tM in Chromatography is the time of exist mobile phase. But, the mobile phase exist together with all the compunds that tasted in the process. So, why in the graph of signal to time we can see a peak separated from all the other ?

thanks, Liat

  • 3
    $\begingroup$ You may need to study basics of chromatography. $\endgroup$
    – Poutnik
    Commented Jan 28, 2021 at 10:36
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    $\begingroup$ "time of exist"—do you think the mobile phase disintegrates afterwards? "all the compunds that tasted in the process"—sounds like degustation, not chromatography. $\endgroup$
    – andselisk
    Commented Jan 28, 2021 at 10:42
  • $\begingroup$ Welcome to Chemistry Stack Exchange! Please take a moment to familiarize yourself with the norms of our online forum. Adding an example may increase the likelihood of a positive response here. $\endgroup$
    – z1273
    Commented Jan 29, 2021 at 2:39
  • $\begingroup$ chemistry.stackexchange.com/q/63595/102629 $\endgroup$
    – cngzz1
    Commented Jan 30, 2021 at 10:24

1 Answer 1


The dead time "disturbance" $t_M$, is not a proper peak per se in HPLC. It is a disturbance recorded by the detector when the injected sample solvent reaches the detector.

When you open your garden hose tap, it takes some finite time when you start seeing water out of the other end of the hose. Dead time is exactly like that. It is the time it takes for the sample solvent to travel from the injector, column and finally to the detector. This is the of time of exit as you call it.

Everything must elute after the dead time, because nothing can travel faster than the unretained sample solvent. There are a few exceptions but accept this as a fact for most of the cases.


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