Chapter 5, Problem 5.30P

(a)

Interpretation Introduction

Interpretation:

The response factor for the analyte has to be calculated.

Concept Introduction:

Response factor,

In chromatography, a response factor is well-defined as the proportion between the concentration of a compound being analyzed and the response of the detector to that compound.

$\frac{\text{signal from analyte}}{\text{concentration}\text{of}\text{analyte}}\text{=F}\left(\frac{\text{signal from standard}}{\text{concentration}\text{of}\text{standard}}\right)$

$\frac{{\text{A}}_{\text{X}}}{\left[\text{X}\right]}\text{=F}\left(\frac{{\text{A}}_{\text{S}}}{\left[\text{S}\right]}\right)$

$\begin{array}{l}\text{where,}\\ {\text{A}}_{\text{X}}\text{=signal}\text{from}\text{analyte}\\ \left[\text{X}\right]\text{=concentration}\text{of}\text{analyte}\\ \text{F=}\text{response}\text{factor}\\ {\text{A}}_{\text{S}}\text{=signal from}\text{standard}\\ \left[\text{S}\right]\text{=}\text{Concentration}\text{of}\text{standard}\end{array}$

Answer to Problem 5.30P

The response factor for the analyte is $\text{0}\text{.1684}$.

Explanation of Solution

Given,

In chromatographic analysis, a solution has an analyte concentration of $\text{3}\text{.47mM}$ and a standard concentration of $\text{1}\text{.72}\text{mM}$ gave peak areas $\text{3}\text{473}\text{and}\text{10}\text{222}$ respectively. Then addition of $\text{1}\text{.00mL}\text{of}\text{8}\text{.47mM}$ standard concentration was added to $\text{5}\text{.00}\text{mL}$ of unknown $\text{X}$ and then the mixture was diluted to $\text{10}\text{.0}\text{mL}$. The peaks gave form the solution are $\text{5}\text{428}\text{and}\text{4}\text{431}$ for X and S

$\frac{\text{signal from analyte}}{\text{concentration}\text{of}\text{analyte}}\text{=F}\left(\frac{\text{signal from standard}}{\text{concentration}\text{of}\text{standard}}\right)$

$\frac{{\text{A}}_{\text{X}}}{\left[\text{X}\right]}\text{=F}\left(\frac{{\text{A}}_{\text{S}}}{\left[\text{S}\right]}\right)$

$\begin{array}{l}\frac{\text{3}\text{473}}{\left[\text{3}\text{.47}\text{mM}\right]}\text{=F}\left(\frac{\text{10}\text{222}}{\left[\text{1}\text{.72}\text{mM}\right]}\right)\\ \text{F=}\frac{\text{3}\text{473}}{\left[\text{3}\text{.47}\text{mM}\right]}\text{×}\left(\frac{\left[\text{1}\text{.72}\text{mM}\right]}{\text{10}\text{222}}\right)\\ \text{F=0}\text{.1684}\end{array}$

Conclusion

The response factor for the analyte was calculated as $\text{0}\text{.1684}$.

(b)

Interpretation Introduction

Interpretation:

The concentration of $\text{S}\text{(mM)}$ in $\text{10}\text{.0}\text{mL}\text{mixture}$ has to be calculated.

Concept Introduction:

Response factor,

In chromatography, a response factor is well-defined as the proportion between the concentration of a compound being analyzed and the response of the detector to that compound.

$\frac{\text{signal from analyte}}{\text{concentration}\text{of}\text{analyte}}\text{=F}\left(\frac{\text{signal from standard}}{\text{concentration}\text{of}\text{standard}}\right)$

$\frac{{\text{A}}_{\text{X}}}{\left[\text{X}\right]}\text{=F}\left(\frac{{\text{A}}_{\text{S}}}{\left[\text{S}\right]}\right)$

$\begin{array}{l}\text{where,}\\ {\text{A}}_{\text{X}}\text{=signal}\text{from}\text{analyte}\\ \left[\text{X}\right]\text{=concentration}\text{of}\text{analyte}\\ \text{F=}\text{response}\text{factor}\\ {\text{A}}_{\text{S}}\text{=signal from}\text{standard}\\ \left[\text{S}\right]\text{=}\text{Concentration}\text{of}\text{standard}\end{array}$

The concentration of $\text{S}$,

$\left[\text{S}\right]\text{=initial}\text{concentration×dilution}\text{factor}$

Answer to Problem 5.30P

The concentration of $\text{S}\text{=}\text{0}\text{.847}\text{mM}$.

Explanation of Solution

Given,

In chromatographic analysis, a solution has an analyte concentration of $\text{3}\text{.47mM}$ and a standard concentration of $\text{1}\text{.72}\text{mM}$ gave peak areas $\text{3}\text{473}\text{and}\text{10}\text{222}$ respectively. Then addition of $\text{1}\text{.00mL}\text{of}\text{8}\text{.47mM}$ standard concentration was added to $\text{5}\text{.00}\text{mL}$ of unknown $\text{X}$ and then the mixture was diluted to $\text{10}\text{.0}\text{mL}$. The peaks gave form the solution are $\text{5}\text{428}\text{and}\text{4}\text{431}$ for X and S

$\left[\text{S}\right]\text{=initial}\text{concentration×dilution}\text{factor}$

$\left[\text{S}\right]\text{=}\left(\text{8}\text{.47mM}\right)\left(\frac{\text{1}\text{.00}\text{mL}}{\text{10}\text{.0}\text{mL}}\right)$

$\left[\text{S}\right]\text{=}\text{0}\text{.847}\text{mM}$

Conclusion

The concentration of $\text{S}\text{(mM)}$ in $\text{10}\text{.0}\text{mL}\text{mixture}$ was calculated as $\text{0}\text{.847}\text{mM}$.

(c)

Interpretation Introduction

Interpretation:

The concentration of $\text{X}\text{(mM)}$ in $\text{10}\text{.0}\text{mL}\text{mixture}$ has to be calculated.

Concept Introduction:

Response factor,

In chromatography, a response factor is well-defined as the proportion between the concentration of a compound being analyzed and the response of the detector to that compound.

$\frac{\text{signal from analyte}}{\text{concentration}\text{of}\text{analyte}}\text{=F}\left(\frac{\text{signal from standard}}{\text{concentration}\text{of}\text{standard}}\right)$

$\frac{{\text{A}}_{\text{X}}}{\left[\text{X}\right]}\text{=F}\left(\frac{{\text{A}}_{\text{S}}}{\left[\text{S}\right]}\right)$

$\begin{array}{l}\text{where,}\\ {\text{A}}_{\text{X}}\text{=signal}\text{from}\text{analyte}\\ \left[\text{X}\right]\text{=concentration}\text{of}\text{analyte}\\ \text{F=}\text{response}\text{factor}\\ {\text{A}}_{\text{S}}\text{=signal from}\text{standard}\\ \left[\text{S}\right]\text{=}\text{Concentration}\text{of}\text{standard}\end{array}$

Answer to Problem 5.30P

The concentration of $\left[\text{X}\right]\text{=}\text{6}\text{.16}\text{mM}$.

Explanation of Solution

Given,

In chromatographic analysis, a solution has an analyte concentration of $\text{3}\text{.47mM}$ and a standard concentration of $\text{1}\text{.72}\text{mM}$ gave peak areas $\text{3}\text{473}\text{and}\text{10}\text{222}$ respectively. Then addition of $\text{1}\text{.00mL}\text{of}\text{8}\text{.47mM}$ standard concentration was added to $\text{5}\text{.00}\text{mL}$ of unknown $\text{X}$ and then the mixture was diluted to $\text{10}\text{.0}\text{mL}$. The peaks gave form the solution are $\text{5}\text{428}\text{and}\text{4}\text{431}$ for X and S

$\frac{\text{signal from analyte}}{\text{concentration}\text{of}\text{analyte}}\text{=F}\left(\frac{\text{signal from standard}}{\text{concentration}\text{of}\text{standard}}\right)$

$\frac{{\text{A}}_{\text{X}}}{\left[\text{X}\right]}\text{=F}\left(\frac{{\text{A}}_{\text{S}}}{\left[\text{S}\right]}\right)$

$\begin{array}{l}\frac{\text{5}\text{428}}{\left[\text{X}\right]}\text{=0}\text{.1684}\left(\frac{\text{4}\text{431}}{\left[\text{0}\text{.847}\text{mM}\right]}\right)\\ \left[\text{X}\right]\text{=}\frac{\text{5}\text{428}}{\text{0}\text{.1684}}\text{×}\left(\frac{\left[\text{0}\text{.847}\text{mM}\right]}{\text{4}\text{431}}\right)\\ \left[\text{X}\right]\text{=6}\text{.16}\text{mM}\end{array}$

Conclusion

The concentration of $\text{X(mM)}$ in $\text{10}\text{.0}\text{mL}\text{mixture}$ was calculated as $\text{6}\text{.16}\text{mM}$.

(d)

Interpretation Introduction

Interpretation:

The concentration of $\text{X}\text{(mM)}$ in the original unknown has to be calculated.

Concept Introduction:

Response factor,

In chromatography, a response factor is well-defined as the proportion between the concentration of a compound being analyzed and the response of the detector to that compound.

$\frac{\text{signal from analyte}}{\text{concentration}\text{of}\text{analyte}}\text{=F}\left(\frac{\text{signal from standard}}{\text{concentration}\text{of}\text{standard}}\right)$

$\frac{{\text{A}}_{\text{X}}}{\left[\text{X}\right]}\text{=F}\left(\frac{{\text{A}}_{\text{S}}}{\left[\text{S}\right]}\right)$

$\begin{array}{l}\text{where,}\\ {\text{A}}_{\text{X}}\text{=signal}\text{from}\text{analyte}\\ \left[\text{X}\right]\text{=concentration}\text{of}\text{analyte}\\ \text{F=}\text{response}\text{factor}\\ {\text{A}}_{\text{S}}\text{=signal from}\text{standard}\\ \left[\text{S}\right]\text{=}\text{Concentration}\text{of}\text{standard}\end{array}$

Answer to Problem 5.30P

The original concentration of $\left[\text{X}\right]\text{in the unknown is 12}\text{.3}\text{mM}\text{.}$

Explanation of Solution

Given,

In chromatographic analysis, a solution has an analyte concentration of $\text{3}\text{.47mM}$ and a standard concentration of $\text{1}\text{.72}\text{mM}$ gave peak areas $\text{3}\text{473}\text{and}\text{10}\text{222}$ respectively. Then addition of $\text{1}\text{.00mL}\text{of}\text{8}\text{.47mM}$ standard concentration was added to $\text{5}\text{.00}\text{mL}$ of unknown $\text{X}$ and then the mixture was diluted to $\text{10}\text{.0}\text{mL}$. The peaks gave form the solution are $\text{5}\text{428}\text{and}\text{4}\text{431}$ for X and S

The original concentration of $\left[\text{X}\right]$ was two times greater than the diluted concentration

Diluted concentration is

$\begin{array}{l}\left[\text{X}\right]\text{=}\text{6}\text{.16}\text{mM}\\ \text{original concentration}\text{=}\text{2}\left[\text{X}\right]\text{=}\text{2}\left[\text{6}\text{.16}\text{mM}\right]\text{=12}\text{.3}\text{mM}\end{array}$

Conclusion

The original concentration of $\text{X(mM)}$ in the unknown was calculated as $\text{12}\text{.3 mM}$.