Calculating Fertilizer Rates from Nutrient Recommendations
Soil test recommendations are given in lb/ac or kg/ha of nutrients. To determine the fertilizer rate for a particular nutrient, multiply the rate of the desired nutrient by 100 and divide by the percentage of the nutrient in the fertilizer.
Example 1
- Recommended rate of N is 80 lb/ac Using 46-0-0, the rate of fertilizer required is: (80 x 100) / 46 = 174 lb/ac
Example 2
- Recommended rate of P 2O 5 is 40 lb/ac Using 11-52-0, the rate of fertilizer required is: (40 x 100) / 52 = 77 lb/ac 77 lb/ac of 11-52-0 would also supply (11/100) x 77 = 8.5 lb/ac of N.
Example 3
- Recommended rate of K 2O is 15 lb/ac. Using 0-0-60, the rate of fertilizer required is: (15 x 100) / 60 = 25 lb/ac
Converting Fertilizer Prices into Price per unit of Nutrient
The cost of a fertilizer is related to its plant nutrient content. If a nitrogen fertilizer such as 34-0-0 is being purchased, the cost should be about three-quarters that of 46-0-0. When buying fertilizer, one should compare prices on the basis of cost per pound of "actual" nutrient, not the price per tonne of fertilizer material.
Example 1
If urea (46-0-0) costs $367/tonne, the cost per pound of nitrogen (N) is calculated as follows:
- Nitrogen in one tonne (1,000 kg or 2,204 lb) of 46-0-0 (containing 46% N): (46/100) x 2,204 = 1,014 lb Cost per lb of N is: $367/1,014 = $0.362
Example 2
(Based on 11-52-0 at $391/tonne) In order to calculate the cost of phosphate in 11-52-0, the value of nitrogen must first be subtracted.
- Nitrogen in one tonne (1,000 kg or 2,204 lb) of 11-52-0 is (11/100) x 2,204 = 242 lbs The value of nitrogen is 242 x $0.362 = $88 (from example 1, which calculated the value of N to be $0.362/lb) Cost of phosphate per tonne is $391 - $88 = $303 Phosphate in one tonne (1,000 kg or 2,204 lb) of 11-52-0 is: (52/100) x 2,204 = 1,146 lb Cost per lb. of P 2O 5 is : $303/1,146 = $0.264
For further information, contact your MAFRD GO Representative.
