Question

Heptane, decane, and an unknown had adjusted retention times of $12.6\min$(heptane),$22.9\min$(decane), and$20.0\min$(unknown). The retention indexes for heptane and decane are 700 and 1000 , respectively. Find the retention index for the unknown.

Short answer

The unknown has a retention index of 932.

Step-by-step solution

Definition of retention index

• 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.

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

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}$$