Shouldn't mineral NPK fertilizer ingredients add up to 100%?

Question

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):

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

What constitutes the rest?

Answer 1

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

Answer 2

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.