Thanks to Matt Crook for his error-checking and information.
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Suppose you want to make a red flame like the one on the right…
…but you can’t find a vendor for strontium chloride (I couldn’t) and you don’t want to spend in excess of $20 for lithium chloride (I didn’t). For a few bucks, you can get some strontium carbonate ($3.47/lb from skylighter.com) and easily convert it to strontium chloride.
Here’s how.
How to make strontium chloride from strontium carbonate
Tip: Use glass vessels and glass stirring implements because they are non-reactive.
Combine equal parts by weight strontium carbonate and muriatic acid (obtainable at any pool supply store). This assumes a ~33% solution of muriatic acid.
Tip: Don’t have scale? Find yourself a postage scale or a small kitchen scale (for baking).
Tip: What if your solution has a different concentration? Either do the math (explained later) to calculate a new ratio or else dilute your acid to ~33%.
Add the two reagents gradually because they will release energy. They’ll bubble and fizz and heat up. They could bubble over your container if you’re not careful.
Let the reaction between muriatic acid and strontium carbonate run it’s course (may require some stirring). Then let the fluid evaporate. What you’re left with is strontium chloride.
Tip: If you err on the side of too much HCl, it should be okay. It should evaporate. If you err on the side of too much strontium carbonate, it will be safe, but you’ll end up with some carbonate impurity in your chloride. This should not impair your flame colour appreciably.
To get the flame, you might dissolve your strontium chloride in enough methanol so that there are no sediments. Then light the methanol on fire.
The science behind it
Yes, strontium carbonate is preferred for pyrotechnics, but as tressure points out in this chemistry thread, it won’t yield a red flame in methanol. (I expect it requires much greater heat, such as fireworks can produce.) His advice, like mine, is to convert it to the chloride.
First thing to know: muriatic acid is hydrochloric acid (HCl), so the reaction we aim to produce is as follows.
SrCO3 + 2HCl -> SrCl2 + H2O + CO2
The formula says that we need twice as many HCl molecules (molecular weight: 36.46 g/mol) as SrCO3 (molecular weight: 147.63 g/mol). By weight, then we need twice as much SrCO3 as solid HCl (the exact opposite of the molarity ratio: 147.63 / (36.46 * 2) = 2.02).
As mentioned above, however, you’ll probably buy HCl in a stock solution, rather than measuring out a solid and diluting it yourself. I had a 31.45% solution by weight on hand. Water weighs half as much as the acid solute:
H2O molecular weight = 18.02 g/mol
HCl molecular weight = 36.46 g/mol
So in a mix that’s close to 33%, half the molecules are HCl, and half are H2O. That being the case, 10 g of the acid solution has 5 g of HCl and can be mixed with 10 g of SrCO3. I.e. you can mix equal parts by weight of the HCl solution and SrCO3.
Matt went so far as to calculate the enthalpies (the energy required or released) for this reaction and posted them on the Psandboxi at the Pedestrian site (29 Oct 2011). He was right, of course: it was slightly exothermic.
Yes, I spelled ‘Psandbox’ correctly. That’s how it was spelled by whoever set it up (must have been Matt or Jonathan).
I’m glad it worked out for you, Markham.I’m not sure who’s idea it was to call it the Psandbox, but it wasn’t mine. I am, however, responsible for perpetuating that spelling on duck-of-doom.com.