Sciencemadness Discussion Board - Boron extraction from borax

Boron extraction from borax

bonemachine - 27-10-2002 at 13:10

I am looking for a way to extract boron from borax or sodium perborate. Is this possible?

Nick F - 27-10-2002 at 13:15

I'd guess that it'd be possible by adding conc. aq. borax to excess conc. aq. HCl, the boric acid should ppte (I think), and then reducing the boric acid with charcoal.
This will produce the amorphous allotrope; it can also form a VERY hard crystaline form by recrystalisation from molten aluminium (which could also be used for the reduction).
Boron makes a damn good fuel once you get it lit, too.

hmm

bonemachine - 27-10-2002 at 13:35

I got 30% HCL an maybe i try it. Actualy boron is a hell of a good fuell if you consider that it is used with kno3 in airbag compositions for rapid compustion.

-

bonemachine - 27-10-2002 at 13:42

I am also interested in manufacture of b(NO3)3. I had a thought that i could use kno3 as a starting material and maybe borax? Anyone have some experience?

vulture - 27-10-2002 at 13:45

Reducing with Al does not work, as you get AlB¹² (yes twelve), commonly referred to as "square" (quadratic) boron.
Magnesium does work.

Whoa

PrimoPyro - 27-10-2002 at 13:49

AlB12 wouldn't that be um 'good' for thermite fuel?

Is it especially dense?

Nick F - 27-10-2002 at 14:09

AlB₁₂? Wow.
Maybe that's actually what you get when you recrystalise B from molten Al, but I read that it was an allotrope of B.
Yes, it should make a very good fuel...

-

bonemachine - 28-10-2002 at 01:25

Yes that fuel looks very inderesting due to it's high density. What abougt boron nitrate? Can anyone imagine if there is a possible way to synthesise it?

BromicAcid - 16-7-2003 at 16:04

Here are some relevent reactions for involving boron and its oxide.

[Boric oxide or better anhydrous borax, may be reduced by an equivilent amount of metallic calcium or CaH2 The reaction is perfomed in a crucible, 5 - 10% excess CaO being added to moderate the violence of the reaction. Ignition is accomplished by igniting a fuse mixture of sodium peroxide and metalic calcium with magnesium ribbon.]

(a) B2O3 + 3Ca ---> 3CaO + 2B
(b) Na2B4O7 + 6Ca ---> Na2O + 6CaO + 4B
(c) 2B2O3 + 3CaH2 ---> 4B + 3CaO + 3H2O

[When boron trioxide is heated with twice its weight or more of finely powdered magnesium in a tightly covered iron crucible magnesium boride is formed (a), which hydrolyzes in water at 0 C, and liberates H2 (b).]

(a) B2O3 + 6Mg ---> Mg3B2 + 3MgO
(b) Mg3B2 + 6H2O ---> Mg3B2*(OH)6 + 3H2

[Boron trioxide is reduced by magnesium.]

B2O3 + 3Mg ---> 2B + 3MgO

[Hydrochloric acid (or nitric acid) reacts upon magnesium boride to form boron hexahydride. The hydride is an inflammable gas.]

Mg3B2 + 6HCl ---> 3MgCl2 + B2H6

Hope this helps, I know it did for me to fine tune a bit.

BromicAcid - 18-9-2003 at 16:12

I'm re-doing some work with producing boron and I'm going the easier method of aluminum reduction. I figure that I can react the boron produced with chlorine and produce the boron trichloride which would volitize leaving the mass of Al12B and Al2O3 and be collectable. The problem is that I cannot find any resources on hand that tell me how to reduce the boron trichloride to boron except for the reaction with a hot tungsten filiment at elevated temperatues in the presence of hydrogen. So what I'm looking for is a simpiler reaction with boron trichloride to produce elemental boron. Any further help would be appreciated!

Edited to answer my own question for future generations:

2BCl3 + 3Zn ----> 3ZnCl2 + 2B (900 C)

[Edited on 1/11/2004 by BromicAcid]

chemoleo - 18-9-2003 at 16:24

on an other note, I doubt that B(NO3)3 is produced easily, as this would be the nitric acid anhydride of boric acid (B(OH)3). In other words, this is NOT a salt of a hydroxide, like with Ca(OH)2 etc!

[Edited on 19-9-2003 by chemoleo]

BromicAcid - 15-12-2004 at 11:14

Quote:

I am also interested in manufacture of b(NO3)3. I had a thought that i could use kno3 as a starting material and maybe borax? Anyone have some experience?

I was reading through a book "Preparative Inorganic Reactions Vol 1",specifically an article by C.C. Addison and N. Logan when I came across mention of B(NO3)3 and immediately thought of this thread:

Quote:

Chlorine nitrate is a yellow liquid, melting point -107C., boiling point (by extrapolation) +18C. The chlorine atom may be considered to carry a partial positive charge Cl+ - ONO2-, so that in its reactions the compound is a ready source of nitrate ions. A further possible advantage of chlorine nitrate lies in the fact that its reactions are not complicated by the NO+ or NO2+ ions which are inevitably present in N2O4 or N2O5.

Because of its low melting point, reactions with chlorine nitrate, e.g.,

TiCl4 + 4ClNO3 ---> 4Cl2 + Ti(NO3)4

can be carried out conveniently at the temperature of solid carbon dioxide (-80C.) and chlorine, with excess chlorine nitrate, can be removed in vacuum at this temperature. The nitrates B(NO3)3 (-78C.), Al(NO3)3 (-7C.), and Sn(NO3)4 (-60C.) are said to be prepared in this way at the temperatures given.

JohnWW - 16-12-2004 at 17:06

B(NO3)3 would be a covalent nitrate; this class of compounds are highly explosive, like cellulose nitrate. Besides, because it is also electron-deficient, it would be polymeric, with O bridges between B atoms.

Reduction with Al powder

Magius - 25-12-2004 at 07:04

Does anyone know what temperature the reduction of boric acid by aluminum would start at, and if the reaction would be exothermic?

The reaction between Calcium and sodium tetraborate that Bromic mentioned seems interesting. Could the use of HCl be skipped the tetraborate be reduced directly, looking something like this?

3Na2B4O7 + 13Al ---> 3Na2O + 6Al2O3 + AlB12

Would there be anyway to seperate the Al2O3 from the AlB12? This fuel looks very interesting for a themite reaction.

cyclonite4 - 25-12-2004 at 08:06

To seperate Al2O3 and AlB12, try mixing with a stoichiometric amount of NaOH, to form the tetrahydroxyaluminate ion, which is soluble in water. I'd give you the stoich now, but i'm about to goto sleep and cant be bothered figuring it, shouldn't be to hard for you.

[Edited on 25-12-2004 by cyclonite4]

S.C. Wack - 25-12-2004 at 09:31

The Al and Mg reactions all seem to use a large amount of S as well, so I thought that it ought to be mentioned.

cyclonite4 - 25-12-2004 at 19:09

@S.C. Wack: By 'S' do you mean sulfur, because I can't see one mention of the word sulfur

. What do you mean by 'S'?

BromicAcid - 25-12-2004 at 21:12

For what purpose do you want the AlB12, it's fairly inert to everything.

Quote:

Does anyone know what temperature the reduction of boric acid by aluminum would start at, and if the reaction would be exothermic?

I believe that even the reaction of boric acid with aluminum once initiated would be exothermic. When I did a thermite type reaciton between boric oxide and aluminum the oxide was made by dehydration of boric acid and probably still contained an appreciable precentage of it, the mix itself upon heating would not ignite, but when a very small area of it was mixed with KClO4 that part quickly caught fire and the heat spread through the rest of the reaction mass.

S.C. Wack - 26-12-2004 at 00:11

So you can't see one mention of the word sulfur - this means that you haven't looked at the preparations of B (from Mg) or AlB12 in Brauer's Handbook of Preparative Inorganic Chemistry, Schlessinger's Inorganic Laboratory Preparations, or Biltz's Laboratory Methods of Inorganic Chemistry. Well, I have, thus my earlier post. It seems to be preferred.

JohnWW - 26-12-2004 at 13:46

Are those on the forum's FTP?

S.C. Wack - 26-12-2004 at 16:19

Alchemist uploaded the English version of Brauer to the FTP. It was available as a torrent at roguesci. I might make a better one, because quality (and in places legibility) was sacrificed for speed. Mephisto did an excellent job with the original German set and it is in his folder. I did Schlessinger (which only mentions the Al exp) myself, it isn't pretty and my last few uploads were done much better.

I didn't think that the whole Biltz, Hall, and Blanchard book was worth doing entirely and I'm not uploading a partial book. The details there for these experiments (both use S and give no alternative) are in part contained in the related prep of Si (though not as abbreviated as with Schlessinger), so I'll give Brauer's:

Boron
I. According to Moissan, very impure amorphous boron, containing
about 80-90% B, is obtained by the reaction of B2O3 with magnesium.
According to Kroll the optimum yields are obtained as follows: A
fireclay crucible, appoximately 20 cm. high and 16 cm. in diam-
eter, is painted with a paste of ignited MgO and sintered MgCl2 and
dried in a low-temperature oven. A mixture of 110 g. of B2O3,
115 g. of Mg shavings (the use of Mg powder frequently leads to
explosive reactions) and 94 g. of powdered S is placed in the cruc-
ible. The reaction is started with an ignition pellet, after which it
proceeds vigorously. After the mixture is cooled, it is extracted
in water and then in dilute HCl for a week. The residue is treated
several times by heating with HF and HCl, washed with water and
dried in vacuum at 100C. The yields are variable, with a maximum
of 46%

AlB12
A mixture of 50 g. B2O3, 75 g. of S and 100 g. of Al (all the
reagents must be dry) is reacted in a fireclay crucible. After
cooling, the melt is removed from the crucible and pulverized, and
water is added. After elutriation of the slag, the reduced particles
are sorted out from the residue, separated as far as possible from
the slag, and treated with concentrated HCl until a brilliant black
crystalline residue remains. The latter is treated with 40% HF in a
Pt crucible, washed with water and left in HCl until gas evolution
ceases. It is then filtered, washed and dried.

Preparation of Boron Trioxide

I am a fish - 23-3-2005 at 11:05

I am planning to make boron trioxide from boric acid.

When heated at 100°C, boric acid decomposes into metaboric acid:
H3BO3 → HBO2 + H2O

When heated further, metaboric acid decomposes into boron trioxide:
2HBO2 → B2O3 + H2O

However, I don't know the conditions for the second reaction. Does anyone know what temperature and duration of heating is required to reliably complete the conversion?

chemoleo - 23-3-2005 at 11:43

Didn't Bromic do a reaction with it somewhere? I think he also tried a thermite.

Anyway, I also made it a while ago, from boric acid - I just heated it until no more bubbling would occur (under a bunsen). This then produced a glassy mass that could be pulled into superthin threads. As far as I remember it is also slightly irritant to the nose (that is, the dust).

I am a fish - 24-3-2005 at 14:08

I made some boron trioxide today. I obtained a glass-like substance. However, it certainly couldn't be pulled into threads. When I tried, it merely shattered. I wonder what was different between the procedures we used.

Edited to add:

Chemoleo: What temperature was your boron trioxide at when you pulled it into threads? I previously assumed that you had let it cool to room temperature.

[Edited on 25-3-2005 by I am a fish]

12AX7 - 24-3-2005 at 19:06

Well, when molten. Like glass. Matter of fact, borax is essentially a lower-melting, somewhat softer analog of silica, which also forms glass, pure.

Tim

TheBear - 8-4-2005 at 08:28

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]

12AX7 - 8-4-2005 at 08:38

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

S.C. Wack - 8-4-2005 at 10:47

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

TheBear - 8-4-2005 at 12:20

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]

BromicAcid - 8-4-2005 at 19:02

Boron trichloride and tribromide can be reduced with zinc at high temperature, but more readily handleable then hydrogen gas in my opinion.

chemoleo - 8-4-2005 at 19:29

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]

TheBear - 9-4-2005 at 00:18

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]

chemoleo - 9-4-2005 at 11:50

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.