woelen
Super Administrator
Posts: 7977
Registered: 20-8-2005
Location: Netherlands
Member Is Offline
Mood: interested
|
|
Sodium molybdate, ratio Na2O : MoO3?
I purchased some MoO3 from eBay:
http://cgi.ebay.nl/MOLYBDENUM-TRIOXIDE-Mo66-4-High-grade-mat...
This is very impure material, it is green and on further analysis it is clear that this contains a lot of iron, but no other transition metal in more
than trace amounts. I purified this material and now I have a beautiful white crystalline solid, which easily dissolves in water. It is a form of
sodium molybdate.
My question is what I exactly have.
I did the following:
Take all 100 grams of the material and add approximately 400 ml of distilled water. Swirl to make a green suspension.
In a separate beaker, take 100 ml of water and dissolve a lot of 99% NaOH.
Next, add the concentrated solution of NaOH very slowly and swirling the solution all the time, over the course of an hour or so. Add so much of the
solution that all but a tiny fraction (maybe 1 gram) of the green powder dissolves. I did not want to have an excess of NaOH added.
When the green powder dissolves, then a brown flocculent suspension is obtained. When the liquid is swirled and then allowed to stand, the green
material sinks to the bottom in seconds, while the brown flocculent solid remains floating in the liquid for a much longer time. In this way I could
easily determine whether all of the MoO3 is dissolved or not.
I then vacuum filtered the liquid in a Buchner funnel with ultra fine paper. When this is done, an almost clear light yellow/brown liquid is obtained
and more than 1.5 mm of chocolate brown solid remained in the filter (90 mm diameter).
The light yellow liquid then was put in a long tube and allowed to stand for a few days. A thin layer of very fine solid was deposited on the bottom
of the thin cylinder, appr. 5 mm of very fine flocculent solid. Above this, a completely clear very light greenish/yellow liquid was obtained.
This clear liquid was drawn into a syringe and put in a big glass dish. The last 1 cm of liquid (with 5 mm clear liquid and 5 mm flocculent
precipitate) was discarded.
Appr. 10 ml of liquid was discarded, could not be filtered due to the extremely fine solid particles, the rest was evaporated to dryness on a warm
place (40 C or so, in the glass dish for more than 7 days). Finally, a white, perfectly dry, non-hygroscopic, solid was obtained.
I know that what I have is almost pure molybdate, it tests negatively for iron with conc. thiocyanate. It is white, so other first row transition
metals are not in the solid.
The problem is that molybdate is not a single entity, but these can be very complicated. The only thing I know of my solid is that it is something
with empirical formula Na2O.xMoO3.yH2O, with unknown x and y. It might be Na2MoO4.2H2O, but I'm not sure, I did not measure the amount of NaOH I
added, I only know that the amount of NaOH added is just not enough to dissolve all MoO3.
I already read a lot in my books about molybdenum, but this does not solve my problem. If there is someone, who knows more about the values of x and
y, then that would be nice. I did not yet label the bottle with the molybdate, because the formula is not known to me.
The material is very nice though for experiments. It dissolves very easily, leaving the liquid somewhat opalescent. A tiny amount of NaOH makes the
liquid completely clear (and colorless).
The white solid dissolves in water very easily
[Edited on 31-5-09 by woelen]
|
|
|
panziandi
Hazard to Others
Posts: 490
Registered: 3-10-2006
Location: UK
Member Is Offline
Mood: Bored
|
|
I would hazard a guess as to it being Na2MoO4.2H2O, if you are worried I suppose you could reform the trioxide from it and then react
stoichiometrically with sodium hydroxide by a standard procedure, and then recrystalise it? You could make the trioxide from molybdenum disulphide by
roasting and then compare there "molybdates" you obtain.
Could you analyse the product gravimetrically?
M.W. dihydrate is 241.95 g/mol and that of the anhydrous form is 205.92 g/mol. Perhaps weigh out 0.01mol of your molybdate and carefully heat until
you get a constant weight (in theory should have decreased to 2.059g) that way it will be pretty suggestive that you have Na2MoO4.2H2O
You could even precipitate out Mo and/or Na from a known mass of the molybdate and find the mass of ppt, then work out the ratio of each in the
compound etc and deduce formula?
|
|
|
JohnWW
International Hazard
Posts: 2849
Registered: 27-7-2004
Location: New Zealand
Member Is Offline
Mood: No Mood
|
|
With an excess of NaOH, the molybdate formed would be the monomeric species Na2MoO4; see http://en.wikipedia.org/wiki/Sodium_molybdate . This is the stuff, as the dihydrate, used in fertilizer as a molybdenum trace element supplement,
and as a corrosion inhibitor.
Crystallization from more acid solutions with less Na+ or other univalent cation present would result in bridged polymeric species, similarly to
dichromate, trichromate, and pyrosulfate, and to polymeric tungstates. The reagent- and analytical-grade "molybdate" sold for spectrophotometric
analytical purposes t determine phosphate in water, taking advantage of the intense yellow color formed by reaction with phosphate (but silicate,
arsenate, stannate, plumbate, and antimonate interfere), is a polymeric species, ammonium heptamolydate, (NH4)6Mo7O24ยท4H2O; see http://en.wikipedia.org/wiki/Ammonium_heptamolybdate .
|
|
|
woelen
Super Administrator
Posts: 7977
Registered: 20-8-2005
Location: Netherlands
Member Is Offline
Mood: interested
|
|
I did an experiment using an analytical balance and pure MoO3 and NaOH which is 98...99% pure (the balance being water).
The experiment is done as follows:
- Take 10 ml of distilled water and put in a small beaker (25 ml)
- Put the beaker with the water in it in an analytical balance and tare at 0.
- Immediately add a spatula full of MoO3 and read the increase of weight --> this shows 546.5 mg of MoO3
- Retare again (setting readout to 0) and immediately add some NaOH --> this shows 139 mg of NaOH
- Swirl for several minutes
- Retare again and add a tiny amount of NaOH and immediately measure again and then swirl for a few minutes (this is repeated all over again, until
the MoO3 has dissolved almost completely (just a weak opalescence is left).
The final result is 258.5 mg of NaOH added.
The retaring all over again is necessary, because of evaporation of water. With such a sensitive balance you can see the weight going down at a rate
of 100 ug per second or so.
With these weights, 1 mole of MoO3 is dissolved by adding 1.7 mole of NaOH. When the impurity of the NaOH and some absorption of water vapor from air
is taken into account, then the ratio will become somewhat lower, somewhere between 1.65 and 1.7.
This ratio is close to 3 : 5, so I think that in reality one has the following:
3MoO3 + 5NaOH --> ??
The ratio Na2O : MoO3 in the final product is 5 : 6. The product is not Na2MoO4, because in that, the ratio of Na2O : MoO3 is 6 : 6. My final product
is somewhat richer in MoO3 than Na2MoO4.
Another nice observation is that the solution I obtained has a pale yellow color with a faint green hue, while the starting products are purely white
NaOH and pure MoO3. The liquid I obtained from the impure green MoO3 also had this same yellow color.
What could the formula be of the solid material which I have now? Or did I obtain a mix of polymolybdates?
|
|
|
watson.fawkes
International Hazard
Posts: 2793
Registered: 16-8-2008
Member Is Offline
Mood: No Mood
|
|
Quote: Originally posted by woelen  | | What could the formula be of the solid material which I have now? Or did I obtain a mix of polymolybdates? |
Have you measured the pH? That would give an indication about how much oxygen from the NaOH is still in solution in the form of
hydroxyl ions.
Also, the Wikipedia page on molybdates is informative about solution dynamics.
|
|
|
kmno4
International Hazard
Posts: 1495
Registered: 1-6-2005
Location: Silly, stupid country
Member Is Offline
Mood: No Mood
|
|
Something better than Wikipedia 
http://dx.doi.org/10.1021/ja01239a508
|
|
|
woelen
Super Administrator
Posts: 7977
Registered: 20-8-2005
Location: Netherlands
Member Is Offline
Mood: interested
|
|
@watson.fawkes: The pH of the solution is somewhere between 9 and 10, not stongly alkaline at all.
With a ratio 5 : 6 of Na2O and MoO3 a possible formula could be 2Na2MoO4.NaHMoO4.xH2O. When the ratio is 4 : 6 then the formula could be
Na2MoO4.NaHMoO4.xH2O.
I also want to determine x. I poured the liquid in a petri dish, rinsing final drops with a little distilled water and putting that in the petri dish
as well. let the liquid evaporate on a dry place in a petri dish, until I get a totally dry powdery solid in the petri dish. I determine the weight of
the dish + solid. Then I remove the solid, rinse the petri dish and dry the petri dish. Then I weigh again. The difference is the weight of the solid.
I know how much MoO3 I added and how much NaOH, then I can determine the amount of water of crystallization. That also may give some information about
the structure, because water can be present as H2O, but also in the form of on oxo-group, which is converted to two hydroxo groups (O(2-) + H2O -->
2OH(-)). In this way, one can have stuff like NaHMoO4, where the real structure would be Na-O-M(=O)2-OH. In a polymolybdate species, there may be
multiple OH groups.
@KMnO4: Your source might be better than Wikipedia, but the majority of the members cannot access this source and if that restriction is taken into
account as well, then to my opinion Wikipedia is a better source 
Could you provide us with the full PDF?
[Edited on 4-6-09 by woelen]
|
|
|
watson.fawkes
International Hazard
Posts: 2793
Registered: 16-8-2008
Member Is Offline
Mood: No Mood
|
|
At pH 10,
10 ml of water has only 1 mg of OH(-) ion, or .055 mmol. You've got 3.80 mmol MoO3 and 6.46 mmol NaOH. Since OH(-) translates one-to-one to H(+) in
the crystal, the amount of NaHMo{x}O{y} species in your crystal is minimal, less than half a molar percent. The relevant reaction is thus:
Mo{x}O{y}(k-) + OH(-) + Na(+) --> Mo{x}O{y+1}((k+2)-) + H(+) +
Na(+)
Clearly the molybdate species aren't balanced to actual ratios in this equation. It does show, however, that crystallation favors higher oxygen
content == less polymerization.
So I think you've got a mixture of molybdates in the crystal. The general equation for your reaction is thus:
x MoO3 + (2y-6x) NaOH --> Mo{x}O{y}((2y-6x)-) + (2y-6x) Na(+) +
(y-3x) H2O
This equation is relatively simple because all the molybdates have the Mo in oxidation state +6. If you had to account for mixed oxidation states, the
number of variables would be greater. Your observed ratio 1.70 corresponds to the ratio (2 y/x - 6 ) in the above equation. I'm looking for a
proportionally blended reaction, so I want the two stoichiometric ratios for possible molybdates immediately above and below. Those molybdates are
MoO4(2-), ratio 2, and Mo2O7(2-), ratio 1. All others have ratios less than 1, so I can discount them. I'll write down the reactions for these two for
clarity.
MoO_3 + 2 NaOH --> MoO4(2-) + 2 Na(+) + H2O
2 MoO_3 + 2 NaOH --> Mo2O7(2-) + 2 Na(+) + H2O
Taking a 5/3 ratio for Na/Mo, that's 4 parts of the first equation and 1 part of the second. Under the 5/3 assumption, the crystal is
4(Na2(MoO4)).Na2Mo2O7.
If you stick with a 17/10 ratio, though, it comes out 14 parts MoO4 and 3 parts Mo2O7. The heights of the integer ratios just keep going up from here,
though.You'll need to do some crystallography to keep going in this vein.
Almost certainly, though, there's some NaHMoO4 in there. There's likely some Na2Mo7O24 as well. I'd rather not hazard a guess about these.
|
|
|
woelen
Super Administrator
Posts: 7977
Registered: 20-8-2005
Location: Netherlands
Member Is Offline
Mood: interested
|
|
Yes, I indeed think that I have a mix of compounds and that my material is somewhere between Na2MoO4 and NaHMoO4. The real structure probably is very
complicated.
I now also determined the weight of the dry powder after evaporating to dryness very well. The total weight of the dry powder is 875 mg (+/- 5 mg).
So, summarizing: 546.5 mg of MoO3, 258.5 mg of NaOH and 875 mg of dry solid after evaporation.
Now, I did the computations again, and I have found that the amount of water in the final product is a VERY sensitive function of the ratio of MoO3
and Na2O. The problem is that the purity of the NaOH is not known well. The bottle says 98...99%, the rest being water, but it also could have
absorbed some water from the air, before it fell into the solution. So, the ratio can be 1 : 1.7, leading to a 3 : 5 ratio of MoO3 and NaOH, but it
could also be 1 : 1.5. I can determine upper and lower bounds, but I need to do a little more math before final formula-ranges can be given.
|
|
|
sparkgap
International Hazard
Posts: 1234
Registered: 16-1-2005
Location: not where you think
Member Is Offline
Mood: chaotropic
|
|
Hope this hels you, woelen.
sparky (~_~)
Attachment: na2moo4xh2o.pdf (375kB)
This file has been downloaded 657 times
"What's UTFSE? I keep hearing about it, but I can't be arsed to search for the answer..."
|
|
|
|
![[*]](images/xpblue/default_icon.gif){kind=icon}
