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Topic: Solubility of Unknown  (Read 12121 times)

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Offline prncess23

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Solubility of Unknown
« on: October 05, 2007, 02:56:42 AM »
Hello everyone, hopefully you guys can help me on this.

1. Via experiment i figured out that both Acetone & water and acetone and hexane are miscible. Question is: How can you explain these results, given that water and hexane are immiscible?

2. We are given a test tube containing two immiscible liquids and a solid organic compound that is dissolved in one of the liquids. The identity of the two liquids are told to us as well as the solid, but we don't know the relative positions of the liquids or in which liquid is the solid dissolved.

Find out the positions of the liquids and in which liquid the solid is dissolved in first a) without doing an experiment (use density) and b) experimentally.

a) the test tube i got had: methylene chloride and water as the liquids and fluorene as the solid.
The densities are:
methylene chloride : Density. 1.3255 g/cm³, slightly soluble in water and miscible with most organic solvents.
water : 1.0 g/cm³
fluorene: 1.202 g/cm³, insoluble in water, soluble in organic solvents.

from that i know what is on top (water) and what is on bottom (methylene chloride), and that fluorene is dissolved in methylene chloride since it is insoluble in water.

However, the problem is that now we have to figure that out experimentally and write out a procedure. I don't know how i would do that, which is where i need your help. Remember, this is the first lab in organic chemistry we did and we aren't advanced yet...it has to be fairly simple. the lab book says something about separating the two layers with a Pasteur pipet and evaporating the liquid. Which ever one has crystals left after evaporation will determine that the solid was dissolved in that specific liquid...right? can i do that??

Thanks.



Offline prncess23

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Re: Solubility of Unknown
« Reply #1 on: October 05, 2007, 03:10:09 AM »
oh yea i forgot to mention one thing.

the tube has a little bit of the solid on the top of the liquid...dont know if that means it was dissolved in water...and since it is floating on the top and water is on the top of the test tube and fluorene is insoluble in water, that might explain why the solid is floating on the top...

Offline saidinstouch

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Re: Solubility of Unknown
« Reply #2 on: October 05, 2007, 03:47:35 AM »
I'm still trying to recall my chemistry, so don't take my answer as the end all, but I will try to help for part 2.

For the experiment, you have to remember that two immiscible liquids wont mix (oil/water) and will instead partition into two layers.  The book gave you the basic solution.  The interface between the two layers is something that is surprisingly defined.  If you want to separate the two layers, take a pipette and remove a majority of the top layer carefully by keeping the tip away from the partition.  After you have a majority of the layer removed you even more carefully and slowly move the pipette closer to the partition and you can actually remove most of the top layer without getting any of the bottom layer in your pipette.  Place the top layer in a petri-dist (or other dish to allow evaporation) and then repeat this process for the bottom layer.  This is a fairly common technique from what I remember, so get used to doing it, though like I said I am sure I have forgotten some things since I was last in the lab.  Hopefully you get the general idea and if there are details missing you can fill them in.

Offline russellm72

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Re: Solubility of Unknown
« Reply #3 on: October 05, 2007, 07:54:08 AM »
Yes separate the two layers into two seperate test tubes. Now if you place the test tube containing the lower layer (DCM) in a warm water bath the DCM will quickly evaporate and you should be left with a nice solid....Fluorene.

You can do the same with the water layer but clearly this would take longer and you will have no residue. Well nothing like what should be left from the DCM evaporation.

Have you heard of TLC?

If so, then TLC both the layers against Fluorene as a standard. This way you can tell in which the compound is dissolved and you can be reasonably confident of it's identity. Clearly for your level this is enough. At a more advanced stage you'd need more techniques to verify the identity of the component e.g. NMR or easier and more apt for you melting point.

You see easy peazy. :)

R.

Offline Door 218

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Re: Solubility of Unknown
« Reply #4 on: October 06, 2007, 03:03:15 AM »
1. Via experiment i figured out that both Acetone & water and acetone and hexane are miscible. Question is: How can you explain these results, given that water and hexane are immiscible?

Polarity decreases in the order of H20 > CH3COCH3 >> C6H8, so does that offer an explanation?

Offline prncess23

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Re: Solubility of Unknown
« Reply #5 on: October 06, 2007, 08:52:28 PM »
so even though there is solid on the top of the water, it is still dissolved in the DCM right?

Offline Door 218

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Re: Solubility of Unknown
« Reply #6 on: October 07, 2007, 02:51:10 PM »
"so even though there is solid on the top of the water, it is still dissolved in the DCM right?"
I doubt any DCM positioned above water because it's significantly more dense, so any fluorene found on the water layer is some that didn't dissolve in DCM.

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