# Carbon to Nitrogen Ratio (C:N) In compost

This is a simplified guide for anyone interested in calculating the Carbon to Nitrogen Ratio. If you’re looking for an even simpler version, check out our guide on creating a compost heap.

Organic matter contains both carbon and nitrogen, but in varying proportions. As organic matter decomposes, the ratio of carbon to nitrogen increases over time.

To achieve the recommended ratio of 30:1 for your compost pile, you can use trial and error, or you can understand the mathematical concept behind it. However, if you’re not interested in the math, don’t worry about it.

It’s important to note that our formula is simplified for home composting use. For more accuracy, you should consider factors such as moisture content, bioavailability of carbon and nitrogen, and lignin content in the materials. For this guide, we will estimate the ratios of carbon to nitrogen.

In this example, we will round the grass ratio up to 20:1 and use 40:1 for the leaves ratio, with frequent rounding during the calculations. The equations provided will demonstrate how to set up the problems.

Here are some examples provided:

- Determining the C:N ratio for a 2-input pile
- Calculating the weight of materials needed for a 2-input pile
- Determining the C:N ratio for a 3-input pile
- Calculating the weight of materials needed for a 3-input pile

## Determining C:N ratio (2-input pile):

You have 5 pounds of grass clippings with a C:N ratio of 20:1. You have 5 pounds of leaves with a C:N ratio of 40:1. You have a total of 10 pounds of material, with 50% being grass and 50% being leaves.

To determine the C:N ratio, you need to multiply the percentage of grass by the C:N ratio of grass and add that to the multiplication of the percentage of leaves by the C:N ratio of leaves.

So, it’s (50% × 20/1) + (50% × 40/1) = 10 + 20 = 30.

This ratio can also be written as 30/1 or 30:1.

Therefore, the C:N ratio of the mixture is 30:1.

## Determining materials needed by weight (2-input pile):

Suppose you have 5 pounds of rotted manure with a C:N ratio of 25:1, and you want to know how much corn stalks with a C:N ratio of 60:1 to add to get the optimum 30:1 ratio.

To begin, let’s set the unknown variable, the weight of cornstalks needed, equal to “W,” and the total weight of the pile equal to “T.”

The percentage of the total pile by weight represented by cornstalks will be W/T. Therefore, the weight of the rotted manure will be (T – W) = 5 lbs. The percentage of the total pile by weight represented by manure will be 5/T.

Next, we can fill these variables and the known 30:1 target ratio into the equation from the first example. That is, we multiply the percentage of manure by the C:N ratio of manure, add the multiplication of the percentage of cornstalks by the C:N ratio of cornstalks, and set it equal to a ratio of 30:1.

(5/T × 25/1) + (W/T × 60/1) = 30/1

Now we can perform basic math to reduce the equation:

125/T + 60W/T = 30

125 + 60W = 30T

since T = (5 + W), 125 + 60W = 150 + 30W

30W = 25

W = 25/30 = 0.83 lbs. of cornstalks required

To check, we can put back the values of W and T into the original equation.

W = 0.83, T = 5.83, Percentage of total weight in manure = 5/5.83 = 86%, 100-86 = 14% for cornstalks.

(86% × 25/1) + (14% × 60/1) =

21.5 + 8.4 = 29.9 (not 30 due to rounding)

Thus, we need 0.83 pounds of cornstalks to achieve the optimum 30:1 ratio, which will result in a ratio very close to 30:1.

## Determining C:N ratio (3-input pile):

You have a total of 12 lbs. of material, consisting of 5 lbs. of grass clippings (20:1), 5 lbs. of leaves (40:1), and 2 lbs. of rotted manure (25:1). Out of the total weight, 42% is grass, 42% is leaves, and 16% is manure.

To calculate the C:N ratio of the mixture, multiply the percentage of grass by the C:N ratio of grass, add the multiplication of the percentage of leaves by the C:N ratio of leaves, and add the multiplication of the percentage of manure by the C:N ratio of manure.

(42% × 20/1) + (42% × 40/1) + (16% × 25/1) = 8.4 + 16.8 + 4 = 29.2

The C:N ratio of the mixture is 29.2:1.

## Determining materials needed by weight (3-input pile):

To achieve an optimum 30:1 ratio, you have 5 pounds of rotted manure (25:1) and 4 pounds of grass clippings (20:1), and you want to determine how much corn stalks (60:1) in weight you should add.

Let the unknown variable, weight of cornstalks needed, be “W.” Let the total weight of the pile be “T.” The % of the total pile by weight represented by cornstalks will be W/T.

Therefore, the weight of the rotted manure will be (T – W – 4) = 5 lbs. The % of the total pile by weight represented by manure will be 5/T.

Therefore, the weight of the grass will be (T – W – 5) = 4 lbs. The % of the total pile by weight represented by grass will be 4/T.

Using the equation from the other examples and substituting the values of the variables and the known 30:1 target ratio, multiply the % of manure by the C:N ratio of manure, add the multiplication of the % of grass by the C:N ratio of grass, and add the multiplication of the % of cornstalks by the C:N ratio of cornstalks.

(5/T × 25/1) + (4/T × 20/1) + (W/T × 60/1) = 30/1. Perform basic math to reduce the equation.

125/T + 80/T + 60W/T = 30 205 + 60W = 30T Since T = (5 + 4 + W), 205 + 60W = 270 + 30W 30W = 65 W = 65/30 = 2.2 lbs. of cornstalks required.

To check, plug in the values back into the original equation: W = 2.2; T = 11.2; Percent of total weight in manure = 5/11.2 = 45%; Percent of total weight in grass = 4/11.2 = 36%; 100 – 45 – 36 = 11.2% for cornstalks.

(45% × 25/1) + (36% × 20/1) + (19% × 60/1) = 11.3 + 7.2 + 11.4 = 29.9 (not 30 due to rounding).

Thus, you need 2.2 lbs. of cornstalks.