Sciencemadness Discussion Board - Experiment report: Cobalt Molybdate, amazing colors and a collloidal mystery

Experiment report: Cobalt Molybdate, amazing colors and a collloidal mystery

MidLifeChemist - 20-10-2020 at 06:54

REPORT: Cobalt Molybdate + bonus experiment

I'll admit, I wasn't originally planning on doing an experiment report on CoMoO4, it was a simple precipitation reaction (inspired by a beautiful photo from SM user Bezaleel), and I didn't think it would warrant a full report. But I found many things about this experiment to be interesting, some of which will allow me to tie in to my colloidal mystery that I referred to in the subject.

REACTANTS:

I added 3.5 grams of Cobalt Sulfate ($9 for 114 grams from a pottery store) to 40ml of distilled water in a 100ml beaker, with my magnetic stirrer on at 500RPM. Not sure how much hydration it had, maybe the heptahydrate?). It formed a pretty pinkish-red solution.

In a separate beaker, I dissolved 4.5 grams Sodium Molybdate Dihydrate (227 grams for $9, Alpha Chem) in a small amount of distilled water and got a clear solution. I used what I believed to be an excess of Na2MoO4, based on a recommendation from Bezaleel. Anyways, I thought it couldn't hurt.

OH, THOSE COLORS!

I poured the Sodium Molybdate into the CoSO4 solution while the magnetic stirring was turned on. Instantly a gorgeous purple/violet precipitate formed, which took over the entire beaker. Seeing this color in motion was quite a sight. But the real magic happened when I turned off the stirring. The CoMoO4 quickly settled into the bottom 2/3rd of the beaker, leaving a pink-red solution in the top third. I brought out all the family members to witness this spectacle and they were quite impressed. Of course, it could be repeated over and over - turning on the stirring filled the beaker with a light purple, and when I turned it off, two layers of dramatic colors appeared.

CHEM MYSTERY #1

But why was the solution pinkish-red? I thought I had used an excess of Sodium Molybdate. Hmmmm. So after removing the magnetic stirrer from the beaker using a steel rod (part of my stand clamp - bought on eBay), I filtered the precipitate using gravity filtration through a brown coffee filter, and all the solution passed through the filter within 30 seconds. I gave the beautiful lavender-violet precipitate a quick wash, and put it aside to dry on top of two paper towels on a 150mm KarterSci watch glass ($7 for pack of ten), still on the coffee filter. I was looking forward to taking a beautiful photo of this compound - which as you will see later, was harder than I thought.

Now I thought I would precipitate out more CoMoO4 from the pinkish-red solution (the filtrate) by adding more Na2MoO4. A sprinkled a little in, but nothing. Nada. No precipitate. Hmmm. So then, I poured in a little Sodium Hydroxide. I saw a flash of blue, then quickly the solution turned brownish-gray, with a fine particle precipitate forming.

A DIFFERENT FILTERING EXPERIENCE

Again, I used gravity filtering with my brown coffee filter for this greyish precipitate - which I had a very small amount of, compared to the amount of CoMoO4. What took 1 minute to filter before, now took over 30 minutes! The water just did not want to go through my coffee filter, with this fine substance clogging it up. In addition, some of it even passed right through the filter! This reminded me of my Iron Hydroxide colloidal experience from last week, which we'll visit in a moment.

TESTING THE GREY-BLACK PRECIPITATE

I let the greyish-black precipitate dry, and the next day I wanted to test it a little. It was very non-reactive to HCl, only dissolving a little bit after some time. There was a somewhat of a pinkish-red color, but it was still mostly greyish-black. I thought I even smelt a whiff of Chlorine. Was my mystery substance Co2O3? Was it Co3O4? It reminded me of what I got when I baked Cobalt Hydroxide in the oven for a little bit - a fine, black non-reactive powder. If it did have Co+3 in it, is it possible it could have oxidized a little Cl- to Cl2?

BACK TO THE COLLOIDAL MYSTERY

Having the Cobalt Oxide pass through my filter reminded me of my experiments with FeCl3 from last week. I had a concentrated solution of FeCl3, and I wasn't sure of the Molarity. I thought I would try to get a rough estimate of the concentration by "titrating" it with a NaOH solution to see how much FeCL3 it took to neutralize it, but it didn't go as planned, and Iron chemistry is complicated (not the best idea!). But I had better luck performing a back-titration with dilute ammonia.

[EDIT] To be 100% clear, I'm not recommending back titrating with Ammonia as a method to determine Fe+3 concentration, and certainly not over a method more accurate like iodometry - as Texium was kind enough to point out. I was simply trying it out to see if I could get the Fe+3 concentration in a ballpark range.

I decided to try again with 10% Ammonia (from Ace Hardware) and do a small test. I poured the Ammonia into a beaker, and slowly added some of my FeCl3 solution. The reaction was beautiful - clumps of brown fluffy Iron Hydroxide (or FeOOH) appeared, with the remaining solution staying clear. As I added more FeCL3, more and more regions of beautiful fluffy hydroxide appeared, and I thought I would keep adding FeCl3 until I saw it had stopped reacting with the Ammonia. However, I accidently added a little too much. No worries, it was nice to filter the precipitate. I gave the solution a quick stir, prepared by Buchner funnel for vacuum filtration, and poured it in.

But guess what - IT ALL PASSED THROUGH THE FILTER. Every drop of it. But I could see the precipitate in the solution, but it looked different from before, it was more fine and dispersed throughout the solution. Had I produced colloidal Iron Hydroxide? I researched it, and it seemed possible.

So how to recover my Iron? Ok, I took the filtrate and added more Ammonia. Quickly, the fluffy precipitate from before came back, and the entire solution became very thick with mushy Iron Hydroxide. This was quite the opposite of a colloid - this was a thick, hydrated, messy precipitate. It easily filled my 110ml porcelain buchner funnel, and I had to filter it three times, giving my a giant plate-full of gunky Iron Hydroxide which even after three days does not seem to have completely dried. And it looks nothing like the pretty reddish-colored Iron Hydroxide powder I saw someone post here last week. The filtrate was clear, which told me I had Ammonia in excess and all of the FeCl3 had reacted. This had seriously gunked up my buchner funnel, but a dilute hydrochloric acid soak cleaned it up nicely.

PHOTOGRAPHIC CHALLENGES - PHOTOGRAPHING PURPLE

Bezaleel had shared a very nice photo of his CoMoO4, and I also wanted something for posterity. I tried a few photos with a few different devices. My eyes said "purple", but the image said "blue". Blue? why were my photos blue? I quick google search showed that cameras have trouble capturing this particular hue of lavender/purple. No worries, time to use my D810 in raw mode with the 105mm macro lens with VR turned on, and do some lightroom adjustments. Lightroom is a little bit of a resource hog, but no worries, I got the job done by adjusting the blue hue and the photo now does a much better job of showing the colors of CoMoO4.

Cobalt Molybdate sitting in my porcelain mortar.

This is the corrosion on my stand clamp, after briefly dipping it into the Cobalt / Molybdate solution to retrieve my magnetic stirrer. Perhaps I should have studied my activity series better...

ENDING

So that's the end of my long experiment report. Although the chemistry here is simple, I find it fascinating. And I can't recommend the CoMoO4 experiment enough - both reactants are easily and cheaply purchased, at least in the USA.

That's all for now. As always - thoughts, comments, tips, questions, and constructive criticisms are all highly appreciated. Or just be a lurker!

[Edited on 21-10-2020 by MidLifeChemist]

Mateo_swe - 20-10-2020 at 07:37

That is a beautiful intense color.
I have read is that hydroxides is generally hard to filter.

Bedlasky - 20-10-2020 at 08:12

You cannot precipitate more molybdate from solution because of formation of [Co(MoO4)6]10- complex. (NH4)4H6[Co(MoO4)6] can be made from ammonium heptamolybdate and cobalt chloride and crystallized out. If you want to precipitate most of the molybdate, you must add excess of cobalt chloride. Excess of sodium molybdate leads to formation of more soluble complex. For hexamolybdatocobaltate formation is better mildly acidic solution, but it obviously forms also in neutral solution. Look at my article:

https://colourchem.wordpress.com/2020/08/27/molybdenum-chame...

Sodium molybdate heptahydrate? If I know, sodium molybdate is sold as dihydrate.

Greyish black precipitate is probably CoO(OH) which oxidize HCl to chlorine.

Red FeO(OH)? I never heard about it. There are two forms of anhydrous Fe2O3 - black and red. But FeO(OH) precipitated from solution is always brown.

Determination of FeCl3 by titration with NaOH is probably stupidest thing I ever heard. FeCl3 isn't simple acid like HCl. If you want to determine Fe3+ concentration, use iodometry or gravimetric determination as Fe2O3.

[Edited on 20-10-2020 by Bedlasky]

MidLifeChemist - 20-10-2020 at 09:27

Quote: Originally posted by Bedlasky

Sodium molybdate heptahydrate? If I know, sodium molybdate is sold as dihydrate.

Greyish black precipitate is probably CoO(OH) which oxidize HCl to chlorine.

Red FeO(OH)? I never heard about it. There are two forms of anhydrous Fe2O3 - black and red. But FeO(OH) precipitated from solution is always brown.

Determination of FeCl3 by titration with NaOH is probably stupidest thing I ever heard. FeCl3 isn't simple acid like HCl. If you want to determine Fe3+ concentration, use iodometry or gravimetric determination as Fe2O3.

Thank you for the information on the complexes and for the link.

My Na2MoO4 is the dihydrate, heptahydrate was a mistype.

>> is probably stupidest thing I ever heard.

Some people should listen more and talk less. Anyways, if you know how much of a base is required to precipitate all of a metal salt out of solution, you can approximate how much metal salt you had. And it is much simpler than the methods you recommended, although certainly less accurate.

[Edited on 20-10-2020 by MidLifeChemist]

Texium - 20-10-2020 at 10:39

Quote: Originally posted by Bedlasky

Determination of FeCl3 by titration with NaOH is probably stupidest thing I ever heard. FeCl3 isn't simple acid like HCl. If you want to determine Fe3+ concentration, use iodometry or gravimetric determination as Fe2O3.

Jeez, a little bit harsh there. Plenty of far stupid-er things have been said on this forum!

You are right that it wouldn't work though. Precipitating iron with base gives a whole mess of hydrous hydroxide/oxide, and there's no way of knowing how many moles were precipitated, since it wouldn't even be clear how many moles of base would need to be added to affect precipitation. Iodometry is a good method. All you need is KI, starch indicator, and a standard thiosulfate solution. You would need to standardize the thiosulfate solution prior to titrating the unknown using a solution containing a known amount of Fe(III).

MidLifeChemist - 20-10-2020 at 11:14

Quote: Originally posted by Texium (zts16)

You are right that it wouldn't work though. Precipitating iron with base gives a whole mess of hydrous hydroxide/oxide, and there's no way of knowing how many moles were precipitated, since it wouldn't even be clear how many moles of base would need to be added to affect precipitation. Iodometry is a good method. All you need is KI, starch indicator, and a standard thiosulfate solution. You would need to standardize the thiosulfate solution prior to titrating the unknown using a solution containing a known amount of Fe(III).

Texium, thanks for the advice. I do have KI and Sodium Thiosulfate, I probably also have some starch, and I have no doubt iodometry would be more accurate.

However, I do think that the Fe+3 concentration also can be determined by reacting with Ammonia. If you know how many moles of Ammonia (let's say X moles) was used to react with the FeCl3 solution to precipitate out all of the Iron, then the numbers of moles of FeCl3 you had in that solution was X/3. This is assuming you need 3 moles of NH4OH for every mole of Fe(OH)3 or FeOOH (or the various hydrated forms), which I believe is a good assumption, but I'd be happy to hear arguments contradicting that.

unionised - 20-10-2020 at 13:37

Quote: Originally posted by Bedlasky

Determination of FeCl3 by titration with NaOH is probably stupidest thing I ever heard.
[Edited on 20-10-2020 by Bedlasky]

I have heard much worse ideas.
The easy way would be via back titration.
You could also do the titration directly with a pH probe- though it might need a lot of cleaning later.

Lion850 - 20-10-2020 at 14:59

Midlifechemist thanks for the report. I’ll add prep of cobalt molybdate to the ever growing list of experiments I would like to do.

Texium - 20-10-2020 at 18:32

Quote: Originally posted by MidLifeChemist

However, I do think that the Fe+3 concentration also can be determined by reacting with Ammonia. If you know how many moles of Ammonia (let's say X moles) was used to react with the FeCl3 solution to precipitate out all of the Iron, then the numbers of moles of FeCl3 you had in that solution was X/3. This is assuming you need 3 moles of NH4OH for every mole of Fe(OH)3 or FeOOH (or the various hydrated forms), which I believe is a good assumption, but I'd be happy to hear arguments contradicting that.

The method you propose is not impossible, but it would be very difficult to do accurately. It's very hard to tell when all of the iron is precipitated. You would need to stir the solution as you add ammonia, to make sure that the ammonia is reacting with all the iron and isn't just floating around near the top. That would disturb the precipitate, which would make it impossible to see if new precipitate was forming. So you'd have to add it a couple drops at a time, stir, wait for the precipitate to settle, add a couple more drops. Even if you managed to not overshoot it, trying to spot the last bit of precipitate forming would be a lot more difficult than seeing the sudden change from dark blue to pale yellow that would be observed as the end point of iodometry. With the concentrations that you'd likely be measuring, it might actually be impossible to see the endpoint. The titration with ammonia would also be thrown off if there's any free acid in the solution. A clear solution of iron(III) necessarily contains some acid. At or above neutral pH, it will start precipitating. Iodometry, on the other hand works best with an acidic solution.

Bedlasky - 20-10-2020 at 20:57

MidLifeChemist: Sorry, I don't mean to insult you. I just mean that this isn't really good idea, not that you are stupid. I am sometimes too much straightforward, so again, I am sorry.

Determination of Fe(III) by NaOH isn't possible. FeO(OH) or Fe(OH)3 is only very simplified formula. When you neutralize [Fe(H2O)6]3+, you end up with monomeric [Fe(H2O)3(OH)3]. But this monomeric species also undergo polymerization which change number of OH- in complex (I don't remember if number is higher or lower, I read about it long time ago, in that document there were also eqautions for polymerization of Fe(OH)3 or Cr(OH)3 and sadly I cannot find it). So it really isn't that simple how it looks. Btw. Ferric hydroxide precipitate can trap some NaOH inside it, adsorb NaOH or even react with NaOH (this is reason why in gravimetry is ferric hydroxide precipitated using dilute ammonia - heating will remove any ammonia or ammonium ions and water).

woelen - 20-10-2020 at 23:30

First I want to add that your write-up is an example of excellent home chemistry. This is the kind of things we really like to see over here at sciencemadness!

The estimation of iron with hydroxide is not accurate, as others have pointed out. It, however, can be used for getting a quick and rough estimate. If you just are interested in a rough estimate it may be good enough. One way to make it somewhat more accurate is indeed using ammonia for precipitation and after filtering heating of the precipitate to 200 or so. This drives off any excess ammonia and also most water, leaving behind (more or less) pure Fe2O3. By weighing that, you get a rough estimate (an upper bound) of the amount of iron in your sample.

_________________________________________________________________

Quote:

MidLifeChemist: Sorry, I don't mean to insult you. I just mean that this isn't really good idea, not that you are stupid. I am sometimes too much straightforward, so again, I am sorry.

@Bedlasky: Thumbs up!

Bedlasky - 21-10-2020 at 01:13

MidLifeChemist: Back to your CoMoO4. I remember that I posted some photos of metal molybdates/heteropolymolybdates on this forum. Look at this:

https://www.sciencemadness.org/whisper/viewthread.php?tid=15...

You can see that complex molybdates have quite a different colours than normal molybdates. I have in PC some writeup about preparation of these compounds, but it's not finished yet because I tried to prepare Fe(III) and V(IV) salts, but yields are terrible from some reason. But I'll soon publish it and post on forum. Mn, Cr and Ni salts can be quickly and easily made in good yield and have beautiful colours (I especially love colour of ammonium hexamolybdatonickelate).

Btw. I read that anhydrous CoMoO4 is green in colour (your CoMoO4 is monohydrate). You can try heat carefully small sample of it in test tube and see if it changes its colour.

[Edited on 21-10-2020 by Bedlasky]

valeg96 - 21-10-2020 at 01:14

If you liked the fast settling of precipitate, a similar phenomenon is observed with cobalt thiocyananate mercurate. It's a deep blue powder that settles very rapidly and (if in slight defect of Hg) leaves a pink Co(II) solution behind. Thing can be repeated countless times and it's always amusing. I have the sodium and ammonium molybdates but never stumbled upon anything, really. I might try this one though.

Quote: Originally posted by Bedlasky

I have in PC some writeup about preparation of these compounds, but it's not finished yet because I tried to prepare Fe(III) and V(IV) salts, but yields are terrible from some reason. But I'll soon publish it and post on forum.

I'll be waiting for that writeup, and probably try them as well. I'm getting bored of Cu/Ni/Co complexes here. If you want (and if I have the chemicals!) I could try the syntheses that aren't working well with you.

[Edited on 21-10-2020 by valeg96]

[Edited on 21-10-2020 by valeg96]

MidLifeChemist - 21-10-2020 at 04:48

@unionised- thank you for your comments; yes, back titrating with an excess of ammonia is what I had planned, I think I get within 20% of the true Molarity with this method

@lion850 - you are very welcome!

@Texium - thanks for your reply. What you have described is messy,but back titrating with an excess of Ammonia is much easier and I think can get you in the ballpark (say within 20%) , and would allow the entire solution to be pushed in the basic ph range. However, I would certainly never debate that iodometry would be much more accurate.

@bedlasky - no need to apologize, I don't think anyone here takes offense to anonymous virtual replies

I have also read about the complex chemistry of Iron Hydroxide and it's various forms, I have seen no evidence to suggest that the 3 to 1 mole ratio between NH4OH and Fe+3 ions would not hold true with a slight excess of 10% Ammonia within an acceptable tolerance of error for my purposes. And by washing the precipitate in the filter and doing the 2nd part of the back titration on the filtrate to measure the amount of excess Ammonia, I don't think there would enough "trapped" Ammonia to throw off the results too much. But no one is debating that iodometry would be much more accurate.

@woelen - thanks for your kind reply, and for your suggestion of heating the precipitate to drive off all of the water, I have done similar procedures with copper and cobalt hydroxides to try to measure my yields. According to this paper, physically bound water is driven off at 130C, and structurally bound water is driven off at 275C. 275C is just under the max temperature of my oven, so that might work. https://www.degruyter.com/downloadpdf/journals/pac/67/11/art...

@bedlasky - I look forward to your writeups on the Mo complexes, especially since I have easy access to Mo, Mn, Ni, Cr and Co salts and would love to give these complexes a try. I'll gently heat some of the CoMoO4 and see if I get a green anhydrous version.

@valeg96 - thanks for your reply. Cobalt thiocyananate mercurate certainly sounds very interesting.

woelen - 21-10-2020 at 06:25

Quote: Originally posted by valeg96

If you liked the fast settling of precipitate, a similar phenomenon is observed with cobalt thiocyananate mercurate. It's a deep blue powder that settles very rapidly and (if in slight defect of Hg) leaves a pink Co(II) solution behind. Thing can be repeated countless times and it's always amusing. I have the sodium and ammonium molybdates but never stumbled upon anything, really. I might try this one though.

This one indeed is beautiful. I have written a web page about it:

https://woelen.homescience.net/science/chem/exps/cobalt_thio...

Texium - 21-10-2020 at 06:52

Quote: Originally posted by MidLifeChemist

@Texium - thanks for your reply. What you have described is messy,but back titrating with an excess of Ammonia is much easier and I think can get you in the ballpark (say within 20%) , and would allow the entire solution to be pushed in the basic ph range. However, I would certainly never debate that iodometry would be much more accurate.

Fair enough, I guess you’re just one of those people who has to actually try stuff and witness that it won’t work instead of taking someone else’s word for it. Nothing wrong with that though!

At the very least, please test out the method that you’re going to use on at least two Fe(III) solutions of known molarity so that you can quantify your margin or error and know whether it’s worth using the method on your unknown. If you don’t do that you’ll have no idea whether your measurement of the unknown is actually within 20%. Without comparing to a control, your experimental results will be meaningless.

MidLifeChemist - 21-10-2020 at 08:09

Quote: Originally posted by Texium (zts16)

I'm happy to take someone's word for it if they make a good argument. But so far, I've not seen a good argument for why a back-titration with an excess of dilute ammonia would have an error of greater than 20%. I've only seen good arguments for why it is less accurate than other methods, which I concur with. And yeah, I'm more than happy to try things out for myself - that is how we will all learn from each other.

I agree with 100% that this method should be tested before being promoted to others. When I mentioned it in my initial report, I did not intend to promote it to other people who may need a more accurate measurement than I need (in fact, I don't really need it at all). Perhaps I should have been more clear about that, but then we would all miss out on what I think is an interesting discussion.

Texium - 21-10-2020 at 09:18

Back titration would be better than direct titration with ammonia, which is what I thought you were initially suggesting. However, you are still working with a couple of assumptions that aren't trivial:

Quote: Originally posted by MidLifeChemist

I don't know how much ammonia would be trapped in the precipitate, but I would feel uneasy assuming that it wouldn't be a factor. Also, ammonia is volatile, and the more that you manipulate the solution, including the filtering step, the more that you will have evaporate.

Here's another one: is there also going to be cobalt present in your unknown solution? If you aren't absolutely certain that there isn't, then you can throw the ammonia method out entirely, since cobalt complexes with ammonia. Cobalt(II) is not responsive to iodometry though, so that's another point in favor of it.

Also, I'm not talking about promoting it to others, I'm just saying that if you want to know whether your results are good or not, you'll have to go through the process of a responsible analytical chemist and perform a controlled test before you use it on your unknown. Whatever titrant you use though, whether it's ammonia for back-titrtation, or sodium thiosulfate for iodometry, keep in mind that you'll need to standardize it prior to running the titrations. Assuming you have 10% ammonia out of the bottle doesn't cut it for analytical work, and could introduce enough error to render your results meaningless. You'll have to standardize the acid that you use to back titrate the ammonia as well. So the ammonia method actually requires more standardizing and overall titrations than iodometry.

By no means am I trying to discourage you, I just want to be sure that you're being thorough.

[Edited on 10-21-2020 by Texium (zts16)]

MidLifeChemist - 21-10-2020 at 09:37

Quote: Originally posted by Texium (zts16)

By no means am I trying to discourage you, I just want to be sure that you're being thorough.

[Edited on 10-21-2020 by Texium (zts16)]

All great advice, thank you and it is much appreciative.

Lion850 - 22-10-2020 at 02:39

I had a go at cobalt molybdate today but a slightly different route. I have a 500g bottle of "molybdic acid", I think old stock, and wanted to use that. I dissolved some molybdic acid in a sodium hydroxide solution aiming to get 15g of sodium molybdate in solution and then slowly added a solution of cobalt acetate. This gave the ppt pictured below, the color is very similar to the color in Midlifechemist's photo above.

The stoichiometry was a bit off, as the filtrate of the final solution (to separate the purple precipitated product) was still too red to my liking, thus it seems not all the cobalt acetate was consumed. When I did the stoichiometry I used MoO3 in the equation because I did not know the molybdic acid hydrate but I forgot to take into account that the assay of MoO3 in the molybdic acid is only 90%. See label below. The label also says "contains ammonium hydrogen molybdate" which explains the ammonia smell each time some molybdic acid was added to the sodium hydroxide solution. Why is the ammonium hydrogen molybdate present? Is it an intermediate when it was made or maybe for stabilizing purposes?

EthidiumBromide - 22-10-2020 at 12:45

Quote: Originally posted by Lion850

Why is the ammonium hydrogen molybdate present? Is it an intermediate when it was made or maybe for stabilizing purposes?

To quote from Wikipedia

Quote:

Ammonium dimolybdate (ADM) is an intermediate in the production of molybdenum compounds from its ores. Roasting typical ore produces crude molybdenum(VI) oxides, which can be extracted into aqueous ammonia, affording ammonium molybdate. Heating solutions of ammonium molybdate gives ADM. Upon heating, solid ammonium dimolybdate decomposes to molybdenum trioxide

Dissolving in ammonia is quite commonly used to extract and purify metal compounds form ores, such as in the case of uranium, with ammonium diuranate.

[Edited on 22-10-2020 by EthidiumBromide]

DraconicAcid - 22-10-2020 at 13:11

So if I understand this correctly, mixing a cobalt source with a molybdate will give some cobalt molybdate as a purple precipitate, and will also form some of the cobaltomolybdate complex ion, which can be isolated as the ammonium salt? So two products from one reaction?

Lion850 - 22-10-2020 at 20:47

Hi DraconicAcid there was indeed a minute bit of a pinkish salt apart from the purple. And I did wonder whether this was due to the fact that there was a wee bit of ammonia going around. But I don’t know enough so can just speculate.

If Bedlasky still follows this thread he can probably clarify.

itsallgoodjames - 28-10-2020 at 09:08

Wow, that's a really bright and vibrant purple. It's a shame it's as toxic as it is, it'd probably make a really nice purple dye.

Bedlasky - 28-10-2020 at 09:24

Quote: Originally posted by DraconicAcid

So if I understand this correctly, mixing a cobalt source with a molybdate will give some cobalt molybdate as a purple precipitate, and will also form some of the cobaltomolybdate complex ion, which can be isolated as the ammonium salt? So two products from one reaction?

I never do it from one solution. I prepared these compounds separately. In filtrate after CoMoO4 there is just small amount of complex, most of Mo end up as CoMoO4.

Sodium molybdate + slight excess of cobalt salt --> cobalt(II) molybdate

Slight excess of ammonium heptamolybdate + cobalt salt --> ammonium hexamolybdatocobaltate(II)

Ammonium heptamolybdate forms acidic solutions which is important for preparation of hexamolybdatocobaltate. In neutral solution there is a formation of cobalt molybdate.

Quote: Originally posted by itsallgoodjames

Wow, that's a really bright and vibrant purple. It's a shame it's as toxic as it is, it'd probably make a really nice purple dye.

It isn't that much toxic. But metal molybdates are slightly soluble, not much, but still. Cobalt phosphate is cheaper and much less soluble purple dye, similar in colour.

itsallgoodjames - 28-10-2020 at 14:57

Quote: Originally posted by Bedlasky

Quote: Originally posted by itsallgoodjames

Wow, that's a really bright and vibrant purple. It's a shame it's as toxic as it is, it'd probably make a really nice purple dye.

It isn't that much toxic. But metal molybdates are slightly soluble, not much, but still. Cobalt phosphate is cheaper and much less soluble purple dye, similar in colour.

Still not something you'd probably want to be exposed to constantly though

MidLifeChemist - 31-10-2020 at 21:22

Quote: Originally posted by Bedlasky

Btw. I read that anhydrous CoMoO4 is green in colour (your CoMoO4 is monohydrate). You can try heat carefully small sample of it in test tube and see if it changes its colour.

[Edited on 21-10-2020 by Bedlasky]

I tried heating a sample a couple different ways, including with my propane torch. No change at 300C. I finally got it to change to grayish-blue color with high heat, but after an hour it changed back to purple. I never saw a green color. If there is a green anhydrous version, I was not able to make it.

Bedlasky - 1-11-2020 at 13:23

I read it on atomistry. But obviously it isn't true.