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All 21 posts   Subject: Solvent Miscibility Chart   Please login to post   Down

(Chief Bee)
08-09-02 17:20
No 343523
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      Solvent Miscibility Chart
(Rated as: excellent)

Solvents are said to be miscible if the two components can be mixed together in all proportions without forming two separate phases.

For example, ethanol is miscible with water. Alcoholic beverages (or laboratory solutions of ethanol) comes in all strenghts between 0.5% and 99.5%, and regardless of composition water and ethanol never separates into two phases. On the other hand, toluene is immiscible with water in all proportions - neither one drop of water in a beaker of toluene or one drop of toluene in a beaker of water will dissolve and mix, and if you try to mix 50ml of each in a 100ml beaker, the toluene will just happily sit on top of the water in its own separate layer regardless of how much you threaten the mixture, it simply doesn't care.

If you mix two miscible solvents with a third, things can become a little complicated and unpredictable, especially if the third solvent is miscible with one, but immiscible with the other of the first two. The result will also vary with the ratios between the solvents and the temperature. Better make sure you do not try to make use of solvent mixtures with more than two components if you want completely predictable results.

Another similar table, with additional data:
(Hive Bee)
08-09-02 20:42
No 343586
User Picture 
      Very cool! Thanks Rhodium!     

Very cool! Thanks Rhodium!

The above post is purely fictional. Any resemblance to "real-life" is purely coincidental.
(Hive Bee)
10-27-02 23:51
No 373381
User Picture 

helpful Rhodium. TY

10-31-02 16:39
No 375023
      What if?     

What if somthing has a lesser density than toluene,but is miscible with water.Do you get three layers of separation?
(Title Addict/Eraser)
10-31-02 16:48
No 375029
      such as...     

jetson pondered this same question just the other day but hasn't yet to try it.  not sure if the two are miscible(tetra and xylene)but he thinks that h2o tetrachloroethylene and xylene without shaking/stirring and added xylene last would seperate into three layers.

the devil is so lonelymad
(Synaptic Self-Mutilator)
10-31-02 18:43
No 375075
      Three layers, yeah
(Rated as: excellent)

If all three are insoluble in each other you get three layers. (eg mercury, toluene and water) Then if you take a piece of gold, a stone, a piece of HDPE plastic and a piece of wood, drop them in, and you'll see the gold sink to the bottom, the stone float on the mercury, the HDPE plastic float on the toluene and the wood float on the water. smile
(Hive Bee)
11-01-02 01:27
No 375225
User Picture 
      Three layers with different densities     

Suppose Billy Bob has three solvents (x,y, and z). Two of them are miscible with each other (x and y), and the other is immiscible in either one (z). Of the two solvents that are miscible with each other, x is denser than y. Z is also denser than y.

Supposing you can put all three solvents in a container without allowing x and y to mix, the jar would be stacked into three layers. It seems to me that solvents x and y will never be able to mix with one another because of solvent y seperating the two layers (because no matter how many times you turn it upside down, the middle layer will remain in the middle). Is this a fair assumption?

Also, suppose there is an equal volume of each solvent. If you were able to mix solvents x and y after the above experiment, where will the combined solvents sit in relation to solvent z? Will solvent z be under or over the mixture of solvent x and solvent y? Do you simply average the densities of x and y to determine the density of the two mixed togethor?

When you meet a swordsman, draw your sword: Do not recite poetry to one who is not a poet.
(Hive Bee)
11-01-02 06:21
No 375300
User Picture 
      I'm glad you brought that up     

The thing about emulsions... Let's assume the emulsion aka frothy bubbly shit has two solvents instead of three (for simplicity's sake). If solvent x has honey in it, but it has formed an emulsion with solvent y, how would one go about seperating them into two layers to aid in the extraction of honey?

I was thinking of adding a third solvent which is miscible with solvent x but not solvent y in order to change the density of solvent x (hopefully making it better able to sink below solvent y). Would this help to break up the emulsion, or would this just complicate things? If not, what other ways can somebody use to break up an emulsion?

When you meet a swordsman, draw your sword: Do not recite poetry to one who is not a poet.
(Title on BackOrder)
11-01-02 06:31
No 375303
User Picture 
      an easy way     

If the layer you don't want contains a lot of water, you can add a saturated NaCl solution to weight down the water layer and separate the two.  If you are wanting to keep the water layer instead, you can usually add a NP solvent of lower density to separate the emulsion. 

If at first you don't succeed, try, try again. Then give up. There's no use being a fool about it.
08-11-03 10:11
No 452951

Slight deviation, please forgive.
Can any non-polar solvent be used to extract the honey? Why is some of the experimentals is diethyl ether/toluene chosen in preference to chlorinated solvents such as DCM and CCl3 for end extractions?

Is solubilty what factors in here or can Et20/toluene be happily substituted with DCM?
(Chief Bee)
08-12-03 19:01
No 453146
User Picture 
      Factors influencing choice of solvent
(Rated as: excellent)

Solubility factors play a role, as in any given extraction you want the solvent to extract as much of your product as possible, but as little as possible of the impurities (and when washing a solution its the other way around, you want only the impurities to be removed, not any of your product). To be able to predict which solvent that has the most optimal properties for a given extraction, you need a lot of experience, so don't despair if you are unable to do this.

Another factor which plays a role is the boiling point of the solvent, you want it to be relatively low, as for it to be easily removed by distillation/evaporation, it is therefore you often see diethyl ether and DCM being used in journal articles. The density is also of interest, sometimes you want the solvent to become the bottom layer (all chlorinated solvents) and sometimes the upper layer (most other non-polar solvents). The density should also be as different to water as possible, so that it easily floats or sinks, otherwise emulsions are likely to occur. The solvent should of course not react with any of the substances present, and it should naturally be immiscible with water (or separate layers will not develop).

Remaining factors include flammability (preferably non-flammable), toxicity (preferably non-toxic to both the chemist and the environment), cost/availability (it should be as cheap as possible), water solubility (it is usually desirable if the solvent is of low solubility in water, and that water has a low solubility in the solvent), and finally the tendency of the solvent to form emulsions (which is very undesirable).

Of course, you cannot have everything you ask for, so it is always a trade-off between all the above factors when a solvent is chosen.

The chlorinated solvents are in general very good solvents for everything, so they will extract everything that is not very very polar from an aqueous solution (including impurities), therefore, if you have a relatively impure aqueous solution you want to extract your product from, then it is desirable to use a less powerful solvent, such as ether or sometimes toluene. Ideally, you should not use a more powerful solvent than necessary, you should rather extract 3-4 times with a less powerful one than  only 1-2 times with a powerful one). If you have a relatively pure solution (where you mainly want to remove organic or inorganic salts and other water-soluble substances from your product), then there is every reason to choose a powerful non-polar solvent (like DCM), as to minimize the amount of solvent necessary to retrieve all of your product.

Sometimes, for example in those cases where your product is of relatively low solubility in organic solvents, you may need to use a powerful solvent anyhow, despite having a relatively impure reaction matrix. In those cases you cannot help extracting a lot of byproducts together with your desired product, but there is few lab mistakes as unnecessary as to use a too weak solvent (or extracting too few times, or with a too small solvent volume) so that you fail to extract the entire amount of product, leaving some of it behind in a phase that might be thrown away before discovering your error. Therefore, always make sure that you use a powerful enough solvent, and that you extract with a large enough volume of it* as to extract all of your product.

If you look at the polarity index at you can get a rough estimate how powerful the solvent is (the higher the polarity index (but still insoluble in water), the more powerful the solvent is). Acetone is very polar, but is miscible with water, so it cannot be used. The most polar water-insoluble solvents include ethyl acetate, chloroform and dichloromethane. The least polar ones include Pentane/Hexane/Cyclohexane and Petroleum Ether, and Benzene/Toluene/Xylene and Diethyl ether lies somewhere inbetween.

* Always divide the extraction solvent into several portions (most often three portions, used in nearly any literature procedure). It is significantly more efficient to extract a solution three times with 100 mL of a solvent each time, than it is to extract once with 300 mL! See the full explanation in Post 453799 (Rhodium: "Theory of Extraction", Newbee Forum).
08-13-03 10:52
No 453295
      solvent choice     

Just wanna say thanks. This is the last hurdle before I get under way and just wanted to get the facts straight.

That PDF is awesome smile

Rated: excellent - for surewink
(Hive Bee)
08-15-03 16:42
No 453756
      Factors influencing choice of solvent     

That was such a helpful post Rh.
Just beautiful. I probably would have asked for several of the answers presented.
(Chief Bee)
08-15-03 22:15
No 453799
User Picture 
      Theory of Extraction
(Rated as: excellent)

Here is a clarification to the section about extraction portion size (the last, and the third paragraph from the bottom) I just added to Post 453146 (Rhodium: "Factors influencing choice of solvent", Newbee Forum)

Theory of Extraction

"Several small extractions are better than one big one." Doubtless you've heard this many times, but now I'm going to try to show that it is true.

By way of example, let's say you have an aqueous solution of oxalic acid, and you need to isolate it from the water by doing an extraction. In your handbook, you find some solubilities of oxalic acid as follows: 9.5 g/100 g in water; 23.7 g/100 g in ethanol; 16.9 g/100 g in diethyl ether.

Based on the solubilities, you decide to extract into ethanol from water, forgetting for the moment that ethanol is soluble in water and that you must have two insoluble liquids to carry out an extraction. Chagrined, you forget the ethanol and choose diethyl ether.

From the preceding solubility data we can calculate the distribution, or partition coefficient, for oxalic acid in the water-ether extraction.

This coefficient (number) is just the ratio of solubilities of the compound you wish to extract in the two layers. Here,

which amounts to 16.9/9.5, or 1.779.

Imagine you have 40 g of oxalic acid in 1000 mL water, and you put that in contact with 1000 mL ether. The oxalic acid distributes itself between the two layers. How much is left in each layer? Well, if we let x g equal the amount that stays in the water, 1.779x g of the acid has to walk over to the ether. And so

Wt of oxalic acid in ether = (1000 mL)(1.7 g/mL) - 1779x g
Wt of oxalic acid in water = (1000 mL)(x g/mL) = 1000x g

The total weight of the acid is 40 g (now partitioned between two layers)


2779x g = 40 g
x = 0.0144


Wt of oxalic acid in ether - 1779(0.0144) g = 256 g
Wt of oxalic acid in water = 1000(0.0144) g = 14.4 g

Now, let's start with the same 40 g of oxalic acid in 10 mL of water, but this time we will do three extractions with 300 mL ether. The first 300 mL portion hits, and

Wt of oxalic acid in ether = (300 mL)(1.779x g/mL) = 533.7x g
Wt of oxalic acid in water = (1000 mL)(x g/mL) = 1000x g

The total weight of the acid is 40g (no partitioned between two layers), and

1533.7x g = 40g
x = 0.0261


Wt of oxalic acid in ether = 533.7(0.0261) = 13.9 g
Wt of oxalic acid in water = 1000(0.0261) g = 26.1 g

That ether layer is removed, and the second jolt of 300 mL fresh ether hits, and

Wt of oxalic acid in ether = (300 mL)(1.779x g/mL) = 533.7x g
Wt of oxalic acid in water = (1000 mL)(x g/mL) = 1000x g

But here we started with 26.1 g of acid in water (now partitioned between two layers) and

1533.7x g = 26.1 g
x = 0.0170


Wt of oxalic acid in ether = 533.7(0.0170)g = 9.1 g
Wt of oxalic acid in water = 1000(0.0170) = 17.0 g

Again, that ether layer is removed, and the third jolt of 300 mL fresh ether hits, and

Wt of oxalic acid in ether = (300 mL)(1.779x g/mL) = 533.7x g
Wt of oxalic acid in water = (1000 mL)(x g/ mL) = 1000x g

But here, we started with 17.0 of acid in water (now partitioned between two layers) and

1533.7x = 17.0 g
x = 0.011


Wt of oxalic acid in ether = 533.7(0.011) g = 5.87 g
Wt of oxalic acid in water = 1000(0.011) g = 11.0 g

(They don't quite add up to 17.0 g - I've rounded them off a bit.)

Let's consolidate what we have. First, 13.9 g, then 9.1 g, and finally 5.87 g of oxalic acid, for a total of 28.9 g acid extracted into 900 mL of ether. OK, that's not far from 25.6 extracted once into 1000 mL of ether. That's because the distribution coefficient i fairly low. But it is more.
That's because several small extractions are better than one large one.

Reference: James W. Zubrick, "The Organic Chem Lab Survival Manual", 4th Ed., John Wiley & Sons, 1997.

Edit: It has been shown in Post 464879 (Chromic: "More fun with math", Chemistry Discourse) that the above equations can be used to derive the fact that for optimum extraction effiacy with a given amount of solvent, all the extractions should be of the same size, so extracting something with 3x50mL is better than with either 40+50+60 mL or 60+50+40 mL.
10-21-03 15:51
No 465981
      What about amount per extraction?     

So its obvious from the fine theoretical work posted here by our fearless leader that three EQUAL extractions is the way to go.  But is there a rule for calculating how much solvent to use in each extraction as it relates to the mass of the material that is to be extracted and as well the total mass of the solution being extracted?  

To make it a more practical example, could someone illustrate the required amount of DCM to use to recover the finished honey from say a 2.5 KG NaBH4 post reduction solution (flooded workup) ?
(Chief Bee)
10-22-03 19:03
No 466165
User Picture 
      calculating the solubility product     

could someone illustrate the required amount of DCM to use to recover the finished honey from say a 2.5 KG NaBH4 post reduction solution (flooded workup)?

You cannot determine it beforehand without knowing the partition coefficient for MDMA between DCM and aqueous alcohol (and I doubt anyone has tabulated that anywhere). It will also differ if you have more or less salts dissolved in the solution, as well as if you have used methanol, ethanol or isopropanol as the alcohol, and how much water you flooded it with.

But - if you adhere to a standard protocol, and measure how much product you get in the first, second and third extraction, you can calculate the solubility product as in Post 464877 (Chromic: "Okay, more guessing", Chemistry Discourse) and then from that you can work out the optimal number of extractions, and how large each of them should be to retrieve 99.5% of all your product.
(Hive Bee)
11-24-03 02:24
No 472760
      Re: Properties of solvents     

Critical compilation of scales of solvent  parameters.

           Part I. Pure, non-hydrogen bond donor solvents
Pure Appl. Chem.,          Vol. 71, No. 4, pp. 645-718, 1999
           It has long been known that solvents often affect chemical reactivity, this involving, e.g., the shift of the position of chemical equilibria (thermodynamic aspect) as well as significant changes in reaction rate constants (kinetic aspect). Physical properties, particularly the frequencies and intensities of transitions in IR, UV-visible, fluorescence, NMR and ESR spectroscopies are also known to be affected by solvents.

 These phenomena are consequences of differences in the solvation of reagents and products (thermodynamic effects) or reagents activated complexes (kinetic effects). Differential solvation of species in the ground and excited states accounts for the spectral  phenomenology indicated above. Differences in solvation of a given solute  in two different solvents determine the size of the corresponding partition coefficient.

We're  all in this world together,
07-11-04 05:30
No 518657
      regarding miscable solvents, and A/B     

First, I want to say thank you to all of you guy's here. I thought when SWIM started his present endeavour, he could just go to Rhodium's (MOST EXCELLENT) site, pick a synth, use clean glass, and get the job done..... HAHAHAH.. was SWIM wronng! And he would not have been anywhere as far along as je is at solving most of SWIMS problems if i hadnt been using this excellent hive, and availing myself of your freely shared knowledge. You have my gratitude, and my appreciation..
Now, for my very first question,ever posted at the HIVE, I would put this forward:
SWIM was cleaning up his lab, and had a whole bunch of solvent filled mason jars, that he is SURE have some honey in them (MDMA). Problem is...some of the so,lvents are mixed... The one that most concerns him is the one with water and tone.... Now this is my question... If SWIM wanted to extract whatever honey was there...(from a water/acetone mix) would it work to use a solvent that would dissolve acetone and honey, but not water..(Xylene??) and tehn basify the mix, pushing the amine in to the Xylene and out of the water, or should i add xylene, then acidify it, and extract via the water?
Thanx for reading this!
WYLD at heart in the WYLDERNESS

"To enjoy life, you must take big bites; moderation is for monks." R.A.Heinlein as L.Long
07-11-04 16:49
No 518712
      just evap     

acetone bp 56C - remove with distillation
drive off remaining water to yield resultant product.
if it is ungassed then use dcm to extract - evap it and start again with clean dry acetone.
i think you need to be a little more systematic in your approach dude
(Hive Bee)
07-29-04 19:19
No 522521
User Picture 
      Rubber products chemical resistance     

What rubber hose, gasket material to choose with different solvents see the chemical resistance chart....

Just hold on to the thread...that keeps us going
09-19-04 21:56
No 532213
      Personal experience     

On the three layer theme, from personal experience:

Suppose you have benzaldehyde salted out of water, ie two layers: bottom benzaldehyde (denser than water, practically insoluble in water), top water with NaCl and other inorganic salts - in my case, mostly acetates and (hydro)sulfites. Very carefully adding xylene (lighter than water) as not to disturb the liquids indeed causes three layers to form. But a slight shake of the flask and the benzaldehyde layer "connects" to the top xylene and starts to slowly migrate. If you add enough xylene (or toluene, ether or what have you) you might suck the bottom benzaldehyde in a minute. Btw this is a hell of a funny experiment to watch laugh

Cheers and good luck with your extractions

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