Carbon and Nitrogen ContentsOrganic matter is not solely carbon or solely nitrogen. All live organic matter has some of both elements, in varying proportions. After death, the ratio of carbon to nitrogen increases over time. The following examples explain how to compute the recommended 30:1 ratio of Carbon to Nitrogen in a compost pile. Most composters use trial and error to attain a good ratio, but it is helpful to some to understand the mathematical concept. If you don't want to understand the mathematical concept, don't worry about it. You can go to the Klickitat County Compost Calculator and get estimate volumes for your pile. Please be aware that our formula is simplified for use by home composters. To be truly correct, you must factor in the moisture content, bioavailability of carbon and nitrogen, and lignin content in the materials. We will instead use the simpler method of considering estimated ratios of carbon to nitrogen. We have provided a reference which gives Carbon to Nitrogen ratio (C:N) estimates for organic matter. If the C:N ratio is not in the reference, we were unable to find one. For purposes of this example, we are going to round the grass ratio up to 20:1, and use 40:1 for the leaves ratio. We will also use rounding frequently in performing the math. These equations illustrate how to set up the problems. The examples given are:
Example of determining C:N ratio (2input pile):
You have 5 pounds of grass clippings (20:1). Multiply the % of grass by the C:N ratio of grass, add the multiplication of the % of leaves by the C:N ratio of leaves. (50% × 20/1) + (50% × 40/1)= 10 + 20 = 30 > which in fraction notation is 30/1 or 30:1. The C:N ratio is 30:1.
Example of determining materials needed by weight (2input pile):You have 5 pounds of rotted manure (25:1) and you want to know how much corn stalks (60:1) in weight to add to get the optimum 30:1 ratio.
Set the unknown variable, weight of cornstalks needed, to equal "W". Fill these variables and the known 30:1 target ratio into the equation from the first example. In other words, multiply the % of manure by the C:N ratio of manure, add the multiplication of the % of cornstalks by the C:N ratio of cornstalks, equal to a ratio of 30/1. (5/T × 25/1) + (W/T × 60/1) = 30/1 > Perform basic math to reduce the equation.
125/T + 60W/T = 30
To check, put back into original equation:
(86% × 25/1) + (14% × 60/1) =
Example of determining C:N ratio (3input pile):
You have 5 pounds of grass clippings (20:1). Multiply the % of grass by the C:N ratio of grass, add the multiplication of the % of leaves by the C:N ratio of leaves, add the multiplication of the % 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 > which in fraction notation is 29.2/1 or 29.2:1.
Example of determining materials needed by weight (3input pile):You have 5 pounds of rotted manure (25:1) and 4 pounds of grass clippings (20:1) and you want to know how much corn stalks (60:1) in weight to add to it to get the optimum 30:1 ratio.
Set the unknown variable, weight of cornstalks needed, to equal "W".
Therefore, the weight of the rotted manure will equal (T  W  4) = 5 lbs.
Therefore, the weight of the grass will equal (T  W  5) = 4 lbs. Fill these variables and the known 30:1 target ratio into the equation used in the other examples. In other words, 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, add the multiplication of the % of cornstalks by the C:N ratio of cornstalks, equal to a ratio of 30/1. (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
To check, put back into original equation:
(45% × 25/1) + (36% × 20/1) + (19% × 60/1) = = 2.2 lbs. of cornstalks are required

