4.6: Chromatography I

A laboratory is using the GLC chromatography system described in Methods to separate a series of alcohols. The chromatogram has the following appearance: Solvent front, 0.11 min.; Width of elution peaks at baseline absorbance: methanol, 0.1 min.; ethanol, 0.15 min.; isopropanol, 0.20 min.; n-propanol, 0.25 min.; acetone, 0.17 min. The elution times of the major peaks are listed in the figure legend. The laboratory decides to determine if acetone is sufficiently well separated from ethanol.

QUESTIONS

1. Is the separation (i.e. resolution) of acetone and ethanol adequate?
2. For quality control purposes, the HETP of the column is calculated weekly and compared to the initial value. When the HETP decreases to 70% of the initial value, the column is taken out of service. This column’s initial HETP was 0.083 cm/theoretical plate. Is the current HETP acceptable?

Questions to Consider

1. What parameters are needed to calculate resolution?
2. Calculate k, the capacity factor, for each of the alcohols using Equation 5-6 on p. 117.
   Calculation: $$k = \frac{V_{R} - V_{O}}{V_{O}} = \frac{t_{R} - t_{O}}{t_{O}}$$
3. Calculate the alpha factor for ethanol and acetone. (Equation 5-7, p. 117)
   Calculation: $$\text{Alpha} = \frac{k_{2}}{k_{1}}$$
4. Calculate the number of theoretical plates for the internal standard (Equation 5-3, p. 114).
5. Using the calculated and measured parameters, now calculate the resolution, R, using equations 5-2 and 5-10.
6. Presume that the column is 200 cm long. What is the HETP? Use Equation 5-4 for this calculation.

Answer

1. Since an R greater than 1.25 (p. 112) is considered adequate, by either calculation the resolution between acetone and ethanol is sufficient.
2. Since the column’s efficiency has only decreased by 13%, (0.083/0.094), the HETP is acceptable, and the column can remain in service.

Answers to Questions to Consider

1. Resolution between two compounds can be calculated by either Equation 5-2 (p.112) or Equation 5-10 (p. 116). Equation 5-2 is a general resolution equation which can be used to calculate resolution directly from the chromatogram using the distances (or time) between the peaks and the peak width. In equation 5-10, resolution is calculated using the theoretical plates of the column, a selectivity factor, and a capacity factor.
2. The capacity factor is calculated from the relationship: $$\begin{split} k^{\prime} &= \frac{t_{R} - t_{O}}{t_{O}} \\ &= \frac{0.98 - 0.11}{0.11} = 7.90\; methanol \\ &= \frac{1.33 - 0.11}{0.11} = 11.1\; ethanol \\ &= \frac{2.12 - 0.11}{0.11} = 18.3\; isopropanol \\ &= \frac{2.88 - 0.11}{0.11} = 25.2\; n-propanol \end{split}$$
3. $$\begin{split} Acetone\; k^{\prime} &= \frac{1.55 - 0.11}{0.11} = 13.1 \\ Alpha &= \frac{k_{2}^{\prime}}{k_{1}^{\prime}} = \frac{13.1}{11.1} = 1.19 \end{split}$$
4. Number of theoretical plates $$N = 16 \left(\dfrac{t_{R}}{W}\right)^{2} = 16 \left(\dfrac{2.88}{0.25}\right)^{2} = 2123$$
5. Using Equation 5-2: $$\begin{split} R &= \frac{d_{2} - d_{1}}{0.59(W_{2} + W_{1})} \\ &= \frac{1.55 - 1.33}{0.5(0.15 + 0.17)} \\ &= \frac{0.22}{0.16} \\ &= 1.37 \end{split}$$or 5-10 $$\begin{split} R &= \left(\dfrac{N^{0.5}}{4}\right) \left(\dfrac{\alpha - 1}{\alpha}\right) \left(\dfrac{k_{2}^{\prime}}{(1+k_{2}^{\prime})^{2/3}}\right) \\ &= \left(\dfrac{2123^{0.5}}{4}\right) \left(\dfrac{1.19 - 1}{1.19}\right) \left(\dfrac{13.1}{(1+11.0)^{2/3}}\right) \\ &= 2.02 \end{split}$$
6. For a 200 cm column: $$HETP = \frac{200}{2123} = 0.094\; cm/theoretical\; plate$$HETP = total length of stationary phase/# of theoretical plates