ChemNEWBIE1320 - 5-9-2013 at 17:14
Hey everyone, I'm sure this is a very simple problem, but I am having an issue solving it as I am working on a pre-lab for a steam distillation
experiment I am getting ready to run.
Basically, what I'm trying to find is the amount of water required to steam distill a TOL/benzil sol'n. I have tried several times to get the correct
answer.
Using Dalton's law, we can assume that
molH2O/molorganic material=pH2O/porganic material.
The partial pressure of the organic material must be equal to
760mmHg-pH2O.
And the weight of H2O req'd per g of organic material is
Wt. H2O/g organic material=(MWH2O*pH2O)/(MWOrganic Material(760mmHg-pH2O))
I have a number in mind. I used the bp of the azeotrope of TOL/H2O (@ ~84°C) and the corresponding pH2O (@ ~416.8 mmHg). I also
have added MW of TOL (92.14 g/mol) and MW of benzil (210.23 g/mol) in the denominator of the second equation.
Should I be using a different number for the respective constituent's p @ T°C such as 25°C? Should I exclude MW of benzil from the
equation? Anyone have any ideas or am I just completely lost here? Thanks in advance!
-Newbie
{repost from original in "Organic" section}
bfesser - 5-9-2013 at 17:32
No problem. I removed the duplicate topic. (You can always do this yourself by hitting <img src="./images/xpblue/edit.gif" /> and checking the
<strong>! Delete this message !</strong> box.)
ChemNEWBIE1320 - 5-9-2013 at 17:39
thanks!
DJF90 - 6-9-2013 at 01:48
It'd help if you included the procedure for the experiment, so we can see if one or both of the components are supposed to be steam distilled...
http://d.web.umkc.edu/drewa/Chem321L/Handouts/Lab2Distillati...
It appears the steam is just to remove toluene. Using the example in the text it should be easy enough to work out what you want to know. Bp
toluene-water azeotrope is 84.1 *C, so using the table in that document, vapor pressure of water at that temperature is ca. 416.8 mmHg. As 760 mmHg is
atmospheric, that means that the vapor pressure of toluene at 84.1*C is ca. 760-416.8 = 343.2 mmHg.
Using the equations given in the text, moles water/moles toluene = press. water/ press. toluene = 416.8/343.2 = 1.21
So for each mole of toluene that distills, 1.21 moles of water also passes over. Converting that to a mass ratio...
(18x1.21)/(1.00 x 92.14) = 21.78/92.14 = 0.24g water required per gram of toluene.
As its an approximation, and benzil is apparently non-volatile with steam, it does not feature in any of the calculations.
ChemNEWBIE1320 - 6-9-2013 at 06:01
Thanks DJ, I very much appreciate the explanation. I guess initially I was thinking that the presence of benzil in the distilling flask alone would
contribute to a change in pressure which is what was hanging me up. I did work the problem excluding benzil's MW and got 0.24 g h2o/g TOL. Thanks
again
Newbie