Making a Mg supplement (is this calc correct?)

akirsonis

New member
I am trying to make a magnesium supplement with magnesium chloride hexahydrate (MgCL2 . 6H20). The MW of the compound is 203.302 g/mol and the MW of Mg is 24.305 g/mol. So in 203.302 g of the compound there should be 24.305 g of Mg2+, correct?

I want the solution to have a [Mg2+] of 100,000 ppm ( equivalent to Brightwell)

I am trying to make the solution in 5 gallon batches and this is how I am calculating the amount of magnesium chloride to add, but the amount I come up with seems absurdly large. Perhaps someone can find the error in my logic.

(5 Gallons)(3.785 L / 1 Gallon)(100,000 mg Mg2+ / 1 L)(1 g Mg 2+ / 1000 mg Mg2+)(203.302 g Magnesium chloride hexahydrate / 24.305 g Mg2+) = 15,830 g of the magnesium chloride hexahydrate.

15,830 g (1 kg / 1000 g)(2.2 lbs / 1 kg) = 34.83 pounds of magnesium chloride hexahydrate

So I need almost 35 pounds of magnesium chloride hexahydrate to make 5 gallons of a solution that is 100,000 ppm Mg2+? That seems like a very large amount.

Randy, what am I doing wrong?
 

disc1

-RT * ln(k)
That calculation is correct. Keep in mind that about half of the weight of that powder is water.

I don't know if you can put that much in without help. Bright wells product may have a chelator or some other trick that helps him get more in there. Low pH will help too.
 

akirsonis

New member
Thanks for the confirmation. We will see how hard it is to dissolve. I believe my formula is 836 g/L and the solubility of magnesium chloride hexahydrate is 1,570 g/L.
 

akirsonis

New member
Also, assuming this is possible to make without any tricks to help with solubility, I would add the magnesium chloride to my 5 gallon container and fill to the 5 gal mark with water, rather than weighing out my magnesium chloride and adding 5 gallons of water, correct?
 

disc1

-RT * ln(k)
With that much water in the crystal I'd bring it up in a little less than 5ga and get it mostly dissolved then top it up to 5ga.
 
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