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TheBear
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Did an experiment at school today (free hands  )
B2O3 was produced by heating B(OH)3, the B2O3 was powdered using mortar and pestle (lost half of it that way :p). Then we mixed it up with Mg powder
and put it in a cruicible with a cap. Everything was heated until it began to melt, then the cap was removed and the flame was directed into the
cruicible using a blowpipe. Everything reacted in a flash (I was very starteled since I was expecting a much more peaceful reaction  ). What was left was a black and crusty material. To this 5M HCl acid was added
whereupon small explosions were heard (I was asuming H2 from unreacted Mg).
After breaking up the pieces in the beaker we were left with a suspension of this black mass. The black mass was filtered out. On a flame colour test
the black mass did not colour the flame green.
All amounts of chemicals were calculated from this assumption:
B2O3 + 3Mg ---> 2B + 3MgO
MgO + 2HCl ---> MgCl2 + H2O
Apparently things are more comlicated than this (I've realised after reading this thread). My questions to you are:
What do you think this black crusty material is?
And where did out boron go? 
I'm now thinking we might lost it in the form of B2H6? (The small explosions and all?)
However I can not figure out what the black crusty material is for something when it does not colour our flame green. (might be that it still contains
boron).
Note: HCl solution of black crusty material had a very funky™ smell.
[Edited on 8-4-2005 by TheBear]
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12AX7
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| Quote: | Originally posted by TheBear
Note: HCl solution of black crusty material had a very funky™ smell. |
Did it smell like, uhh crap I have no comparison. Well how about this-
Roast a sulfate with carbon. Say, gypsum. Makes CaS. Add HCl, makes H2S. I say it smells like slag, because...
Take all your aluminum slag from your meltings (surely everyone has a home foundry?  ), melt and add salt until a soupy black mess is obtained, pour ingots. Black stuff is aluminum oxide in salt. Smash up to reveal a
horrible stink that smells just like sulfides in acid (but with sulfides nowhere to be found??).
Tim
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S.C. Wack
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From the Handbuch der Präparativen Chemie:
Nach der Methode von Moissan verfährt man folgendermassen: Reines, im Platintiegel durch öfteres Umschmelzen sorgfältig von Wasser befreites
Bortrioxyd wird mit reinstem, besonders von Kieselsäure und Eisen freiem Magnesium im Verhältnis 3 B2O3 auf 1 Mg, was ungefähr einem Drittel der
berechneten Menge an Magnesium entspricht, sehr fein pulverisiert und in einem bedeckten hessischen Tiegel in den vorher auf Rotglut erhitzten
Perrotschen Ofen gebracht. Nach ungefähr 4—5 Minuten geht die Reaktion vor sich und dabei entwickelt sich eine solche Hitze, dass der Tiegel
weissglühend wird. Der nach dem Erkalten sich meist leicht loslösende Kuchen zeigt eine äussere, wenig tiefe, schwarze Schicht, während der
innere, mehr oder weniger blasige Teil eine braune Farbe besitzt. Die ganze Masse ist durchsetzt von den weissen Teilen des Borates. Man entfernt
sorgfältig die schwarze Schicht und behandelt die pulverisierte braune Masse mit viel siedendem Wasser und reiner Salzsäure. Schliesslich wird
fünf- oder sechsmal mit konzentrierter Salzsäure ausgekocht. Der Rückstand wird mit destilliertem Wasser gewaschen und mit einer lOproz.
alkoholischen Lösung von Kalilauge gekocht, wieder gewaschen und mit zur Hälfte verdünnter Flußsäure im Platingefäss mit Platinrückflusskühler
gekocht. Das nochmals sorgfältig gewaschene und im Vakuum über Phosphorpentoxyd getrocknete Produkt ist braungefärbt, enthält kein Wasser, keinen
Wasserstoff und keine Borsäure und besteht aus 93,97—95,00% Bor, 2,28—4,05% Magnesium und 1,18—1,60% Unlöslichem. Um den geringen Gehalt an
Magnesiumborid zu entfernen, wird das Pulver mit Bortrioxyd gemischt und nochmals in derselben Weise behandelt; dann ist es etwas heller gefärbt und
besteht aus 98,30% Bor. 0,37% Magnesium und 1,18% Unlöslichem. Der schwarze, unlösliche Rückstand besteht hauptsächlich aus Bornitrid. Der aus den
Verbrennungsgasen herrührende Stickstoff lässt sich in der Weise unschädlich machen, dass man den Tiegel, in welchem die Reaktion vor sich geht, in
einem anderen Tiegel vollständig mit einer Mischung von Titandioxyd und Kohle umgibt. Das unter dieser Vorsichtsmassregel erhaltene Pulver zeigt
einen Gehalt von 99,2—99;6% Bor. Der aus den Verbrennungsgasen bei letzterer Vorsichtsmassregel aus dem umgebenden Kohlenpulver herrührende
Kohlenstoff zeigt sich als Borcarbid in der äusseren schwarzen Schicht des Reaktionskuchens, welche deshalb sorgfältig entfernt werden muss. Diese
Verunreinigung durch die Verbrennungsgase lässt sich ganz umgehen, wenn man die Reduktion der Mischung in einem Porzellanschiffchen und einer
Porzellanröhre, ausführt unter Anwendung einer stickstoffreien Wasserstoffatmosphäre. Das so erhaltene Produkt ist sehr rein; doch sind die
Ausbeuten gering.
Reinstes amorphes Bor wurde in jüngster Zeit durch Reduktion von Borchlorid mit Wasserstoff im Hochspannungslichtbogen dargestellt.
Eigenschaften: Das nach der Methode von Moissan dargestellte Bor ist ein graubraunes Pulver. Es entzündet sich an der Luft bei 700°.
Moissan (in French of course):
Attachment: moissan.pdf (453kB)
This file has been downloaded 971 times
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TheBear
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Ahh.. Seems like there's a decent chance that what I have is still boron. But the only way one will be able to get the shiny metal will be by
making the trichloride (poisonous as *** if I remember correctly) and react it with hydrogen?
Though I have my doubts since I'm not sure wheter the little piece of black material (perhaps darkbrown depending on how you look at it) actually
caught fire during the flame test. The propane torch reaches more than 700 *C right?
Anyone have any suggestions for any analytical tests to determine what this consists of (no I don't have access to NMR  ).
Another thougt: during the reaction the contents gets extremely hot and if my memory isn't failing me I remember MgO changing into a very
unreactive form at extreme temperatures, is this the reason for why such long boiling with strong acids is necessary?
[Edited on 8-4-2005 by TheBear]
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BromicAcid
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Boron trichloride and tribromide can be reduced with zinc at high temperature, but more readily handleable then hydrogen gas in my opinion.
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chemoleo
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TheBear, did you get it translated with Babelfish or something? From the reference of S.C.Wack, it is possible that the black residue is boron nitride
('Der schwarze, unlösliche Rückstand besteht hauptsächlich aus Bornitrid' ) , which should be found on the surface of the reaction
mainly.
This black residue is removed, to expose the comparatively brown (B2O3-bubbly) boron within. Flame test is not an answer by the way, i.e. Cu metal
won't give a green colour even though its salts will. It's a question of ionisation temperatures, which determine the absorption spectra.
I.e., BaSO4 will require more heat to preduce a green flame, than Ba(ClO3)2 will.
Acc. to S.C.Wacks ref, one would desire 3 B2O3 to 1 Mg, which is then heated to red heat. The reaction commences itself. Afterwards, the black outer
layer is removed, and the inner boiled with hot water/HCl (5-6 times). It can be cleaned further with HF.
In fact, this prep sounds very similar to a silicon prep I posted somewhere.
[Edited on 9-4-2005 by chemoleo]
Never Stop to Begin, and Never Begin to Stop...
Tolerance is good. But not with the intolerant! (Wilhelm Busch)
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TheBear
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No babelfish, after 6 years of inefficient german lessons in school I actually understand the language pretty well. What I meant was that I could
accept calling the "black material" dark brown.
And I don't think it's likely that atmospheric nitrogen could have reacted with more than a small part of the reactants.
But what I can't see is why to use 3 B2O3 for every Mg, I added the other way around (3 Mg for every B2O3) but I guess it's neccessary to
get free boron. (Does it form an alloy with Mg in accordance to earlier posts in this thread if "my" method is used?)
[Edited on 9-4-2005 by TheBear]
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chemoleo
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Oh, the reason why they use an excess of B2O3 vs Mg is because they are trying to avoid an excessively energetic reaction, keep the amount of side
products down (i.e. Mg3B2) by restricting Mg, and possibly to increase the ease of purification.
Never Stop to Begin, and Never Begin to Stop...
Tolerance is good. But not with the intolerant! (Wilhelm Busch)
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