[alt-photo] Re: molarity problem

[EJ Photo]

Joe, It is best to look at the compound as a unit and parts. It might look
like your mixing the abbreviations in your question which can also leads
problems, in some cases, it looks like mol = molecule and others you are
referring to a mole. 

there was a citing in an earlier post as to the magic number, Avogadro's #
of 6.022 x 10 the 23rd. that is a mole or number abbreviated mol ( without .
)

molecule = mol. (with .) 

Just trying to help those following along. With all the different email
programs, etc . can jump around.
 
http://www.cas.org/products/print/ca/standabbrevacro.html


And Alberto pointed out, careful of rounding errors. 
 

Eric Neilsen
Eric Neilsen Photography
4101 Commerce Street, Suite 9
Dallas, TX 75226
 
www.ericneilsenphotography.com
skype me with ejprinter
www.ericneilsenphotography.com/forum1
Let's Talk Photography
 

-----Original Message-----
From: alt-photo-process-list-bounces at lists.altphotolist.org
[mailto:alt-photo-process-list-bounces at lists.altphotolist.org] On Behalf Of
Joseph Smigiel
Sent: Friday, July 23, 2010 10:04 PM
To: The alternative photographic processes mailing list
Subject: [alt-photo] molarity problem


On Jul 23, 2010, at 7:25 PM, Terry King wrote:

>
>
> 'Molarity' has it's place ...


OK.  Let me see if I have this straight using a different process.   
In wetplate collodion some of the commonly used salts are Cadmium  
Bromide (CdBr2) and Cadmium Iodide (CdI2).  I'd like to get away from  
using cadmium since it is so toxic and substitute something like  
Potassium Bromide (KBr) and Potassium Iodide (KI) salts.  The halogen  
combines with the silver in the Silver Nitrate (AgNO3) sensitizing  
bath to form Silver Bromide (AgBr) and Silver Iodide (AgI) and  
therefore the proper balance of halogen salts is important.

So, to maintain proper halogen content when substituting the metals,  
I'd need to know the following:

KBr is 119 gm/mol

KI is 166 gm/mol
CdBr2 is 272 gm/mol

CdI2 is 366 gm/mol



There would be 2 mol Potassium and 2 mol Bromide in 2 gm KBr.

There would be 1 mol Cadmium and 2 mol Bromide in 2 gm CdBr2.  (is  
that 1:2 mol ratio correct for the compound?)

Then, if I have:

2 gm KBr  * (1 mol KBr /119 gm KBr) ~ 0.0168 mol KBr

and

0.0168 mol KBr * (1 mol Br/ 1 mol KBr) * (1 mol CdBr2 / 2 mol Br) *  
(272 gm CdBr2/ 1 mol CdBr2) ~ 2.285 gm CdBr2

2 gm KBr would contain as much bromide as 2.285 gm CdBr2.

Therefore the substitution would be ~ 1 gm KBr / 1.14 gm CdBr2.   
(Similarly, the substitution between KI and CdI2 would be ~1 gm KI /  
1.10 gm CdI2.)

If a formula called for 1.0 gm CdBr2, I would need to substitute ~  
0.88 gm KBr.


Could a chemist on the list verify the above for me?  Thanks.

Joe


p.s.,  Since KBr is essentially insoluble in alcohol or ether, and  
collodion deteriorates with very little water content, and the grain  
alcohol used already has 5% water content, making percentage salt  
solutions for this process doesn't make sense.  At least I don't see  
where it would be helpful.
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