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The relative retention for two compounds in gas chromatography is 1.068 on a column with

Chapter 22, Problem 22-F

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QUESTION:

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

(a) Find the unadjusted relative retention \((\gamma)\) for the two compounds.

(b) What length of column will separate the compounds with a resolution of 1.00?

(c) The retention time for air \(\left(t_{\mathrm{m}}\right)\) is 2.00 min. If the number of plates is the same for both compounds, find \(t_{\mathrm{r}}\) and \(w_{1 / 2}\) for each peak.

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

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QUESTION:

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

(a) Find the unadjusted relative retention \((\gamma)\) for the two compounds.

(b) What length of column will separate the compounds with a resolution of 1.00?

(c) The retention time for air \(\left(t_{\mathrm{m}}\right)\) is 2.00 min. If the number of plates is the same for both compounds, find \(t_{\mathrm{r}}\) and \(w_{1 / 2}\) for each peak.

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

ANSWER:

Step 1 of 4

The given values in the question and their related equations are shown below:

The height of the plate is given as 0.520 mm.

The equation for the relative retention is given as follows:

\(\begin{aligned}
(\alpha) & =\frac{t_{r 2}}{t_{r 1}} \\
& =\frac{t_{r 2}-t_{m}}{t_{r 1}-t_{m}} \\
& =1.068 \\
\frac{t_{r 2}-2}{t_{r 1}-2} & =1.068 \ldots (1)
\end{aligned}\)

Where \(\alpha\) is the retention factor

Similarly, the retention factor(k) is given as follows

\(\begin{aligned}
k & =\frac{t_{r 1}-t_{m}}{t_{m}} \\
& =\frac{t_{r 1}-2}{2}=5.16
\end{aligned}\)

This implies

\(\begin{array}{l}
t_{r 1}=2=10.32 \\
t_{r 1}=12.32 \ldots \ldots .(2) \\
t_{r 1}^{\prime}=10.32 \ldots \ldots .(3)
\end{array}\)

From equation no.3 and 4

\(\begin{array}{l}
\frac{t_{r 2}-2}{10.32}=1.068 \\
t_{r 2}=13.02 \ldots \ldots(4) \\
t_{r 2}=11.02 \ldots \ldots .(5)
\end{array}\) 

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