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Simple solution
January 5, 2012 12:25 PM   Subscribe

This is pretty straightforward algebra/chemistry which is why I'm embarrassed to ask my co-workers to help me so...

If I have the dilution ranges of a series of solutions of MSG, how can I figure out, in grams, how much MSG I need in 1L of water to make the stock solution? I get that:

x M =y moles/z L

Let's say my stock would be 1.0x10(-1)M. And the molar mass of MSG is 169.111g/mol.

Given the above:

0.1M (msg) = 0.1moles/1.0 liter

and converting moles to grams = 0.1 moles x 169.111g/mol = 16.9g of MSG/L

Is that it? Or did I miss some element? It seems too simple but it's been a while.
posted by Katine to Science & Nature (8 answers total) 4 users marked this as a favorite
 
You seem to have it. 16.9 grams, it is.
posted by BevosAngryGhost at 12:38 PM on January 5, 2012


The general formula are:

Concentration (molarity) = number of moles / Volume

number of moles = mass / gram molecular mass

So:

C = n/V

C = (m/GMM)/V

Rearrange for mass:

m = C * GMM * V

(dimentionality: [mass] = [mol]/[vol] * [mass]/[mol] * [vol])

Plugging your numbers in:
C = 0.1 mol/L
GMM = 169.11 g/mol
V = 1L

=> m= 0.1 * 169 * 1 = 16.9 g
posted by bonehead at 12:41 PM on January 5, 2012


Thank you!
posted by Katine at 12:50 PM on January 5, 2012


This is tough to do not on a napkin. Heh. Here's an everyday way of thinking about it.

I always mentally set 1 M equal to (grams/mol)/1L.
So for MSG and it's MW of 169.1, I say "1 M = 169.1 g/ 1 L" (or, as often, 169.1 g / 1000 ml).
I also turn all my decimals into fractions and take care of any pesky powers of 10.
Usually that means converting to thousandths, but we'll ignore that for now.
(to me 1.0x10(-1)M = 1/10 M = 0.1 M = 100 mM)


Then I set equivalent fractions. Generically, that's
(desired molarity)*(1 M in grams per liter) = (grams needed for stock)/(volume of stock desired)


Here, that looks like this:
(1/10)*(169.1 g / 1 L) = x / (1 L)
Solving for x yields 16.9 g.

The advantage of thinking of it this way is that it makes sense conceptually for nice values like 1L of a 0.1 M solution but it still works for non-intuitive values

Say you need, 3 L of a 375 mM solution. Using MSG, that would look like this:
(375/1000)*(169.1 g / 1 L) = x / (3 L)
Now, solving for x = 3 * .375*169.1 g = 190.23 g MSG.

Conversely, if you ran only had a 5 g bottle of MSG and just wanted to make as much 0.1 M stock as you could, the formula would look like this:
(1/10)*(169.1 g / 1 L) = 5 g / x
Solving for x here gives you
(16.91 g) * x = 5 g * (1 L)
x = 5/16.91g * 1L => x = 0.295 L or 295 ml.

This is pretty much what bonehead said above, but typed up in the way this molecular biologist thinks about on a daily basis. I swear it looks very neat and orderly when written as proper fractions instead of this typed mess. Between this and civi=cfvf, you could now do most of my job.
posted by maryr at 2:33 PM on January 5, 2012 [1 favorite]


If I have the dilution ranges of a series of solutions of MSG, how can I figure out, in grams, how much MSG I need in 1L of water to make the stock solution? I get that:
Just a little nitpick. It is not 16.9g per litre of water; it is 16.9g per litre of the solution (i.e. the final volume). To make a litre of the solution, you will likely use slightly less than 1 litre of water. This matters in particular for solutions at high concentrations, where the differences between the two can be dramatic.
posted by kickingtheground at 11:05 PM on January 5, 2012


For solutions of 1M or less, the common rule of thumb is to treat the solute behaviour as ideal and not worry about density effects. It's measurable, but usually less than the uncertainty of measurement of weight in either the balance or the glassware. Class A glassware is only 1% to contain, after all.

Solutions can go either way anyway, expanding or contracting. Sugar strongly contracts an aqueous solutions, for example. The standard barkeepers' simple syrup is 2 pints of dry sugar in 1 pint of water, yielding, more or less, 1 pint of syrup. Such non-ideal behaviour really only become important for solutions of highly soluble materials in concentrations greater than a percent or two of their saturation values.

Short answer: for dilute solutions, you don't sweat fugacities.
posted by bonehead at 6:33 AM on January 6, 2012


blah: class A is +/- 0.1%.
posted by bonehead at 6:36 AM on January 6, 2012


The usual method, or rather, the method I was taught, in molecular biology is to make your solution in say 950ml of H2O as measured in a a beaker, then bring it up to volume in a graduated cylinder. That was you don't go over a liter by adding your solute AND you get a more precise measurement - beaker marks very widely.
posted by maryr at 8:06 AM on January 6, 2012


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