1. Four Steps in Fertilizer Calculations
Determine the area
Determine the rate per area
Determine the amount of nutrient (area X rate)
Convert amount of nutrient to amount of fertilizer

2. Two Basic Types of Calculations (nutrient vs. fertilizer)
You have a given amount of fertilizer and you need to calculate how much nutrient is in it
You have a given amount of a nutrient to apply and you need to calculate how much fertilizer to use

3. Memorize and Understand This!
Weight of fertilizer X % of Nutrient = Weight of nutrient
# of urea X 46% N = # of N
Weight of fertilizer = (Weight of nutrient)/(% of nutrient)
# of urea = (# of N)/(46% N)

4. Converting From Fertilizer to Nutrient :
How many pounds of N are in a 50# bag of ammonium nitrate (35-0-0)?
50# fert X 0.35 = 17.5 #N
50# fert X (0.35#N/#fert) = 17.5 #N

5. Converting From Nutrient to Fertilizer :
How many pounds of ammonium nitrate (35-0-0) do you need to give you 5# of N?
x# fert X 0.35 = 5 #N
x# fert X (0.35#N/#fert) = 5 #N
x# fert = (5 #N)/(0.35#N/#fert) = # fertilizer

6. Example 1 The area of a home lawn is 15,000 ft2
You want to apply 1#N per 1000 ft2 using ammonium sulfate (21-0-0)
Total amount of N = (1#N/K) X (15K) = #N
Total amount of fertilizer = 15#N/0.21 = # ammonium sulfate

7. Example 1, more difficult
You want to apply 1#N per 1000 ft2 to the same lawn, using 75% IBDU and 25% urea
15#N X 75% = 11.25#N from IBDU (31%N)
x# IBDU = (11.25#N)/(0.31#N/#IBDU) = 36.3# IBDU
15#N X 25% = 3.75#N from urea (46%N)
x# urea = (3.75#N)/(0.46#N/#urea) = 8.2# urea

8. Example 2, Acres 1 acre = approx. 43,000 ft2 = 43K
1#N/K X (43K/acre) = 43#N/acre
How much ammonium nitrate (33-0-0) would you buy for 75 acres if you want to apply a total of 4#N/K during the next year, in 4 separate applications? It comes in 50 # bags. How many bags would you order?

9. Continued 1#N/K = 43#N/acre 4#N/K = ?#N/acre = 172#N/acre
For 75 acres, need: 172#N/acre X 75 acres = 12900#N
x # AN = (12900#N)/(0.33#N/#fert) = #AN
39091 #AN/(50 #AN/bag) = 782 bags

10. Example 3
You are growing 25 acres of tall fescue and plan on applying a total of 4 #N on the following schedule: 2# N as MU (39-0-0) on May 1 1# N as on Mar # N as on Oct. 1
Question 1: How much of each fertilizer do you need to order?

11. Question 1 Calculations:
Will apply 2#/K of both fertilizers
(2#N/K)(43K/acre)(25 acres) = 2150#N
(2150 #N)/(0.39 #N/# MU) = 5513 # MU
(2150 #N)/(0.26 #N/# ) = 8269 # fert

12. Example 3 continued
You also want to apply 1.5 # actual P/1000 and 3 # actual K/1000 ft2.
Question 2: How much actual P and K are applied in the applications?

13. Question 2 calculations
There are a couple ways to approach this.
Figure out how much fertilizer is applied per 1000 ft2:
(8269 # fert)/(25 acres) = 331 # fert/acre (331 # fert/acre)/(43K ft2/acre) = 7.7 #fert/K
Now calculate how much actual P and K are applied in 7.7 # fert/1000 ft2.

14. Continued... 7.7 # fert/1000 ft2 (26-4-8)
For P: (7.7 #fert)(.04 #P2O5/#fert) = 0.3# P2O5/1000 ft2
(0.3# P2O5)(0.44# P/#P2O5) = 0.14 # P/1000
To apply a total of 1.5 actual P/1000, need 1.36# P/1000 ( ) from some other source. Let’s use super phosphate (0-20-0).

15. Continued... Need 1.36 #P/1000 from 0-20-0 for 25 acres
How much total P? (1.36#P/K)(43K/acre)(25 acres) = 1462 #P
How much is needed? First calculate the amount of P2O5 equal to 1462 #P: (1462#P)/(0.44#P2O5/#P) = 3323 # P2O5
How much is needed?: (3323 # P2O5)/(0.20# P2O5 /# fert) = #

16. Continued... Now figure out how much K must be applied:
Remember we applied 7.7 #26-4-8/1000
How much actual K is applied in 7.7 # fert/1000 ft2? (7.7#fert/1000)(0.08#K2O/#fert) = 0.62# K2O
(0.62# K2O)(0.83#P/# K2O) = 0.51# K/1000

17. Continued...
To apply a total of 3 actual K/1000, need 2.5# K/1000 (3-0.5) from some other source. Let’s use KCl (0-0-60)
How much extra K is needed? (2.5#K/1000)(43K/acre)(25 acres) = 2688 #K
Convert from actual K to K2O: (2688#K)/(0.83#K2O/#K) = 3238 # K2O

18. Continued...
How much is needed?: (3238#K2O)/(0.60#K2O/#fert) = 5397 # of KCl