Calculating Fertilizer Costs
Guide A133
R.P. Flynn, Associate Professor, Agricultural Science Center at Artesia
Charles Siepel, Hidalgo County Program Director and County Extension Agent
College of Agriculture, Consumer and Environmental Sciences
New Mexico State University
Quick Facts
 The guaranteed analysis of a fertilizer includes the percentages of nitrogen, phosphorus, potassium, and other plant nutrients present in quantities large enough to conform to state law.
 Guaranteed analysis must be given for every fertilizer material sold in New Mexico.
 The cost per pound of nitrogen (N), phosphorus (as P_{2}O_{5}), or potassium (as K_{2}O) is calculated using the total cost and the nutrient percentage in the fertilizer. Cost per pound of nutrient should be the major criteria in determining which fertilizer source to use.
 If more than one plant nutrient is contained in a fertilizer, the cost of one or more nutrients must be assumed. This cost is subtracted from the total fertilizer cost, and the residual cost is used for determining the cost per pound of the nutrient in question.
 When liquid fertilizers (solutions or suspensions) are priced by the gallon, the cost and the density of the material as well as the percent of the nutrient in the fertilizer must be known to complete the cost calculation. If the liquid fertilizer is priced by the ton, the calculations are similar to dry fertilizer materials.
 Always use soil testing to determine what fertilizer(s) are needed for your particular soil and crop.
The 1978 New Mexico Fertilizer Act requires that the guaranteed analysis of every fertilizer material sold in the state be given in a clearly legible and conspicuous form. The guaranteed analysis provides the percentages of nitrogen (N), available phosphorus (expressed as percentage P_{2}O_{5}), watersoluble potassium (expressed as percentage K_{2}O), and other nutrients in quantities that conform to state law.
The Fertilizer Act defines a commercial fertilizer as any substance that contains one or more recognized plant nutrients, is used for its plant nutrient content, and is designed for use or claimed to have value in promoting plant growth (except unmanipulated animal and vegetable manures, marl limes, limestone, wood ashes, gypsum, and other exempt products). A fertilizer material means a commercial fertilizer that either: 1) contains important quantities of no more than one of the primary plant nutrients—nitrogen, phosphoric acid, and potash; or 2) has approximately 85% of either nitrogen, phosphoric acid, or potash in the form of a single compound; or 3) is derived from a natural source and processed in such a way as to purify and concentrate the plant nutrient.
Although the guaranteed analysis expresses phosphorus and potassium on the oxide basis (P_{2}O_{5} and K_{2}O) these plant nutrients occur in the fertilizer as mixtures of different chemicals. For example, the chemical formula for diammonium phosphate (DAP) is (NH_{4})_{2}HPO_{4}. DAP has a guaranteed analysis of 18460 expressed as 18% N, 46% P_{2}O_{5}, and 0% K_{2}O—but it actually contains no P_{2}O_{5}. The use of the oxide expression for plant nutrient content is a carryover from early practices when chemists ignited fertilizer samples and weighed the oxides. Soil test recommendations for P and K additions to soil have been corrected for this oxide expression.
To calculate the cost per pound of elemental P or K, the guarantee must be changed from P_{2}O_{5} to P and K_{2}O to K. (No conversion is necessary for N because it is already expressed on an elemental basis.) These conversions are provided in table 1. Use table 2 to convert between English and metric units.
Table 1. Conversion factors for P and P_{2}O_{5} and K and K_{2}O.
Convert column 1 to column 2, multiply by: 
Column 1  Column 2  Convert column 2 to column 1, multiply by: 
2.29  P  P_{2}O_{5}  0.437 
1.21  K  K_{2}O  0.836 
Table 2. Englishmetric conversions.
Convert column 1 to column 2, multiply by: 
Column 1  Column 2  Convert column 2 to column 1, multiply by: 
0.454  pound (lb)  kilogram (kg)  2.205 
0.907  ton  metric ton (Mg)_{}  1.102 
3.78  gallon (gal)  liter (L)  0.265 
Fertilizer Containing a Single Nutrient
Calculating the cost per pound of a nutrient in a fertilizer containing a single element is relatively simple. Some examples are provided below.* Similar procedures can be used for any fertilizer containing one plant nutrient.
Example A. Urea, CO(NH_{2})_{2}, has a guaranteed analysis of 4600 and has cost an average of $229 per ton (2,000 lb) since 1997. What is the cost per pound of N?
First, calculate the pounds of N in the fertilizer:
2000 lb fertilizer × 0.46 = 920 lb of N
Next, calculate the cost per pound of N
$229 ÷ 920 lb N = $0.25/lb N.

First, calculate the pounds of P_{2}O_{5} in the fertilizer:
2,000 lb fertilizer × 0.46 = 920 lb P_{2}O_{5}.
Next, calculate the cost per pound of P_{2}O_{5}:
$261 ÷ 920 lb = $0.28/lb P_{2}O_{5}.

First, convert pounds of P_{2}O_{5} to units of P (refer to table 1 for conversion factor):
920 lb P_{2}O_{5} × 0.437 = 402 lb P.
Next, calculate the cost per pound of P:
$261 ÷ 402 lb P = $0.65/lb P.
Mixed Fertilizers
Mixed fertilizers contain more than one nutrient. An example is granulated diammonium phosphate (18460). Although mixed fertilizers supply more than one nutrient, fertilizers should be mixed to meet the specific needs of the crop in question based on sound soil sampling and analysis. Always test your soil using good techniques prior to fertilization (see NMSU Extension Guide A114, Test Your Soil).
Example D. Diammonium phosphate (18460) has cost an average of $269 per ton in the Mountain Region. Calculate the cost of the P_{2}O_{5} in this fertilizer.
For this determination, the cost per pound of N from example A ($0.25/lb) can be used; then the cost of the P_{2}O_{5} (the one that is of greatest interest) can be calculated.
What is the cost of the P_{2}O_{5} in 18460 using the value of N from urea (see example A)?

First, calculate the pounds of N and P_{2}O_{5} in a ton of fertilizer:
2,000 lb fertilizer × 0.18 = 360 lb N;
2,000 lb fertilizer × 0.46 = 920 lb P_{2}O_{5}.
Next, calculate the portion of the total fertilizer cost that can be attributed to N:
$0.25/lb N × 360 lb N = $90.
To calculate the total cost of P_{2}O_{5} in the fertilizer, subtract the cost of N from the total cost of the fertilizer:
$269 − $90 = $179.
 $179 ÷ 920 lb = $0.19/lb P_{2}O_{5}.
The cost for P_{2}O_{5} can then be compared to other
P_{2}O_{5} sources. Notice that the cost of P_{2}O_{5} in 18460
was substantially less than the cost of P_{2}O_{5} in 0460
(see example B).
Example E. Calculate the cost of elemental P from 18460 based on example D.
 First, convert pounds of P_{2}O_{5} to pounds of P (from example D, refer to table 1 for conversion factor):
920 lb P_{2}O_{5} × 0.437 = 402 lb P.
Then calculate the cost per pound of P:
$179 + 402 lb P = $0.45/lb P.
Example F. Calculations based on a “plant food approach” lump N and P together for making value calculations that spread the cost of both elements over each other. Care must be taken when approaching fertilizer value this way. Each element must be expressed on an elemental basis before its value can be calculated. What is the cost of total plant food in 18460?
 First, because 18460 supplies two nutrients, both must be expressed on an elemental basis so their “plant food” totals can be summed. In one ton of 18460, there are 360 lb of elemental N (2,000 × 0.18) and 402 lb of elemental P (2,000 × 0.46) for a total of 762 lb of “plant food”. The cost of 18460 has cost an average of $269 per ton (table 3). The value of 18460 is then $0.35/lb (from $269 ÷ 762 lb) of “plant food”(both N and P).
Table 3. Fertilizer prices as reported by the USDA for the mountain states of Colo., N.M., Wyo. and Mont.†
Fertilizer  Analysis  Year  
N�P_{2}O_{5}�K_{2}O  1997  1998  1999  2000  2001  ave  
Dollars per ton  
Ammonium nitrate  33.500  233  199  178  180  262  210 
Ammonium phosphate  10340  263  270  258  254  269  263 
Ammonium sulfate  2100  197  196  177  175  202  189 
Ammonium phosphate sulfate  16200  216  215  208  196  210  209 
Anhydrous ammonia  8200  313  261  218  234  421  289 
Diammonium phosphate  18460  289  283  273  248  254  269 
Monoammonium phosphate  11520  286  278  268  257  255  269 
Muriate of potash  0060  162  167  173  169  173  169 
Triple superphosphate  0460  274  267  262  248  254  261 
Urea  4600  272  199  181  199  296  229 
Urea ammonium nitrate‡  3200  223  197  172  191  264  209 
† Agricultural Prices 2001 Summary. USDA National Agricultural Statistics Service. July 2001. ‡ Fertilizer cost reported from Southwestern states (Ariz., Calif., Nev. and Utah). 
Solution or Suspension Fertilizers
When determining the cost per pound of nutrients in liquidbased fertilizers that are priced by the gallon, the density of the material must be known. When the liquid fertilizer is priced on a weight basis (cost/pound or cost/ton), the calculations are similar to those used to determine the nutrient cost of dry fertilizer materials. Most liquids are priced on a cost per pound basis.
Example G. A hypothetical zinc (Zn) chelate costs $6/gal. It has a density of 11.2 lb/gal and contains 6% Zn. What is the cost per pound of Zn?
 First, find pounds of Zn per gallon of solution:
11.2 lb/gal × 0.06 = 0.67 lb Zn/gal.
Next, calculate the cost of Zn:
$6/gal ÷ 0.67 lb/gal = $8.96/lb Zn.
 The steps to solving this problem are exactly like those given in example A. First, calculate the pounds of N in the fertilizer:
2,000 lb fertilizer × 0.32 = 640 lb N.
Next, calculate the cost per pound of N:
$209 + 640 lb N = $0.33/lb N.
Summary
Every fertilizer material sold in New Mexico must contain a guaranteed analysis that consists of at least three numbers that indicate the percent N, percent P_{2}O_{5}, and percent K_{2}O. Any other nutrients that are guaranteed also must be listed on the label. The cost per pound of N, P_{2}O_{5}, and K_{2}O in a fertilizer with only N, P, or K can be calculated using the cost per ton of fertilizer and the percent N, P_{2}O_{5}, or K_{2}O in the material. To calculate the cost per pound of elemental P or K, a factor must be used to convert percentage P_{2}O_{5} to percentage P, and percentage K_{2}O to percentage K (table 1).
For mixed fertilizers (those with more than one plant nutrient), the cost per pound of one or more nutrients that could replace the nutrient found in the mix must be used. The contribution of these nutrients to the cost of the mix is subtracted from the total mixed fertilizer cost. The residual cost is then used to determine the cost per pound of the nutrient in question.
For liquidbased fertilizers (solutions or suspensions) that are priced by the gallon, cost calculations require the price per gallon of material, the density of the liquid, and the percent of the nutrient present.
The cost per pound of nutrient should be the major criteria in determining which fertilizer material to use. Most NP_{2}O_{5} K_{2}Obearing fertilizers perform equally well when applied properly. Other factors, such as ease of handling, safety considerations, and ease of integration into a grower’s production program, also influence which fertilizer material is the best to use.
Always test your soil prior to fertilization.
*The cost provided for all of the following examples are used for illustrative purposes and do not necessarily reflect what the cost per pound of the fertilizer should be.
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Revised and electronically distributed January 2003, Las Cruces, NM.