NMSU: Fertilizer Cost Calculations
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Calculating Fertilizer Costs


Guide A-133
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 P2O5), or potassium (as K2O) 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 P2O5), water-soluble potassium (expressed as percentage K2O), 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 (P2O5 and K2O) these plant nutrients occur in the fertilizer as mixtures of different chemicals. For example, the chemical formula for diammonium phosphate (DAP) is (NH4)2HPO4. DAP has a guaranteed analysis of 18-46-0 expressed as 18% N, 46% P2O5, and 0% K2O—but it actually contains no P2O5. The use of the oxide expression for plant nutrient content is a carry-over 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 P2O5 to P and K2O 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 P2O5 and K and K2O.

Convert column 1
to column 2,
multiply by:
Column 1 Column 2 Convert column 2
to column 1,
multiply by:
2.29 P P2O5 0.437
1.21 K K2O 0.836

Table 2. English-metric 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(NH2)2, has a guaranteed analysis of 46-0-0 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.

Example B. Superphosphate (0-46-0) has cost an average of $261 per ton in the Mountain Region, which includes New Mexico. What is the cost per pound of P2O5?
    First, calculate the pounds of P2O5 in the fertilizer:
    2,000 lb fertilizer × 0.46 = 920 lb P2O5.

    Next, calculate the cost per pound of P2O5:
    $261 ÷ 920 lb = $0.28/lb P2O5.

Example C. What is the cost per pound of P in the superphosphate from example B? Notice in example B that there was 920 lb of P2O5 in one ton of superphosphate. Converting P2O5 to P allows for the cost per pound of P to be found.
    First, convert pounds of P2O5 to units of P (refer to table 1 for conversion factor):
    920 lb P2O5 × 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 (18-46-0). 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 A-114, Test Your Soil).

Example D. Diammonium phosphate (18-46-0) has cost an average of $269 per ton in the Mountain Region. Calculate the cost of the P2O5 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 P2O5 (the one that is of greatest interest) can be calculated.

What is the cost of the P2O5 in 18-46-0 using the value of N from urea (see example A)?

    First, calculate the pounds of N and P2O5 in a ton of fertilizer:
    2,000 lb fertilizer × 0.18 = 360 lb N;
    2,000 lb fertilizer × 0.46 = 920 lb P2O5.

    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 P2O5 in the fertilizer, subtract the cost of N from the total cost of the fertilizer:
    $269 − $90 = $179.
Finally, calculate the cost per pound of the P2O5:
    $179 ÷ 920 lb = $0.19/lb P2O5.

The cost for P2O5 can then be compared to other P2O5 sources. Notice that the cost of P2O5 in 18-46-0 was substantially less than the cost of P2O5 in 0-46-0 (see example B).

Example E. Calculate the cost of elemental P from 18-46-0 based on example D.

    First, convert pounds of P2O5 to pounds of P (from example D, refer to table 1 for conversion factor):
    920 lb P2O5 × 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 18-46-0?

    First, because 18-46-0 supplies two nutrients, both must be expressed on an elemental basis so their “plant food” totals can be summed. In one ton of 18-46-0, 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 18-46-0 has cost an average of $269 per ton (table 3). The value of 18-46-0 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�P2O5K2O 1997 1998 1999 2000 2001 ave
Dollars per ton
Ammonium nitrate 33.5-0-0 233 199 178 180 262 210
Ammonium phosphate 10-34-0 263 270 258 254 269 263
Ammonium sulfate 21-0-0 197 196 177 175 202 189
Ammonium phosphate sulfate 16-20-0 216 215 208 196 210 209
Anhydrous ammonia 82-0-0 313 261 218 234 421 289
Diammonium phosphate 18-46-0 289 283 273 248 254 269
Monoammonium phosphate 11-52-0 286 278 268 257 255 269
Muriate of potash 0-0-60 162 167 173 169 173 169
Triple superphosphate 0-46-0 274 267 262 248 254 261
Urea 46-0-0 272 199 181 199 296 229
Urea ammonium nitrate‡ 32-0-0 223 197 172 191 264 209
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 liquid-based 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.
Example H. Urea ammonium nitrate (UAN) solution (32-0-0) has cost an average $209 per ton in the Southwestern states (table 3). What is the cost per pound of N?
    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 P2O5, and percent K2O. Any other nutrients that are guaranteed also must be listed on the label. The cost per pound of N, P2O5, and K2O in a fertilizer with only N, P, or K can be calculated using the cost per ton of fertilizer and the percent N, P2O5, or K2O in the material. To calculate the cost per pound of elemental P or K, a factor must be used to convert percentage P2O5 to percentage P, and percentage K2O 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 liquid-based 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 N-P2O5- K2O-bearing 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.