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GC Problem

Calculate:
1. Corrected retention times for air and 3 alcohols
2. Specific retention volumes for 3 alcohols
3. Distribution constant for 3 alochols
4. Capacity factor for 3 alcohols
5. The average number of theoretical plates and the average height of a theoretical plate for the column

Given:
1. Column length: 1.1m ID = 2mm
2. Weight of stationary phase = 1.4g /density = 1.02g/ml
3. Average volumetric flow in column = 30.2ml/min
4. Inlet pressure = 26.1psi above ambient press
5. Room temperature = 21.2C column temp = 102C

RETENTION TIMES:
air = 0.3 min
propanol = 2 min
butanol = 4.2 min
hexanol = 8 min

PEAK WIDTHS:
propanol = 0.2min
butanol = 0.4 min
hexanol = 0.8min

Solution Preview

Retention times are corrected for the amount of time needed for a nonadsorbent analyte to be detected, in this case air. All retention times need to be corrected by subtracting 0.3 minutes.

air = 0 min
propanol = 1.7 min
butanol = 3.9 min
hexanol = 7.7 minutes.

The ...

Solution Summary

Retention times are corrected for the amount of time needed for a nonadsorbent analyte to be detected, in this case air. The retention volume is given as the amount of mobile phase to pass befor the peak max is achieved: flowrate x retention time. Distribution constants for the three alcohols is the retention time divided by the peak width (time in stationary phase/ time in mobile phase). Capacity factors for the alcohols is the retention time divided by the dead time. The number of theoretical plates N is N = 16 (T/W)^2.

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