= Solution
Complete oxidation of $10^{12}\,\mathrm{kg}$ of carbon adds $n_C=10^{12}/(12\times10^{-3})=8.33\times10^{13}\,\mathrm{mol}$ of <carbon dioxide>. With atmospheric <mass> about $5.1\times10^{18}\,\mathrm{kg}$ and mean air molar mass $28.97\times10^{-3}\,\mathrm{kg\,mol^{-1}}$, there are about $1.76\times10^{20}\,\mathrm{mol}$ of air. The dry-air mole fraction, approximately the volume fraction, changes by
$$
\boxed{\Delta C\simeq10^6\frac{8.33\times10^{13}}{1.76\times10^{20}}\simeq0.47\,\mathrm{ppmv\ per\ GtC}.}
$$
It is the carbon inventory that counts: oxidation adds oxygen to each molecule but does not change the number of carbon-containing molecules. Using $5.15\times10^{18}\,\mathrm{kg}$ instead changes the estimate by only about one percent. The atmospheric constants are standard Earth estimates, not additional data read from a missing plot.https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html[terrestrial atmosphere: total mass and mean molecular weight]{ref}
Back to article page
