 Solution

Complete oxidation of $10^{12}\mathrm{kg}$ of carbon adds $n_C=10^{12}/(12\times 10^{-3})=8.33\times 10^{13}\mathrm{mol}$ of carbon dioxide. With atmospheric mass about $5.1\times 10^{18}\mathrm{kg}$ and mean air molar mass $28.97\times 10^{-3}\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$, there are about $1.76\times 10^{20}\mathrm{mol}$ of air. The dry-air mole fraction, approximately the volume fraction, changes byIt is the carbon inventory that counts: oxidation adds oxygen to each molecule but does not change the number of carbon-containing molecules. Using $5.15\times 10^{18}\mathrm{kg}$ instead changes the estimate by only about one percent. The atmospheric constants are standard Earth estimates, not additional data read from a missing plot.^[ref]