Question

How much magnesium sulfate and magnesium chloride will be needed to be dissolved in 1 gallon of water to contain 47,000 ppm magnesium, 70,000 ppm sulfate, and 86,000 chloride? in grams ? and in mL?

This is what I did, but it is not making sense. I need 332 g of MgSO4 and 438 g MgCl2 that does not equal 789 g. Where are the other 19 grams coming from?

1 gallon = 3.78541 L

1 ppm = 1 mg/L

70,000 ppm sulfate = 70,000 mg/L sulfate

So, sulfate concentration in 3.78541 L (1 gal) = 70,000 x 3.78541 = 264,978.7 mg is 70,000 ppm

1 MgSO4 has 1 Mg2+ and 1 sulfate

So, MgSO4 required to get 70,000 ppm SO4^2- in 1 gal water = 264,978.7 mg x 120.37 g/mol/96.06 g/mol = 332,037 mg = 332 g

For 86,000 ppm Cl = 86,000 mg/L x 3.78541 L = 325,545.26 mg to make 86,000 ppm in 1 gallon

So, MgCl2 required to get 86,000 ppm Cl- in 1 gal water = 325,545.26 mg x (95.2 g/mol/2)/35.4 g/mol = 437,739 mg = 438 g

Mg2+ concentration required = 47,000 ppm = 47,000 mg/L = 47,000 x 3.78541 = 177,914.27 mg in 1 gallon MgSO4 + MgCl2 = 1 Mg (from sulfate) + 1 Mg (from MgCl2) = 88957.15 mg each

So for 47,000 ppm total Mg, MgSO4 + MgCl2 required = 88957.15 mg x 120.37 g/mol/24.3 g/mol + 88957.15 mg x 95.2 g/mol/24.3 g/mol = 440,558.41 mg (MgSO4) + 348,435.33 mg (MgCl2) = 788,994 mg = 789 g