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The definition of dead time ($t_\mathrm m$) is mentioned as the time taken for elution of an unretained species. It is also mentioned that the retention time ($t_\mathrm r$) is sum of the time analyte spends in mobile ($t_\mathrm m$) and stationary phases ($t_\mathrm s$). 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, $t_\mathrm m$)?

Both seems independent of each other. An analyte's time in the different phases is dependent on its affinity while the unretained species takes the same amount of time no matter what.

Please explain how they are same.

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  • $\begingroup$ The said 'duplicate' differs in the fact that it is with regard to the associated formula while the gravity of this question lies in what tm means. $\endgroup$ – Zam May 26 '19 at 3:36
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The analyte spends the same time in the mobile phase as it tooks for the mobile phase to pass through the column.

An analyte's time in the different phases is dependent on it's affinity.

The time $t_\mathrm{m}$ is given.

The ratio of an analyte's time in the different phases is dependent on it's affinity and is equal:

$$\frac{t_\mathrm{s}}{t_\mathrm{m}}=K_\mathrm{D} \cdot \frac{V_\mathrm{s}}{ V_\mathrm{m}} $$.

where

  • $t_\mathrm{s}, t_\mathrm{m}$ are time is stationary and mobile phase
  • $K_\mathrm{D}$ is the phase distribution constant
  • $V_\mathrm{s}, V_\mathrm{m}$ are volumes of stationary and mobile phase.
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    $\begingroup$ I couldn't understand your first sentence. $\endgroup$ – Zam May 26 '19 at 4:41
  • $\begingroup$ The analyte spends the same time $t_\mathrm{m}$ in the mobile phase as the carrier(mobile phase) in the column. $\endgroup$ – Poutnik May 26 '19 at 4:42
  • $\begingroup$ Poutnik, I am afraid your first statement is not correct. The partition ratio determines how much time an analyte spends in the mobile phase and the stationary phase. $\endgroup$ – M. Farooq May 27 '19 at 17:57
  • $\begingroup$ In fact, both, as the first ( affinity, or distribution constant) being an intensive parameter, the second ( partition ratio ) an extensive parameter. Note also I have said determines, not equals to. And also, I i have followed the OP reasoning. $\endgroup$ – Poutnik May 27 '19 at 18:08
  • $\begingroup$ @M._Farooq Whenever the analyte is in the mobile phase, it moves at the same speed as the carrier. The time to go from start to finish (not counting the time stuck in the stationary phase) is the same. Different analytes spend more or less time in the stationary phase, so $t_m$ is constant and $t_s$ is variable, down to zero for unretained species. $\endgroup$ – Karsten Theis May 27 '19 at 21:30

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