}\right]}\frac{1.00}{0.2320\mathrm{M}} \\ =\mathrm{F}\left(\frac{0.913}{0.265\mathrm{M}}\right) \\ \mathrm{F}=1.253 \\ 2,3-\mathrm{dimethyl}-2-\mathrm{butanol}\mathrm{has}\mathrm{a}\mathrm{response}\mathrm{factor}\mathrm{of}1.253. \end{gathered}$$

Analyte concentration $\left[X\right]$

(b)

It is necessary to compute the areas of Pentanol and 2,3-dimethyl-2-butanol.

Peaks' area:

The formula below can be used to compute the area of a Gaussian peak.

Gaussian peak area $=1.064\times \mathrm{peak}\mathrm{height}\times {\mathrm{w}}_{1/2}$

$$\begin{gathered} \mathrm{here}{\mathrm{w}}_{1/2}=\mathrm{width}\mathrm{at}\mathrm{half}\mathrm{height} \\ \mathrm{pentanol}\mathrm{and}2,3-\mathrm{dimethyl}-2\mathrm{butanol}\mathrm{areas}\mathrm{were}\mathrm{calcuated}. \\ {\mathrm{pentanol}}^{\text{'}}\mathrm{s}\mathrm{territory}\mathrm{is}158{\mathrm{mm}}^2 \\ 2,3-\mathrm{dimethyl}-2\mathrm{butanol}\mathrm{has}\mathrm{a}\mathrm{surface}\mathrm{area}\mathrm{of}164{\mathrm{mm}}^2. \\ \mathrm{The}\mathrm{size}\mathrm{of}\mathrm{the}\mathrm{compounds}\mathrm{is}\mathrm{calculated}\mathrm{by}\mathrm{measuring}\mathrm{the}\mathrm{height}\mathrm{and}\mathrm{width}\mathrm{of} \\ \mathrm{the}\mathrm{buildings}{\mathrm{w}}_{1/2}\mathrm{in}\mathrm{millimetres}. \\ \mathrm{For}\mathrm{Pentanol},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{relative}\mathrm{peaks}\mathrm{is}\mathrm{computed}\mathrm{as}\mathrm{follows}: \\ \mathrm{Height}=40.1\mathrm{mm} \\ {\mathrm{w}}_{1/2}=3.7\mathrm{mm} \\ \mathrm{Area}=1.064\times \mathrm{peak}\mathrm{height}\times {\mathrm{w}}_{1/2} \\ \mathrm{Area}=1.064\times 4040.1\mathrm{mm}\times 3.7\mathrm{mm} \\ \mathrm{Area}=158{\mathrm{mm}}^2 \\ 2-\mathrm{dimethyl}-2\mathrm{butanol}(2,3-\mathrm{dimethy}-2-\mathrm{butanol}) \\ \mathrm{Give}: \\ =77.0\mathrm{mm} \\ \mathrm{Height}{\mathrm{w}}_{1/2}=2.0\mathrm{mm} \\ \mathrm{Area}=1.064\times \mathrm{peakheight}\times {\mathrm{w}}_{1/2} \\ \mathrm{Area}=1.064\times 77.0\mathrm{mm}\times 2.0\mathrm{mm} \\ \mathrm{Area}=164{\mathrm{mm}}^2 \\ \mathrm{The}\mathrm{area}\mathrm{of}2,3-\mathrm{dimethly}-2\mathrm{butanol}=164{\mathrm{mm}}^2 \end{gathered}$$

(c)

Given,

2,3-dimethyl-2-butanol has a large surface area$=164{\mathrm{mm}}^2$

Factor of response =1.253

Pentanol concentration = 93.7 mM

Area of Pentanol$=158{\mathrm{mm}}^2$

Let [X] be the 2,3-dimethyl-2-butanol concentration

$$\begin{gathered} \frac{{\mathrm{A}}_{\mathrm{x}}}{\left[\mathrm{X}\right]}=\mathrm{F}\frac{{\mathrm{A}}_{\mathrm{S}}}{\left[\mathrm{S}\right]} \\ \frac{164}{\left[\mathrm{X}\right]}=1.253\left(\frac{158}{93.7\mathrm{mM}}\right) \\ \left[\mathrm{X}\right]=77.6\mathrm{mM} \\ \mathrm{The}\mathrm{amount}\mathrm{of}2,3-\mathrm{dimethyl}-2-\mathrm{butanol}\mathrm{present}=77.6\mathrm{mM} \end{gathered}$$