Chemistoftheelements

Yes, I've figured since that its was some tungstic oxide hydrate. Wasn't at all sure at the time, though.

It surprises me just how lasting these names are, and even chemical manufacturers occasionally perpetuate this confusion, as I’ve witnessed with a bottle of “tungstic acid”.

This admitedly isn't the best example, but it concludes on a probable structure, i.e. there still remains an uncertainty. However this may well be explained by the difficulty of producing material which is sufficiently crystalline and uniform in composition, as your second link hints, thanks. I take your point about the molecular formulas, but the H3AsO4. 1/2H2O in the wikipedia link I took to imply a compound with As single bonded to 3 OH groups, and double bonded to an O, analogous to an orthophosphoric acid unit, apart from the ½ H20 bit. I read it this way as wikipedia phrases it as the "hemihydrate form"(of arsenic acid). The structure As2O5.4H2O which I suggested is hydrous As2O5, i.e with no OH groups. This highlights the confusion, and part of my frustration, that hydrous acids and hydrous oxides are banded together with the same name- cf "perrhenic acid": http://en.wikipedia....Perrhenic_acid.

Precisely. But that would be a challenge....you thinking of giving it a crack?

For an example of what I mean: http://www.nrcresearchpress.com/doi/abs/10.1139/v72-106 if XRD etc, cannot answer the question of the structure of this compound, what can? Also, although I have limited knowledge of this sort of thing, the structure H3AsO4.½ H2O (http://en.wikipedia.org/wiki/Arsenic_acid) seems a little strange to me; is a water molecule shared between two H3AsO4 units? Do the XRD experiments on this compound (if any have been done) rule out As2O5.4H2O ?

There seems to be much uncertainty of the structure of the oxoacids of certain elements, particularly metalloids, but also of metals such as molybdenum, tungsten (i.e. "molybdic" and "tungstic" acids), and such terms as "perrhenic acid" and "rhenium heptoxide" are frustratingly used interchangeably. Orthosilicic acid certainly exists in solution, but when crystallized from aqueous solution under standard conditions, no-one really knows what happens- perhaps the acid polymerizes, perhaps a hydrous silica is produced. An example of this uncertainty can be illustrated by the compound "Arsenic acid" when crystallized; wikipedia* talks of the structure H3AsO4 ½ H2O, but how do we know that the structure isn't actually the stoichiometrically identical As2O5.4H2O? What are the limitations of Raman spectra, X-ray crystallography and other techniques for determining the structure of those substances which result from crystallization from aqueous solution of such acids? What my question is is this: How can we be sure of the structure of these substances who's structures have so far proved so elusive? * http://en.wikipedia.org/wiki/Arsenic_acid

All I know is that when I've neutralised Fe(II)+ containing solutions with aqueous ammonia, I've gotten a precipitate. Whether that precipitate is Fe(OH)2 or whether it's a hydroxide- ammonia complex, I don't know- the observation that some ammonia- containing complexes are soluble whilst others aren't makes it difficult to tell, and a frustration for me. I'd like to be able to predict theoretically how some complexes (and not just ammonia- containing ones) dissolve in water under standard conditions. Anyway, here's very likely what the ammonia- nickel chloride complex which I mentioned is: http://www.periodictable.com/Elements/028/index.html It's about the fifth picture from the bottom.

Thank you, that was interesting and informative. Yes, it's insoluble enough for the NiCl2 to be amost completely recoverable, as I recall. You can drive the ammonia off in a fume cupboard, which was, infact, part of the practical. I wasn't sure of the formula for a long time, but it's a very striking experiment, and the kind of demonstration which could be used more widely in schools and colleges to encourage people to participate in chemistry.

Hi, I remember performing a practical once whereby ammonia solution was added to a solution of Nickel(II) chloride, with a purple precipitate appearing. This has always stuck with me as an example of an insoluble complex, and noting that many other ammonia complexes are soluble, it has now got me wondering why some complexes of any kind dissolve whilst others don't. Could anyone who has more theory please recommend to me something which would help me understand why some complexes dissolve and others don't in water under standard conditions? I realise this may be a bit involved, so I was thinking of a textbook or an area of inorganic chemistry which covers this? Many thanks.

Thanks. Briefly, withought trying to break out into another thread, could you recommend something which would help me understand why some complexes dissolve and others don't ( e.g. Cu2+ ammonia complexes versus Ni2+ ammonia complexes), in water under standard conditions? Many thanks.

"And chemistoftheelements, would you be in a position to repeat your experiments with stirling silver and pure silver and with warmed 0.88M ammonia in a sealed container?" Hi Greg; after a bit of encouragement from 3 forum members, I've decided to stick around a bit. In answer to you question, not at the moment, but I may have access to a fume cupboard in the not- too distant future- hot ammonia is nasty stuff, so this is essential. What interests me most is that the results so far are strikingly different between fine silver and a high- silver alloy. Maybe it's Ag2S which forms on fine silver, and a mixture of copper compounds on sterling silver, with some silver compounds. As copper is higher up the reactivity series than silver, it may be that it preferentially corrodes and "protects" the silver, in the manner that the zinc in zinc plate on iron will do if this coat is scratched through to the iron. On the solubility of Ag2S, it's always been my understanding that the solubility of a substance per say is not a barrier to forming soluble complexes, but if Ag2S- ammonia complex in water is insoluble, then the tarnish layer on fine silver, if it does consist of Ag2S, will not be removed. For example, TiO2 is extremely insoluble in water, yet is dissolved by hydrofluoric acid, because the complex formed is soluble. Hypervalent_iodine, what's your take on this? Thanks.

In the previous thread, I posted some of my conclusions from an experiment I conducted on the removal of the tarnish formed on 99.99% fine silver, and on the alloy sterling silver. My intention at the time, was to add a little to the discussion based on my experience, not to further an argument. For the sake of science, I will add the results of a second experiment, using the same piece of 99.99%, i.e. fine, silver (as I don’t have a great deal of this available to me), and some more tarnished sterling silver, after treatment with 9% ammonia solution: 1) In the second experiment on the fine silver, there was no further removal of tarnish, and the original, unpublished observation of H2S evolution (which came as a surprise, as there’s no obvious mechanism for this to occur) was not replicated in the second experiment. 2) The sterling silver again, in the second experiment, markedly de-tarnished, to the extent that this could be used as an effective method of at least helping to completely remove the tarnish from sterling silver, at least of it’s more usual composition, in which copper is the alloying element. No H2S evolution took place. The only firm conclusions I reach from the above experiments are as follows: 1) The tarnish layer formed on fine silver is of a different chemical nature of that formed on sterling silver. The species in question were not identified by the author of these experiments. 2) More study is warranted into the differences in the observations between the first and second experiments on fine silver, and into the nature of the tarnish films formed on fine, and sterling, silver. Onto the topic of humility and politelness now, and may I add that after my original post on the topic, which was intended for the purpose of encouraging scientific discussion, I gained the distinct impression that John Cuthber had formed some sort of grudge, as evidenced by his unwarranted behaviour in the Neodymium chloride thread in the Inorganic chemistry section. Although I don’t necessarily condone the methods which Greg used in this thread, it is clear that he was provoked by the same behaviour in the previous, now closed, thread on silver tarnishing, as I was in the Neodymium chloride thread. I can’t say I blame him, as I also have a low tolerance for arrogance and lack of humility. It has been demonstrated by further discussions in this thread (after the conversation between Greg and John Cuthber, that Greg is ammenble to civilised discussion with others who use a more civil approach; it is largely dependent on the manner in which any correcting and commenting is done, and this applies to most people. Please refer to my point about politeness on page 2 of the Neodymium chloride thread in Inorganic chemistry). I joined this forum for the sake of discussing chemistry, and I really am not interested in arguing with people who cannot grasp the basic rules of social interaction. Regretfully for me at least, I’m out of here.

OK, I goofed whilst numbering the points- there are infact 9 of these! Of course, if Pr(IV) reduces instantly, i.e. is extremely unstable relative to Pr(III), then you will get only one precipitate, containing both metals. If this is the case, which it may well be, then you are no worse off than you were to begin with.

Hi elementcolloctor1, yes, I know the feeling, it can be a little hard to keep track of these things. I think we're getting a little confused her, so I think it might be helpfull if I re- post a conversation we had about 3 days ago. Note that as far as I can tell, whether the 8- point method I suggest works is a matter, I think, of whether you have any Pr in there in the first place, and just how long Pr(IV) will hang around in acid solution; you MIGHT, might be lucky and get a small amount of Pr(IV) precipitate, but as John Cuthber has rightfully pointed out, it is pretty unstable. Here was our ealier conversation: elementcollector1, on 31 December 2011 - 12:45 AM, said: Hah, speaking of sodium bicarbonate, I just lost a large beaker to the stuff. Poor thing has a massive chunk missing. Anyway, I know oxalic acid is used as wood bleach, so I'll look into that. I'd like to go all the way to Nd and maybe Pr metals if I could, but we'll see. Process of getting to Nd from neomagnets, in short: 1) Dissolve in acid. 2) Add oxalic acid or salt to precipitate insoluble oxalates. 3) Ignite in air to produce oxides. (for separation of Pr) 4) Somehow convert the Pr6O11 that formed into PrO2. 5) (for separation of Pr) Place in 5% acetic acid? I read below on Sciencemadness that PrO2 is insoluble in 5% acetic acid, and apparently Nd2O3 is. Elementcollector1, 1) Happy new year! 2) To answer this and your latest question: There’s some good logic to this, but I’m not sure that this would work, for several reasons: 1) At the oxide stage, the Pr would be in solid solution in the Nd2O3, and probably being only a minor constituent , would be prevented from oxidising to higher than Pr(III). In other words, the Nd(III) would stabilise the Pr(III), in the same way that Th(IV) stabilises U( IV) in high ( and low) U thorianites. 2) Oxidation states of Pr higher than 3 would be unstable in acid solution- think CeO2, so adding acetic acid would probably cause what little Pr in a >III oxidation state to revert pretty quickly to PrIII and go into solution. 3) Lanthanoid (and related oxides) don’t easily dissolve in acid once they’ve been ignited. Working on a theory that whether a metal oxide dissolves or not depends on the ionic radius, the coordination number of the metal, and possibly it’s affinity for oxigen, I produced some Yb2O3, Sc2O3 and Sm2O3 and found that all 3 were very reluctant to dissolve; after ignition at 1000 degrees C (and then cooling), only the Sm2O3 dissolved, but very slowly. If, however you heat them only minimally ( just enough to ignite the oxalate in your case), then you will probably succeed. 4) By using H2O 2, you may end up with a product containing O-O bonds, which would be bad, because such a compound may prove thermally dangerously unstable. You would have to work on very small quantities to find this out, or alternatively oxidise with something else. What I would suggest is the following, which circumvents problem 1, but not the others, so it’s not guaranteed to work: 1) Dissolve in acid 2) Add H2C2O4 to precipitate Nd and Pr and separate them from Fe, etc. Add NH3 to raise pH to increase precipitation if necessary, or dilute with water. 3) Gently heat precipitate to create Nd/Pr oxides. 4) Dissolve oxides in H2SO4, NOT HCl, as this is reducing. 5) Oxidise this resulting solution to produce Pr2 O(SO4)3(?), a higher basic sulfate of praseodymium which will be insoluble. 6) Filter. You have now recovered Pr. 7) Ignite your P2 O(SO4)3 7)To remaining solution, add NH3 to precipitate Nd(OH)3. 8) Ignite the Nd(OH)3. From here, you’ll have to figure out a way to reduce the Pr and Nd oxides. Good luck, and let us know how you get on. In the mean time, I’ll try to figure out how to create subscripts!

Hi elementcollector1 "Please explain. When did the oxide crystals form again?" This is at the stage after you ignite your oxalate, which you were wondering how you should convert, in terms of it's praseodymium content, to PrO2, and you suggested H2O2..<br style="mso-special-character:line-break"> <br style="mso-special-character:line-break"> You wrote: “ Hah, speaking of sodium bicarbonate, I just lost a large beaker to the stuff. Poor thing has a massive chunk missing. Anyway, I know oxalic acid is used as wood bleach, so I'll look into that. I'd like to go all the way to Nd and maybe Pr metals if I could, but we'll see. Process of getting to Nd from neomagnets, in short: 1) Dissolve in acid. 2) Add oxalic acid or salt to precipitate insoluble oxalates. 3) Ignite in air to produce oxides. (for separation of Pr) 4) Somehow convert the Pr6O11 that formed into PrO2. 5) (for separation of Pr) Place in 5% acetic acid? I read below on Sciencemadness that PrO2 is insoluble in 5% acetic acid, and apparently Nd2O3 is.”<br style="mso-special-character:line-break"> <br style="mso-special-character:line-break"> Then later on: “My main problem with this is how to convert the Pr6O11 to PrO2. Would hydrogen peroxide work? Excess heating? From the same forum as earlier, members have mentioned that on ignition, praseodymium oxalate turns brown, but I'm not sure if this is PrO2 (in which case, yay!) or something else. Nd (III) cannot be oxidized any further by igniting in air, so that's safe.” This was before I came up with my prodedure suggestion. Good luck! "<br style="mso-special-character:line-break"> <br style="mso-special-character:line-break"> " This bit, of course, is complete nonsense. I think this sort of thing is happening because I'm copying from microsoft word before I post.

Indeed reality may well prove impolite. On the second point, if you were referring to this: "1) At the oxide stage, the Pr would be in solid solution in the Nd2O3, and probably being only a minor constituent , would be prevented from oxidising to higher than Pr(III)" then this was at the stage when elementcollector1 was contemplating oxidising the mixed Nd/ Pr oxides with H2O­2. I was thinking that if the Pr was the less abundant constituent by some way, then, being in solid solution, it would be "shielded"- any oxidant would be in contact mostly with Nd oxide, and only the Pr oxide which was right at the surface could be oxidised. The more Pr in the mix, I figure, would allow more oxidation, and this method may become more feasible, especially with fine grounding. Sorry about the irregular line spacing, I'm trying to figure out why it's doing this. P.S. when i say "right at the surface", I mean of the crystals/particles of mixed Nd/Pr oxide.

It was late, I meant that the rest of the lanthanides can’t be oxidised to +4 or greater. I am here to discuss where the practice and theory of chemistry meet, which can only be done by experimenting, otherwise it isn’t much fun, and you don’t learn as much, even if what you are attempting doesn’t work. If I am wrong on anything, I will gladly be corrected, but on the other forums of which I am a member, this has always been conducted in a friendly and constructive manner, even by the occasional leading experts who graciously give their time to these forums- this is a first for me. Politeness costs nothing and often, I find, produces more rewarding discussions. I really hope that this discussion is not turning into a confrontation, as I am not interested in this, not my game. Thank you in advance for your cooperation.

The starting point for this project is a magnet, composed of, and this a a presumption, Pr as well as Nd, along with Fe, etc. The Pr and Nd by this presumption are intentionally there, and presumably the rest of the rare earth's aren't, so this is a completely seperate issue, is it not, to dealing with separating all of the rare earths? If you mean that the commercial extraction processes for the rare earths typically don't take out Pr with Ce, could this not be something to do with the influence of the so many other lanthanoids as are present in rare-earth ores, which aren't capable of being oxidised to a 3+ or higher oxidation state, with which the Pr may be carried over? Your point is valid only if the Ce is taken out first, then the rest of the rare earth elements are treated sepparetely. Lastly, I've said all along that this is project is full of uncertainties, many of which I've highlighted, if you read the whole discussion from when I entered it, so I don't understand your last quip, if that's how it was intended.

This is what I feared may be the case, in the same manner Pb(IV) is reduced to Pb(II) in acid solution, so Pr(IV) may be to Pr(III). But 2 questions: If you ignite Pr in air, you get an oxide containing a fair amount of Pr(IV), and I've always considered the highest oxidation state of metals which can be attained by oxidation in air a measure of the stability of that oxidation state in acid solution. PbO2, for example, is way too oxidising to be able to be formed by heating Pb in air, and not surprisingly, PbO2 dissolves in acids. Wouldn't Pr be stable enough to stay in solution in acid at least partially as Pr(IV)? Secondly, Ce(IV) is pretty oxidising as well, but CeO2 doesn't seem to disolve in HCl, (which it would be expected to oxidise, and, becoming itself reduced to Ce(III)2O3, which dissolves in this medium much more readily). If the observations for CeO2 can be extrapolated to Ce(IV) in acid solution, would this not also apply, albeit to a lesser degree, with Pr? Thanks.

Good. Before I forget, and this important point I neglected earlier, you may find that the pH range for Pr precititation and that of Nd overlap. By this I mean (and don’t take these figures literally- they’re possibly way out and just to illustrate a point) your Pr(IV) may start precipitating about pH2 and be completely precipitated by 5, whist for Nd(III) it may be 4-7. As you add your NH3 ( which you do in many steps to stop, stir and observe), you will will hopefully reach a point where the precipitate begins to change colour, due to the Nd beginning to drop out of solution (precipitate). Filter as soons as this happens, and not after. You’ve got Pr as free of Nd as you’re going to get using this method ( if it works). Back to your solution and precipitate now; you should find that as you’ve added more NH3 in a handful of steps, each time your precitiate has changed colour slightly. Eventually, with more NH3, the colour will no longer change. This will hopefully tell you that the Pr is now out of solution. Filter now and save the precipitate for a future experiment, as it’s Pr mixed with Nd and useless for this experiment. Add further NH3, gradually and stepwise, and if no colour change occurs, you can neutralise completely (take it to pH 7), and you have your Nd, largely as Nd(OH)3 if you’ve neutralised all the way. P.S. the formula for your basic Pr sulfate is probably more like Pr2(OH)2(SO4)­3­­.

Yes, and this is even desirable, as it makes for a gentler means of raising pH. It’s what I’ve always used. I have a feeling that you may not need to use the NH3, as I think your praseodymium will precipitate at a fairly low pH. Just make sure you dissolve all your mixed Nd/Pr oxides before you oxidise, as then you’ll know that then you have a solution which is too acidic to precipitate Nd, but not necessarily Pr when the solution is oxidised to crease Pr(IV). As for sheet metal, check out steel beer cans, or food cans, although the latter are a little thicker and would be a bit more difficult to cut. All you’re looking to do is to ignite your precipitate to a bright red heat, so you may have what you need already; do a few experiments, and with steel cans if this doesn’t work, noting whether the steel oxidises and flakes too much, in which case you may have to use stainless. I have only college- level chemistry, and all this advice comes from my work, current study, and hobby experience, so in a sense, my practice is ahead of my theory, so don’t take my (or anyone’s for that matter) word as fact- be cautious and always experiment on a small scale first. The lanthanoids are a fascinating group of elements which deserve more attention, which are as least equally as frustrating, so good luck, keep us informed and don’t give up! P.S. The colour of your Pr(IV) precipitate may not be the same as that of pure Pr(IV) in solution, so that it may not be brown. Think of NiCl2, which is grass- green as a solid and in solution, but when NH3 is added to an aqueuous solution containing this species, you get a very purple precipitate of NiCl2­- NH3 complex! This is not due to the NH3 as such, but to the resultant coordination number of the Ni atom. What I'm saying is that it may be better to look for colour CHANGE as you adjust the pH, rather than colour per say. P.S. 2, when you add H2C2O4, to precipitate your Nd and Pr, if you use any NH3­, then do so very cautiously- you don’t want to raise the pH enough to precipitate Fe or any other contaminents. H2C2O4 ­should give you the advantage here, as it should precititate Nd and Pr in quite low pH (acidic) solution, so NH3 ­may not be needed.

It’s not the rate of precipitation, but the amount of precipitate. The pH at which precipitation occurs depends on the species in question. In electopositive metals like Pr, which tend to form oxides and hydroxides with no acidic properties, precipitation will be promoted by alkalinity ( i.e. higher pH). In the case of the stage where you’re trying to precipitate the Pr2O(SO4)3, the same principle applies. But because the Pr will be in a higher oxidation state, it will be less basic, and will precitate at a lower pH, i.e. the solid will be more stable against acid. Think of the example of Sn- it has 2 common oxidation states, with Sn(II) predomination, except when it comes to oxides, when Sn(IV) is easily produced.; Sn(IV)O2 easily precipitates in acid conditions, whereas Sn(II)O requires more alkaline conditions, i.e. a higher pH. As Nd(III) is more basic than Pr(IV), your Pr will precipitate from your Nd/Pr in H2SO4 solution at a lower pH, i.e. in more acidic conditions, than your Nd will. You may find, though, that your solution is so acidic that even the Pr won’t precipitate, hence the requirement to raise the pH by adding NH3 (and it must be NH3, not, for example NaOH). You don’t want to raise the pH (make your solution more alkaline) so much, though, that your Nd precipitates out as well. I’m not sure what colour/color Pr(IV) is (this will depend anyway on the individual compound due to differences in coordination number), but from my own experiments, as you raise the pH, your Nd will begin to precipitate not as Nd(OH)3 but as Nd2O(SO4)2, which is orange (I began with 99.9% Nd metal, so there’s no possibility of contamination), so this may, and I stress may, serve as an indication of when to stop. A crucible and a blowtorch will be a good way to ignite your precipitates. When you heat your Pr2O(SO4)3, be sure to do this outdoors, as you will produce SO3! You may find you need to heat strongly here- to 10000C at least. Perhaps using small pieces of sheet metal would be better for this one, and heat underneath (I’ve used Ti on a number of occasions, but with any metal, you have to be carefull it doesn’t flake and contaminate the material you’re trying to ignite) as the crucible may conduct too much heat away. The Ca method of reducing your oxides was the way I had in mind, too. I’ve not done this myself, and have often wondered how you avoid contaminating the metal you’re trying to produce with Ca. I guess it depends on getting your stoichiometry exactly right, and performing in a inert atmosphere. Good luck! P.S. The bit about the greater stability of higher oxides towards acids seems to apply only to those which are not strongly oxidising. For example, when you treat Pb in HNO3, you might think that you’d produce a passivating layer of Pb(IV)O2 which would prevent the Pb from dissolving, but this doesn’t happen- the Pb dissolves, and you can produce Pb(NO3)2 this way. However, it’spossible that Pb(IV)O2 forms at least temporarily, as this reaction takes ages. Pb(IV) is very strongly oxidising. After doing a bit of research, Ce(IV) seems to be a lot more stable, (albeit still oxidising), and this quite possible applies to Pr(IV) as well, which is encouaging for the sake of your endevour.

Elementcollector1, 1) Happy new year! 2) To answer this and your latest question: There’s some good logic to this, but I’m not sure that this would work, for several reasons: 1) At the oxide stage, the Pr would be in solid solution in the Nd2O3, and probably being only a minor constituent , would be prevented from oxidising to higher than Pr(III). In other words, the Nd(III) would stabilise the Pr(III), in the same way that Th(IV) stabilises U( IV) in high ( and low) U thorianites. 2) Oxidation states of Pr higher than 3 would be unstable in acid solution- think CeO2, so adding acetic acid would probably cause what little Pr in a >III oxidation state to revert pretty quickly to PrIII and go into solution. 3) Lanthanoid (and related oxides) don’t easily dissolve in acid once they’ve been ignited. Working on a theory that whether a metal oxide dissolves or not depends on the ionic radius, the coordination number of the metal, and possibly it’s affinity for oxigen, I produced some Yb2O3, Sc2O3 and Sm2O3 and found that all 3 were very reluctant to dissolve; after ignition at 1000 degrees C (and then cooling), only the Sm2O3 dissolved, but very slowly. If, however you heat them only minimally ( just enough to ignite the oxalate in your case), then you will probably succeed. 4) By using H2O 2, you may end up with a product containing O-O bonds, which would be bad, because such a compound may prove thermally dangerously unstable. You would have to work on very small quantities to find this out, or alternatively oxidise with something else. What I would suggest is the following, which circumvents problem 1, but not the others, so it’s not guaranteed to work: 1) Dissolve in acid 2) Add H2C2O4 to precipitate Nd and Pr and separate them from Fe, etc. Add NH3 to raise pH to increase precipitation if necessary, or dilute with water. 3) Gently heat precipitate to create Nd/Pr oxides. 4) Dissolve oxides in H2SO4, NOT HCl, as this is reducing. 5) Oxidise this resulting solution to produce Pr2 O(SO4)3(?), a higher basic sulfate of praseodymium which will be insoluble. 6) Filter. You have now recovered Pr. 7) Ignite your P2 O(SO4)3 7)To remaining solution, add NH3 to precipitate Nd(OH)3. 8) Ignite the Nd(OH)3. From here, you’ll have to figure out a way to reduce the Pr and Nd oxides. Good luck, and let us know how you get on. In the mean time, I’ll try to figure out how to create subscripts!

Elementcollector1, it's likely that the free acid, H2C2O4 would be better than the sodium salt, as using the latter, you may get a double- salt or a Na2C2O4-Nd complex, and removing the sodium may prove difficult. If you need to raise the pH at any point, try adding NH3, but try this in a small scale first, so you don't ruin the whole lot if it doesn't work!

Well, you’re braver than me; I hope your Pb does the job with your Am! As for U and Th, they’re amoung the primordials, i.e. have extremely long half- lives- over a million times longer than the most stable isotope of Am, so in theory shielding wouldn’t be necessary for gram quantities, but sealing in glass would be recommendable to prevent Rn escape. In these 2 cases, it’s probably advisable to go for minerals, as getting the elements would be in many cases be quite difficult, I would think.