91Ó°ÊÓ

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

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

where,

$\left[{\mathrm{A}}_{\mathrm{x}}\right]$signal from analyte

$\left[\mathrm{X}\right]$concentration of analyte

$\mathrm{F}$response factor

${\mathrm{A}}_{\mathrm{s}}$signal from standard

$\left[\mathrm{S}\right]$Concentration of standard

A solution was prepared by mixing $5.00\mathrm{ML}$of unknown element $\mathrm{X}$with $2.00\mathrm{mL}$of solution.

It has $4.13\mathrm{μ²\mu }$of standard element $\mathrm{S}/$millilitre and it is diluted to $10.0\mathrm{mL}$

Using standard mixture to determine the response factor.

We knew that $\left[\mathrm{X}\right]=\left[\mathrm{S}\right]$and the ratio of signals ${\mathrm{A}}_{\mathrm{x}}/{\mathrm{A}}_{\mathrm{s}}\mathrm{is}1.31$

Mixture of unknown plus standard, the concentration of $\mathrm{S}$is

$\left[\mathrm{S}\right]=\left(4.13\mathrm{μ²\mu }/\mathrm{mL}\right)\left(\frac{2.00}{10.0}\right)=0.826\mathrm{μ²\mu }/\mathrm{mL}$

For the unknown mixture $\mathrm{F}=\frac{{\mathrm{A}}_{\mathrm{x}}/{\mathrm{A}}_{\mathrm{s}}}{\left[\mathrm{X}\right]/\left[\mathrm{S}\right]}$

role="math" localid="1654931185928" $$\begin{gathered} 1.31=\frac{0.808}{\left[\mathrm{X}\right]/\left[0.826\right]} \\ \left[\mathrm{X}\right]=0.509\mathrm{μ²\mu }/\mathrm{mL} \end{gathered}$$

The concentration of $\mathrm{X}$diluted from 5.00 to $1.00\mathrm{mL}$in the mixture with $\mathrm{S}$,

The original concentration of$\mathrm{X}=\left(\frac{1.0}{5.0}\right)\left(0.509\mathrm{μ²\mu }/\mathrm{mL}\right)=1.02\mathrm{μ²\mu }/\mathrm{mL}$