Question 747436
You would make a diluted solution of the product.  This diluted solution would allow you to add an accurate amount of this product to your quantity of water needing treatment.  


Look at this proportion.


{{{2/10000=n/500}}}
{{{n=2*500/10000}}}
{{{n=1000/10000}}}
{{{highlight(n=0.10)}}} ounces.


Maybe this 0.10 ounce is too small to weigh or maybe it is enough.  One way is to use grams.  {{{0.10*(1/16)*454= 2.84}}} is how many grams.  WHAT TO DO?
Depends on how much you can dissolve in what amount of water.  As example, you want this 2.84 grams of product; and you could choose to dissolve 20 grams of it in 500 ml. of water.  


Then, you take a quantity of this solution, this much:
{{{2.84*500/20 = 71}}} ml. of this solution.


So, based on my example combined with your question, you could first dissolve 20 grams of your product in 500 ml. of water; then take 71 ml. of this solution and add it to your 500 gallons of water that you want to treat with the product.  The addition of the 71 ml. of water with the dissolved product would be negligible.


Note the assumption you mean as you said, "ounces" and that 1 pound is 16 ounces.