Making a 12% Magnesium Chloride solution

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Can specific gravity be ignored?
I'm fixing to make a 12% magnesium chloride solution, 12% referring to weight. The rest of the solution will be comprised of water. I'm looking to use 4 oz. of water in total.

Four ounces of water weighs about 114.4 grams. Twelve percent of 114.4 grams is about 13.73 grams.

I'll be using magnesium chloride flakes. A data sheet from the manufacturer states that a 12% solution of their magnesium chloride in water has a specific gravity of 1.105.

My question is, do I need to make use of the specific gravity information? Or can I just mix four ounces of water with 13.73 grams of magnesium chloride flakes?

I'd appreciate any assistance, thanks in advance.
 
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Thanks for the catch, I just realized that I had calculated the weight of the water based on "weight-ounces," not "volume-ounces."

So the correct weight for four (volumetric) ounces of water is 118.294 grams. I also realized that preparing a 12% solution is a bit trickier than just multiplying 118.294 grams by 12 percent to get the weight of the magnesium chloride.

Well long story short I came up with a weight of about 16.13 grams of magnesium chloride to obtain a 12% solution for four ounces of water.

I appreciate your reply which got me on the right track.
 
Just to help clarify or confirm, "w/w" or weight per weight, is mass of the solute to mass of the finished solution. "Ounces" means the fraction of one pound (a force; not a mass) cut into sixteen equal parts and taking ONE of those parts. "Fluid Ounce" is a volume measure. For this, a density may be needed.

What quantity of water are you expecting to use? Four OUNCES or four FLUID OUNCES?
 
Not absolutely sure if I missed some detail about needed facts of conditions; but I took a couple conversion ratios from some un-tracked online sources.


Online found these....
70 deg. F, water density 0.99802 g/cc
1 fl. oz. is 29.5735 ml.



If one wants 12% w/w Magnes. Chloride using 4 fl. oz. of water,

4(floz)*29.5735(ml/floz)=118.294 ml. of water

The "mass" of water for that,

118.294(ml)*0.99802(g/ml)=118.060 grams of water


How much of the salt to use, x.

x/(x+118.060)=0.12

x=0.12(x+118.060)

x=0.12x+(0.12)(118.060)

0.88x=14.1672

x=14.1672/0.88

x=16.099
You would report or use this as 16 grams.


post-note: I assumed the magnesium chloride to be anhydrous.
 
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w/w% = mass of solute/mass of solution

Note, the denominator is not just the mass of the solvent, but mass of solute + mass of solvent. In the dilute limit, especially with compounds of low molar mass, you can approximate the denominator it as mass of solvent. But not applicable here.

Since you are preparing a 12% solution and not a 12.0% or 12.00% solution, approximate the mass of water as its volume (4 oz = 118 ml => 118 g water). Weigh it out in a beaker on a scale unless you have a good high volume graduated cylinder. Or use 120 ml of solvent if precision isn't too important.

You want 12% MgCl2 by mass. Be conscious of its hydration number. Unless you bought it from a commercial chemical vendor, it most likely is in a hydrated form unless otherwise specified. An anhydrous sample will become hydrated over time, so if it's old, also don't assume anhydrous.

Let x be the mass of sample needed in grams. The equation becomes

x/(x+118 g) = 0.12

Solve for x to get your mass and keep in mind that x g of anhydrous MgCl
2 and its hydrates yield drastically different actual Mg2+ concentrations. To standardize your solutions, use molarity instead. Two separate 1.0 M aq. solution of anhydrous MgCl2 and MgCl2 • 6 H2O contain the same amount of Mg2+.

Hope it helps.
 
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Thanks guys for the info
 
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