C
csmith
Guest
I'm in need of assistance with reverse math. To follow is my equation used to figure liquid dosing. My PO4 level is a bit low (about 2 ppm) so for my next batch I'm upping it a bit. I'm in need of assistance with the reverse math of this.
Equation:
[(atomic mass of compound / mass of desired nutrient) * desired ppm] / (x) mL dosed for each 1 L of water column * 1000 (to convert the liter bottle to mL) / 1000 (to convert mg to grams)
K-H2-P-O4
K - 39.0983
H - 1.00794 *2
P - 30.973762
O - 15.9994 *4
0.5 mL dosed for every L of water column
Desired ppm of PO4 is 4
So...
{[39.0983 + (1.00794 * 2) + 30.973762 + (15.994 * 4)] / [30.973762 + (15.9994 * 4)] * 4} / 0.5 (removed the *1000 / 1000 for obvious reasons)=
{[39.0983 + 2.01588 + 30.973762 + 63.976] / [30.973762 + 63.976] * 4} / 0.5 =
[(136.063942 / 94.949762) * 4] / 0.5 =
11.46407861454145
So roughly 11.46 grams of KH2PO4 gives me 4 ppms of PO4 per 0.5 mL dosed in each L of water column from a 1000 mL stock. I've got that part. What I need to know is how to figure what 11.46 grams of KH2PO4 gives me as it concerns K. The following equation is as far as I can get. Desired ppm replaced with an (x) as it's now the variable. I've also changed the mass of desired nutrient to that of K.
{[39.0983 + (1.00794 * 2) + 30.973762 + (15.994 * 4)] / [39.0983] * (x)} / 0.5 = 11.46407861454145
Can anyone help me out here? I'm lost.
Equation:
[(atomic mass of compound / mass of desired nutrient) * desired ppm] / (x) mL dosed for each 1 L of water column * 1000 (to convert the liter bottle to mL) / 1000 (to convert mg to grams)
K-H2-P-O4
K - 39.0983
H - 1.00794 *2
P - 30.973762
O - 15.9994 *4
0.5 mL dosed for every L of water column
Desired ppm of PO4 is 4
So...
{[39.0983 + (1.00794 * 2) + 30.973762 + (15.994 * 4)] / [30.973762 + (15.9994 * 4)] * 4} / 0.5 (removed the *1000 / 1000 for obvious reasons)=
{[39.0983 + 2.01588 + 30.973762 + 63.976] / [30.973762 + 63.976] * 4} / 0.5 =
[(136.063942 / 94.949762) * 4] / 0.5 =
11.46407861454145
So roughly 11.46 grams of KH2PO4 gives me 4 ppms of PO4 per 0.5 mL dosed in each L of water column from a 1000 mL stock. I've got that part. What I need to know is how to figure what 11.46 grams of KH2PO4 gives me as it concerns K. The following equation is as far as I can get. Desired ppm replaced with an (x) as it's now the variable. I've also changed the mass of desired nutrient to that of K.
{[39.0983 + (1.00794 * 2) + 30.973762 + (15.994 * 4)] / [39.0983] * (x)} / 0.5 = 11.46407861454145
Can anyone help me out here? I'm lost.