91影视

• The retention index of a sample component is a number calculated by interpolation (usually logarithmic).
• And relating the sample component's adjusted retention volume (time) or retention factor to the adjusted retention volumes (times) of two standards eluted before and after the sample component's peak.

It is necessary to compute the unknown's retention index.

Retention index of Kovats:

For linear alkanes, the retention index, I, equals 100 times the number of carbon atoms.

I=800 for an Octane molecule.

For a Nonane molecule,$\mathrm{l}=900$

Using the calculation, the retention index of a compound eluted between Octane and Nonane will be between 800 and 900,

$$\begin{gathered} \mathrm{l}=100\left[\mathrm{n}+\left(\mathrm{N}-\mathrm{n}\right)\frac{{{\mathrm{logt}}_{\mathrm{r}}}^{\text{'}}\left(\mathrm{unknown}\right)-{{\mathrm{logt}}_{\mathrm{r}}}^{\text{'}}\left(\mathrm{n}\right)}{{{\mathrm{logt}}_{\mathrm{r}}}^{\text{'}}\left(\mathrm{N}\right)-{{\mathrm{logt}}_{\mathrm{r}}}^{\text{'}}\left(\mathrm{n}\right)}\right] \\ \mathrm{The}\mathrm{number}\mathrm{of}\mathrm{carbon}\mathrm{atoms}\mathrm{in}\mathrm{a}\mathrm{smaller}\mathrm{alkane}\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{n}. \\ \mathrm{N}\mathrm{represents}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{carbon}\mathrm{atoms}\mathrm{in}\mathrm{a}\mathrm{bigger}\mathrm{alkane}. \\ {\mathrm{t}}_{\mathrm{r}}\left(\mathrm{n}\right)=\mathrm{smaller}\mathrm{alkane}\mathrm{retention}\mathrm{time}\mathrm{has}\mathrm{been}\mathrm{altered} \\ {{\mathrm{t}}^{\text{'}}}_{\mathrm{r}}\left(\mathrm{N}\right)=\mathrm{Alkane}\mathrm{retention}\mathrm{time}\mathrm{has}\mathrm{been}\mathrm{altered}\mathrm{for}\mathrm{larger}\mathrm{alkanes}. \\ \mathrm{Calculate}\mathrm{the}\mathrm{unknown}\text{'}\mathrm{s}\mathrm{retention}\mathrm{index}. \\ \mathrm{Given}, \\ \mathrm{Heptane}\mathrm{retention}\mathrm{time}\mathrm{was}\mathrm{adjusted}\mathrm{to}12.6\mathrm{minutes}. \\ \mathrm{Decane}\mathrm{retention}\mathrm{time}\mathrm{has}\mathrm{been}\mathrm{adjusted}\mathrm{to}22\mathrm{minutes}. \\ \mathrm{Unknown}\mathrm{retention}\mathrm{time}\mathrm{was}\mathrm{adjusted}\mathrm{to}20.0\mathrm{minutes}. \\ \mathrm{Heptane}\mathrm{retention}\mathrm{index}=700\mathrm{Decane}\text{'}\mathrm{s}\mathrm{retention}\mathrm{index}\mathrm{is}1000. \\ \mathrm{The}\mathrm{unknown}\text{'}\mathrm{s}\mathrm{retention}\mathrm{index}\mathrm{is}\mathrm{determined}\mathrm{as}, \\ \mathrm{l}=100\left[\mathrm{n}+\left(\mathrm{N}-\mathrm{n}\right)\frac{{{\mathrm{logt}}_{\mathrm{r}}}^{\text{'}}\left(\mathrm{unknown}\right)-{{\mathrm{logt}}_{\mathrm{r}}}^{\text{'}}\left(\mathrm{n}\right)}{{{\mathrm{logt}}_{\mathrm{r}}}^{\text{'}}\left(\mathrm{N}\right)-{{\mathrm{logt}}_{\mathrm{r}}}^{\text{'}}\left(\mathrm{n}\right)}\right] \\ \mathrm{l}=100\left[7+(10-7)\frac{\log (20.0)-\log (12.6)}{\log (22.9)-\log (12.6) \\ }\right] \\ \mathrm{l}=100\left[7+\left(3\right)\frac{1.3010-1.1003}{1.3598-1.1003}\right] \\ \mathrm{l}=100\left[7+\left(3\right)\frac{0.2007}{0.2595}\right] \\ \mathrm{l}=932 \\ \mathrm{The}\mathrm{unknown}\text{'}\mathrm{s}\mathrm{retention}\mathrm{index}\mathrm{is}932. \\ \mathrm{The}\mathrm{unknown}\text{'}\mathrm{s}\mathrm{retention}\mathrm{index}\mathrm{was}\mathrm{calculated}\mathrm{and}\mathrm{determined}\mathrm{to}\mathrm{be}932. \end{gathered}$$