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!

  • $\begingroup$ @immx Your question is similar to this post chemistry.stackexchange.com/questions/717/… . $\endgroup$ – Eka Aug 30 '13 at 3:53
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    $\begingroup$ @Eka I'm sorry but I don't agree... I appreciate the "homework policy" you're directing me to; this isn't "homework" and I'm not asking for an answer on a plate. I think I should have mentioned that I don't see how molarity can be greater than the molarity of a pure liquid. $\endgroup$ – Louis Maddox Aug 30 '13 at 5:57
  • $\begingroup$ @Eka I followed the link to the post you mentioned and read your answer. You might consider revising it (see the [comment I left under the question] (chemistry.stackexchange.com/questions/717/…) $\endgroup$ – Silvio Levy Jul 31 '14 at 0:34

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.


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