$\beta$-mercaptoethanol molarity I'm writing up a practical and $\beta$-mercaptoethanol was used, with no concentration given. I've found I don't need the BME molarity since I'm giving percentage (v/v) concentration, however I'm slightly confused as to why there are different values for its concentration.

Sigma-Aldrich provide BME concentration 14.3 M (pure liquid), yet the openwetware.org protocol for SDS sample buffers gives a molarity of 14.7 M. I can't see precisely how this is derived from its density, or why there'd be different molarities floating around.

Not that it matters for my writing here, but since I don't know the molarity I'm just a bit puzzled now as to what it may have actually been, and would rather understand now than have to work it out further down the line. Any answers greatly appreciated!
From the Aldrich data:
 $\rho = 1.114\ \frac{\mathrm{g}}{\mathrm{mL}}\ \text{at}\ 25\ \mathrm{^\circ C}$

Converted to a more convenient unit:
 $\rho = 1.114\ \frac{\mathrm{g}}{\mathrm{mL}} \cdot \frac {1000\ \mathrm{mL}}{1\ \mathrm{L}} = 1114\ \frac{\mathrm{g}}{\mathrm{L}}$

Given the molar mass (also from Aldrich) of:
 $M = 78.13\ \frac{\mathrm{g}}{\mathrm{mol}}$

We learn that one liter has:
 $1114\ \frac{\mathrm{g}}{\mathrm{L}} \cdot \frac{\mathrm{mol}}{78.13 \ \mathrm{g}} = \frac{1114}{78.13}\ \frac{\mathrm{mol}}{\mathrm{L}} = 14.26\ \frac{\mathrm{mol}}{\mathrm{L}}$

I can't speak to why your other source has a different number.