Thank you for working that out for me. I follow how you derived conc. in bottles. I'm not sure how you derived mass percent. Is that chemistry work? Essentially calculating the mass of K in the total K2SO4 dissolved in the bottle? I follow how you came up with the values of K, NO3 etc in mg/mL using the mass percent extraction. I don't follow how you made the conclusion highlighted in bold. On a side note, in my research I've discovered that dosing the high light recommendation for GLA PPS Pro yields similiar PPMs of NPK and micros that EI reccomends:
According to GLA if I dose 2mls of macros/10gal and .2mls of micros/10 gal my results will be comparable to EI and as follows:
This limits water column nutrient levels to 28 ppm NO3, 2.8 ppm PO4, 36 ppm K, 2.8 ppm Mg, 0.28 ppm Fe(TE).
Allow me to start from the end... be careful on the wording existing on GLA site ... "limits" is the important word. That does not mean it actually adds that much. What does it mean, under what conditions is this limited ? You have to ask GLA... If you look at the math a 2ml dose in 10g would add ~2ppm NO3, 0.2ppm PO4 etc Doesn't look like EI to me.
Glad I could help with the calculations.
The mass percent is based on the atomic weight of the molecule/ion/element, so yes a chemistry thing. Take water for example :
Total atomic weight 2*H+1*O = 2*1+1*16=2+16=18
The mass percent of O in the water molecule is then 88,9% (16/18*100). This then can also be applied at a larger scale meaning that from 100kg water, O would be responsible for 88.9kg. I am only looking at the K from the K2SO4 and treating it as if I could weigh it alone. In reality it is not possible to weight a ion from the molecule but we can still calculate its weight at a theoretical basis. As the K+ would fully dissolve (hopefully) , from K2SO4, we can then calculate its concentration.
Next. How I got from a bottle having a conc. of NO3 ... 39.99 mg/mL to 30mL of solution in 10G giving NO3 ... ~31,7 mg/L
NO3 ... 39.99 mg/mL means that :
1ml of solution has 39.99mg
30ml of solution have 30*39.99=1199.7mg
If we add this amount in 1L we would get a conc of 1199.7mg/L.
The same amount in 37,9L (10G) would result in 1199.7/37.9=31.65mg/L
Well actually I just used a formula but this gets you trough the thought process.
Any questions please ask.