[alt-photo] Re: Stoichiometry for the nonscientist

[Katharine Thayer]

On Jul 29, 2010, at 1:26 AM, David Ashcraft wrote:

> I pulled this definition off the internet:
> Molar concentration or molarity is most commonly in units of moles  
> of solute per liter of solution. For use in broader applications,  
> it is defined as amount of solute per unit volume of solution, or  
> per unit volume available to the species, represented by lowercase  
> c:[2]
>
> Here, n is the amount of the solute in moles,[1] N is the number of  
> molecules present in the volume V (in litres), the ratio N/V is the  
> number concentration C, and NA is the Avogadro constant,  
> approximately 6.023 × 1023 mol−1.
> Or more simply: 1 molar = 1 M = 1 mole/litre.
>
> I do not have a background in chemistry and have no idea what this  
> means.  Several posts have been made that state how easy it is to  
> understand (dumb me), I have no idea.  Surely someone could share  
> the knowledge and explain simply so we all could understand.
> At the start of this I didn't care, but now feel like I need to know.

I'd guess that came from Wikipedia. (You can see I don't have a lot  
of respect for the "encyclopedia anyone can edit." ) Wherever it came  
from,  it's not a very helpful explanation.  I see that since then,  
you've started figuring it out from Loris's examples, but just in  
case you or someone else would still like a simpler explanation of  
molarity and stoichiometry, here's my attempt at it:

Chemical reactions,  like the ones that create most photographic  
images,  are usually expressed in equation form, showing the number  
of molecules of each substance that combine to form a molecule of the  
reaction product(s).  The question for the practicing photographer  
is, how does the reaction equation translate into how much you need  
of each of the reactants (ingredients)?    Stoichiometry is the fancy  
word we use for the calculations that express the molecules in the  
chemical equation in terms of how many grams we need of each of the  
reactants, and a mole (the molecular weight of each of the substances  
in grams) is the magic measure that translates molecules into grams.

It's an amazing and remarkable fact of chemistry that no matter what  
kind of chemical we have, a mole of that compound contains exactly  
the same number of molecules as a mole of a different chemical  
compound.  (You don't really need to know what that number is, though  
it's been mentioned a couple of times in this thread, and was  
mentioned in the article you referred to; it's Avogadro's number,  
6.02 x 10^23;  the only thing you need to understand is that a mole  
always contains the same number of molecules, no matter what it's a  
mole of, and that's why we use moles to figure equivalencies of  
material in chemical equations.)

As Loris explained,  to calculate how much of a substance equals a  
mole, you add up the atomic weights (I guess they call it "mass"  
these days rather than "weight" which is how I learned it) for each  
of the elements in the material;  the sum, in grams, equals a mole of  
the substance.  So, to use  a very simple example, carbon has an  
atomic mass of 12, so 12 grams of carbon is a mole.  For sodium  
chloride, NaCl, the atomic mass of Na is 23, the atomic mass of  
chlorine is 35;  a mole of NaCl is 23+35=58 grams.   (You can get the  
atomic mass of each of the elements in a molecule from the periodic  
table, from a chemistry book or from the internet; they are pretty  
easy to find, and in fact you can usually find the molecular mass of  
the compound already calculated without having to add up the masses  
of the individual atoms yourself.)    A 1 molar solution is one mole  
in a liter of solution.

That's really all there is to it,  at least to a general  
understanding of what it's about and why molarity is important.  I'm  
one of those people who wants to understand how the processes I use  
work, and I do have a background in chemistry, so it seems natural to  
me to go into these things.  But I can also agree with Diana that if  
you have a recipe that works and if you have no background or  
interest in chemistry, you can make fine prints without ever  
understanding the chemistry at this level,  just as it's possible to  
drive a car without having an understanding of internal combustion or  
to heat up leftovers without knowing how a microwave works.  Each to  
his own, is what I've always said on this list, for... goodness,  a  
dozen years now.