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Author: Subject: Converting percent (%) solution to molarity (M)
BromicAcid
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sad.gif posted on 17-6-2004 at 19:15
Converting percent (%) solution to molarity (M)


If you are lucky enough to find an OTC product that lists the contents more often then not they are in % composition. I used to think I could do the conversion fairly well, however during my recent excursion into HBr manufacture my figures for the molarity of the solution (48%) thus resulting were off by an unacceptable amount.

Therefore I ask three questions, beginning questions, assume I have no prior ability:

How would I find the molarity of a 48% solution of HBr in H2O? (Yes, I know the answer but I want to see how others found it).

How would I find the molarity of a 6% NaOCl solution? (I know this one too but wanted a solid dissolved in water to make this question more complete).

And to round off the common states of matter, the molarity of a 70% isopropyl alcohol solution in water.

And what's up with the w/w and w/v aspects. Seriously, I never learned any of this in chemistry, everything was always in molarity and it was unheard of to have a percent solution, therefor my current techniques are a mishmash of internet help sites that somehow produce reasonable numbers.

Yes, I have googled my butt of on this one, and because this is in the beginnings section I'm hopping, no, pleading, for a bit of spoon feeding :o




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DDTea
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[*] posted on 17-6-2004 at 20:01


I have found that percentages are done by mass and not volume; at least from my experience with 30% Nitromethane. So, naturally, it's just making percentage calculations for a given mass... For example, 30% of 100 g of NM solution is 30 g... At least, such calculations worked for the Chloropicrin synthesis, so I'm assuming I've done this properly. If, however, it is done by volume, naturally you will find the percent of a given volume, then look up the density of the given chemical to find the mass.
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Magpie
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[*] posted on 17-6-2004 at 21:18
molarity vs %


Samosa is correct but I will approach this a little differently:

Molarity is g-moles per liter of solution.

For consumer products gases and solids are usually reported in wt%. For liquids in solution it will likely be vol%.

Assuming, for the sake of argument, that HBr is a liquid, then molarity would be:

molarity = [(vol% HBr/100)(vol of soln, mL)(sp. gr. of HBr)/(MW of HBr)]/[vol of soln, L]

- this assumes no volume change on mixing.

For 6% NaOCl,

molarity = [(wt% NaOCl/100)(wt of soln, g)/(MW HBr)]/[vol of soln, L].

Hope this helps and that I didn't just add to the confusion.




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[*] posted on 18-6-2004 at 08:09


chemists tend to like
molarity.....moles per litre of SOLUTION
-good for volumetric chemistry

physists like
molality.... moles per kilo (in the case of water 1l) of SOLVENT
-good if you are dealing with masses, not volumes.


for concentrations up to about 3-5M there really isnt much difference between the two......the 6% bleach soln. will almost be exactly the same in molality as molarity. However for the 70 % IPA its actually more sensible to deal with mole fractions/ molality.
1kilo of solution contains
700 g IPA, 300g water
3.333x700 g IPA: 1kg H2O
xmoles IPA: ymoles H2O

The problem with the convertion to molarity is you will need the density of the solution.... and thats a function of concentration. Mole fraction and molal is a far easier way to go in all but volumetric chemsitry.
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The_Davster
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[*] posted on 18-6-2004 at 16:45


When calculating the molarity of the 6% sodium hypochlorite solution why did you divide by the molar weight of HBr, or is that a typo?:P
So a 10.8% sodium hypochlorite solution would work out to what? I need this so I can run the stoich and calculate my yield for the haloform reaction.

Edit: I actually got off my lazy ass and got some numbers. I did two trials because I was using a 50mL graduated cylinder for volume measurements(could not find my 10mL pipett).
Trial 1 v=10.0 mL m=11.4g
Trial 2 v=1.00mL m=10.7g
ave mass per 10 mL=12.05g...hmm those numbers are too different mabey I should have shook the bleach up first?

so c=((10.8%)/(100%))(12.05g))/((74.44g/mol)(0.0100L))
c=1.75mol/L
Does that sound right to everyone?

[Edited on 19-6-2004 by rogue chemist]
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[*] posted on 18-6-2004 at 18:10
molarity calculation


rogue chemist:

1) You divide the weight of the solute (HBr or NaOCl) in grams by the molecular weight. This gives you the number of g-moles that you have. This is necessary no matter what the solute.

2) Your calculation looks good to me.




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[*] posted on 18-6-2004 at 18:12
meh


That was a typo! sorry.



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[*] posted on 18-6-2004 at 19:29


The only problem I face in this regard is that not all chemical suppliers state the r.d. of the solution on the labels, printing only the % composition. Therefore, when I try to look up the r.d.'s of the solutions (with known % compositions) from chemfinder, they very from site to site! eg. one site listed 70% HNO<sub>3</sub> as 1.37 while another claims it is 1.42. My experimental observations agree with neither- 1.25!

Does anyone know any site which has an index of the r.d's of some common reagents along with their respective % compositions?
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[*] posted on 19-6-2004 at 08:31
r.d.?


I have never heard the term "r.d." but I assume it means density.

Since I am "old school" I go to Perry's "Chemical Engineer's Handbook" for densities of common reagents. I believe the CRC handbook also has such tables. Then there is the ultimate "International Critical Tables" if you have access to a good technical library.

There is a thread on this I believe. Search on Tom Haggen. He was looking for such a table on nitric acid IIRC.




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[*] posted on 19-6-2004 at 19:11


r.d.= relative density. r.d. is the ratio of the mass of "x" volume of a solution to "x" volume of pure water at 14.5<sup>o</sup>C.

Thanks for the references- I'll look them up.
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[*] posted on 7-1-2014 at 03:49


I had this exact same problem several years ago, and I asked my science teacher how to work it out. I got this formula:
Conc. moles = <u>known molarity x unknown percentage</u>
____________________known percentage

So to calculate the concentration of a 28% solution of sulphuric acid:
<u>18.4 x 28</u> = 5.26M
__98

It does, however, require the knowledge of the percentage concentration of a known molar concentration.

[Edited on 7-1-2014 by eidolonicaurum]

[Edited on 7-1-2014 by eidolonicaurum]




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Zyklon-A
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[*] posted on 7-1-2014 at 06:42


Posted in 2004....



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[*] posted on 9-1-2014 at 01:18


It may be of use to future people who look at it.



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[*] posted on 9-1-2014 at 03:59


my first thought is:
48% HBr means that in 100g of the acid there are 48g of HBr gas
6% NaOCl means that in 100g of solution there is 6g NaOCl
70% IPA means that in 100g of the soln. there are 70g IPA
from that you can calculate the molarity.

but after reading the replies, I'm confused... :o






all above information is intellectual property of Pyro. :D
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BromicAcid
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[*] posted on 9-1-2014 at 04:30


The real problem is that for concentrated solutions the density is affected and there is no accurate rule of thumb. Concentrated ammonia solution is 0.88 g/mL whereas concentrated hydrogen chloride solution is 1.19 g/mL. Without knowing the density it can be difficult to convert to molarity because molarity by definition requires you to know moles per liter.

So eidolonicaurum your equation would not account for a much more or much less concentrated solution because the density is going to change so wildly but for solutions of similar concentration it would work because using the reference you would have a nearby starting point.




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blogfast25
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[*] posted on 9-1-2014 at 06:12


Quote: Originally posted by Pyro  
from that you can calculate the molarity.

but after reading the replies, I'm confused... :o




Yes, but not without knowing what the volume of 100 g of solution is, i.e. what is the density of the solution.

A) Divide the 100 g by the density in g/cm<sup>3</sup> and multiply by 1000 to get L (dm<sup>3</sup> as we're all supposed to call it now).

B) Divide the % of substance by its molar mass to obtain moles.

Divide number of moles by number of L, i.e. A) by B) to get moles dm<sup>-3</sup> i.e. M.


[Edited on 9-1-2014 by blogfast25]




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[*] posted on 9-1-2014 at 07:29


ah, ok. thanks for clarifying!



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[*] posted on 10-1-2014 at 11:26


Quote: Originally posted by blogfast25  
Divide number of moles by number of L, i.e. A) by B) to get moles dm<sup>-3</sup> i.e. M.




Ooopsie: that should have been:

"Divide number of moles by number of L, i.e. B) by A) to get moles dm<sup>-3</sup> i.e. M."




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