1. Reading A Fertilizer Label Topic 2035 Melissa M. Fowler

2. Fertilizer r Any material that is added to the soil that supplies one or more of the required minerals for plants.

3. Fertilizer Make-up r Nitrogen (N) r Phosphorus (P) r In the form of phosphate (P 2 O 5 ) r Potassium (K) r In the form of potassium oxide (K 2 O) r Micronutrients r Fillers

4. Reading the Label r Labeled by amount of N, P 2 O 5, and K 2 O r Listed as %N - % P 2 O 5 - % K 2 O r Example: 10-25-30 r 10% Nitrogen r 25% Phosphate r 30% Potassium Oxide

5. Practice Problems r Tell what percentage of components are N, P 2 O 5, and K 2 O? r 10-20-10 r 10% N and K 2 O r 20% P 2 O 5 r 0-0-60 r 0% N and P 2 O 5 r 60% K 2 O

6. Pounds of Nutrient r Determined by multiplying the percentage of nutrient by the total pounds of fertilizer being analyzed. r Example: 50 pounds of 10-20-35 r Pounds N = 50 x 10% = 5 pounds N r Pounds P 2 O 5 = 50 x 20% = 10 pounds P 2 O 5 r Pounds K 2 O = 50 x 35% = 17.5 pounds K 2 O

7. Practice Problem r Determine the pounds of N, P 2 O 5, and K 2 O in 200 pounds of 12-10-8. r Pounds N = 200 x 12% = 24 pounds N r Pounds P 2 O 5 = 200 x 10% = 20 pounds P 2 O 5 r Pounds K 2 O = 200 x 8% = 16 pounds K 2 O

8. Available P r Determined by multiplying pounds of P 2 O 5 by the phosphorus/phosphate conversion factor r Conversion factor =.44 r Atomic weight of P 2 / atomic weight of P 2 O 5 r Atomic weights: P=31, O=16 r (31 x 2) / (31 x 2 + 16 x 5) =.44

9. Example Problems r Determine the available P in 40 pounds of P 2 O 5. r 40 x.44 = 17.6 pounds P r Determine the available P in 100 pounds of 15-12-13. r Pounds P 2 O 5 = 100 pounds x 12% = 12 pounds P 2 O 5 r Pounds P = 12 x.44 = 5.28 pounds P

10. Practice Problems r Determine the available P in 15 pounds of P 2 O 5. r 15 x.44 = 6.6 pounds P r Determine the available P in 1 ton of 10-25-30. r Pounds P 2 O 5 = 1000 pounds x 10% = 100 pounds P 2 O 5 r Pounds P = 100 x.44 = 5.28 pounds P

11. Available K r Determined by multiplying pounds of K 2 O by the potassium potassium oxide conversion factor r Conversion factor =.83 r Atomic weight of K 2 / atomic weight of K 2 O r Atomic weights: K=39, O=16 r (39 x 2) / (39 x 2 + 16) =.83

12. Example Problems r Determine the available K in 40 pounds of K 2 O. r 40 x.83 = 33.2 pounds K r Determine the available K in 100 pounds of 15-12-13. r Pounds K 2 O = 100 pounds x 13% = 13 pounds K 2 O 5 r Pounds K = 13 x.83 = 10.79 pounds K

13. Practice Problems r Determine the available K in 45 pounds of K 2 O. r 45 x.83 = 37.35 pounds K r Determine the available K in 500pounds of 17-8-10. r Pounds K 2 O = 500 pounds x 10% = 50 pounds K 2 O r Pounds K = 50 x.83 = 41.5 pounds K