- - AGRICULTURE CORE CURRICULUM - - (CLF6000) Advanced Core Cluster: ORNAMENTAL HORTICULTURE (CLF6350) Unit Title: ELEMENTS NECESSARY FOR PLANT GROWTH ____________________________________________________________________________ (CLF6354) Topic: READING THE Time Taught in Year(s) FERTILIZER LABEL 1 Hour 3 and 4 ____________________________________________________________________________ Topic Objectives: Upon completion of this lesson the student will be able to: Learning Outcome #: (H-4) - Read the directions on a fertilizer label and demonstrate proper dilution and application techniques. (H-5) - Calculate the content of N-P-K in a fertilizer container from information on the package. (H-15) - Calculate the amount of fertilizer needed for an acre of crop using a selected nitrogen source. (H-4, - Calculate the cost per pound of plant nutrient in a ton 5, 15) of fertilizer. Special Materials and Equipment: Sample fertilizer labels References: Hartmann, H., Kofranek, Anton M., Rubatzky, Vincent E., & Flocker, William J. (1988). PLANT SCIENCE: GROWTH, DEVELOPMENT, AND UTILIZATION OF CULTIVATED PLANTS (2nd ed.). Englewood Cliffs, NJ: Prentice-Hall. WESTERN FERTILIZER HANDBOOK (7th ed.). (1985). Danville, IL: Interstate Printers and Publishers. Evaluation: Quiz by instructor TOPIC PRESENTATION: READING THE FERTILIZER LABEL A. A fertilizer may contain one or more of the three primary nutrients: nitrogen, phosphorus, and potassium. 1. A container of fertilizer may also contain some secondary or micronutrients. 2. A little observation and study will help you understand the meaning of information shown on a fertilizer container (label). B. Each bag of commercial fertilizer carries a label stating the analysis of its contents. 1. This analysis is represented by three figures, for example, 21-7-14. This three-number designation is called a grade. a. The first figure is the percent nitrogen by weight, in this case, 21 pounds per 100 pounds of fertilizer. b. The second figure is phosphorus, specifically, 7 percent phosphoric acid (P2O5). c. The third figure represents potassium, specifically, 14 percent potassium (K2O). 2. Another way of viewing the above information is that a ton of 21-7-14 fertilizer contains: a. 420 pounds of nitrogen. b. 140 pounds of phosphorus (P2O5). c. 280 pounds of potassium (K2O). d. the total of all nutrients is 840 pounds. e. The remaining 1160 pounds consists of other elements in the formulation. These include oxygen, sulfur, magnesium, calcium, and others. NH4NO3 or ammonium nitrate is only 33.5 to 34.0 percent N for the following reason: 1) It has 2 atoms of N with atomic weight of 14 each --> 28 2) It has 3 atoms of O with atomic weight of 16 each --> 48 3) It has 4 atoms of H with atomic weight of 16 each --> 3 ___ 79 4) 28 as a percent of the total of 79 = 35.4%. 5) Because the prilling granulation process works better with a slightly acidic side, more nitric acid (which, dissolved in water, results in nitrate, NO3) is present. Thus, a slightly lower percent (33.5 to 34.0 percent) is made rather than the 35.4%. 3. The fertilizer ratio is the relative portion of each of the primary nutrients. a. As an example, a 12-12-12 grade is a 1:1:1 ratio, and, as in the sample used above, a 21-7-14 grade is a ratio of 3:1:2. b. A zero in a grade indicates that a particular element is not included in the fertilizer. Example: 1) The grade designation for ammonium nitrate is 34-0-0. This means that ammonium nitrate contains 34 percent nitrogen and that there is no phosphorus (P) present, and there is no potassium (K) present in that formulation. C. Here is an example of how we can calculate the amount of a given fertilizer that we must apply to get the desired amount on each acre. Let's look at a realistic problem: 1. We plan to apply 80 lbs. of actual nitrogen per acre of almonds. We have chosen to use ammonium sulfate. How many pounds of ammonium sulfate must we apply to obtain 80 lbs. per acre? a. Ammonium sulfate (NH4)2SO4 = 21% nitrogen. b. Since the N content of ammonium sulfate is 21% we divide 21 into the 80 lbs. of N we want per acre. 1) 8000 divided by 21 equals 380.95 lbs. of ammonium sulfate 2) Answer 380.95 pounds of ammonium sulfate applied per acre will give us 80 pounds of actual nitrogen per acre. **************************************************************************** Note: Had we chosen to use ammonium nitrate (33.5% nitrogen) we would divide by 33.5. If we had chosen to use urea (46% nitrogen) we would divide by 0.46 For estimates on nitrogen use, it takes approximately: 5 pounds of ammonium sulfate to yield 1 pound of nitrogen. 3 pounds of ammonium nitrate to yield 1 pound of nitrogen. 2 pounds of urea to yield 1 pound of nitrogen. **************************************************************************** __________________________________________________________ ACTIVITY: 1. Give each student a fertilizer label. Have them calculate the number of pounds of nitrogen in that specific container. 2. Use examples similar to the problem given in the text of this topic. Have students calculate the amount of specific fertilizers needed to obtain a given amount of nitrogen desired. __________________________________________________________ D. Here is an example of how to calculate the cost per pound of a nutrient. 1. The cost of ammonium sulfate is $120/per ton. a. Ammonium sulfate (NH4)2S04 = 21% nitrogen. b. A ton of fertilizer contains 2000 pounds. c. Since the ton or 2000 pounds is 21% nitrogen we multiply 2000 x 21 and find it contains 420 pounds N/ton. d. To find the cost of each pound of nitrogen we divide the cost per ton by the number of pounds of nitrogen. For example, $120/per ton divided by 420 (lbs. of nitrogen) gives $0.2857 per pound of nitrogen. e. For urea at $230/per ton, the cost would be $0.25 per pound of nitrogen. 2000 x 0.46 = 920 pounds nitrogen per ton. $230 divided by 920 gives $0.25 per pound of nitrogen. 1/6/91 PJK/clh #%&C