Aluminium ChloroSulphate Crystal

Quote: Originally posted by aga

No experiment is useless. That's like saying a negative experience is useless. What happens is that people Learn at least something. Even if it is just how to get carbon off the inside of an RBF. I distilled weeds with suphuric acid. Now i know not to poke the glass for hours, just get some pirahñas in. Where i'm at, any experiment, useless or otherwise gives me useful experience.

While my product is drying, here’s some more ‘random theorising’ on AJ’s scheme:

"2 AlCl2(OH) (aq) + MgSO4 (aq) ---?---) Mg(OH)2 (s) + Al2(SO4)Cl4"

1. The Ksp of Mg(OH)2 is 5.6 x 10^-12, that means that from a 1 M Mg²⁺ solution Mg(OH)2 only starts precipitating from about pH > 8.4. But at that pH, Al³⁺ cannot stay in solution either.

2. Existence/availability of AlCl2(OH): although there is a CAS number for this formula, precious few sources stock it. Even arguably the largest inventory in the world, Sigma Aldrich, does not stock it. I suspect it only really exists in solution, not as a pure substance. Some datasheets I found indicate an BP of 100 C: that clearly refers to a solution, not an ionic solid.

3. Assume I’m wrong on all this and that Mg(OH)2 precipitates (but not Al). A solution that contains Al3+, sulphate and chloride in the molar ratios of 2/1/4 would then be left behind. But it’s by no means certain that something like “Al2(SO4)Cl4” would crystallise out: whatever crystallises out would have the lowest solubility limit of all possible ones.

[Edited on 9-6-2014 by blogfast25]

1. Semi-quantitative determination of sulphate:

The previous titrations showed the sample solution to be about 0.05 M in Al, so if the empirical formula of the material is AlClSO4 then this solution must also be 0.05 M in sulphate and 0.05 M in chloride.

50 ml of 0.05 M sample solution was placed in a beaker. 50 ml of 0.05 M MgSO4 was prepared and placed in a second beaker. Then 100 ml of 0.1 M Ba(NO3)2 solution was prepared.

50 ml of the barium nitrate solution was added to both the beakers. Left: the sample solution, right: the MgSO4 solution:



The precipitate is the highly insoluble BaSO4, which forms in both solutions in roughly equal amounts (visual assessment).

This is unequivocal proof the material contains sulphate ions in its lattice.

2. Semi-quantitative determination of chloride:

The BaSO4 suspension of the sample solution was filtered on Buchner to remove all BaSO4 and the filtrate placed in a beaker.

About 100 ml of 0.05 M NH4Cl was prepared in another beaker.

To both was added about 50 ml of silver nitrate (AgNO3) 0.1 M. In both cases a comparable amount of insoluble AgCl precipitated. Left: sample solution after filtering and addition of AgNO3 solution, right: NH4Cl 0.05 M after addition of the same amount of AgNO3 solution:



This is unequivocal proof the material contains chloride ions in its lattice.

3. Yield determination:

Based on the amount of product (6.85 g) obtained, assuming it is AlClSO4.6H2O and the 20 g of ASH I started from, the Actual Yield would be about 40 %. I think this could be improved by reducing the amount of HCl 37 % used.

4. Recrysallisation experiment:

The remaining 3.49 g of product were subjected to a recrystallisation test. The material was mixed with 5.4 g of water. This ratio amounts to a mol fraction of AlClSO4 of about 0.033 which is comparable to that of hot, saturated ASH solutions.

The mixture was heated and the product dissolved quickly and to a clear solution. It’s now cooling to check if it will crystallise out.

5. Could this be a double salt of aluminium sulphate and aluminium chloride?

A double salt with the composition Al2(SO4)3.AlCl3.18H2O would correspond to the formula AlClSO4.6H2O, so it’s a possibility.

6. Another batch:

Another batch has now been prepared, this one from 40 g commercial ASH and 80 ml HCl 37 %, for further experimentation.

[Edited on 15-6-2014 by blogfast25]

Quote: Originally posted by The Volatile Chemist

You can make so many double salts, it seems unlikely they'd have a listing for every single one.

If a 'pure substance' is a stable compound there is absolutely no reason NOT to attribute an identifier like CAS or EINECS to it, although I'm uncertain about the protocol. A double salt is a pure substance in its own right.

Whether the compound is actually useful or whether there are many similar ones is neither here nor there.

Your argument would be like a biologist saying: 'pff, there are that many species of beetles, why catalogue them all?' (it's a career ending opinion, BTW! )

My jury is finally out (at least more or less) on the composition of the Al chloride sulphate double salt.

The chloride content was determined using Mohr’s titration: titration of chloride with silver nitrate and potassium chromate as indicator.

Because this titration has to be performed at pH 6.5 to 9, a pH at which aluminium would precipitate as Al(OH)3, the aluminium had to be eliminated prior to the titration, as follows.

3.33 g of dried product (from the second batch) was accurately weighed (1 mg) and dissolved in 100 ml of water. A perfectly clear solution was obtained. About 2 ml of 33 w% ammonia solution was then very gradually added to a pH of about 8. The aluminium precipitates as Al(OH)3.

This was filtered off with a Buchner filter and the filter cake washed repeatedly with small amounts of water until the total filtrate was about 200 ml. This was quantitatively transferred into a 250.0 ml volumetric flask and diluted to the mark.

The silver nitrate solution was first standardised against recrystallised sodium chloride.

The titration yielded a w% chloride of 12.7 w%. Combined with the previously obtained value of Al w% = 9.4 w%, from these values can be deduced that the molar Al/Cl ratio of the original material is 0.995, in other words 1 (one). And from the neutrality requirement (Al³⁺ + Cl^- + SO₄^2- = electrically neutral) can be deduced that the molar ratio Al/Cl/SO4 = 1/1/1.

From this the water content could also be estimated and the formula of the material to be AlClSO₄.7H₂O but I’m fairly sure that because of the indirect determination of water, there is some error on that 7. If it was actually 6, the overall formula would be Al₂(SO₄₃.AlCl₃.18 H₂O.


[Edited on 21-6-2014 by blogfast25]

I've been invited by a German university researcher to co-author an academic paper on the synthesis, analysis and characterisation of this compound. His own research confirms the composition I claimed and also found the hydration (per mole of Al) to be 7, as I found.

If all goes ahead as planned I'll report on where this 'child of ScienceMadness' can be found.