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close this book Soils, Crops and Fertilizer Use
close this folder Chapter 9: Using chemical fertilizers
View the document What are chemical fertilizers?
View the document Are chemical fertilizers appropriate for limited-resource farmers?
View the document An introduction to chemical fertilizers
View the document Common chemical fertilizers and their characteristics
View the document The effect of fertilizers on soil pH
View the document Fertilizer salt index and "burn" potential
View the document Basic application principles for N, P, and K
View the document Fertilizer application methods explained and compared
View the document Troubleshooting faulty fertilizer practices
View the document Getting the most out of fertilizer use: crop management as an integrated system
View the document Understanding fertilizer math

Understanding fertilizer math

There's a surprising amount of math devolved in using chemical fertilizers. This section covers the following useful fertilizer math skills:

• Converting fertilizer recommendations from an N-P2O5-K2O basis to the actual kind and amount of fertilizer needed.

• Selecting the most economical fertilizer.

• Mixing fertilizers.

• Determining how much fertilizer is needed per area, per Plant, and Per length of row.

• Converting fertilizer dosages from weight to volume.

USE THE METRIC SYSTEM!: It greatly simplifies fertilizer math and most other calculations. Even if your country doesn't use metrics, it's well worth your while to use it for calculation purposes. Here's how to quickly convert some common non-metric units into metric (see Appendix A also):

lbs./acre X 1.12 = kg/ha; 1 lb. = 0.454 kg = 454 g

1 acre = about 4000 sq. meters (actually 4048 m2) 1 manzana (Latin America) = 7000 sq. meters

4" = 10 cm

8" = 20 cm

12" = 30 cm

16" = 40 cm,

18" - 45 cm

24" = 60 cm

30" = 75 cm

32" = 80 cm,

36" = 90 cm

40" = 100 cm


As explained in Chapter 9, fertilizer recommendations aren't always given in terms of actual kind and amount of fertilizer. Instead, technical brochures and soil testing labs often give recommendations in terms of the amount of N, P2O5, and K2O needed per hectare. In this case, it's up to you and the farmer to determine what kind and amount of actual fertilizer is needed per hectare to match this recommendation. Let's run through a practice problem:

PROBLEM 1: A farmer's cooperative has just received the following fertilizer recommendation for a one hectare tomato field.








At transplanting




1st sidedressing at 30 days



2nd sidedressing at 60 days



3rd sidedressing at 90 days



Suppose the local ag supply store has the following fertilizers available. What kind will be needed, how much of each, and what will the cost be?

Fertilizers Available

Cost per 50 kg Sack












STEP 1: Let's begin with the 40-80-40 transplanting recommendation. The first thing to do is to look at the ratio of N:P2O5:K2O and then look for a fertilizer with a similar ratio. The 40-80-40 figure has a ratio of 1:2:1. Look at the fertilizer list and you'll see that 10-20-10 is the only one with a 1:2:1 ratio, so it's the one that' 8 needed.

STEP 2: There are 2 ways to find out how much 10-20-10 is needed to supply the 40 kg N, 80 kg P2O5, and 40 kg K2O needed for the hectare:

a. You know that each 100 kg of 10-20-10 supplies 10 kg of N, 20 kg of P2O5, and 10 kg of K2O. Therefore 400 kg would supply 40-80-40.

b. The second way is to divide the percentage of N, P2O5 or K2O in the 10-20-10 into the respective amount of N, P2O5, or K2O needed. Let's do this using N:

40 kg N needed = 40 kg = 400 kg 10-20-10 needed 10% N in the fertilizer 0.10

Note that you would get the same answer using P2O5 or K2O so it's only necessary to do this division once.

STEP 3: Now what about the N sidedressings of 30 kg actual N each? In this case, choosing the right fertilizer is easy, since the 20-0-0 fertilizer is the only one containing just N. To find out how much 20-0-0 will be needed to supply the 30 kg of N needed for a sidedressing, use one of the 2 methods in STEP 2 as follows:

a. You know that each 100 kg of 20-0-0 supplies 20 kg of N. 200 kg would supply 40 kg N. It would take 150 kg of 20-0-0 to supply 30 kg (i.e. 150 X 20% = 30)

b. Divide 20% into 30 kg which gives you 150 kg.

Therefore, 3 sidedressing of 150 kg 20-0-0 each will be needed for a total of 450 kg.

STEP 4: You've determined that 400 kg of 10-20-10 and 450 kg of 20-0-0 are needed for the hectare, so you can now calculate the cost:

400 kg 10-20-10 at $16/100 kg =

$64 450 kg

20-0-0 at $12/100 kg =


$118 TOTAL

You won't always be able to fit a recommendation exactly, because the right type of fertilizer may not be available locally. At any rate, you don't have to be exact, since soil tests and recommendations aren't 100% accurate anyway. But, do try to get within 10-25% of the amounts recommended. There's nothing wrong with having to apply more P than needed in order to supply enough K or vice-versa; P won't leach, and K is fairly immobile. However, avoid putting too much N on at planting or leaching losses may be high.

Let's look at a situation where the fertilizers don't exactly fit the recommendation (Problem 2):

PROBLEM 2: Soil test results recommend that Fatou fertilize her maize field as follows:


kg per hectare





At planting




At knee high stage



Given the following fertilizers, how much and what kind will be needed per hectare?

Fertilizers Available







STEP 1: Let's begin with the planting recommendation of 30 kg N, 50 kg P2O5, and 40 kg K2O. That's a 3:5:4 ratio (or 1:1.7:1.3). None of the available fertilizers has this ratio, but 12-24-12 is the closest with a 1:2:1: ratio.

STEP 2: Let's figure out how much 12-24-12 is needed to supply the 30 kg of initial N:

30 kg N needed / 12% N in fertilizer = 250 kg 12-24-12

250 kg of 12-24-12 per hectare would supply 30 kg N, 60 kg P2O5 and 30 kg K2O. This falls short on K2O. by about 25% but runs over on P2O5 about 20% This is still satisfactory. Now what would happen if we tried to supply the exact amount of P2O5 (50 kg) using 12-24-12?:

50 kg P2O5 needed / 24% P2O5 in fertilizer = 208 kg 12-24-12

208 kg/hectare of 12-24-12 supplies 25-50-25 which is about 20% less N and 40% less K2O. than called for.

The 3rd option is to see how much 12-24-12 it would take to supply the exact amount of K2O (40 kg) called for:

40 kg K2O needed / 12% K2O in fertilizer = 333 kg 12-24-12

333 kg of 12-24-12/hectare supplies 40-80-40 which is about 30% more N and 60% more P2O5 than called for at planting. You could adjust for the extra N by applying less in the sidedressing, but there's no way to compensate for the 30 kg extra P2O5 applied. True, some of this excess will be available to future crops, but at the expense of having to buy about 33% more 12-24-12 compared to the 250 kg rate.

Thus, of the 3 options, the first one of applying 250 kg of 12-24-12 is best.

STEP 3: Now for the N sidedressing. The 33-0-0 fertilizer (ammonium nitrate) is the only straight N source, so it's the one to use. Calculate the amount needed to supply the 50 kg of N as follows:

50 kg N needed / 33% N in fertilizer = 150 kg 33-0-0 needed


You can't compare a 14-14-14 and 10-30-10 fertilizer on the basis of cost per unit of nutrient for 2 reasons:

• A 1:1:1 ratio fertilizer may be better suited than a 1:3:1 ratio or vice-versa, depending on the soil and the crop.

• N, P2O5 and K2O. don't necessarily cost the same per kg of nutrient.

However, you can compare straight fertilizers having just one of the "Big 3", such as urea (45-0-0) vs. ammonium sulfate (20-0-0), or single superphosphate (0-20-0) vs. triple superphosphate (0-48-0). You can also compare NP or NPK fertilizers having the same ratio, such as 10-20-10 and 12-24-12.

When comparing several sources of the same nutrient as to cost, what counts is the cost per kg of nutrient, not the cost per sack. Let's run through a practice problem:

PROBLEM 3: Which of the 3 N fertilizers below is the most economical source of N, other considerations aside?


% N

Cost per 50 kg sack




Ammonium nitrate



Ammonium sulfate



SOLUTION: Although ammonium sulfate has the lowest cost per sack, it's not necessarily the cheapest. The real test is the cost per kg of N. Here's how to calculate it:

UREA: A 50 kg sack contains 22.5 kg of N (50 kg x 45%)

$18.00 / 22.5 kg N = $0.80 per kg of N

AMMONIUM NITRATE: A 50 kg sack contains 16.5 kg of N.

$15.84 / 16.5 kg N = $0.96/kg of N

AMMONIUM SULFATE: A 50 kg sack contains 10.5 kg of N.

$11.76 / 10.5 kg N = $1.12/kg of N

This makes urea the cheapest source of N. Usually, the fertilizer with the highest content of the nutrient will be the most economical due to lower shipping costs per unit of actual nutrient. However, this isn't always the case.

Other factors may be important aside from the cost per kg of nutrient. Although it's the most economical (in this case), urea is a very highly concentrated source of N; farmers unfamiliar with urea may over-apply it and waste money or injure their crops. As for ammonium sulfate, it's often the most costly per kg of N, yet it might be the best choice for a sulfur-deficient soil, unless another sulfur-bearing fertilizer were used at planting time. On the other hand, ammonium sulfate is considerably more acid forming in its longterm effect on soil pH than either urea or ammonium nitrate (see Table 9-1). Ammonium nitrate is a quicker-acting N source than ammonium sulfate or urea, because half of its N is already in the mobile, nitrate form. It might be the best choice where a crop is showing N deficiency symptoms (see Appendix E) or where sidedressing has been delayed.


There are cases where it's necessary to mix 2 or 3 different fertilizers together in order to obtain the nutrient ratio needed to suit a recommendation. For example:

PROBLEM 4: Suppose that the extension office recommends the following fertilizer rates for cabbage at planting time:

kg per hectare







The local ag supply store has the following fertilizers on hand:



0-45-0 (triple superphosphate)

Is it possible to meet the 40-80-40 recommendation by mixing 2 or more of these together? If so, what proportions are needed, and what is the resulting fertilizer formula?

SOLUTION: The 40-80-40 recommendation has a 1:2:1 ratio. The 15-15-15 provides NPK in a 1:1:1 ratio. What's needed is to increase the amount of P by adding some 0-45-0 fertilizer. The easiest way to calculate the amounts needed is to set up a worksheet as follows:





100 kg 15-15-15


15 kg

15 kg

15 kg

X kg 0-45-0


0 kg

15 kg

0 kg

100 + X kg


15 kg

30 kg

15 kg

This worksheet helps visualize the problem. It shows that in order to end up with a 1:2:] N:P2O5:K20 ratio, we need to combine 100 kg of 15-15-15 with enough 0-45-0 to add 15 extra kg of P2O5 To figure out how much 0-45-0 is needed, divide 15 by 45%:

15 kg P2O5 needed / 45% P2O5 = 33 kg 0-45-0

Now let's fill in the worksheet:





100 kg 15-15-15


15 kg

15 kg

15 kg

33 kg 0-45-0


0 kg

15 kg

0 kg

133 kg


15 kg

30 kg

15 kg

This shows that mixing 100 kg of 15-15-16 with 33 kg of 0-45-0 will produce 133 kg. of a fertilizer with a 1:2:1 ratio.

Determining the true formula of the mix: At first glance, it would seem that the formula of the mixture is now 15-30-15, but it isn't! What you've made is 133 kg of fertilizer containing 15 kg N, 30 kg P2O5, and 15 kg K2O. But, fertilizer formulas are based on nutrient content in percent (i.e. kg of N, P2O5, and K2O per 100 kg of fertilizer). Here's how to derive the true formula:

True formula = 15-30-15 / 1.33 = 11.25-22.5-11.25

CAUTION!: Not all Fertilizers can be Mixed

• Lime in any form should not be mixed with ammonium N fertilizers or urea. It will cause loss of N as ammonia gas.

• Lime should also not be mixed with any chemical fertilizer containing P, because it may convert some of the P into an insoluble, unavailable form.


Fertilizer recommendations are usually given on a per hectare (or per acre) basis. However, such figures are of little use unless you know how to determine the following:

• How much actual fertilizer is needed, given the size of the particular field?

• If the fertilizer will be applied using an LP (localized placement) method, how much fertilizer is needed per plant if the hole or half-circle method is used, or how much per length of row if it's banded? (These 2 application methods were covered earlier in this chapter.)


BLANK BOXES = Fertilizers which can be mixed.

BOXES WITH AN "X" = Fertilizers which can be mixed only shortly before use.

BOXES WITH A "O" = Fertilizers which cannot be mixed for chemical reasons.

EXAMPLES: Ammonium sulfate should not be mixed with lime.

Urea can be mixed with single or triple superphosphate shortly before use.


For Large Fields: Measure the field's dimensions and calculate the area. If its shape is not rectangular, you may have to divide it up into triangles and rectangles and determine the area of each. (The area of a triangle equals 1/2 the base X the height.)

PROBLEM 5: Suppose soil tests recommended applying 250 kg/ha of 16-20-0 to grain sorghum at planting time. How much is needed for a field measuring 40 x 80 meters?

SOLUTION: One hectare = 10000 sq. meters

The field's size = 3200 sq. meters (40 x 80)

3200 sq. meters X 250 kg/ha / 10000 sq, meters = 80 kg of 16-20-0 needed

For Small Plots: The metric system has some very handy shortcuts. A very useful one is:


In other words, to convert from kg/ha to g/sq. meter, just drop a zero and change kg to grams!! Here's why it works:

100 kg/ha = 100,000 grams/hectare

100,000 grams / 10,000 sq. meters = 10 grams/sq. meter

PROBLEM 6: If the extension service recommends broadcasting 10-30-10 at 600 kg/ha for nursery seedbeds when no compost or manure are available, how many grams of 10-3010 would be needed for a nursery seedbed measuring 1 X 5 meters?

SOLUTION: 600 kg/ha = 60 g/sq. meter

1 x 5 meters = 5 sq. meters

5 sq. meters x 60 g/sq. meter = 300 grams of 10-30-10 needed


NOTE: The calculations below are based on open fields with evenly-spaced rows running across them. Where "intensive" gardening is used (beds with alleyways around them), another factor needs to be considered. We'll cover this after explaining the open-field calculations.

If using the half-circle or hole method of placement, the farmer will need to know how much fertilizer is needed per plant (or group of plants if they're in "hills"). There are several ways of doing this, but most people agree that the following method is the simplest:

PROBLEM 7: Angelita is planning to plant a field of maize with rows 90 cm apart. She'll plant 3 seeds per hole with 60 cm between holes, using a planting stick. The extension office recommends applying 18-46-0 at 150 kg/ha. If she uses the hole method of fertilizer placement, how many grams of 18-46-0 should each seed group receive?

SOLUTION: 150 kg/ha = 15 g/sq. meter

0.9 m X 0.6 m = 0.54 sq. meters of space belonging to each plant group.

0.54 X 15 g/sq. meter = 8.1 grams of 18-46-0 per plant group

NOTE: As you see, it's not necessary to know the field's size in order to arrive at the above answer. All that's needed is the rate per hectare and the in-row and between-row spacings. Of course, you need to know the field's area (or the total number of plants) to figure out how much fertilizer to buy.

PROBLEM 8: A communal garden project has run out of manure and is about to transplant cabbage on a field measuring 20 x 20 meters. The local extension office recommends applying 16-20-0 at 250 kg/ha using the half circle method. How much fertilizer should each plant receive if the rows are 60 cm apart with 40 cm between plants in the row?

SOLUTION: 250 kg/ha = 25 g/sq. meter

0.6 m X 0.4 m = 0.24 sq. meters space occupied by each plant

0.24 X 25 g/sq. meter = 6 grams of 16-20-0 needed per plant.


NOTE: The calculations below are based on open fields with evenly-spaced rows running across them. Where "intensive" gardening is used (beds with alleyways around them), another factor needs to be considered. We'll cover this after explaining the open-field calculations.

When banding fertilizer, farmers need to know how much to apply per meter of row length (or per row). As with per-plant dosages, there are several ways of calculating this, but the simplest and quickest method is shown below:

PROBLEM 9: Suheyla is about to apply an N sidedressing to her grain sorghum field. The recommendation is 200 kg/ha of 21-0-0 (ammonium sulfate). The rows are spaced 90 cm apart, and she plans to apply the fertilizer in a band running down the middle of each row. How many grams of 21-0-0 should be applied per meter of row length?


STEP 1: 200 kg/ha = 20 g/sq. meter

STEP 2: All the fertilizer in a meter of row length will be confined in a band. If you can calculate the area belonging to that one meter of row length, you can figure out the dosage per meter:

Area belonging to 1 meter of row length = 1 meter of length X 0.9 m of width = 0.9 sq. meters

STEP 3: 0.9 sq. meters X 20 g/sq. meter = 18 g of 21-0-0 meter.


All the above dosage calculations were based on the open-field system of crop spacing where the rows are spaced equally across the field. (Both systems are explained and illustrated in Chapter 4.) However, if you use the same assumption when calculating dosages for intensively-grown crops (bed-and-alley system) you'll end up significantly shortchanging the plants on fertilizer. Here's why:

• In the intensive system, vegetables are grown under very close spacings within beds (flat, raised, or sunken) that are separated by alleyways used for all foot and equipment traffic.

• Since virtually all root growth takes place in the soil within the beds, no fertilizer (or water) should be applied to the alleys.

• The fertilizer recommendation per hectare is the same for both systems, BUT this means that the dosage per plant, per meter of row length, and per planted area (i.e. beds only) will be higher in the intensive system to make up for the fact that no fertilizer is applied to the alleyways.

• Another way of explaining this is that plants grown under the intensive system are spaced much more closely than when grown on an open-field basis. Because of this, more fertilizer is needed per sq. meter of actual bed. Since alleyways aren't fertilized, the amount of fertilizer per hectare ends up the same for both systems.

NOTE: You may think that water rates need to be similarly increased per sq. meter, but not so. That's because the high plant densities under bed-and-alley cropping shade more of the soil surface and thus lower evaporation losses of water; this helps compensate for the increased usage caused by the higher plant density. Also, the plant leaves shade each other more, which lowers transpiration (actual plant usage).

Let's go over how to calculate fertilizer dosages for bed-and-alley cropping.

PER AREA (Bed-and-Alley System)

In the open-field system, 100 kg per hectare equals 10 grams per sq. meter. Now, in the intensive system you have 2 types of area: bed area (planted area) and alley area. This means that 100 kg/ha of fertilizer doesn't work out to 10 g/sq. meter of actual bed area. If you applied 10 g per sq. meter to all the bed area, you'd end up applying much less than 100 kg/ha, because of not fertilizing the alley area which can equal about 30-40% of the total area.

Let's run through a practice problem on how to adjust for this:

PROBLEM 10: Akbar has 10 beds each measuring 1 x 10 meters; they're separated from each other by alleyways 60 cm wide on all 4 sides. He is told to apply 12-24-12 at 300 kg/ha at planting and wants to know how much fertilizer to buy.


STEP 1: 300 kg/ha = 30 g/sq. meter of total area (beds + alleys)

STEP 2: Determine the total area (beds + alleys) occupied by Akbar's plots:

It's accurate enough to assume that each 1 x 10 meter bed is separated from another by a 60 cm alleyway on each of its 4 sides.

Therefore each bed along with its associated portion of alley (half the width of each alley) measures 1.6 x 10.6 meters which equals 17 sq. meters.

10 beds x 17 sq. meters (bed + alley) = 170 sq. m

STEP 3: Calculate the amount of 12-24-12 needed for all the beds, based on the total area involved.

170 sq. m x 30 g/sq. m = 5100 g = 5.1 kg of 12-24-12 needed

STEP 4: Calculate the amount needed per bed:

5.1 kg are needed but will be applied only to the beds themselves, not alleys.

5100 grams = 510 grams 12-24-12 needed per bed 10 beds

Now you can see how much difference there is in dosages. If you had based the dosage on 30 g/sq. m and used bed area alone, each bed would receive 300 grams of fertilizer (10 x 30) instead of the 510 grams it really deserves!

PER PLANT (Bed-and-Alley System)

In this case, the easiest way to calculate the upwardly-adjusted rate is to count the number of plants on a bed and then divide that into the amount of fertilizer needed per bed as we did for Akbar's plot above.

PROBLEM 11: Suppose Akbar is planning to transplant cabbage on the beds above. He'll run 3 rows down the length of each bed with 40 cm between rows and 40 cm between plants in the rows. How much 12-24-12 should each cabbage transplant receive if he plans to use the half-circle method and the same rate per hectare (300 kg)?


STEP 1: Find how how many cabbage plants will fit on each bed:

25 plants will fit in each row (24 spaces with 40 cm between them, with the first and last plant being 20 cm from the bed's end). 75 total plants/bed. (See Figure 9-3.)

FIGURE 9-3: A 1 x 10 m bed can accomodate 75 cabbage plants spaced 40 cm x 40 cm.

STEP 2: Find the dosage per plant:

In the above problem, we determined that Akbar needs 510 g of 12-24-12 per bed.

510 g = 6.8 g 12-24-12 per plant 75 cabbage plants

Now, let's compare this dosage to that obtained from using open-field system math calculations as in Problems 7 and 8 a few pages back:

300 kg/ha = 30 g/sq. m

0.4 m (40 cm) X 0.4 m = 0.16 sq. m of space belong to each plant

0.16 x 30 g/sq. m = 4.8 grams (too little)

If Akbar applied 4.8 g per plant, each bed would receive only 360 g (instead of 510 g), which would work out to about 212 kg/ha instead of the recommended 300 kg/ha. (If each bed occupies 17 m2 (including alleyway area), there would be about 588 beds in a hectare; 588 x 360 g = 212 kg.)


In this case, the simplest method is to find the amount of fertilizer needed per bed as in Problem 10 and divide this by the number of rows per bed.

PROBLEM 12: Suppose Akbar decides to plant leaf lettuce on the 10 beds in Problem 10 g in rows 20 cm apart running the short way (i.e. 1 meter long rows). Using the same fertilizer rate (300 kg/ha of 12-24-12, how much should be applied per meter of row if the band method is used?


STEP 1: Find out how many rows will fit on each 1 x 10 m bed:

50 rows with 49 row spaces each of 20 cm will fit on a 1 x 10 m bed. Each of the 2 end rows will be 10 cm in from the bed's edge.

STEP 2: Determine the dosage per meter of row (i.e. one row in this case):

From Problem 10, we know that 510 grams are needed per bed, so:

510 grams = 10.2 g of 12-24-12 per 50 rows one meter of row length

Again, let's compare this intensive system dosage with that obtained by using open-field calculations as in Problem 9:

300 kg/ha of 12-24-12 = 30 g/sq. meter

0.2 m (20 cm) X 1 meter = 0.2 sq. meters space belonging to each meter-long row

0.2 sq. meters x 30 g/sq. m = 6 g of 12-24-12 per sq. m (too low)





Grams of Fertilizer Equal to:



100 cc(ml)

1 Level Tablespoonful (15 cc)

Ammonium sulfate

108-120 g

16-18 g

Ammonium nitrate

85 g

13 g



Ammonium nitrate

100 g

15 g




75-79 g

11-12 g


98-104 g

15 g


93-108 g

14-16 g

Potassium chloride

100-120 g

15-18 g



Single superphosphate

109-11, g

16-18 g

Triple superphosphate

100-112 g

15-17 g

Most other NP and

93-110 g

14-16.5 g

NPK fertilizers


As shown by the above dosage problems, the amount of chemical fertilizer needed per plant or per meter of row is surprisingly small, usually ranging from 5-30 grams. To assure accuracy and cost-effectiveness, farmers should not attempt to estimate such small amounts. However, since few farmers or gardeners have easy access to accurate scales, it's very helpful if to convert the fertilizer dosage from weight to volume. This doesn't mean simply converting grams to cubic centimeters, either. The dosage should be given in terms of a commonly available volume measure such as a:

• Juice can

• Tuna fish can

• Bottle cap lid

• Match box

• Spoon size commonly used in the area

This can be done by using a gram scale (check the post office or a pharmacy) to measure the densities of the common fertilizers available and comparing them to water (1 gram = 1 cc or 1 ml). Then you can measure the volume of commonly available containers like those above and calculate how many grams of fertilizer they hold.

Fertilizers vary a lot in their density, depending on type, brand, and moisture content. If no scales are available, use Table 9-6.

Here's a practice problem for converting fertilizer weight to volume:

PROBLEM 13: How many grams of urea would a 120 cc juice can hold?

SOLUTION: 100 cc of urea weighs 74-79 grams.

(120 cc / 100 cc) X 74-79 g = 89-95 g of urea in one juice can