fraction collection

[calistra]

Hi! I have problems when collecting fractions using the fraction collector. I calculated delay volume based on tubing dimensions, it should be 48 µl, flow rate is 0.4 ml/min. Peaks that I want to collect elute at 65.340, 66.946 and 68.258 min, and have width of 0.75, 0.65 and 0.70 min, respectively. Problems is I have recovery of 50% when I check the fractions. Can I somehow improve my recovery? My idea was that maybe because of software processing etc. fraction collector starts to collect fractions later.
Thank you for your help

[DJ]

You're gonna first have to tell us what system, software you're using, how you determined recovery, etc

[Peter Apps]

If the widths you use are at half height then 50 % recovery (in half the peak volume) is pretty good accuracy.

Peter

[calistra]

If the widths you use are at half height then 50 % recovery (in half the peak volume) is pretty good accuracy.

Peter

Sorry I forgot to write: widths are calculated at the baseline.

[calistra]

You're gonna first have to tell us what system, software you're using, how you determined recovery, etc

I am using Dionex HPLC system with Chromeleon software. Column is HILIC with 3 µm particles (150x4.6 mm) and detector I´m using is fluorescent detector.

I collected fractions, injected them again on the column and calculated peak area in the collected fraction and in the original sample from which fractions were collected.

[Peter Apps]

We need details of how you did the calculation of recovery - what data did you use, and what was your 100% recovery expectation of peak area from the collected fraction.

Peter

[calistra]

I just extracted Chromeleon data, so for example for the first peak area under peak (RFU*min) is 100.3060 (this is for the peak that is collected). After its reinjection area is 56.476 what is 56.3% of this 100.3 that had original peak.

[HPLCaddict]

In the collected fractions your analytes will be in a lower concentration as in the original sample, because they are diluted during the chromatographic process. Therefore, if you inject the same amount of collected fraction as with the original sample, the peak area must be lower than before, even if you have 100% recovery.

[Peter Apps]

Your injection volume is one of the many details that you do not provide, but let's assume that it is 150 ul. At the flow rates you give, presuming that you collect for the whole of the peak width, your fraction volume is 300 ul - so your substance comes out of the column in double the volume it went in, and your collected fraction has a concentration one half of that of the original solution. Then you inject 150 ul of the fraction, which contains half as much substance as the first injection, and get a peak that has half the area of the original peak. You then calculate that you have 50% recovery.

Peter

[calistra]

Your injection volume is one of the many details that you do not provide, but let's assume that it is 150 ul. At the flow rates you give, presuming that you collect for the whole of the peak width, your fraction volume is 300 ul - so your substance comes out of the column in double the volume it went in, and your collected fraction has a concentration one half of that of the original solution. Then you inject 150 ul of the fraction, which contains half as much substance as the first injection, and get a peak that has half the area of the original peak. You then calculate that you have 50% recovery.

Peter

I injected 100 ul. And I collected fraction that has as you said volume of about 300 ul. Then I dried the sample down to have again the same volume as the injected sample-to 100 ul. And then I injected whole volume.

[Peter Apps]

Classic - you post on Sunday saying you have a problem, but we do not get the critical details of your procedure until Wednesday.

How are you injecting the whole of the 100 ul of your reconstituted fraction ? If you are doing it by valve with a 100 ul loop then you have way too little volume to fill the loop properly. In fact, if I remember correctly, loading a loop with its nominal volume does leave about half the expected volume in the loop and would account for your 50% recovery.

How repeatable is the 50% ?, is it the same for every fraction ?, and how many times have you tried it ?

Peter

[calistra]

It´s 100 ul loop. I measured it 3 times and it was always round 50%. How so you then suggest to measure it more accurately?

[Peter Apps]

You cannot accurately fill a 100 ul loop if you only have 100 ul of liquid. You might do better to dilute your fraction to 500 ul so that the loop is completely and consistently filled, and then take the dilution into account when you calculate recovery.

Peter

[calistra]

OK, thank you!

[lmh]

... and don't believe that a 100uL loop is really 100uL unless someone swears it! Some loops are sold with enormous tolerances.
... and beware of calculated delay volumes when working out how far behind your detected signal the fraction collector actually is. It can be very hard to calculate exactly what the volumes are in every part of the system, and my limited experience of working on collected fractions is that quite often I find people's compounds of interest in the fraction adjacent to the one they swore would contain it!