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### Topic: Mole fraction calculation  (Read 19845 times)

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#### betty87

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• Mole Snacks: +0/-0 ##### Mole fraction calculation
« on: March 19, 2008, 12:04:55 PM »
I obtained a top fraction and bottom fraction from a distillation column containing water and methan ol. I know the density of the mixture and of the two components, but have no data on the volume. I need to find the mole fractions of each of the components in each of the mixture. This is what I have come up with:

Mass mixture = mass methanol + mass water
m =pV
pV mixture = pV methanol + pV water
For the top fraction:
0.7954 Vmixture = 791.2 Vmethanol + 1000 Vwater
mass = no of moles x molecular weight

So for methanol
Mass = 791.2 Vmethanol / 32 = 24.7V moles
And water
mass = 1000 Vwater / 18 + 55.6Vmoles

But this can't be right as the top fraction would be mainly methanol, and also my calculation for the bottom fraction comes out the same as this. I am totally stuck!

Any help would be greatly appreciated!

#### Borek ##### Re: Mole fraction calculation
« Reply #1 on: March 19, 2008, 12:15:40 PM »
What is pV?

It is possible to convert density to mole fraction using density tables. CASC does it automatically. Without density tables it is impossible.
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#### betty87

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• Mole Snacks: +0/-0 ##### Re: Mole fraction calculation
« Reply #2 on: March 19, 2008, 12:19:34 PM »
rho V as in density X volume

Yeah i thought it was impossoble too. This is definitely the only information we have access to, so I'm stumped.

#### Borek ##### Re: Mole fraction calculation
« Reply #3 on: March 19, 2008, 12:29:14 PM »
If you are not allowed to use density tables, whole thing doesn't make sense to me, as dependence between density and composition is not linear. Perhaps you are expected to look for this data by yourselves?
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#### betty87

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« Reply #4 on: March 19, 2008, 12:40:53 PM »
Well I suppose we must have to look for it, although nothing was mentioned to us. If the relationship between density and composition isn't linear, then would it be appropriate to use interpolation to find the composition from the tables?

#### Borek ##### Re: Mole fraction calculation
« Reply #5 on: March 19, 2008, 01:23:05 PM »
Linear interpolation between points in the table will work. That's assuming the table has a little bit more than density of pure water and density of pure methanol Say every 5%. Could be even 10% will do. A lot depends on the substance, but methanol has very flat and smooth density vs composition plot.
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#### betty87

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• Mole Snacks: +0/-0 ##### Re: Mole fraction calculation
« Reply #6 on: March 19, 2008, 06:29:20 PM »
Right I'll do that then!
Thank you for your *delete me*!

#### eugenedakin

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• Gender: • My desk agrees with the law of entropy ##### Re: Mole fraction calculation
« Reply #7 on: March 19, 2008, 08:31:40 PM »
Here is the way I see it:

0.7954*M(mix) = 0.7912*M(methanol)+1.000*M(water)

M(mix) = M(methanol) + M(water) : assume the total mass is 100 (just a number)

100 = M(methanol) + M(water)
M(methanol) = 100 - M(water)

substituting 100 into the first equation:

79.54 = 0.7912*M(methanol)+1.000*M(water)

combine the two equations:

79.74 = 0.7912*(100-M(water)) + 1.000*M(water)
79.74 = 79.12 - 0.7912M(water) + 1.000M(water)
0.62 = 0.2088M(water)
water = 2.97

substitute the answer in on of the above equations:

100 = M(methanol) + 2.97

M(methanol) = 97.03

The answer I have is 97.03 % (or 0.9703 mole fraction) is methanol, and 2.97% (or 0.0297 mole fraction) is water.

I hope this helps Eugene

There are 10 kinds of people in this world: Those who understand binary, and those that do not.

#### Borek ##### Re: Mole fraction calculation
« Reply #8 on: March 20, 2008, 04:20:33 AM »
Eugene: volumes are not additive, while your approach gives approximate answers, it also gives false idea that that's a general way of doing it - it is not.
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#### eugenedakin

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• Gender: • My desk agrees with the law of entropy ##### Re: Mole fraction calculation
« Reply #9 on: March 20, 2008, 07:56:00 PM »
Hello Borek,

Yes, you are quite correct.  The approach I gave is an approximation.  In a laboratory, my method is probably too general for most applications.  In the oilfield, this is accurate enough (commonly referred to as OFCE, or Oilfield Close-Enough).

As an example in the oilfield, a blend (methanol and water) in a tank quite often contains a high level of contamination (1% crude oil, probably sand, clay and other possible contamination).  Most calcuations such as this frequently would like to know what the freeze point is, or the affinity to absorb more water.

Again, you are right when talking about chemistry within, lets say, pharamceutical areas, in the oilfield, this is accurate enough.

Sincerely,

Eugene
There are 10 kinds of people in this world: Those who understand binary, and those that do not.

#### Rabn

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« Reply #10 on: April 02, 2008, 03:18:14 AM »
If you are distilling the mixture why not stick a thermocouple or thermometer before the distillation condensation section of the set up. THere should be a definite gap in time between the methanol vaporizing and the water vaporizing.  You will likely get an azeotrope, but knowing the the percentage would allow you to find the desired answer.