Felab
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Lithium halides
Lately I have beeng trying to make lithium halides (bromide and iodide) because I want an ethanol soluble halide salt.
I have made lithium iodide by reacting iodine with zinc to make zinc iodide and then reacting that with lithium carbonate to make lithium iodide.
It worked but I had trouble knowing how much lithium carbonate to add, because if I added too litle I would have zinc iodide contamination and if I
added too much I would have lithium carbonate contamination. I decided to add an excess of lithium carbonate and then recrystalise the product in
acetone. I filtered the solution of lithium iodide , evaporated it, and I dissolved it in acetone.
The solution made two different crystals, one was needle like and the other was cubic (what you expect from lithium iodide). I think the needle like
crystals where zinc iodide, but how could I test?
Also, lithium iodide is soluble in acetone when cold to a certain extent, so when trying to filter the crystals the solution was too thick so it
didn't pass through the filter. It ended up absorbing moisture from the air and turning liquid.
Any clues on a better solvent for recrystalising lithium iodide or another way to purify it?
Also, I want to make lithium bromide too by the same procedure.
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ninhydric1
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Just use ideal stoichiometric quantities.
ZnI2 + Li2CO3 -> 2 LiI + ZnCO3
It is a 1:1 molar ratio of the reactants. ZnI2 has a molar mass of 319 g/mol while Li2CO3 has a molar mass of 74 g/mol. So the mass to mass ratio ZnI2
to Li2CO3 is 319:74 or about 4.3 g ZnI2 per 1 g Li2CO3.
This should give you a relatively pure product of LiI solution and ZnCO3 as long as you filter and process the resulting products correctly. No need
for additional recrystallizations.
Additionally, ZnCO3 can be recycled to make more ZnI2 (with HI), which can be used again (if needed).
The philosophy of one century is the common sense of the next.
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Felab
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Quote: Originally posted by ninhydric1  | Just use ideal stoichiometric quantities.
ZnI2 + Li2CO3 -> 2 LiI + ZnCO3
It is a 1:1 molar ratio of the reactants. ZnI2 has a molar mass of 319 g/mol while Li2CO3 has a molar mass of 74 g/mol. So the mass to mass ratio ZnI2
to Li2CO3 is 319:74 or about 4.3 g ZnI2 per 1 g Li2CO3.
This should give you a relatively pure product of LiI solution and ZnCO3 as long as you filter and process the resulting products correctly. No need
for additional recrystallizations.
Additionally, ZnCO3 can be recycled to make more ZnI2 (with HI), which can be used again (if needed). |
My problem is that I don't know the exact concentration of the zinc iodide solution so I will always have some Li2CO3 contamination if I do it
stoichiometrically.
If I had HI I would react it with the lithium carbonate directly.
[Edited on 9-2-2019 by Felab]
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ninhydric1
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How do you make your ZnI2? If you use a fixed quantity of Zn and I2 you should know your theoretical yield of ZnI2 and base your calculations off
that.
The philosophy of one century is the common sense of the next.
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Felab
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| Quote: Originally posted by ninhydric1 | | How do you make your ZnI2? If you use a fixed quantity of Zn and I2 you should know your theoretical yield of ZnI2 and base your calculations off
that. |
I tried to do so but during transfer steps and filtrations you loose some of the solution.
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Felab
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| Quote: Originally posted by Felab | | Quote: Originally posted by ninhydric1 | | How do you make your ZnI2? If you use a fixed quantity of Zn and I2 you should know your theoretical yield of ZnI2 and base your calculations off
that. |
I tried to do so but during transfer steps and filtrations you loose some of the solution. |
Also, when I added the zinc to the iodine, some white powder remained undissolved. I suspect it is zinc oxide but what is clear is that my reagents
aren't pure, which makes everything more difficult.
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fusso
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Is I2 the limiting reagent? If yes then you can use it to work out other masses.
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Felab
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I did, the problem is that the purity of my reagents is unknown and the loses during transfer steps and filtrations.
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I am a fish
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If the crystals have visible structure, they're (probably) quite pure.
If you're simply trying to tell which is which, the flame test ought to do it. Assuming you have an excess of zinc iodide, the lithium iodide will
turn the flame crimson, and the zinc iodide won't.
If both contain lithium, you may have an excess of lithium carbonate, in which case you can tell them apart by adding an acid.
Edit:
Another problem that occurs to me is sodium contamination, especially if you're using cheap lithium carbonate from a pottery supplier. The cubic
crystals could be sodium iodide.
[Edited on 15-2-2019 by I am a fish]
1f `/0u (4|\\| |234d 7|-|15, `/0u |234||`/ |\\|33d 70 937 0u7 /\\/\\0|23.
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j_sum1
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My eyes must be deceiving me!
5 and a half years and you're back.
Good to see you. 
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Tsjerk
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Welcome back Fish!
I would go at this problem by first using an excess zinc so there is no iodine left, then use an excess of lithium carbonate and extract you lithium
iodide with ethanol as neither lithium carbonate nor zinc carbonate dissolve in ethanol.
You could do it all one pot, evaporate the water (I guess you do it in water?), and extract with ethanol. Only lithium iodide should be in the
ethanol.
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Felab
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| Quote: Originally posted by I am a fish |
Another problem that occurs to me is sodium contamination, especially if you're using cheap lithium carbonate from a pottery supplier. The cubic
crystals could be sodium iodide.
[Edited on 15-2-2019 by I am a fish] |
A small sodium iodide contamination isn't a problem for me. Also i did a flame test on the lithium iodide and it turned the flame a deep red, without
any signs of yellow or green.
I think I will do what Tsjerk suggests.
Also, as this procedure should work for lithium bromide if you substitute the iodine with bromine I will try that one first since I haven't ampoulled
my bromine and I don't want to loose it all before I get to making it.
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Felab
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Lithium Bromide
Hi. Today I made some lithium bromide by reaccting zinc with bromine and reaccting the resulting zinc bromide with lithium carbonate. It worked
perfectly, yielding a white, fluffy powder that was very deliquescent, turned a flame deep red and reaccted with silver nitrate to produce silver
bromide.
When I dissolved the lithium bromide in ethanol to purify it, the ethanol turned violet, and then the color shifted back to a very slight yellow when
I evaporated all the solvent. I suspect that the violet colour appeard because tribromides in ethanol are violet, but I am probably wrong. Any ideas
on that?
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