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Author: Subject: Figuring out ratio of binary mixtures with known specific gravity
Douchermann
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[*] posted on 19-6-2009 at 12:07
Figuring out ratio of binary mixtures with known specific gravity


Hey guys, sorry I couldn't word the title better. I have a solution of nitromethane and methanol. It's not the 8% azeotrope, because the SG of that should be 0.812g/ml and the solution I have has an SG of 0.881g/ml. I'm trying to figure out a formula to solve this easily without knowing the percentage of either. I did a guess and check, and the closest I felt like getting is 74.25% methanol / 25.75% nitromethane. However, the guess and check was painful and monotonous. Since the variables are in a fixed ratio, I was wondering if anyone knew of, or could figure out a formula. I tried this:

(X*0.7918) + (Y*1.1371) = 0.881

but by plugging in zero's for variables and solving for the opposites, none of the data would match up. I know double variable equations are possible, I just cant remember if it's doable with simple algebra. Anyone have any pointers or ideas?

By the way, Methanol = 0.7918g/ml Nitromethane = 1.1371g/ml




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Nicodem
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[*] posted on 19-6-2009 at 12:29


You can not calculate the composition of a liquid from the specific gravity since the two compounds interact (dipole, Van der Waals, and other interactions).
You can only determine the composition if you have tabular data or an equation made from empiric data (these are usually polynomial equation while you presented a linear one; so as you see, you see it too simple).

Edit: I think you missed that Y=1-X in your conception of the problem. But anyway, as I said, the solution is not in a linear equation.

[Edited on 19/6/2009 by Nicodem]
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Douchermann
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[*] posted on 19-6-2009 at 13:33


You're right, i completely forgot about interactions. However, I could still get in the ball park, could I not? Even if not from an equation, just a linear graph. Error within the 5% range is acceptable.



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Nicodem
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[*] posted on 20-6-2009 at 08:18


I'm afraid the error using a linear correlation would be more than 5%.
Anyway, if you still have not solved your equations, these are my results: X=0.742 which makes Y=0.258.
Maybe you could use other properties instead. While tabular data or an empirical equation for the density of MeOH/MeNO2 mixtures might be difficult to obtain, it is quite likely that you can find a boiling point tabular data or an equation for this mixture. Chemical engineers like to publish this sort of factographic stuff. I suggest you consult the literature. You might be lucky. Or else just separate the two compounds.
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kmno4
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[*] posted on 20-6-2009 at 08:43


Set (partial) of densities for MeNO2 - MeOH can be found here:
Bulletin of the Chemical Society of Japan
Vol.43 , No.6(1970)pp.1634-1642
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Nicodem
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[*] posted on 20-6-2009 at 10:45


Kmno4, is there anything that you can not find? :D
Douchermann, the absolute error between a linear calculation value and empirical value is 5.9% (which is 7.9% relative error). If you are satisfied with that, then you can use a linear equation (but the error depends on composition as well).




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JohnWW
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[*] posted on 20-6-2009 at 12:51


The International Critical Tables, of which there are uploads in the References section (if not, I will re-upload them), should have the densities of binary mixtures of various mutually soluble liquids over the range of mixtures.
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kmno4
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[*] posted on 21-6-2009 at 00:49


Quote: Originally posted by JohnWW  
The International Critical Tables, should have the densities of binary mixtures of various mutually soluble liquids over the range of mixtures.

Should ?
Does it mean that you do not know if these Tables contain these data ?
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JohnWW
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[*] posted on 21-6-2009 at 01:07


The Tables do have the densities of many binary mixtures, and of some ternary mixtures, of miscible liquids; along with phase diagrams and refractive indices and some other properties. Have you found them yet?
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Nicodem
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[*] posted on 21-6-2009 at 06:53


Quote: Originally posted by Nicodem  
Douchermann, the absolute error between a linear calculation value and empirical value is 5.9% (which is 7.9% relative error). If you are satisfied with that, then you can use a linear equation (but the error depends on composition as well).

Ups... I missed that the table list compositions in mol fractions not volume fractions, so the error I gave above is completely wrong. The actual difference between the value obtained from your equation and the empirical one reported in Bull. Chem. Soc. Jap. is very small. The relative error is only 0.84%.




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