magnetar68
New member
25 yeas ago I took chemistry as a freshman in college. Well, I seemed to have forgotten the basics :headwally:. In order to ensure I understood the math in the dosing calculators online, I tried to do the same calculation by hand, but I was unsuccessful. I realize this is pretty basic stuff, but I need a little assistance:
The question involves Randy's Magnesium dosing solution. I have two types of magnesium compounds: Magnesium Chloride Hexahydrate and Magnesium Sulfate Heptahydrate. I want to mix them in a ~10:1 ratio MgCl/MgSO4 (7 1/4 cups MgCl and 3/4 cups MgSO4) into 1 gallon of water. From this solution, I want to dose 100ppm of Mg into a 120gallon fish tank. The calculator online gives the answer of 967.7ml of solution.
I assume I need to know how much Mg is in all of this powder being dissolved into the 1 gallon of water. Now, I know the molar mass of both MgCl2*6(H2O) and MgSO4*7(H2O) as that calculation is trivial from the atomic weights of the atom counts. This seems to give the % Mg for each of these by mass:
MgCl2*6(H2O) is 203.27 g/mol, Mg is 24.31 g/mol, so 11.957% Mg (24.31/203.27)
MgSO4*7(H2O) is 246.42 g/mol, Mg is 24.31g/mol so 9.863% Mg (24.31/246.42)
While I know the molecular weight in g/mol I need to know g/cm3 to get a volumetric weight rather than a molar mass. Randy cites these in his article and provides the source of the datasheets. I assume these bulk densities depend on the specific nature of the powders and must be determined by experiment:
MgCl2*6(H2O): 0.85 g/cm3
MgSO4*7(H2O): 1.05 g/cm3
H2O: 1.00 g/cm3
The directions for this solution are to use 7 1/4 cups or MgCl and 3/4 cups of MgSO4. 7.25 cups is 1715.26 ml (cm3) and 0.75 cup is 177.44 ml (cm3). Also, I know 1 gallon is 3.785 liters of distillled H20. This means my solution has:
0.85 g/cm3 * 1715.26 cm3 = 1457.90g MgCl2*6(H2O)
1.05 g/cm3 * 177.44 cm3 = 186.31g MgSO4*7(H2O)
1.00g/cm3 * 3785.00 cm3 = 3785.00g H2O
Since MgCl2 is 11.95% Mg, I have 174.33g Mg from the MgCL2
Since MgSO4 is 9.86% Mg, I have 18.38g Mg from the MgSO4
This means my 1 gallon of solution has a total of 192.71g Mg.
120G of water is 454.25 liters or 454,250ml, but assuming a specific gravity of 1.025g/ml, then I have 454250ml*1.025g/ml which is 465,606.25g of seawater.
100ppm by weight would be 46.56g of Mg, so I would need 46.56/192.71 or 24.16% of the 1 gallon of solution.
I am not sure I have all of that right, but here is where I know I made a mistake. I have 3785ml of water with some stuff in it. I think it has 192.71g of Mg. 24.16% would be 914.51ml, but the answer in the online calculators is 967.7ml.
One idea is that I need to account for the H20 added by the hydrates. This means that 1457.9g*53%=775.06g H20 and 186.31g*51%=95.32g H2O for an additional 870.38g of H2O from the hydrates. But if I add this this 870ml to the 3785ml in the gallon for a total of 4665.38ml of H20, then 24.16% of this is 1124.81ml which is too high.
I would really like to understand how these calculations are done, so sorry for the long post.
The question involves Randy's Magnesium dosing solution. I have two types of magnesium compounds: Magnesium Chloride Hexahydrate and Magnesium Sulfate Heptahydrate. I want to mix them in a ~10:1 ratio MgCl/MgSO4 (7 1/4 cups MgCl and 3/4 cups MgSO4) into 1 gallon of water. From this solution, I want to dose 100ppm of Mg into a 120gallon fish tank. The calculator online gives the answer of 967.7ml of solution.
I assume I need to know how much Mg is in all of this powder being dissolved into the 1 gallon of water. Now, I know the molar mass of both MgCl2*6(H2O) and MgSO4*7(H2O) as that calculation is trivial from the atomic weights of the atom counts. This seems to give the % Mg for each of these by mass:
MgCl2*6(H2O) is 203.27 g/mol, Mg is 24.31 g/mol, so 11.957% Mg (24.31/203.27)
MgSO4*7(H2O) is 246.42 g/mol, Mg is 24.31g/mol so 9.863% Mg (24.31/246.42)
While I know the molecular weight in g/mol I need to know g/cm3 to get a volumetric weight rather than a molar mass. Randy cites these in his article and provides the source of the datasheets. I assume these bulk densities depend on the specific nature of the powders and must be determined by experiment:
MgCl2*6(H2O): 0.85 g/cm3
MgSO4*7(H2O): 1.05 g/cm3
H2O: 1.00 g/cm3
The directions for this solution are to use 7 1/4 cups or MgCl and 3/4 cups of MgSO4. 7.25 cups is 1715.26 ml (cm3) and 0.75 cup is 177.44 ml (cm3). Also, I know 1 gallon is 3.785 liters of distillled H20. This means my solution has:
0.85 g/cm3 * 1715.26 cm3 = 1457.90g MgCl2*6(H2O)
1.05 g/cm3 * 177.44 cm3 = 186.31g MgSO4*7(H2O)
1.00g/cm3 * 3785.00 cm3 = 3785.00g H2O
Since MgCl2 is 11.95% Mg, I have 174.33g Mg from the MgCL2
Since MgSO4 is 9.86% Mg, I have 18.38g Mg from the MgSO4
This means my 1 gallon of solution has a total of 192.71g Mg.
120G of water is 454.25 liters or 454,250ml, but assuming a specific gravity of 1.025g/ml, then I have 454250ml*1.025g/ml which is 465,606.25g of seawater.
100ppm by weight would be 46.56g of Mg, so I would need 46.56/192.71 or 24.16% of the 1 gallon of solution.
I am not sure I have all of that right, but here is where I know I made a mistake. I have 3785ml of water with some stuff in it. I think it has 192.71g of Mg. 24.16% would be 914.51ml, but the answer in the online calculators is 967.7ml.
One idea is that I need to account for the H20 added by the hydrates. This means that 1457.9g*53%=775.06g H20 and 186.31g*51%=95.32g H2O for an additional 870.38g of H2O from the hydrates. But if I add this this 870ml to the 3785ml in the gallon for a total of 4665.38ml of H20, then 24.16% of this is 1124.81ml which is too high.
I would really like to understand how these calculations are done, so sorry for the long post.