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?
 
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.
 
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.
 
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?
 
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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