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This question actually comes from backyard composting. When choosing materials to compost the ideal ratio of Carbon to Nitrogen of the raw materials to make compost is 30:1. There are basically no materials that are 30:1 naturally, so you have to work with what you have. For instance, grass clippings are 20:1 and wood chips are 400:1.

My question is how to you determine the quantity $x_1$ of a material with a $C:N$ ratio $r_1$ as a function of the quantity $x_2$ of a second material with $C:N$ ratio of $r_2$ where the sum of the two materials will be $r_{1+2}$, in this case $30:1$?

Randy
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2 Answers2

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If you have $x_2$ with $\frac CN=r_2$, you get $\frac {x_2r_2}{1+r_2}$ of $C$ and $\frac {x_2}{1+r_2}$ of $N$ and similarly with subscripts $1$. The ratio of the mix is then $r_{1+2}=\frac {x_2r_2(1+r_2)+x_1r_1(1+r_1)}{x_2(1+r_2)+{x_1(1+r_1)}}$ Now solve for $x_1$ and you have what you need.

Ross Millikan
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Another approach could be used considering the composition of the two elements to be mixed. In the first component (we shall use it a s the reference) the fraction of $C$ is given by $\frac {a_1}{a_1+b_1}=\frac {r_1}{r_1+1}$ ($r_1=\frac {a_1}{b_1}$) and the fraction of $N$ is given by $\frac {b_1}{a_1+b_1}=\frac {1}{r_1+1}$. The same for the second component (change $1$ to $2$).

Now consider that you mix $1$ quantity of the first component (this is your $x_1$) and a quantity $x$ of the second component (this is your $x_2$). You then have $(1+x)$ of mixture which contains $$C=\frac {r_1}{r_1+1}+x\frac {r_2}{r_2+1}$$ and $$N=\frac {1}{r_1+1}+x\frac {1}{r_2+1}$$ and you want that the ratio of this last quantities to be equal to $r_3$. So, $$r_3=\frac{\frac {r_1}{r_1+1}+x\frac {r_2}{r_2+1}} {\frac {1}{r_1+1}+x\frac {1}{r_2+1} }$$ from which can be extracted $$x=\frac{({r_2}+1) ({r_3}-{r_1})}{({r_1}+1) ({r_2}-{r_3})}$$

Claude Leibovici
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