(a) If retention times are 1.0 min for CH4, 12.0 min for octane, 13.0 min for unknown, and 15.0 min for nonane, find the Kovats retention index for the unknown. (b) What would be the Kovats index for the unknown if the phase ratio of the column were doubled? (c) What would be the Kovats index for the unknown if the length of the column were halved

Find: Kovats retention index (I) for the unknown

In this problem, use the formula for Kovats retention index (I), as shown below:

$\mathrm{l}=100\left[\mathrm{n}+\left(\mathrm{N}-\mathrm{n}\right)\frac{{{\log }_{\mathrm{r}}}^{\text{'}}\left(\mathrm{unknown}\right)-{{\log }_{\mathrm{r}}}^{\text{'}}\left(\mathrm{n}\right)}{{{\log }_{\mathrm{r}}}^{\text{'}}\left(\mathrm{N}\right)-{{\log }_{\mathrm{r}}}^{\text{'}}\left(\mathrm{n}\right)}\right]$

$$\begin{gathered} -\mathrm{n}=\mathrm{no}.\mathrm{of}\mathrm{carbon}\mathrm{atoms}\mathrm{in}\mathrm{the}\mathrm{smaller}\mathrm{alkane} \\ -\mathrm{N}=\mathrm{no}.\mathrm{of}\mathrm{carbon}\mathrm{atoms}\mathrm{in}\mathrm{the}\mathrm{larger}\mathrm{alkane} \\ -{{\mathrm{t}}_{\mathrm{r}}}^{\text{'}}\left(\mathrm{n}\right)=\mathrm{adjusted}\mathrm{retention}\mathrm{time}\mathrm{of}\mathrm{the}\mathrm{smaller}\mathrm{alkane} \\ -{{\mathrm{t}}^{\text{'}}}_{\mathrm{r}}\left(\mathrm{N}\right)=\mathrm{adjusted}\mathrm{retention}\mathrm{time}\mathrm{of}\mathrm{the}\mathrm{larger}\mathrm{alkane} \end{gathered}$$

The smaller alkane is octane with 8 carbon atoms while the larger alkane is nonane with 9 carbon atoms.

Compute for the adjusted retention time compound using the formula below:

${{\mathrm{t}}^{\text{'}}}_{\mathrm{r}}={\mathrm{t}}_{\mathrm{r}}-{\mathrm{t}}_{\mathrm{m}}$

Retention time for octane:

$\mathrm{t}{\text{'}}_{\mathrm{r}}\left(\mathrm{n}\right)=12.0-1.0=11.0\min$

Retention time for nonane:

${{\mathrm{t}}^{\text{'}}}_{\mathrm{r}}\left(\mathrm{N}\right)=15.0-1.0=14.0\min$

Retention time for unknown:

${{\mathrm{t}}^{\text{'}}}_{\mathrm{r}}\left(\mathrm{unknown}\right)=13.0-1.0=12.0\min$

Compute for the Kovats retention index I:

$$\begin{gathered} \mathrm{l}=100\left[8+\left(9-8\right)\frac{\log \left(12.0\right)-\log \left(11.0\right)}{\log (14.0)-\log (11.0)}\right] \\ =836.08 \end{gathered}$$

$\mathrm{New}\mathrm{phase}\mathrm{ratio}{\mathrm{\beta }}^{\text{'}}=2\mathrm{\beta }$

Thus, the new retention factor will become:

${\mathrm{k}}^{\text{'}}=\frac{\mathrm{K}}{{\mathrm{\beta }}^{\text{'}}}=\frac{\mathrm{K}}{2\mathrm{\beta }}=\frac{1}{2}\mathrm{k}$

The retention factor k in terms of the retention time is expressed as:

$$\begin{gathered} \mathrm{k}=\frac{{\mathrm{t}}_{\mathrm{r}}-{\mathrm{t}}_{\mathrm{m}}}{{\mathrm{t}}_{\mathrm{m}}}=\frac{{{\mathrm{t}}_{{\mathrm{r}}_{}}}^{\text{'}}}{{\mathrm{t}}_{\mathrm{m}}} \\ {{\mathrm{t}}^{\text{'}}}_{\mathrm{r}}={\mathrm{kt}}_{\mathrm{m}} \end{gathered}$$

The new retention factor is:

${\mathrm{k}}^{\text{'}}=\frac{{{\mathrm{t}}^{\text{'}}}_{\mathrm{r}}\left(\mathrm{new}\right)}{{\mathrm{t}}_{\mathrm{m}}}=\frac{1}{2}\mathrm{k}$

$$\begin{gathered} \mathrm{The}\mathrm{new}\mathrm{adjusted}\mathrm{retention}\mathrm{time}{{\mathrm{t}}^{\text{'}}}_{\mathrm{r}\left(\mathrm{new}\right)}\mathrm{is}: \\ {{\mathrm{t}}^{\text{'}}}_{\mathrm{r}\left(\mathrm{new}\right)}=\frac{1}{2}{\mathrm{kt}}_{\mathrm{m}}=\frac{1}{2}{{\mathrm{t}}_{\mathrm{r}}}^{\text{'}} \end{gathered}$$

Thus, values for the adjusted retention time of the compounds will be halved.

Compute for the Kovats retention index I:

$\mathrm{l}=100\left[8+(9-8)\frac{\log {\displaystyle \frac{12.0}{2}}-\log \left[{\displaystyle \frac{11.0}{2}}\right]}{\log \left({\displaystyle \frac{14.0}{2}}\right)-\log \left({\displaystyle \frac{11.0}{2}}\right)}\right]=836.08$

No change in Kovats retention index.

The length of the column will not affect the retention time; thus, the retention index of the unknown will not change.

I=836.08

a) I=836.08

b) I=836.08, no change

c) No change