I have granules of $\ce{Al2O3}$ and granules of $\ce{SiO2}$. They look almost identical and I have mixed them up. Is there a simple chemical test that would allow me to differentiate between these granules?
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1$\begingroup$ HF will etch the SiO2, but not harm the Al2O3 (unless heated). Note that HF is potentially nasty stuff. Optical measurements, including Raman, would do it as well. X-ray fluorescence would as well. Rutherford backscattering too... $\endgroup$– Jon CusterCommented Jun 13, 2017 at 19:05
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$\begingroup$ I have some HF... I am thinking of trying this with proper precautions in a fume hood, etc. After a while at room temperature I could expect SiO2 granules should shoe some kind of disintegration? I was under the impression that Al2O3 would also react with HF. $\endgroup$– sTr8_StrugginCommented Jun 13, 2017 at 19:22
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$\begingroup$ Certainly sapphire is resistant to HF unless heated well above room temperature. SiO2 will dissolve pretty quickly. $\endgroup$– Jon CusterCommented Jun 13, 2017 at 19:36
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1$\begingroup$ What about separating based on density? For corundum, I see 3.95 g/cc, and quartz is 2.65 g/cc. Find a liquid with a density between those two and you're good to go. $\endgroup$– Todd MinehardtCommented Jun 13, 2017 at 19:37
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1$\begingroup$ @ToddMinehardt Yup, flotation is the way to go - unless the granules are small enough for stable dispersion to be an issue. $\endgroup$– StianCommented Jun 13, 2017 at 19:42
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4 Answers
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Although not a chemical method, using flotation in bromine to separate your mixture will work at room temperature. Unlike the chemical methods given in other answers, this physical method of separation does not destroy or otherwise degrade the components of your mixture, so you can recover them.
The density of bromine at 20 $^\circ$C is 3.1 g$\cdot$cm$^{-3}$; the densities of $\ce{Al2O3}$ and $\ce{SiO2}$ at the same temperature are 3.95 g$\cdot$cm$^{-3}$ and 2.65 g$\cdot$cm$^{-3}$, respectively. The $\ce{SiO2}$ will float and the $\ce{Al2O3}$ will sink.
As noted in the comment thread above, this method will work unless the particle size is too small.
And as noted in the comments below, sodium polytungstate solution (with a density of 3.1 g$\cdot$cm$^{-3}$) is an alternative to bromine.
Because there are likely a large number of substances with a density in the range required to separate your mixture, the suggestions here represent a small number of those choices.
answered Jun 13, 2017 at 19:53
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$\begingroup$ Bromine is not very practical for sink-float analysis. I'd rather suggest sodium polytungstate solution. Sigma-Aldrich offers it with the same density as bromine. sigmaaldrich.com/catalog/product/sigma/71913 $\endgroup$ Commented Jun 14, 2017 at 23:06
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$\begingroup$ @aventurin - Thanks for the info, I've edited my answer to include it. $\endgroup$ Commented Jun 14, 2017 at 23:14
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1$\begingroup$ Bromine has the extra feature of the possibility of separating the separator from his life... $\endgroup$– StianCommented Jun 15, 2017 at 13:49
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Acids can differentiate $\ce{Al2O3}$ and $\ce{SiO2}$. There is a famous reaction called etching of glass where $\ce{SiO2}$ reacts with hydrofluoric acid $\ce{HF}$. $\ce{Al2O3}$ also reacts with $\ce{HF}$ but it requires heat. This test somehow differentiates them but we need to determine which substance is which by reacting with an acid which reacts with only one oxide. Strong acids like $\ce{HCl, H2SO4}$ or will suffice.
It reacts with acids to form corresponding aluminium salts(sulfate, chloride)
$$\ce{Al2O3 + 3H2SO4 ->[\Delta] Al2(SO4)3 + 3H2O}$$
$$\ce{Al2O3 + 6HCl ->[\Delta] 2AlCl3 + 3H2O}$$
Silicon(IV) oxide does not react with any of the above acid. Rather silicon(II) oxide reacts with sulfuric acid or nitric acid to produce silicon(IV) oxide.
answered Jun 13, 2017 at 19:36
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$\begingroup$ $\ce{\alpha Al2O3}$ does neither react with sulphuric nor with hydrochloric acid. $\endgroup$ Commented Jun 14, 2017 at 18:28
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$\begingroup$ @aventurin aluminum metal does not react with any of the acids due to passivation layer of Al2O3 but since Al2O3 is present here in powdered form, surface area increases and reaction of acids is possible. $\endgroup$ Commented Jun 14, 2017 at 18:50
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$\begingroup$ The question is not about aluminium metal. My comment was about alpha aluminium oxide which is known to be insoluble in sulphuric and hydrochloric acid. Btw, I'm not sure if granules is the same as powder. $\endgroup$ Commented Jun 14, 2017 at 20:04
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Give a fine powdered mixture of your substance and $\ce{CaF2}$ a lead crucible. Add $1~\pu{ml}$ of concentrated sulfuric acid and warm gently.
If your substance is $\ce{SiO2}$ then gaseous $\ce{SiF4}$ will form:
$$\ce{2 F- + H2SO4 -> 2 HF + SO4^{2-}} \tag{1}$$ $$\ce{SiO2 + 4 HF -> SiF4 + 2 H2O} \tag{2}$$
A moistened black filter paper placed on the cap of the crucible will show white $\ce{SiO2}$ when dried (reaction (2) reverted).
As an alternative to the black filter paper you could place a water drop hanging at a black glass rod above the small hole in the crucible's cap. You will see silicic acid at the drop if $\ce{SiO2}$ is present.
answered Jun 13, 2017 at 21:35
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$\begingroup$ I believe they are mixed in the same container. $\endgroup$– EquinoxCommented Jun 14, 2017 at 3:18
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A classical method is to digest the oxide in a sixfold excess of an eutectic mixture of sodium and potassium carbonate at a temperature of about 800°C in a nickel, iron or platinum crucible.
$$\ce{Al2O3 + Na2CO3 -> 2 NaAlO2 + CO2}$$ $$\ce{SiO2 + Na2CO3 -> Na2SiO3 + CO2}$$
$\ce{Al2O3}$ and $\ce{SiO2}$ are distinguished by extracting the cooled melt with concentrated hydrochloric acid. $\ce{AlCl3}$ gets dissolved while silicic acid is insoluble.
The procedure can also be used to separate $\ce{Al}$ and $\ce{Si}$ from a mixture of $\ce{Al2O3}$ and $\ce{SiO2}$.
answered Jun 13, 2017 at 21:06