A user posted this on my board and it was a bit over my head so asking your adivce on his post:
Well, I must be very bored today so I did some calculation to verify the recommended dosage claim in the EI dosing method:
Verifying recommended EI dosing
NO3 percentage in KNO3 Calculation
Atomic mass for individual atoms (derived from periodic table)
K molecular mass = 39.0983
NO3 molecular mass = 14.0667 + (15.994 * 3) = 63.0649
Total molecular mass for KNO3 = 101.1632
K % in KNO3 molecule = 39.0983 / 101.1632 = 38.65%
NO3 % in KNO3 molecule = 63.0649 / 101.1632 = 61.35%
Now, for the sake of calculation, let us assume the powder we get from local hydroponics store is 100% pure. This means that 1g of the powder will contain about 61.35% NO3.
Since we have obtained the amount of NO3 in KNO3, we can now move on to the next step of the calculation.
Lets review the recommended KNO3 amount for the stock solution:
40g of KNO3 to 500ml water
By using the formula for calculating PPM:
x PPM = ( mass of KNO3 / mass of water ) * 10^6
x PPM = (40g / 500g) * 10 ^ 6 note: 1mL of water = 1g
x = 80000ppm
As shown in above calculation, we now know the stock solution is an 80,000ppm KNO3 solution.
The EI method suggest that we add 10mL of this stock solution to 100L of water would give us a value of 5ppm. Now lets verify this claim:
10 / 500 * 40 = 0.8g of KNO3 solid in this 10mL solution
(0.8g / 10g + (100L * 1000g/L)) * 10 ^ 6 = approx 8ppm of KNO3
However, each gram of KNO3 contains only 61.35% of NO3. Thus, we need to revise our formula:
((0.8g * 0.6135) / 10g + (100L * 1000g/L)) * 10 ^ 6 = 4.9075 ppm
This above result is extremely close to the posted dosing recommendation. We can thus conclude that adding 10mL of the stock solution to 100L of water would INDEED give you around 5ppm of nitrate (NO3).
With this calculation, we can now tailor the dosage of any nutrient to any tank size : ) yay!
(I know there's a program which calculates the amount, but I feel more confident to dose the amount once I know exactly how the numbers are derived.)