semiconductive
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Posts: 287
Registered: 12-2-2017
Location: Scappoose Oregon, USA.
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Mood: Explorative
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Recrystallization, solubility of glacial acids.
I'm just doing some physical chemistry experiments at the moment, and am curious about what is more likely to recrystallize in a non-water system.
I'll describe what I've done, my reasoning, and I would appreciate any comments pointing out where I likely have made a mistake.
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I bought some sodium saccharine, which turns out to carry water of crystallization.
Saccharine is a weak acid, so I think the sodium might be removable by exposure to a stronger acid. At the same time, by boiling it in an acid and
refluxing I figure the water could be driven off.
So, I tried an experiment
roughly 1CC dry measure of NaSaccharine, in 4CC's glacial acetic acid.
I heated it until it fully dissolved using a soldering iron (25W) in a glass test tube.
The grains of sodium saccharine cause bubbling/boiling until all of it is dissolved. Once dissolved, the boiling point rapidly rises and bumping
occurs. Can easily erupt out of test tube.
When I allow to cool, roughly the same amount of crystals form (volume wise) as initially added as Na:Sacc. I think this means that the solubility of
saccharine in glacial acetic acid is low, when cold.
Acetic acid will absorb water, but the molecular weight of acetic acid + water is lower than that of acetic acid bonded to saccharine; so I expect the
acetic acid to co-boil with water leaving a continuously stronger solution of saccharine in acetic acid.
When re-heating, a second time, the solution does not boil during the dissolution of the crystals. I think this is a good sign that the water of
crystallization, originally in the saccharine, has transferred into the acetic acid. I boil the system down until only 1CC of liquid remains. Then
I add 2cc's of fresh glacial acetic acid again.
Hopefully, the remaining solution is now very close to anhydrous.
I am not sure how to get the sodium to separate from the saccharine.
I know if I add a stronger acid, such as formic acid, that the sodium ought to prefer the formic acid over the saccharine. But, I am not sure which
will want to go into solution more -- the saccharine, or the sodium formate.
It occurrs to me that MgSO4, cooked to dryness, will not dissolve in glacial acetic acid. Perhaps I could use it as a dessicant, and because it is
highly polar, I would expect it to cause any traces of water or small polar molecules (Na:Formate), to stick to it. Adding 1/2CC of dry powder
sinks to the bottom of the tube. It increases the rate of re-crystallization when cooling the tube, remarkably.
The entire tube rapidly fills with crystals when cooling, and the liquid tends to become colloidal when heated to boiling with MgSO4 present.
I begin adding formic acid drop wise, re-heating and cooling the tube, in particular I begin to heat the top of the tube (rather than the bottom) and
place the bottom of the tube on a cold metal surface in order to encourage crystallization to start from the bottom of the test tube. I also cap the
tube with a rubber bung to reduce heat loss from the top.
After several small additions of formic acid (7 drops/ cycle of heating recrystallizing), I notice that a significant amout of the saccharine no
longer crystallizes out. I have roughly 1/2CC of crystals forming, AND the remaining solution becomes clear after several cycles of heating the
remaining liquid and cooling it again. ( But, if I heat the bottom of the test tube, the entire thing becomes colloidal again. )
So, I have a situation where roughly 1CC of dry saccharine powder now re-crystallizes into 1/2cc of crystals. Upon careful heat cycling while leaving
a few crystals un-dissolved on bottom of tube, I can get the remaining solution to turn clear.
Now, I am not sure what is crystallizing out and what is remaining in solution.
I have been told that during crystallization, substances tend to separate since crystals do not like to include multiple different chemicals.
I am hoping that saccharine (being a bigger molecule, less polar), is more likely to remain in solution than sodium formate will. Would you expect
this to happen, but if not, why not?
Thanks. (Curious.)
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