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Why does ts start just after tm ? Does the analyte starts spending time in the stationary phase only after the mobile phase has passed and reached the detector ? Wouldn't there be an overlap between the two ?

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The column is filled with the carrier (liquid or gas) before the sample is injected. Thus if there is no interaction between the sample and the column, then the fastest that the sample can get to the detector is the dead time denoted by $t_M$ in the diagram. If the sample does interact with the column, then it is retarded by a time $t_S$ shown in the diagram. Thus the total retention time from injection to detection is $t_M + t_S$.

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  • $\begingroup$ That seems more convincing. But then, what does ts mean ? The mentioned definition in the picture is of not much help. $\endgroup$ – Zam May 24 at 16:13
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Most of the separation techniques consist of a chromatographic column, stationary phase, and mobile phase, and use common terminology. Therefore, it is better learn some described in the text book for HPLC (Ref.1) as an overview:

The retention time $t_\mathrm{R}$ solute A can be defined as the time from the injection of the sample to the time of compound elution, taken at the maximum (apex) of the peak that belongs to the specific molecular species A (known or unknown). The retention time indicates how long it takes for a compound A to elute from the column (from the injector to the detector). The retention time of the last peak (if sample contains multiple compounds) in a chromatogram is used to estimate the necessary length of the chromatographic run. In general, for a molecular species A, the retention time can be indicated as $t_\mathrm{R}$(A) and time is usually measured in $\pu{min}$. However, the part "(A)" in the notation is sometimes omitted, yet $t_\mathrm{R}$ is always related to a specific molecular species.

Retention time depends not only on the structure of the specific molecule, but also on factors such as the nature of the mobile and stationary phases, the flow rate of the mobile phase, and dimensions of the chromatographic column. Retention time is usually characteristic for a specific compound in a given separation. For this reason, the retention time is critical in identifying analytes once their retention time is known (e.g., by using standards).

Of particular interest in a separation is the dead time $t_\mathrm{M}$, which is the time a nonretained molecular species needs to elute from the chromatographic column. The dead time is also known as void time or holdup time. The dead time $t_\mathrm{M}$ can be also interpreted as part of the retention time $t_\mathrm{R}$(A) for the analyte A, which the analyte spends in the mobile phase moving through the column (That is the reason for the subscrpt "M" meaning mobile). This parameter is not related to the retention process and depends on the flow rate and physical characteristics of the column (i.e. length, diameter, porosity of stationary phase). The difference between the retention time ($t_\mathrm{R}$) and the dead time ($t_\mathrm{M}$) represents the time the analyte A is retained on the stationary phase ($t_\mathrm{S}$). This difference is indicated as reduced retention time $t_\mathrm{S}$ (or $t'_\mathrm{R}$) and is expressed by the formula: $$t_\mathrm{S}=t_\mathrm{R}-t_\mathrm{M}$$

The value for $t_\mathrm{M}$ is typically obtained as an approximation by using compounds that are very slightly retained, since it can be difficult to find a compound that is not retained at all on a chromatographic column. For example, during HPLC runs, the solvent used for injecting the sample (when different from the mobile phase) can be such a compound, and the retention time of this solvent peak can be taken as dead time.

Reference:

  1. Serban C. Moldoveanu, Victor David, Essentials in Modern HPLC Separations; 1st Edn.; Elsevier, Inc.: Waltham, MA; 2013.
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  • $\begingroup$ Still confused. ts gives the retardation in elution due to the stationary phase. How are we sure that ts is the time spent in stationary phase ? It could include the time spent in the column without contact with stationary phase (in case the column is not entirely filled with stationary phase). Or tr itself would be the time spent in stationary phase (in case the column is entirely filled with stationary phase). $\endgroup$ – Zam May 25 at 4:19
  • $\begingroup$ @Zam: The column is not entirely filled with stationary phase. The solute's time spent in the column without contact with stationary phase is its time spent in the mobile phase because column is filled with stationary and mobile phases only (theoretically). No air bubbles. :-) $\endgroup$ – Mathew Mahindaratne May 25 at 6:34
  • $\begingroup$ But how can time spent by analyte in mobile phase be equal to the time required for an unretained species to pass through (dead time, tm) ? Wouldn't the former be less than the latter ? Because former doesn't take, passage through stationary phase into consideration while the latter does. $\endgroup$ – Zam May 25 at 9:31
  • $\begingroup$ @Zam: If you are taking this class, I assume analytical chemistry or instrumental analysis, you may able to think a little deeper. I can take the horse to water, but I can't make him drink. Sorry. $\endgroup$ – Mathew Mahindaratne May 25 at 21:16
  • $\begingroup$ "The dead time tM can be also interpreted as part of the retention time tR(A) for the analyte A, which the analyte spends in the mobile phase moving through the column (That is the reason for the subscrpt "M" meaning mobile)." I have come across statements with this very same meaning in my class but not reasoning. That's what I am enquiring here. $\endgroup$ – Zam May 26 at 4:57

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