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I did an experiment of gas chromatography which contained the compounds: octane, butanone and propanol. We injected a mixture of the three compounds in the polar column and then a mixture of it in the non polar column. I have certain data which say that the most polar compound of the three is octane,then butanone and the least polar is propanol. As for the boiling points, the highest boiling point belongs to octane,then propanol and the lowest boiling point is for butanone.

I know that the lower the boiling point is, the higher the vapor pressure of the compound and the shorter retention time usually is because the compound will spent more time in the gas phase. I also know that If the polarity of the stationary phase and compound are similar, the retention time increases because the compound interacts stronger with the stationary phase. As a result, polar compounds have long retention times on polar stationary phases and shorter retention times on non-polar columns using the same temperature.

But after seeing the results, the order that these compounds appear are: For the non-polar column:

1)air(I get it)

2)propanol

3)butanone

4)octane

I get about octane but between propanol and butanone how do we know which factor(boiling point or polarity) plays the most important role?

I get confused because if we take into consideration polarity, the retention time of butanone should be bigger but if we take into consideration the boiling point, the retention time for propanol should be bigger.

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  • $\begingroup$ If propanol comes out first in your nonpolar column, then it's obviously the (non-)polarity which plays the bigger role. What's wrong with that? $\endgroup$ – Karl Jan 4 '18 at 18:46
  • $\begingroup$ @Karl I have to understand the concept because I have to find out the order of the compounds for the polar column and I don't get if butanone or proponal will have the bigger retention time. $\endgroup$ – chemistrylove Jan 4 '18 at 22:13
  • $\begingroup$ If the difference in boiling points is not large enough to make a difference in the nonpolar column, then it imo surely won't matter in the polar column. Polar-polar interaction is much stronger than nonpolar interaction. $\endgroup$ – Karl Jan 4 '18 at 23:42

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