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Simple question about dissolving a compound

by JG89
Tags: compound, dissolving, simple
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JG89
#1
May9-14, 05:41 PM
P: 726
Suppose I am trying to dissolve a compound into a minimal amount of solvent (let's assume that this compound is not pure and its impurity is not soluble in the solvent). So I have this compound in a glass and I slowly add the solvent, waiting until my compound is dissolved. How do I know that what isn't dissolved yet is the impurity and only the impurity? Perhaps some of what is not dissolved yet is my compound which I am trying to isolate but I need more solvent to dissolve it. How am I to know?


Also, if after dissolving my compound and the solution is murky, does this mean that not everything is fully dissolved? i.e. I should filter the solution so that it's clear?
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Yanick
#2
May9-14, 08:06 PM
P: 382
I don't know that you can really get a very precise means of determining how much more solvent to use or not. If you are going to do quantitative work then you add small amounts of solvent, perhaps as little as a few drops from a pipette at a time, and then quantitate after.

In general solutions are more or less clear, cloudiness usually means that you haven't really dissolved solute but created a suspension. Filtering or centrifuging and looking for a pellet are relatively simple ways to check.
SteamKing
#3
May9-14, 08:27 PM
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If the impurity is insoluble, it shouldn't matter how much solvent you add: the impurity isn't going to dissolve, ever. That's what 'insoluble' means.

eigenperson
#4
May10-14, 01:14 PM
P: 160
Simple question about dissolving a compound

Ideally the impurity will be much less soluble than the compound you are trying to isolate (let's say >100 times less soluble). If this is the case, then at first, when you add a little bit of solvent, a noticeable amount of the compound will dissolve (so the quantity of solid will decrease noticeably). Once all the compound is dissolved, adding a little bit of solvent will produce almost no change in the amount of solid. That's how you know when to stop. As a bonus, if you overshoot a bit it isn't a disaster because the impurity is so insoluble that you will not end up contaminating the solution very much.

On the other hand, if the compounds have similar solubilities, then it's much more difficult to do this kind of purification. You should choose a different solvent or a different method, unless you have a lot of patience and a good idea of how much solvent will be needed.
lavoisier
#5
May11-14, 04:11 AM
P: 18
I think there's something about this in Vogel's Practical Organic Chemistry.

But probably that was about crystallisations. In a crystallisation your compound of interest is usually in the solid phase; we could say that you want to enrich the solid phase in one substance and leave as much as possible of all the other substances behind in the liquid phase.
In your case you're doing the opposite: you want to dissolve the compound of interest and leave the other substance(s) as 'undissolved' as possible.

So, say that you start from a mass m of solid, containing a percentage p of compound of interest (let's call it P). So the solid contains a mass m*p of P and a mass m*(1-p) of an impurity I.

Now you add a volume V of solvent.
If the solubility of P in the solvent is SP, and the solubility of I is SI, with SI < SP, then (assuming the dissolution of each compound is independent of the presence of the other compounds, and neglecting the change in volume etc):
- mass of P in the liquid phase: min(V*SP,m*p)
- mass of I in the liquid phase: min(V*SI,m*(1-p))
- mass of P in the solid phase: max(0,m*p-V*SP)
- mass of I in the solid phase: max(0,m*(1-p)-V*SI)

So, if the above is correct, you can decide how much solvent to use (V) by setting up the appropriate equation.

If you want to get all of P in the liquid, while dissolving as little I as possible, then:
V = m*p / SP
In this case, after filtration and evaporation, the solid will contain a percentage of P equal to p1:
p1 = SP / ( SP + SI )
The above procedure is worth doing only if p1 > p.
If it's true that the solubility of I is very low compared to the solubility of P, then p1≈1, as already pointed out by SteamKing.

How do you get the solubilities of P and I in the solvent?
Personally I would use LCMS. If you add a small amount of solvent to your starting solid, so that both P and I are at saturation, filter the suspension and inject, the areas of the peaks should give you an indication of the relative solubilities. But this will strongly depend on the type of detector, etc.
If you want to be more precise and/or know the absolute values, you'd have to get pure samples of each substance, do a calibration, etc.

Note that this is an approximate method I would use for standard work in organic chemistry. I'm sure analytical chemists have much more refined ways of doing this.


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