Well, you're going to have to define "respective measurements" for each element, but I'll proceed with some assumptions.
I frequently help out with the math for a scenario that usually goes like this: "I want to know how many grams "g" per liter of compound "c" I need in order to make a solution that will increase the concentration of element "e" by "p" PPM (mg/L) for every "v" mL of solution that I add to my "t" gallon tank."
The only information usually missing from the question is the atomic mass "a" of element e, and the molecular mass "m" of compound c. Once we look those up (Google is your friend), then we can do the following:
g = m / a * p / v * t * 3.7854
The "* 3.7854" is obviously a conversion of tank volume "t" from gallons to liters.
Let's see how this would work for your sodium fluoride supplementation. Since you didn't specify "p" for any of the elements you intend to dose, I'll make the assumption that in this case you want to increase your tank's F by 0.94 mg/L ("PPM" -- from the "-0.94" difference in the report you shared above). So, you have the following values for the variables:
m = 41.98817 (google "sodium fluoride mw")
a = 18.9984 (google "fluorine mw")
p = 0.94 (desired F increase in mg/L, or PPM)
v = 100 (100 mL to achieve the increase -- my assumption)
t = 240 (gallons in tank)
So, g = 41.98817 / 18.9984 * 0.94 / 100 * 240 * 3.7854 = 18.87, meaning a solution containing 18.87 grams of NaF in 1 liter of water will raise the F level of a 240 gallon tank by 0.94 mg/L for every 100 mL that is added.
Let's think this through: Our stock solution will contain 18.87 * 18.9984 / 41.98817 = 8.5399 grams of F per liter. So, 100 mL will contain 0.85399 grams, or 853.99 mg of F. Adding that to 240 gallons, or 240 * 3.7854 = 908.5 liters of water will result in 853.99 / 908.5 = 0.94 mg/L of F.
I leave the other calculations to the reader as an exercise.
Be careful when dealing with compounds that may have different levels of hydration; anhydrous CaCl2, CaCl2*2H2O, etc., each have different values for "m". Also be careful when there is more than one atom of "e" in the compound -- you then have to multiply "a" by the number of atoms present, e.g., for vanadium pentoxide (V2O5), then you have to multiply the "a" value by 2. Also, pay attention to the solubility of the compounds you're working with in water, and stay well below their solubility at 0C.
EDIT: I have made a number of edits to correct a few things since I originally posted this. It think it's correct and needs no further edits now.
Thanks a bunch for for formula and explanation. Gonna take a stab at some mixes.