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In 'Fundamentals of Analytical Chemistry' 9th edition by Skoog, West, Holler and Crouch, there is the following passage (page 865).

The dead time (void time), $t_M$, is the time it takes for an unretained species to pass through a chromatographic column. All components spend at least this amount of time in the mobile phase.

Apologies if it seems a silly question, but how does that make sense? Wouldn't all components spend exactly that amount in the mobile phase? That is, of course, assuming a constant flow rate for the mobile phase. Is this where I have gone wrong?

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  • $\begingroup$ What this passage means is that all components will spend at least that time inside the column. $\endgroup$
    – Variax
    Commented Dec 1, 2016 at 12:59
  • $\begingroup$ @Variax Well that's a given but for most components $$\text{time spent in mobile phase} \neq \text{time spent in the column}.$$ Furthermore, this isn't limited to just this sidenote. It is noted elsewhere in the book as well. $\endgroup$
    – Linear Christmas
    Commented Dec 1, 2016 at 13:19
  • $\begingroup$ @ToddMinehardt So by lower bound you're essentially saying that different unretained species spend a different amount in the mobile phase? For me, that goes against the definition of unretained. Furthermore, the book does later use dead time $t_M$ as a measure of speed for the mobile phase (i.e., pure eluent). $\endgroup$
    – Linear Christmas
    Commented Dec 3, 2016 at 13:31
  • $\begingroup$ @LinearChristmas - Thanks for pointing out a problem with what I wrote. I'd revise my comment to say that dead time is simply the rate of the eluent through the column. Thus, it is the lower bound on time for what can travel through the column, as it represents unimpeded flow (say, 1 mL/min, isocratic), and therefore anything that interacts with the immobile phase will elute at a time greater than the dead time. $\endgroup$
    – Todd Minehardt
    Commented Dec 3, 2016 at 18:21
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    $\begingroup$ @ToddMinehardt ;) I think I finally got it, too. Retained does not necessarily mean entering the stationary phase (hence coming to a "complete stop"). It can also simply slow down which should have been obvious to me from the start... If you write a short answer, I'd be happy to tick it! $\endgroup$
    – Linear Christmas
    Commented Dec 3, 2016 at 18:56

1 Answer 1

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Dead time ($t_{\bf M}$, also called holdup time) is the time it takes for the mobile phase (eluent) to traverse one length of the column, for a given flow rate. Because moieties in the eluent do not interact chemically with the stationary phase, $t_{\bf M}$ is therefore a lower bound on retention times for the species that do interact (chemically) with the stationary phase.

Simply put, anything that does interact with the immobile phase cannot elute faster that the pure mobile phase: retention times will be greater than $t_{\bf M}$.

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    $\begingroup$ Also called void time! :) $\endgroup$
    – Linear Christmas
    Commented Dec 3, 2016 at 19:47
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    $\begingroup$ Yes, it turns out that the world of chromatography is rife with terminology that can be contradictory/incomprehensible/etc. :) $\endgroup$
    – Todd Minehardt
    Commented Dec 3, 2016 at 19:52
  • $\begingroup$ @ToddMinehardt How we measure dead time if the column is always filled with mobile phase? The detector would always get signal from the mobile phase. $\endgroup$
    – Antonios Sarikas
    Commented May 15, 2020 at 16:11

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