Cloumn dimension, bed volume, void volume, sample loading vo

[jduan]

hi, there,

Could anyone give me the definition of bed volume of an HPLC column?

And also, could anyone tell me the relationship between each of these parameters? For example, usually the sample loading volume is xx% of colume dimension or bed volume or.....

Thanks so much!!!

jin

[Uwe Neue]

It is the volume of the empty column, calculated as 3.14*d^2*l/4.

The amount that you can load on a column depends - among other things - on the separation and the purpose. If it s an analytical separation, you want a good peak shape for proper integration. If it is a preparative separation, you can load a higher amount, until the peaks start to run into each other (or even beyond that).

Please clarify the question, and maybe we'll have abetter answer.

[tom jupille]

The bed volume is simply the volume of packing in the column. Since the column is a cylinder its equal to (pi * r^2 * L) where
r = column radius
L = column length

The "void volume" of the column is a more confusing term. Most likely it refers to what is officially called the "dead volume", symbolized by Vm. That is the volume of liquid (mobile phase) in the column. That, of course, depends on the characteristics of teh packing material. For reversed-phase HPLC columns, you can figure that about 60% of the bed volume is liquid, so take the formula above and multiply by 0.6 (remember that this is only an approximation!). [The reason "void volume" is confusing is that it sometimes refers to a cavity at the head of the column bed, which generally causes tailing, broad peaks, etc.]

"Sample loading volume", (more often called "injection volume" for analytical-scale separations), can vary all over the map. For prep or semi-prep work it depends on the selectivity of the system for the product vs contaminants, on solubility in the injection solvent, and on the eluotropic strength of the injection solvent relative to the mobile phase. Bottom line is that it must generally be determined empirically. Once you know how much you can load on a given column, you can scale to different column sizes in proportion to the bed volume.

By the way, one of the down-sides to the new forum engine is that the title lengths are limited. You might want to edit your original post with a shorter title and put the current intended title at the top of the post.

oops!, Uwe beat me and posted while I was composing. That's what I get for being long-winded!

[jduan]

I know the sample injection volume depends on experience a lot. However, I heard that, for analytical column, if the sample is dissolved in a solvent which may destroy the conditioning of the column, the injection volume is better less than 15% bed volume. Is this value generally right?

sorry, all the instruments in my lab are busy for routine sample analysis now. No spare one is used for method development at present. However, I am very interested to get some idea from experts out there,

thanks a lot!!!

jin

[Chris Pohl]

Jin,

If your sample is made up in a "solvent" which is weaker than your mobile phase (i.e. a solvent which allows you're analyte to refocus on the head of the column), you can get away with using sample volumes which are equal to or even greater than the bed volume. However, if you're sample is diluted in mobile phase then you will see significant degradation in chromatographic performance even when the injection volume is 5% of the bed volume. If your sample is made up in a solvent which is a stronger eluent than your mobile phase, injecting even 1% of the bed volume will generally degrade chromatographic performance.

[tom jupille]

First off, it depends whether you're doing analytical or preparative work.

I'll comment for analytical-scale work; if you're doing prep, post again; other contributors to the forum are probably better qualified than I am in that area.

OK, for analytical-scale work, you're concerned primarily with keeping the peaks narrow.
- If your sample is dissolved in the mobile phase, then the rule of thumb is to keep injection volume below 15% of the width of the narrowest peak of interest in your chromatogram. This will give you a barely perceptible increase in peak width (about 5%).
- If your sample is dissolved in something stronger than the mobile phase (for reversed-phase, this would be higher % organic), then you have to cut the injection volume down. How much you decrease the injection volume depends on how much stronger your solvent is. Unfortunately, there are no good rules of thumb here; you have to do the experiments.
- If the sample is dissolved in something weaker than the mobile phase (for reversed-phse, this means lower % organic), then you can increase the injection volume. Again, there is no rule of thumb that I know of. Biggest injection I've ever made was 10mL (on a 150 x 4.6 mm column).

[kiknos]

excuse me, i am wondering how "to keep injection volume below 15% of the width of the narrowest peak of interest in your chromatogram".

try again and again with different injection volumes and then locate the one which helps to work out below 15% of the width of the narrowest peak?

if i misunderstands, sorry! but please kindly help, thanks!

best regards,
kiknos

[jduan]

I think it is the corresponding time of 15% of peak width x flow rate,

jin

[Chris Pohl]

Kiknos,

If I understand you correctly, you're asking the question: how does one goes about determining the maximum allowed injection volume? The simplest strategy is to plot the measured efficiency for your target analyte against the injection volume. If you do this, you should observe a plateau region at low injection volumes where efficiency is independent of injection volume. Following the plateau region, there is generally a relatively steep drop off as the injection volume exceeds the limit of the analytical system. The point at which the efficiency drops by more than 10% is generally considered the limit for analytical work. Alternatively, if you use a 10 microliter injection volume with a 4.6 by 150 mm column, this should be sufficient to be below the above-mentioned limit unless your sample is made in the solvent which is more potent than your mobile phase in which, case as mentioned above, you may need to further reduce the injection volume.

[tom jupille]

Both Chris and jduan are exactly right. Chris's suggestion is the most accurate (but also the most time consuming). The short-cut which I had suggested involves the following steps:

1. Make a small-volume injection (e.g., a few microliters if you are using a standard analytical-scale column).
2. Identify the narrowest peak you wish to quantitate (for an isocratic separation, this is usually the earliest peak)
3. Multiply the peak baseline width (in time units) by the flow rate to obtain the peak baseline width in volume units.
4. Multiply by 0.15 to find the estimated maximum injection volume.

Again, this is a short-cut. Chris's procedure (measure width as a function of volume injected) is more accurate.

[kiknos]

Thank you all so much for your kind replies, Chris, Jin and Tom!

Before the discussion, I had the same idea exactly as Chris. Now I am glad there is some much simpler way to estimate before lots of trials!

One more question:
is that just from experiences or upon some supposition?
is that suitable for all HPLC or limited only for RP-HPLC or for small molecules(as we know even both with RP-HPLC, big molecules elute differently from small molecules elute), etc.?

Thanks again!

Best regards,
kiknos in China

[tom jupille]

As a rule of thumb, it is general. The rationale is the same as for total extra-column volume; there was a discussion earlier: http://www.sepsci.com/chromforum/viewto ... column+vol

In essence, the simplifying assumption is that the only extra-column volume we're concerned with is injection volume. In that case, the maximum injection volume depends on how much of a plate loss you are willing to tolerate. As a "back of the envelope" calculation, it's approximately:

v(inj) = v(col) * [N(col)/N(obs) - 1]^0.5 , where:

- v(inj) is the volume injected
- v(col) is the volume of the peak created by the column only (i.e., for a negligibly small injection)
- N(col) is the plate number for that peak created by the column only
- N(obs) is the plate number you would have with a larger injection volume.

Assuming you are willing to lose 2% of your efficiency (measured as plate count; this is just at the limit of perceptability in most cases), then:
N(col)/N(obs) = 1/0.98 = 1.02
Subtract 1 and take the square root and you get about 0.14 (14%)

In other words, an injection volume which is 14% of the volume of your narrowest peak will cost you about 2% of the plate count for that peak (the limit of perceptibility). If you can tolerate a larger loss in efficiency, then you can inject larger volumes.

As pointed out in the thread I referenced earlier, this is a bit simplistic because we really should be talking about dispersion rather than volume, but it does provide an approximation. It does not depend on the type of chromatography or the size of the molecule, but it is subject to some conditions:
- the peaks in question must be "well behaved" (i.e, no tailing problems, not overloaded, etc.)
- it applies only to isocratic separations
- the sample must be dissolved in the mobile phase (or, at least, a solvent of the same strength as the mobile phase).

[kiknos]

Thank you so much for your kind reply, Tom!