I use a mineral based fertilizer in powder form which - according to the label - should not contain any fiber and is fully soluble in water (I dissolved 2 times the recommended dosage and found no solids remaining in the solution):

NPK fertilizer label

The NPK + Mg + trace elements add up to about 60%.

What constitutes the rest?

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    $\begingroup$ Note that the fertilizer does not actually contain chemical compounds like $\ce{P2O5}$, $\ce{K2O}$, and $\ce{MgO}$. These are just the reference compounds for the calculation of results of the elemental analysis. $\endgroup$
    – user7951
    Nov 2, 2019 at 15:41
  • $\begingroup$ I'm not sure if I can follow. From what I have read so far if chemical compounds are listed after the percentage this does indeed indicate that the fertilizer contains these compounds in the indicated mass percentage. To calculate the percentage of one element one has to calculate the the percentage of that element in the compound using atomic weight. For example with K2O. K = 39u, O = 16u. Total atomic weight = 2*39 + 16. Percentage of potassium is 2*39 / (2*39 + 16) ~ 0.83 ~ 83% of K2O is potassium $\endgroup$ Nov 2, 2019 at 16:12
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    $\begingroup$ You got it wrong. The listed compounds are simply not there, as blunt as that. $\endgroup$ Nov 2, 2019 at 16:27
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    $\begingroup$ @CuriousIndeed You still do not get it. The listed fertilizer composition is NOT it's chemical composition. It is list of selected content, expressed in equivalent amount of respective, historically chosen reference compounds, mostly oxides. It does not mean these oxides are there chemically present. $\endgroup$
    – Poutnik
    Nov 2, 2019 at 17:40
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    $\begingroup$ It means if nitrogen/phosphorus/potassium/magnesium compounds were chemically converted to N2/P2O5/K2O/MgO, the mass of N2/P2O5/K2O/MgO would be 10/20/30/2.6% of the mass of the fertilizer. $\endgroup$
    – Poutnik
    Nov 2, 2019 at 17:47

2 Answers 2


There is a specific convention how the nitrogen, potassium and phosphate concentration is given, and it might be surprising because it mentions substances that are not in the fertilizer, namely nitrogen atoms, $\ce{P2O5}$, $\ce{K2O}$ and $\ce{MgO}$. You have to imagine that all of the nitrogen, phosphorous, potassium and magnesium in your sample reacts to form these species, and then you express the masses of those fictitious products as percentage of the original mass.

$$\text{fertilizer}\ce{-> x N + y P2O5 + z K2O + q MgO}$$

For an example, look at this answer: https://chemistry.stackexchange.com/a/64282 which links to this reference https://en.m.wikipedia.org/wiki/Labeling_of_fertilizer

  • $\begingroup$ According to the link on wikipedia, phosphorus and potassium percentages have to be adjusted according to their portion of mass in the chemical compound. However this is not true for nitrogen. What is the rationale behind this convention? $\endgroup$ Nov 2, 2019 at 16:24
  • $\begingroup$ @CuriousIndeed It’s elemental, my dear Watson ... well, atomic, actually. $\endgroup$ Nov 2, 2019 at 16:30
  • $\begingroup$ But it is listed as nitrate or ammonium..(Nitratstickstoff, Ammoniumstickstoff) $\endgroup$ Nov 2, 2019 at 16:32
  • $\begingroup$ @curiousindeed yes, of the nitrogen, some is in the form of nitrate and some as ammonium, but percentages are based on mass of nitrogen (adds up correctly by just adding percentages). $\endgroup$ Nov 2, 2019 at 17:18
  • $\begingroup$ Is there a specific reason for this convention? I find it pretty confusing... $\endgroup$ Nov 2, 2019 at 17:24

The rest is probably Oxygen in nitrates, Hydrogen in ammonium, and Sulfur in sulfates. Let us make some calculations !

In 1000 g fertilizer, there is : - 74 g N (or 74/14 = 5,28 mol N) as nitrate ion

  • 26 g N (or 26/14 = 1.85 mol N) as ammonium ion
  • 200 g P2O5, or 200/142 = 1.41 mol P2O5, or 2.82 mol P, or 2.82 mol PO4 ion.
  • 300 g K2O, or 300/94 = 3.13 mol K2O, or 6.26 mole K
  • 26 g MgO, or 26/24.3 = 1.07 mole MgO = 1.07 mol Mg.

There must be the same number of moles of positive ions as of negative ions. To start with negative ions, one must triple the number of phosphate and add the double of magnesium plus the number of moles of K. This total gives : 3·2.82 + 5.28 + 2·1.07 = 14.81 mol.

For positive ions, it is more difficult because some H ions are added to PO4^3- ions. Phosphate ions are too much basic to exist in the ground. In neutral mixture, it must be transformed into a mixture of H2PO4^- and HPO4^2-. So, to get the sum of positive ions, one must add 1.85 mol NH4, 6.26 mol K, 2·1.07 mol Mg, and about 1.5 times the number of P for obtaining the number of H atoms bound to PO4 ion. This makes : 1.85 + 6.26 + 2.14 + 1.5·2.82 = 14.48 mol.

The agreement between positive (14.48 mol) and negative (14.81 mol) charges is extremely good ! It is a success. And it is not necessary to add any sulfate ions !...

We may carry out the same calculations with the masses, to see whether the total is 1000 g.

  • 5.28 mol NO3 weighs 327.4 g
  • 1.85 mol NH4 weighs 33.3 g
  • 2.82 mol PO4 weighs 267.6 g
  • 6.26 mol K weighs 244 g
  • 1.07 mol of Mg weighs 26 g
  • 2·2.82 mol H weighs 5.64 g

The total of these masses is 906 g. It is not 1000 g. It shows that there are some non ionic substances in the fertilizer, like clay or organic substances.

  • $\begingroup$ I like your approach but aren't Mg and K positive ions? Could it be that you mixed positive with negative? $\endgroup$ Nov 3, 2019 at 18:52
  • $\begingroup$ Mg and K are elements, occuring in fertilizers in form of ions $\ce{Mg^2+}$ and $\ce{K+}$ $\endgroup$
    – Poutnik
    Nov 3, 2019 at 19:03
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    $\begingroup$ That's why I'm asking. He wrote "To start with negative ions, one must triple the number of phosphate and add the double of magnesium plus the number of moles of K" $\endgroup$ Nov 3, 2019 at 19:05
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    $\begingroup$ Note that there are hydrogen or dihydrogen phosphate. One has to calculate charge balance. $\endgroup$
    – Poutnik
    Nov 4, 2019 at 4:15
  • $\begingroup$ I thought we have to add the positive and negative charges. So why does he add phosphate (PO4 3-) to the positive ions Mg2+ and K+? $\endgroup$ Nov 4, 2019 at 20:24

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