Fertilizer Math

This article originally was written for and published in SportsField Management magazine, the official publication of the SportsTurf Managers Association (STMA) in January 2021, pp. 24-25, 28.

What is the best way to determine how much fertilizer is needed to supply a specific amount of nitrogen (or any other nutrient) per 1,000 square feet?

As a part of properly managing and executing a sound nutrient management plan is making sure that you know how much you need to apply for any of the “Big 3” nutrients, Nitrogen (N), Phosphorus (P) and Potassium (K).

There are actually a number of different ways to properly calculate these, but one the easiest ways is as follows:

- Determine what amount of N, P or K you want to apply. For this explanation, let’s say you want to apply .9# of Nitrogen per 1,000 sq. ft.
- Look at the NPK analysis on the bag of fertilizer and convert them to a more math-friendly decimal form. Let’s say you have a bag with the analysis of 20-5-10, which means there is 20% nitrogen, 5% phosphorus and 10% potassium. You’d simply convert each one to .20 nitrogen, .05 phosphorus and .10 potassium.
- Now take your desired application rate (.9# N) and divide it by the decimal form of the nitrogen (.20). It will look like this: .9 / .20 = 4.5 lbs. of fertilizer
- The calculation answer of 4.5 is the number of pounds of fertilizer that would need to be applied per 1,000 sq. ft to achieve the .9 pounds application rate.
- This same method works for any of the three nutrients.

What is the best method to convert lbs. per 1,000 sq. ft. to lbs. per acre?

This is a pretty easy answer. The most important thing to know for this calculation is how many square feet are in an acre…which is 43,560. Once you have this information, all you need to do is multiply the already known lbs. per 1,000 sq. ft. number by 43.56. Let’s use our previous example to demonstrate how it’s done.

- We already know that we need to apply 4.5 lbs. of fertilizer per 1,000 sq. ft., so we simply multiple 4.5 X 43.56 and come up with 196.02, which is the total pounds of fertilizer needed to cover one acre.

What is the best way to determine the area a bag of fertilizer can cover and how many bags are needed to cover large areas?

There are two of specific things that you need to know to determine how much area one bag of fertilizer will cover. The first item is the weight of the bag itself and the other is the application rate of lbs. per 1,000 sq. ft. You divide the bag weight by the application rate and then multiply the answer by 1,000 to get the coverage area in total sq. ft. Again, let’s use our previous example to illustrate how to do this.

- Most professional fertilizers come in either 40 lb. or 50 lb. bags. So let’s assume that our 20-5-10 fertilizer comes in a 50 lb. bag for this exercise.
- We’ve already determined the application rate as 4.5 lbs. per 1,000 sq. ft.
- 50 / 4.5 = 11.111
- 111 X 1000 = 11,111 total sq. ft. coverage – one bag of 20-5-10

Once you have this information, it’s easy to determine the total number of bags necessary to cover larger areas. All you need to do is take the total sq. footage you want to apply the fertilizer to and divide it by the coverage of one bag to figure out how many bags you would need to have.

- Let’s say you want to treat 84,000 total sq. ft. with our 20-5-10 that we just calculated will treat 11,111 sq. ft per 50 lb. bag.
- 84,000 / 11,111 = 7.56 bags needed

What other factors should go into determining proper application rates?

Proper application rates of any nutrients, macro or micro, will vary considerably and it’s always recommended to perform soil tests at regular intervals to determine what the current levels are of each one to help guide the decisions on what’s needed and what is not. These reports will clearly tell you what you have an abundance of and what is missing. Ultimately all fertilizers feed the soil, which in turn, feed the turf. Knowing what’s happening below the surface is always the best way to determine what type of fertilizer is needed and what analysis makes the most sense for your specific site.

Is there anything I might not have touched upon here, or anything else you would like our readers to know regarding fertilizer selection and application?

It’s important to understand that for P (phosphorus) and K (potassium), the values expressed on the bag are Phosphate (P2O5) and Potash (K2O) and ** not **the elemental forms. For many professionals, correctly managing P and K levels in the soil requires doing a little chemistry and math to convert the numbers to the

**amounts of each element. Here’s a brief breakdown on how to make these conversions.**

*actual**Elemental Phosphorus*

Here’s the easy answer of understanding how much elemental phosphorus is contained in phosphate (P2O5)…43.7%. This is calculated with some very basic chemistry and looking at the atomic weight of each element of the phosphate compound.

- Atomic weights: P is 31 and O is 16. You can see these on any periodic table.
- There are two units of P in phosphate, so 31 X 2 = 61 grams
- There are five units of O in phosphate, so 16 X 5 = 80 grams
- 61 grams + 80 grams = 141 total grams of phosphate
- The amount of P is calculated as: 61 / 141 = .437 or 43.7%

With this information, you can now calculate the amount of elemental P in our 50 lb. bag example of 20-5-10 as follows:

- .437 X 5 = 2.185 lbs of P (initial answer)
- 185 X .5 = 1.0925 lbs of P
- It is necessary to multiply the initial answer by .5 because this is a 50 lb. bag because the phosphate compound is expressed as a “percentage”…meaning per 100.
- If this was a 40 lb. bag, you would multiply the initial answer by .4

*Elemental Potassium*

The very same logic is used when calculating elemental potassium. Here’s the easy answer of understanding how much elemental potassium is contained in potash (K2O)…83.0%.

- Atomic weights: K is 39 and O is 16. You can see these on any periodic table.
- There are two units of K in potash, so 39 X 2 = 78 grams
- There are one units of O in potash, so 16 X 1 = 16 grams

78 grams + 16 grams = 94 total grams of phosphate

- The amount of K is calculated as: 78 / 94 = .830 or 83.0%

With this information, you can now calculate the amount of elemental K in our 50 lb. bag example of 20-5-10 as follows:

- .830 X 10 = 8.30 lbs of K (initial answer)
- 30 X .5 = 4.15 lbs of P
- It is necessary to multiply the initial answer by .5 because this is a 50 lb. bag because the phosphate compound is expressed as a “percentage”…meaning per 100.
- If this was a 40 lb. bag, you would multiply the initial answer by .4