Plants require nitrogen (N), phosphorus (P) and potassium (K), along with other elements known as macro- and micronutrients, to grow well. In the wild, plants send out roots and try to find usable sources of these elements.

In a garden, these nutrients are usually supplied (one way or another) by the gardener via fertilizers. With fertilizers, the gardener adds material that contains the desired elements in forms that are either immediately available to the plant, such as is the case with many nutrient salts, or will break down gradually over time to become available to the plant, as organic materials tend to do.

In either case it is in the best interest of the gardener to have some idea of the nutritional value of the fertilizer being added, both to make sure that they don’t overfeed or insufficiently feed their plants, and to make sure that they aren’t adding several sources of one element and not enough of another.

Nutrient solutions are generally based on true solutions, although the math works out the same even if they are mixtures instead. A solution is made of at least two substances. The majority of the solution will be the solvent (in aqueous solutions, like those used in gardening, the solvent used is water).

The substance dissolved into the water is known as the solute. The amount of solute in a solution determines the solution’s concentration. To put is simply, the more nutrient added to the water, the stronger the solution becomes. Knowing what elements (nutrients) a solute (the fertilizer) has, and at what strength they are in, is important to calculating the final nutrient solution given to the plants. To help with this, fertilizers are marked with an N-P-K listing values to help gardeners get an idea of how much N, P and K is in the bottle or bag.

Parts per hundred is a pretty common way to relate two things, although it is more commonly referred to as percent. For example, a fertilizer with an N-P-K rating of 10-5-14, is made of 10 parts N per 100 parts of fertilizer.

Another way to express that is to say it is 10% N. In the case of N, the N-P-K value listed and the amount of elemental N are the same. For P and K, their N-P-K values are for the oxide forms.

Phosphorus oxide is 43.6 parts per 100 elemental P, so—using the N-P-K value above—multiplying 5 by 0.436 will give the elemental P value of 2.18 parts per hundred, or 2.18%. Potassium oxide is 83% elemental K, so a final N-P-K value of 14 would indicate a fertilizer that is 11.62% K (0.83 x 14 = 11.62).

While knowing that the composition of our solute is 10% N, 2.18% P and 11.62% K tells us some useful information about the proportions of what we are putting into the water to make our solution, concentration is another critical factor.

As this is a pretty potent nutrient, the difference between mixing 1 mg per liter and 1 tsp (5 mg) per liter makes for a very different experience for the plants (this is why I recommend at least reading the recommended feeding rates even if you don’t follow them exactly). Just how big of a difference can be shown with a little more math and the numbers we already have.

1 liter of water weighs 1,000 mg. If we add 1 mg of solute, the total weight becomes 1,001 mg. Since our solute is 1 mg of 10% elemental N, we can calculate the weight of the element: 10% of 1 mg = 0.1 mg.

Since fertilizers are added to in small amounts, and the amount of the desired elements are only a fraction of those small amounts, it is common to use parts per million (ppm) to express how much of each element is in the nutrient solution.

Parts per million is used for concentrations smaller than can be easily expressed in parts per hundred (percent), but larger than those commonly expressed in parts per billion. To calculate ppm, simply divide the weight of element in solute by the total weight of solution and multiply that total by one million. So, using the data from our example for N:

• (0.1 mg N / 1,001 mg) x 1,000,000 = 100 ppm elemental N.

We can also calculate the value of P and K using this same formula.

• (0.0218 mg P / 1,001 mg) x 1,000,000 = 22 ppm elemental P
• (0.1162 mg K / 1,001 mg) x 1,000,000 = 116 ppm elemental K

So, if 1 mg of this nutrient is added per liter, the nutrient solution will have 100 ppm N, 22 ppm P and 116 ppm K—which is in the ballpark for many plants that aren’t actively fruiting. Since plants can survive twice those values, going as high as 2 mg per liter would be reasonable to work up to.

Note: if gallons are easier for you to work/mix in, simply multiply the 1 mg by 3.78541 to get the amount to add per gallon (in this case, 3.78541 mg).

Back to our example, if you were to use 1 tsp per liter (or a heaping tablespoon per gallon) instead of 1 mg, calculating ppm shows how dramatic a difference that makes for your plants. Here is the math:

• 1 tsp = 5 mg
• Weight of N in solute = 0.5 mg (10% of 5 mg)
• Weight of P is solute = 0.109 mg (2.18% of 5 mg)
• Weight of K in solute = 0.581 mg (11.62% of 5 mg)
• Total weight of solution = 1,005 mg (1,000 mg water + 5 mg solute)

So,

• (0.5 mg / 1,005 mg) x 1,000,000 = 498 ppm of elemental N
• (0.109 mg / 1,005 mg) x 1,000,000 = 108 ppm of elemental P
• (0.581 mg / 1,005 mg) x 1,000,000 = 578 ppm of elemental K

This will almost certainly cause overfeeding problems.

Also note that the equation can be reversed to estimate a dose from a desired ppm. If a concentration of 150 ppm of N is the goal (and we’re using our sample solute of 10% N), then:

• 150 ppm = 10% x 1,000,000 x amount to add / 1,000 mg (ish).

This reduces to:

• 150 ppm = 0.1 x 1,000 mg x amount to add.