A question that comes up now and then in our Master Classes is how to figure out the volume of an HPLC column. I’d like to share a couple of rules of thumb that I find very useful for estimating column volume, V_{M}.

## Estimating HPLC Column Volume for an Empty Column

Figuring out the volume inside the column isn’t much of a challenge if the column is empty. Remember back to your high school geometry class that the volume (V) of a cylinder is

V = Πr^{2}h (1)

So for a 150 x 4.6 mm i.d. column, h (height) = L (length) = 150 mm, and r (radius) = 0.5d_{c} (diameter of the column = 4.6 mm), giving (Π)(150)(4.6/2)^{2} = 2493 mm^{2} = 2493 µL ≈ 2.5 mL.

## Estimating HPLC Column Volume for 4.6 mm i.d. Columns

However, 2.5 mL is not the volume of a chromatographic column because it is packed with particles. The spherical particles have spaces between them and are also porous, so they have spaces within them. These spaces represent the volume of the mobile phase or column volume that we are really interested in. For most HPLC columns, this volume represents about 60–70% of the volume of an empty column. Let’s see how we can make a shortcut for this. If we multiply any factor of Equation 1 by 60%, we should get the column volume. For the 4.6 mm i.d. column, if we multiply Πr^{2} by 60%, we get 10 — I always like calculations that involve 5’s and 10’s, because I can do them in my head. So for our 4.6 mm i.d. column we now have:

V_{M} ≈ 10 L (2)

where L is the length in mm and V_{M} is in µL. Usually we are more interested in V_{M} in mL, so we divide by 1000 or Equation (3) can be restated as

V_{M} ≈ 0.01 L (3)

now 1% of 150 = 1.5 mL. We can double-check by multiplying the empty volume, 2.5 mL x 60% = 1.5 mL.

**Adjusting for Other Column Diameters**

Equation (3) is very handy for estimating the volume of a 4.6 mm i.d. column but what do you do if the column is a different diameter? We need a rule of thumb that includes both L and d_{c}. Notice that 0.6 π (d_{c}/2)^{2} = (0.6 π/2) d_{c}^{2} = 0.47 d_{c}^{2}. This is close to 0.5 d_{c}^{2}

V_{M} ≈ 0.5 L d_{c}^{2} (4)

where L and d_{c} are in mm and V_{M} is in µL. We can use this to figure out the volume of a 50 x 2.1 mm i.d. column that might be used in an LC–MS application: V_{M} = 0.5 x 50 x 2.12 = 110 µL or ≈ 0.1 mL.

## Special Case for 2.1 mm i.d. Columns

We can make one more shortcut for the 2.1 mm i.d. column, which is the most popular column diameter other than 4.6 mm. We could use Equation 2 or 3 and adjust for the change in cross-sectional area and get the same result. Since the cross-sectional area is proportional to the ratio of the diameters squared, we get (4.6/2.1)^{2} = 4.8. We’re talking about estimates here, so we can round 4.8 to 5 for easy mental math. This means that the volume of a 4.6 mm i.d. column is about five times as large as the same column in a 2.1 mm i.d. format. Let’s do a quick test of this. A 50 x 4.6 mm column would have a volume V_{M} ≈ 0.01 x 50 = 0.5 mL (Equation 3). A 2.1 mm i.d. version of this column should have 1/5 of the volume, so 0.5 mL/5 = 0.1 mL = 100 µL. This is close enough to the value we calculated from Equation 4 above.

## Conclusion

So now we have a few simple shortcuts to estimate column volume. One for the 4.6 mm i.d. columns (Equation 2 or 3), one for any column (Equation 4), and one specifically for the 2.1 mm i.d. column (Equation 2 or 3 divided by 5). These estimates should be within about 10% or so of the value you measure chromatographically.

*This blog article series is produced in collaboration with John Dolan, best known as one of the world’s foremost HPLC troubleshooting authorities. He is also known for his research with Lloyd Snyder, which resulted in more than 100 technical publications and three books.*