## Wednesday, January 9, 2013

### Molecular Weight Distributions and Size Exclusion Chromatography: Part 1

Size Exclusion Chromatography (SEC) is a commonly used technique to separate molecules dissolved in a solvent on the basis of their size. A sample containing molecules of various sizes is injected at one end of a special SEC column, and monitored at the other end using one or more special detectors.

As an analogy it is useful to think of the column as an obstacle course (a 110 meter hurdle race). Somewhat counter-intuitively, the big fat molecules finish  first, and the small nimble ones finish last. The reason why this happens is the same reason why the tortoise finished first. Despite being slower, he did not pause to examine and explore each nook and cranny.

You can imagine that SEC is a popular technique to get the molecular weight distribution (MWD) of polydisperse polymers. Because of the way in which the calibration process and detectors work, the resulting MWD is usually reported in a somewhat "funky units" of w(log M) versus log M, which looks something like the following.
Note 1: The area under the curve, as reported, is unity.
Note 2: Throughout this discussion, "log" refers to log-base-10, and "ln" refers to log-base-e. In SEC it is more common to use base-10, although as I will show shortly, base-e is more "natural".

Okay, so how is the SEC MWD w(log M) related to quantities we know better such as the number distribution N(M) and the weight distribution W(M)?

To recap, N(M) dM tells us the fraction of molecules with molecular weight between M and M+dM, and W(M) dM tells us the weight fraction of the molecules with molecular weight between M and M+dM. These distributions are normalized which means,

$\int_{0}^{\infty} N(M) dM = \int_{0}^{\infty} W(M) dM = 1$

The first moments of N(M) and W(M) are the number-averaged and weight-averaged molecular weights.

$M_n = \int_{0}^{\infty} M N(M) dM$
$M_w = \int_{0}^{\infty} M W(M) dM$

As you know or expect, N(M) and W(M) contain the same information, and they can be transformed into one other.

$W(M) = \frac{M N(M)}{\int_{0}^{\infty} M N(M) dM} = \frac{M N(M)}{M_n}$

The integral in the denominator ensures the normalization of W(M). The brother of the SEC MWD w(log M) turns out to be closely related to the "next moment"

$w(\ln M) \propto M W(M),$

except that it's normalization is not

$\int w(\ln M) dM = 1,$

but rather,

$\int_{0}^{\infty} w(\ln M) d \ln M = 1.$

This is automatically satisfied for w(ln M) = M W(M), since

$\int_{0}^{\infty} w(\ln M) d \ln M = \int_{0}^{\infty} M W(M) \frac{dM}{M} = \int_{0}^{\infty} W(M) dM.$

What about w(log M)? Aren't we more interested in that whole shebang?

Sure, no problem. We will find out that w(log M) and w(ln M) are just a constant (2.303) apart.

$w(\log M) = c M W(M),$

where c is a constant, and work backwards from the normalization (as in the plot above!)

$\int_{0}^{\infty} w(\log M) d \log M = 1.$

This implies

$\int_{0}^{\infty} c M W(M) \frac{dM}{2.303 M} = 1.$

This implies c = 2.303, and w(log M) = 2.303 w(ln M) = 2.303 M W(M).

We will apply this to a particular problem in the next post.