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A non-polar molecule X is analyzed by TLC chromatography on silica to determine its $R_f$ value.

The first eluent was hexane:EtOAc = $9:1$ and the $R_f$ value was $0.5$.
In the second run, the eluent was changed to hexane:EtOAc = $7:3$. Will the $R_f$ value be smaller or greater than $0.5$?

I am confused because there are two important interactions: because the 2nd system is more polar, it will form weaker interactions with substance X, and the $R_f$ value will decrease; on the other hand, the molar polar eluent will form stronger interactions with stationary phase, so the $R_f$ value will increase because the substance X is more easily eluted.

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2 Answers 2

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The most important interaction is that of the mobile phase with the stationary phase. If you increase the mobile phase’s polarity, that interaction will increase, decreasing the interaction of the solute with the stationary phase. Therefore, a higher ethyl acetate content in the mobile phase will increase $R_f$ values.

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  • $\begingroup$ What if the molecule is extremely non-polar and the interactions with mobile phase are dominant? Is this possible? Could I have a reference for your response? $\endgroup$
    – Uros
    Commented May 29, 2016 at 23:08
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    $\begingroup$ @Uros If the interactions with the mobile phase are dominant, the $R_f$ value in pure hexane will be $> 0.5$. Source: Selling my soul to the chemistry lab for the last five years. $\endgroup$
    – Jan
    Commented May 30, 2016 at 8:25
  • $\begingroup$ Part two: Yes, in these lab years I had my fair share of compounds whose $R_f$ values in pure isohexane were larger than $0.8$ (which technically makes them impossible to isolate in standard silica chromatography). No matter which other mobile phase mixture I used, their $R_f$ value never sank. $\endgroup$
    – Jan
    Commented May 30, 2016 at 8:31
  • $\begingroup$ @Jan The capillary action is different for different solvents. How then we can compare the $R_f$ values? $\endgroup$
    – ado sar
    Commented May 19, 2020 at 16:04
  • $\begingroup$ @adosar We basically cannot predict the $R_f$ value of a compound in a different solvent (i.e. if it has $R_f=0.5$ in $3:1$ that only tells me the $R_f$ value will be lower in $4:1$, I cannot give a specific value; both solvents $\text{hexane}:\ce{EtOAc}$). We can, of course, perform multiple TLC experiments and compare the outcomes which leads to the rule of a more polar ratio of the same solvents always giving a higher $R_f$ value for compounds. $\endgroup$
    – Jan
    Commented May 21, 2020 at 4:52
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As, mentioned by Jan Increasing the polarity of solvent will result in increase in Rf value as the compound in even for polar or nonpolar compound, this is the rule of thumb in most cases, may be except do exist i don't know about it. But for my experience in the laboratory polarity result increase in Rf value.

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