The relative retention for two compounds in gas chromatography is 1.068 on a column with a plate height of 0.520mm.The retention factor for compound 1 is 5.16.

(a) Find the retention factor for compound 2.

(b) What length of column will separate the compounds with a resolution of 1.50 ?

(c) The retention time for air (${\mathit{t}}_{\mathbf{m}}$) is 2.00 min. If the number of plates is the same for both compounds, Find${\mathbf{t}}_{\mathbf{r}}{\mathbf{andw}}_{\frac{\mathbf{1}}{\mathbf{2}}}$for each peak.

(d) If the volume ratio of stationary phase to mobile phase is 0.30,Find the partition coefficient for compound 1.

(a) We know that , Relative retention factor, $\mathrm{\alpha }=\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}$

Where $k_1$are$k_2$ the Retention factor for compound 1 and 2 respectively.

Or,${\mathrm{k}}_2={\mathrm{\alpha k}}_1$

Or, $k_2=1.068\times 5.16$

Or,$k_2=5.51$

(b) Resolution = $\frac{\sqrt{\mathrm{N}}}{4}\frac{\left(\mathrm{\alpha }-1\right)}{\mathrm{\alpha }}\left(\frac{{\mathrm{k}}_2}{1+{\mathrm{k}}_2}\right)$Given Resolution = 1.50 , N = No. of plates

Or, $1.50=\frac{\sqrt{\mathrm{N}}}{4}\frac{\left(1.068-1\right)}{1.068}\left(\frac{5.51}{1+5+51}\right)$

Or, $1.50=\frac{\sqrt{\mathrm{N}}}{4}\times 0.064\times 0.846$ Or, $\sqrt{N}=\frac{4\times 1.50}{0.064\times 0.846}$

Or, $\sqrt{N}=110.82$

Or, $\mathrm{N}={(110.82)}^2=12281.07$

∴Length of the column = H×N where H = height of the plate=0.520mm

=0.520 mm/plate × 12281.07 plate = 6386.15 mm= 6.38m .

(a)As we know ,$\mathrm{k}=\frac{{\mathrm{t}}_{\mathrm{r}}-{\mathrm{t}}_{\mathrm{m}}}{{\mathrm{t}}_{\mathrm{m}}}$ Or,${\mathrm{t}}_{\mathrm{r}}={\mathrm{t}}_{\mathrm{m}}\left(\mathrm{k}+1\right)$

Given${\mathrm{t}}_{\mathrm{m}}=2.00\min$ ${\mathrm{t}}_{\mathrm{r}}^{\text{'}}=2\left(5.16+1\right)=12.32\min$

${\mathrm{t}}_{\mathrm{r}}^{"}=2\left(5.51+1\right)=13.02\min$

Where ${\mathrm{t}}_{\mathrm{r}}^{\text{'}}\mathrm{and}{\mathrm{t}}_{\mathrm{r}}^{"}$are the retention time of compound 1 and 2 respectively.

Again $\mathrm{N}=\frac{5.55{\mathrm{t}}_{\mathrm{r}}^2}{{\mathrm{w}}_{{\displaystyle \frac{1}{2}}}^2}$ Or, ${\mathrm{w}}_{\frac{1}{2}}=\sqrt{\frac{5.55\times {\mathrm{t}}^2}{\mathrm{N}}}$

Now

$$\begin{gathered} {\mathrm{w}}_{\frac{1}{2}}^{\text{'}}=\sqrt{\frac{5.55\times {\mathrm{t}}_{\mathrm{r}}}{\mathrm{N}}}=\sqrt{\frac{5.55\times {\left(12.32\right)}^2}{1228.07}}=0.261\min \\ {\mathrm{w}}_{\frac{1}{2}}^{\text{'}\text{'}}=\sqrt{\frac{5.55\times {\mathrm{t}}_{\mathrm{r}}^{\text{'}\text{'}}}{\mathrm{N}}}=\sqrt{\frac{5.55\times {\left(13.02\right)}^2}{12281.07}}=0.276\min \end{gathered}$$

Where ${\mathrm{w}}_{\frac{1}{2}}^{\text{'}}{\mathrm{andw}}_{\frac{1}{2}}^{\text{'}\text{'}}$are width at half height of compound 1 and 2 respectively.

(d) The relation between Retention factor (k) and Partition Co- efficient (K)

for each solute is : $\mathrm{k}=\mathrm{K}\frac{{\mathrm{V}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{m}}}$ O r, $\mathrm{K}=\mathrm{k}\times \frac{{\mathrm{V}}_{\mathrm{m}}}{{\mathrm{V}}_{\mathrm{s}}}$

Where $\frac{{\mathrm{V}}_{\mathrm{m}}}{{\mathrm{V}}_{\mathrm{s}}}=$Volume ratio of stationary phase to mobile phase

For compound 1 , ${\mathrm{K}}_1={\mathrm{k}}_1\frac{{\mathrm{V}}_{\mathrm{m}}}{{\mathrm{V}}_{\mathrm{s}}}$ Or, ${\mathrm{K}}_1=5.16\times \frac{1}{0.30}=17.2$