Can anyone tell me the formula of Ammonium Molybdate. There seems to be some confusion with different formula coming from different quarters.
Dann2Xenoid - 26-7-2007 at 12:04
My copy of "Mellor's Modern Inorganic Chemistry" 1961 edition, by G.D.Parkes (don't laugh, I've had it all my life) states;
"The molybdates do not as a rule, have simple formulae such as X2MoO4, because of the readiness with which POLYmolybdates are formed. Ammonium
molybdate for instance, is represented by the formula (NH4)6Mo7O24.4H2O, ie. as a derivitive of a hepta-molybdic acid. Made from acting upon
molybdenum trioxide with aqueous ammonia, and allowing the solution to crystallise."
Xenoidagent_entropy - 26-7-2007 at 12:06
the formula for the hydrate is: (NH4)6Mo7O24 * 4H2O
F.W. 1235.86
(Taken directly from the label on a jar of the material from Fisher Scientific Company)Rosco Bodine - 26-7-2007 at 12:10
(NH4)2MoO4
Also
(NH4)6Mo7O24 + 4H2O
ammonium heptamolybdate tetrahydrate
[Edited on 26-7-2007 by Rosco Bodine]woelen - 26-7-2007 at 15:05
And to make things even more complicated, there also is an anhydrous form of (NH4)6Mo7O24, which is sold as 'molybdic acid'. The latter is a slowly
soluble white powder. Once it is dissolved in water, it cannot be distinguished anymore from solutions of the 4-hydrate.dann2 - 26-7-2007 at 17:36
Hello,
Thanks for replys. My Ammonium Molybdate is EBAY stuff.
How would I find out how much Mo I have per gram?
The stuff is white lumps, quite hard, as I have to grind them down with a pestle and mortar to get them to dissolve in water.
Dann212AX7 - 26-7-2007 at 19:01
Gravimetric is easy enough. Dissolve a weighed amount in water, add an excess of something whose molybdate is well-defined and highly insoluble
(barium?), filter, dry and weigh the precipitate. Calculate based on molecular weight.
Alternately, pyrolyze to MoO3 (assumes only volatile ammonium is present), or reduce (in solution or with hydrogen, carbon, etc.) to a lower oxide or
to the metal.
Although... I dig the ammonium molybdate/molybdic acid bit. Amazingly vague for such a common reagent, no?
It is also quite useful for the determination of phosphate and silicate.
Cheers,
O3Xenoid - 27-7-2007 at 03:32
Quote:
Originally posted by dann2
How would I find out how much Mo I have per gram?
The stuff is white lumps, quite hard, as I have to grind them down with a pestle and mortar to get them to dissolve in water.
Dann2
Yeah! I see what you mean!
If you dissolve it in water, and then recrystallise it, you will then have the tetrahydrate for certain!
By my calculations the amount of Mo in the tetrahydrate should be 7 x 95.94 / 1235.36 which is .5436 or 54.36%
Xenoid
[Edited on 27-7-2007 by Xenoid]JohnWW - 27-7-2007 at 08:45
Oh yes - I have some of it sitting on a shelf at home here.dann2 - 27-7-2007 at 19:05
Hello,
It when you add a tiny amount of SnCl2 dissolved in acidified water to A.M. dissolved in slightly acid water (HCl) you get a green ppt. If you use
12% HCl to dissolve both the Ammonium Moly. and the SnCl2 you get no ppt but a nice yellow/orange clear solution.
What is the green ppt?
A.M. is used in the dying business.
Dann2woelen - 28-7-2007 at 09:28
Molybdenum usually occurs in oxidation state +6 in its salts. As such, it forms oxo-anions, which are colorless and your ammonium molybdate is an
example of that.
Molybdenum can, however, fairly easily be reduced to a lower oxidation state (+5, +4, and even +3). When the reduction is only partial, such that most
remains in the +6 oxidation state and a small amount goes to +5, then mixed-oxidation state compounds can be formed. These compounds contain (within
the SAME molecule or ion) molybdenum in oxidation state +5 and +6. Such compounds have intense colors, usually blue, but they can also be black, or
dark green. Search the Internet for 'molybdenum blue' and you'll certainly find information on this.
If more reductor is added, then all molybdenum goes to a lower oxidation state. With strong reductors, you can indeed go to the orange/brown
compounds, which are molybdenum(III), complexed with chloride. Similar results are obtained, when powdered zinc is added to a solution of a molybdate
in hydrochloric acid.
