Section 7: Fertilizer Calculations for Liquid Fertilization Programs
Many commercial fertilizers will provide application rates on the label. However, in some cases the application rate may need to be adjusted or individual mineral salts may be used to prepare a custom fertilization program. In either case, it is important to understand how to calculate fertilizer concentrations. These calculations need only be performed for a liquid fertilization program. Controlled-release fertilizers or slow-release fertilizers are applied at recommended rates per container or cubic yard (as recommended on the labels).
Before working through a few examples, it is important to understand the term parts-per-million (ppm). When mixing two liquids of the same density and containing 100% of the material of interest (pure), a ppm is simply 1 part (active ingredient of the material of interest) to 999,999 parts water. However, the materials used are usually not pure, and the solutions containing the active ingredients have densities different from that of water. Additionally, with fertilizers, solids are usually being added to water. In these cases the calculations are based on weight of active ingredient where 1 unit of weight of the material of interest to 999,999 weight units of water.
1 ppm = 1 mg.L-1
It is also helpful to remember that 1% = 10,000 ppm (100 x 10,000 = 1,000,000). Therefore, a 1% solution is equivalent to 10,000 ppm.
Also remember that these calculations are based on the amount of active ingredient or element of interest not the carrier material.
1. Determine how many mg.L-1 of a 20-20-20 are required to produce a 100-ppm N solution.
100 ppm = 100 mg.L-1
If we were dealing with pure N, we would add 100 mg.L-1 of the fertilizer. However, the material is only 20% N.
Therefore, 100 mg of N / 0.20 mg N per mg fertilizer = 500 mg of fertilizer
500 mg 20-20-20 per liter of water provides 100 mg.L-1 (or ppm) N.
2. Determine how many mg.L-1 of a 20-20-20 are required to produce a 100 ppm K solution.
100 ppm = 100 mg.L-1
If we were dealing with pure K, we would add 100 mg.L-1 of fertilizer. However, the material is 20% K2O, and K2O is only 83% K.
To determine the % K in this fertilizer, we have 0.20 x 0.83 = 0.166 or 16.6% K in the fertilizer
Therefore, 100 mg K / 0.166 mg K per mg fertilizer = 602 mg of fertilizer
602 mg 20-20-20 per liter of water provides 100 mg.L-1 (or ppm) K.
3. In example 1 we determined that 500 mg.L-1 of 20-20-20 provided 100 ppm N. How much P is provided by this solution?
The fertilizer is 20% P2O5.
P2O5 is 43% P
Therefore, 0.20 x 0.43 = 0.086 or 8.6% P
We used 500 mg of the fertilizer (per liter).
500 mg x 0.086 = 43 mg of P
43 mg P/L = 43 ppm P
4. How many mg of 20-20-20 need to be used in a liter of stock solution that when put through a 1:100 injector will provide 100 ppm N?
From example #1, we know that 500 mg 20-20-20/L = 100 ppm N.
1 liter of stock will result in 100 liters of total solution ( 1 liter becomes 100 liters when diluted) when put through the injector. Therefore, 500 mg.L-1 x 100 L = 50,000 mg (50 g) in 1 L liter of stock.
5. How many mg.L-1 of Ca(NO3)2 and KNO3 are required to produce a 100 ppm N solution that provides 75% of the N from Ca(NO3)2 and 25% of the N from KNO3. Assume that both fertilizers are pure.
The molecular weight of Ca(NO3)2 is:
40 + [14 + (16)3]2 = 164
There are 2 N in Ca(NO3)2 so the percentage N is:
28/164 = 0.17 or 17%
The molecular weight of KNO3 is:
39 + [14 + (16)3] = 101
There is 1 N in KNO3 so the percentage N is:
14/101 = 0.14 or 14%
Therefore:
Ca(NO3)2 = 17% N
KNO3 = 14% N
We need a total of 100 mg N.L-1. We can break this into 2 more simple problems by:
We need 0.75 x 100 = 75 mg N from Ca(NO3)2 and 0.25 x 100 mg = 25 mg N from KNO3.
We now have two simply calculation that are similar to the example 1.
75 mg N / 0.17 mg N per mg N in Ca(NO3)2 = 441 mg Ca(NO3)2
25 mg N / 0.14 mg N per mg N in KNO3 = 179 mg KNO3
6. How many mg of 15-5-15 need to be used in 10 liters of stock solution that when put through a 1:100 injector will provide 200 ppm N?
200 ppm = 200 mg.L-1
If we were dealing with pure N, we would add 200 mg.L-1 of the fertilizer. However, the material is only 15% N.
Therefore, 200 mg of N / 0.15 mg N per mg fertilizer = 1,333 mg of fertilizer or 1.33 g of fertilizer per liter.
1 liter of stock will result in 100 liters of total solution (1 liter becomes 100 liters when diluted) when put through the injector and 10 liters of stock solution will become 1000 liters of final fertilizer solution. Thus we need to add enough fertilizer to the stock solution to make 1000 liters.
We needed 1.33 g of 15-5-15 per liter and we are making a total of 1000 liters (final amount of fertilizer after dilution). So, 1.33 g 15-5-15 per liter x 1000 liter = 1,330 g 15-5-15 in the 10 liters of stock.
