LIQUID form:
-know the density of the fertilizer
-how much was applied
-N-P-K-S content
If the farmer applied 10 gallons per acre of 11-34-0 fertilizer, how much N and P are there?
The density of 11-34-0 is 11.7 lbs/gal. It has 11% N and 34% P2O5. To get the lbs of liquid 11-34-0, we must multiply 10 gallons by 11.7 lbs/gal which gives us 117lbs of 11-34-0. To get the amount of N, we multiply 117 lbs by 11% giving us 11.7 lbs of N per acre and if we want to convert it to SI units, we multiply 11.7 by 1.12 (1lb/ac = 1.12 kg/ha). Next, we want to know the amount of P, we get the amount by multiplying 117lbs to 34% since it is the grade of the fertilizer used. It will give us 39.78 lbs but in the form of P2O5. If we want to know the elemental P, we must divide to the 2.29 factor (because P2O5 (141.88g/mol) / P2 (61.94g/mol) =2.290 (Factor) which will give us 17.37 lbs per acre of elemental P. And to convert it to kg per ha we have to multiply it to 1.12 and get 19.45 kg per ha elemental P.
SOLID form:
It is like the liquid form calculation but simpler because you do not need the density of the liquid fertilizer. If we applied 100lbs of granular complete or 14-14-14 fertilizer we just multiply 100lbs by 14% giving us 14lbs of N, P, and K in this case. But remember that P and K are in compound form so to get the elemental form we divide 14lbs by 2.29 and 1.2, respectively. And again, if we want it in kg per ha, multiply by 1.12 the lb/ac.
NOTE: Be mindful of P and K as they are often in a compound form in fertilizers. If it is P2O5, you divide the value by 2.29 to get the elemental P and 1.2 to get elemental K from K2O.
larswrites 