**Convert fertilizer.** Many studies have proven that fertilization affects agricultural yields. In theory, fertilizer is included in the production factors along with labor, capital, and other inputs in farming. Farmers must also know how to calculate fertilizer needs.

Fertilization must be based on the right time and the right dose. Timely fertilization provides information about when the elemental fertilizer is applied, while the right dose explains how much fertilizer should be applied according to the age of the plant.

An example of timely fertilization is that N fertilizer is usually applied when the plants enter the vegetative phase, while fertilizers containing P and K are usually applied when the plants enter the generative phase. If we apply N in the generative phase when the plants are starting the flowering stage, we will find that the plants are too lush and late to flower.

There are various obstacles that we sometimes face when reading information on fertilizer requirements. If we come across information about the need for 200 kg of urea fertilizer per hectare, then it is not a problem if we also find urea in our environment. However, it becomes a problem if the N compound fertilizer available in the farm shop around us is P:K compound fertilizer with a ratio of 15:15:15. Of course, we will count for a moment to convert how much the weight of the compound fertilizer is equivalent to 200 kg of urea.

## Characteristics of Macrochemical Fertilizers

N, P, and K are elements that are needed by plants in small amounts compared to other elements. The presence of these elements can be said to be a must if you want to have healthy and productive plants. The need for these elements can be met by relying on chemical fertilizers such as urea, SP-36, and KCL.

The characteristic of these chemical fertilizers is the elemental content of the fertilizer. For example, urea only contains 45% N, while SP-36 contains 36% P. While KCL contains 60% K elements.

This is a strong reason why the continuous use of chemical fertilizers can damage the soil. Because the chemical fertilizer is only partially used by plants or evaporates into the air, while the carrier or binding element of the main element remains in the soil and is difficult to decompose. Only 45% of the fertilizer is used by plants, as are SP-36 and KCL. Repeated use will make the soil hard and barren.

### Calculating fertilizer needs

Based on the information on the content of elements in chemical fertilizers, we can determine and convert the fertilizer needs for either single fertilizer or compound fertilizer. For additional information, there are compound fertilizers consisting of NPK with a content of 16:16:16, there are also compound fertilizers on the market with an NPK content of 15:15:15. In the example I will give, we are using NPK 15:15:15. Whatever the trademark, there is certainly information in the packaging of the ratio of N P and K elements

How much NPK fertilizer (15:15:15) is equivalent to the N element in 200Kg urea.

The amount of N in urea is = 45/100 x 200 = 90 Kg.

Since the N content in compound fertilizer is 15%, the amount of compound fertilizer that has an N content of 90 Kg is:

100/15 x 90 = 600 kg

So the result is that NPK compound fertilizer (15:15:15) has the same N content as urea which weighs 200 kg.

If you come across a recommendation for the amount of fertilizer with a description of only the N element (not urea) then you no longer need to calculate the amount of N in Urea as in the example above, directly convert from the N element obtained into compound fertilizer.

**For example:**

If the N requirement is 150 kg/ha, then the conversion to NPK fertilizer (15:15:15) is

100/15 x 150 = 1000 kg/ha.

One important thing you should remember is that if you come across a fertilizer recommendation package with several elements at once, then use the smallest number as the calculation to convert into compound fertilizer needs. This is to avoid excess doses of fertilizer that can damage the crop.

**For example:**

The recommendation package for N is 150 kg/ha, P is 100 kg/ha, and K is 80 kg in one fertilizer. Then we use the K requirement value of 80 kg to convert to compound fertilizer.

100/15 x 80 = 533 kg/ha

Meanwhile, the shortage of N and P elements should be fulfilled with a single fertilizer; for example, we fulfill it with urea and SP-36. So the urea and SP-36 we need are:

Urea

The NPK fertilizer used above already contains the element N by

15/100 x 533 kg = 80 kg

So the shortage is: 150 – 80 = 70 Kg

Urea requirement = 100/45 x 70 = 155 kg

SP-36

The NPK fertilizer used above already contains the element P by

15/100 x 533 kg = 80 kg

So the shortage is: 100 – 80 = 20 kg

SP-36 needs = 100/36 x 20 = 55 kg

So the fertilizer we need is 533 kg NPK compound fertilizer, 155 kg urea, and 55 kg SP-36.

You can practice this calculation with other types of fertilizers according to information on their elemental content.

Thank you.

**Update: November 9, 2019**

I received a question from a reader that is quite good. Namely, how to determine the use of fertilizers if the composition of compound fertilizers is not the same. In the example above, the composition of N, P, and K in compound fertilizer is the same, sometimes 16:16:16 or 15:15:15. So what if you encounter compound fertilizer with an NPK ratio of 16:20:25?

An example of the question is the need for 120 kg of N, 150 kg of P, and 90 kg of K. The ratio of compound fertilizer is 16:20:25.

Basically, in the use of fertilizers, we avoid excess doses or adhere to the right dose principle. So the minimum limit of single fertilizer use is determined in determining compound fertilizer, and then the rest uses single fertilizer.

We use several scenarios, namely scenario 1, if it meets the overall N in compound fertilizer. scenario 2 meets the overall P needs in compound fertilizer, and scenario 3 meets the overall K fertilizer needs in compound fertilizer. In each of these scenarios, we determine which scenario does not exceed the specified dose.

In scenario 1, 120 kg of N will be met with 750 kg of compound fertilizer (16% ratio). In the content of 750 kg of compound fertilizer with an NPK ratio of 16:20:25, there are 120 kg of N, 150 kg of P, and 187.5 kg of K. This scenario cannot be used because of the K content. **This scenario cannot be used because K exceeds the dosage.**

In scenario 2, a P of 150 kg will be fulfilled with a compound fertilizer of 750 kg (20% ratio). In the content of 750 kg of compound fertilizer with a NPK 16:20:25 ratio, there are N 120 kg, P 150 kg, and K 187.5 kg. **This scenario also cannot be used because K exceeds the dosage.**

In scenario 3, 90 kg of K will be fulfilled with 360 kg of compound fertilizer (16% ratio). In 360 kg of compound fertilizer with a NPK 16:20:25 ratio, there are 57.6 kg of N, 72 kg of P, and 909 kg of K. The scenario can be used because there are no fertilizers available. **The scenario can be used because no fertilizer exceeds its dosage.**

So scenario 3 can be used with 360 KG of compound fertilizer. then add single fertilizer to meet the unmet N and P in the compound fertilizer.

Furthermore, the calculation of fertilizer needs in pots for home garden plants can be read in the following article: Calculating fertilizer needs in pots.

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