The NPK Ratio
The NPK ratio
Fertilizer packaging contains a set of numbers, for example \(\text{6}\):\(\text{1}\):\(\text{5}\) (see figure below). These numbers are called the NPK ratio, and they give the mass ratio of nitrogen, phosphorus and potassium in the fertilizer.
Examples of (a) organic (compost) and (b) inorganic (industrially made) fertilizers indicating the NPK ratio that will be discussed later. Photos by Kessner Photography on wikipedia and Elvera Viljoen in the UCT Chemical Engineering Chemical Industries Resource pack.
The NPK ratio expresses the content of each nutrient as a percentage of N, P and K in this order. A number in brackets after this ratio indicates the percentage by mass of N, P and K that is present in the fertilizer (what percentage of the total fertilizer is N, P and K). For example:
38% of the total fertilizer is nitrogen, phosphorus or potassium.
\(\%\) N: \(\text{3}\) in every \(\text{9}\) parts of the 38% contains nitrogen (N)
\(\%\) P: \(\text{1}\) in every \(\text{9}\) parts of the 38% contains phosphorus (P)
\(\%\) K: \(\text{5}\) in every \(\text{9}\) parts of the 38% contains potassium (K)
The table below gives an idea of the amounts of nitrogen, phosphorus and potassium there are in different types of fertilizers.
Fact:
A starter fertilizer is usually used when seeds or young plants are planted. It has high concentrations of nitrogen, to boost plant growth, and phosphorus to boost root growth.
|
Description |
Grade (NPK ratio) |
|
Ammonium nitrate |
\(\text{34}\):\(\text{0}\):\(\text{0}\) |
|
Urea |
\(\text{46}\):\(\text{0}\):\(\text{0}\) |
|
Bone meal |
\(\text{4}\):\(\text{21}\):\(\text{1}\) |
|
Seaweed |
\(\text{1}\):\(\text{1}\):\(\text{5}\) |
|
Starter fertilizers |
\(\text{18}\):\(\text{24}\):\(\text{6}\) |
|
Equal NPK fertilizers |
\(\text{12}\):\(\text{12}\):\(\text{12}\) |
|
High N, low P and medium K fertilizers |
\(\text{25}\):\(\text{5}\):\(\text{15}\) |
Table: Common grades of some fertilizer materials.
Depending on the types of plants you are growing, and the growth stage they are in, you may need to use a fertilizer with a slightly different ratio. For example, if you want to encourage root growth in your plant you might choose a fertilizer with a greater ratio of phosphorus in it. The main functions of each nutrient (N, P, K) are given in the table below.
|
Nutrient |
Promotes |
When to use |
|
Nitrogen (N) |
leafy plant growth, faster plant growth |
on lawns and other plants with lots of green leaves |
|
Phosphorus (P) |
strong roots, healthy fruit, blooming |
on flowers and flower beds |
|
Potassium (K) |
disease resistance, growth of fruit |
on fruit bearing plants |
Table: The types of plant growth different nutrients promote.
Some countries express the phosphorus content as \(\text{P}_{2}\text{O}_{5}\) and potassium content as \(\text{K}_{2}\text{O}\). We express the NPK ratio in terms of the elements present as explained above. The rest of the fertilizer (62%) is made up of fillers, such as gypsum, lime and sand. Other micronutrients, such as calcium (\(\text{Ca}\)), sulfur (\(\text{S}\)) and magnesium (\(\text{Mg}\)), are often added to the mixture.
Example: NPK Ratios
Question
Calculate the mass of nitrogen (N), phosphorus (P) and potassium (K) that is present in \(\text{500}\) \(\text{g}\) of industrial fertilizer with a NPK ratio of \(\text{5}\):\(\text{2}\):\(\text{3}\) (\(\text{40}\)).
Determine the mass of the total nutrients in the fertilizer sample.
40% of the sample contains nutrients, therefore:
40% of \(\text{500}\) \(\text{g}\) = \(\text{0.4}\) x \(\text{500}\) \(\text{g}\) = \(\text{200}\) \(\text{g}\) which contains nutrients.
Determine the mass of the specific component in the sample
For every \(\text{5}\) units of nitrogen there are \(\text{2}\) units of phosphorus and \(\text{3}\) units of potassium so the total number of units is \(\text{10}\).
\(\text{5}\) of the \(\text{10}\) units are nitrogen, therefore:
\(\dfrac{5}{10}\times\) \(\text{200}\) \(\text{g}\) = \(\text{100}\) \(\text{g}\) will be nitrogen (N).
\(\text{2}\) of the \(\text{10}\) units are phosphorus, therefore:
\(\dfrac{2}{10}\times\) \(\text{200}\) \(\text{g}\) = \(\text{40}\) \(\text{g}\) will be phosphorus (P).
\(\text{3}\) of the \(\text{10}\) units are potassium, therefore:
\(\dfrac{3}{10}\times\) \(\text{200}\) \(\text{g}\) = \(\text{60}\) \(\text{g}\) will be potassium (K).
Example: NPK Ratios
Question
Calculate the mole ratio of an industrial fertilizer with the NPK ratio of \(\text{5}\):\(\text{2}\):\(\text{3}\) (\(\text{40}\)).
Calculate the mass of N, P and K in a \(\text{500}\) \(\text{g}\) sample of the fertilizer
As we are calculating the mole ratio, any amount of fertilizer can be assumed.
As in the previous worked example the ratio is \(\text{5}\):\(\text{2}\):\(\text{3}\) (\(\text{40}\)), so the mass of nutrients will be \(\text{200}\) \(\text{g}\). Therefore:
The mass of N = \(\dfrac{5}{10}\times\) \(\text{200}\) \(\text{g}\) = \(\text{100}\) \(\text{g}\).
The mass of P = \(\dfrac{2}{10}\times\) \(\text{200}\) \(\text{g}\) = \(\text{40}\) \(\text{g}\).
The mass of K = \(\dfrac{3}{10}\times\) \(\text{200}\) \(\text{g}\) = \(\text{60}\) \(\text{g}\).
Calculate the number of moles for each element in this sample of fertilizer
\(\text{n} = \dfrac{\text{m}}{\text{M}}\)
The molar mass (M) for N = \(\text{14.0}\) \(\text{g.mol$^{-1}$}\)
Therefore n for nitrogen = \(\dfrac{\text{100}{\text{ g}}}{\text{14.0}{\text{ g.mol}}^{-1}}\) = \(\text{7.14}\) \(\text{mol}\)
M for P = \(\text{31.0}\) \(\text{g.mol$^{-1}$}\)
Therefore n for phosphorus = \(\dfrac{\text{40}{\text{ g}}}{\text{31.0}{\text{ g.mol}}^{-1}}\) = \(\text{1.29}\) \(\text{mol}\)
M for K = \(\text{39.1}\) \(\text{g.mol$^{-1}$}\)
Therefore n for potassium = \(\dfrac{\text{60}{\text{ g}}}{\text{39.1}{\text{ g.mol}}^{-1}}\) = \(\text{1.53}\) \(\text{mol}\)
Determine the mole ratio
N:P:K mole ratio = \(\text{7.14}\) : \(\text{1.29}\) : \(\text{1.53}\)
To get the mole ratio divide all the numbers by the smallest number (\(\text{1.29}\)):
\(\text{5.5}\) : \(\text{1}\) : \(\text{1.2}\)
The ratio should be expressed as whole numbers:
\(\text{55}\) : \(\text{10}\) : \(\text{12}\)
Note: If the percentage purity is not given (the number in brackets – e.g. 40 in the worked example above), then you can assume that it is a 100% pure fertilizer and no fillers were added.
As mentioned previously, the NPK ratio is expressed in elemental form, for example the P in the ratio refers to the mass of the element phosphorus (P). However, in some countries the ratios are expressed in compound form: \(\text{N}\):\(\text{P}_{2}\text{O}_{5}\):\(\text{K}_{2}\text{O}\). Nitrogen still refers to the element nitrogen, but phosphorus and potassium refer to the compounds phosphorus pentoxide (\(\text{P}_{2}\text{O}_{5}\)) and potassium oxide (\(\text{K}_{2}\text{O}\)). This example will show you how to calculate the mass or moles when a compound ratio is given.
Example: NPK Ratios in Different Countries
Question
Calculate the moles of phosphorus (P) in \(\text{120}\) \(\text{g}\) of a fertilizer with the \(\text{N}\):\(\text{P}_{2}\text{O}_{5}\):\(\text{K}_{2}\text{O}\) mass ratio of \(\text{4}\):\(\text{3}\):\(\text{8}\) (\(\text{50}\)).
Calculate the mass of nutrients in the sample of fertilizer
50% of the sample contains nutrients.
50% of \(\text{120}\) \(\text{g}\) = \(\text{0.5}\) x \(\text{120}\) \(\text{g}\) = \(\text{60}\) \(\text{g}\) which contains nutrients.
Calculate the mass of \(\text{P}_{2}\text{O}_{5}\)
The \(\text{N}\):\(\text{P}_{2}\text{O}_{5}\):\(\text{K}_{2}\text{O}\) ratio is \(\text{4}\) : \(\text{3}\) : \(\text{8}\). \(\text{4}\) + \(\text{3}\) + \(\text{8}\) = \(\text{15}\). Therefore:
\(\text{3}\) of the \(\text{15}\) units are \(\text{P}_{2}\text{O}_{5}\).
The mass of \(\text{P}_{2}\text{O}_{5}\) = \(\dfrac{3}{15} \times\) \(\text{60}\) \(\text{g}\) = \(\text{12}\) \(\text{g}\).
Calculate the number of moles of \(\text{P}_{2}\text{O}_{5}\)
\(\text{n} = \dfrac{\text{m}}{\text{M}}\)
M(\(\text{P}_{2}\text{O}_{5}\)) = \(\text{2}\) x \(\text{31.0}\) \(\text{g.mol$^{-1}$}\) + \(\text{5}\) x \(\text{16.0}\) \(\text{g.mol$^{-1}$}\) = \(\text{142.0}\) \(\text{g.mol$^{-1}$}\)
Therefore n(\(\text{P}_{2}\text{O}_{5}\)) = \(\dfrac{\text{12}{\text{ g}}}{\text{142.0}{\text{ g.mol}}^{-1}}\) = \(\text{0.0845}\) \(\text{mol}\)
Calculate the number of moles of \(\text{P}\)
For every \(\text{1}\) mole of \(\text{P}_{2}\text{O}_{5}\) there are \(\text{2}\) moles of \(\text{P}\) atoms.
Therefore, n(\(\text{P}\)) = \(\text{2}\) x \(\text{0.0845}\) \(\text{mol}\) = \(\text{0.17}\) \(\text{mol}\)
Optional Activity: NPK ratios
Use the NPK ratio table above if needed, to answer the following questions:
-
Calculate the mass of phosphorus that is present in \(\text{250}\) \(\text{g}\) of bone meal.
-
Calculate the moles of nitrogen that are present in \(\text{200}\) \(\text{g}\) of ammonium nitrate (\(\text{NH}_{4}\text{NO}_{3}\)).
-
Calculate the mass of potassium that is present in \(\text{100}\) \(\text{g}\) of seaweed.
-
Calculate the number of moles of potassium in \(\text{100}\) \(\text{g}\) of a commercial fertilizer with the \(\text{N}\):\(\text{P}_{2}\text{O}_{5}\):\(\text{K}_{2}\text{O}\) mass ratio of \(\text{5}\):\(\text{1}\):\(\text{2}\) (\(\text{40}\)).
This lesson is part of:
Chemistry and the Real World