Douchermann
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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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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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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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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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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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Kmno4, is there anything that you can not find? 
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).
…there is a human touch of the cultist “believer” in every theorist that he must struggle against as being
unworthy of the scientist. Some of the greatest men of science have publicly repudiated a theory which earlier they hotly defended. In this lies their
scientific temper, not in the scientific defense of the theory. - Weston La Barre (Ghost Dance, 1972)
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JohnWW
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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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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.
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Should ?
Does it mean that you do not know if these Tables contain these data ?
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JohnWW
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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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| 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%.
…there is a human touch of the cultist “believer” in every theorist that he must struggle against as being
unworthy of the scientist. Some of the greatest men of science have publicly repudiated a theory which earlier they hotly defended. In this lies their
scientific temper, not in the scientific defense of the theory. - Weston La Barre (Ghost Dance, 1972)
Read the The ScienceMadness Guidelines!
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