Caesar’s Last Breath

23 May, 2025

How many molecules from Caesar’s last breath do we inhale with each breath we take? Shockingly, the answer is about one molecule—we actually do share breaths with Caesar! And, by extension, every breath we take is composed of the previous breaths of everyone who ever lived—Socrates, Lincoln, Einstein, etc. Isn’t that crazy?

This is a classic exercise in Fermi estimation—it even has a book named after it. It’s a beautiful introduction to the power of napkin math, showing how much you can figure out with just a few back-of-the-envelope numbers.

Estimating quantities is a valuable skill—if you’re not convinced, check out Nabeel Qureshi’s blog. It’s easier and more fun than it seems, and I find myself enjoying it more as I realize how often I land within an order of magnitude.

Let’s walk through Caesar’s Last Breath together!

We need to determine

1. the volume of Earth’s atmosphere, and
2. the volume of a breath.

If we assume that a breath diffuses evenly throughout the atmosphere and that these molecules are preserved over time (a reasonable assumption—nitrogen is relatively inert), then we can take

$\text{fraction}=\frac{\text{volume breath}}{\text{volume atmosphere}}$

to find what fraction of Earth’s atmosphere is composed of Caesar’s last breath, and then multiply that by the number of molecules in a breath to get our answer (composition of atmosphere = composition of breath).

Here are a few “anchor” values:

• Earth’s radius is ≈ 6400 km = 6.4×10^6 m
• The dense part of the atmosphere is ≈ 10 km “tall” = 10^4 m
• The atmosphere is mostly O₂ and N₂, both ≈ 30 g/mol (1 mol ≈ 6×10^23 molecules)
• Atmospheric density is ≈ 1 kg/m³
• For more “anchor” numbers to fuel your Fermi estimates, click here

Finding some of these values is itself a Fermi estimation—you might know that Mt Everest is about 10 km high or that planes cruise at that altitude and use it as an estimate for the atmosphere’s thickness. Even better, apply the geometric-mean trick: if the atmosphere definitely extends more than 1 km but probably less than 100 km, guess $\sqrt{1\times {10}^2}=10$ km—that’s the right order of magnitude.

Atmospheric volume
The volume of a sphere is $V=\frac{4}{3}\pi r^3$. Approximating $\frac{4}{3}\pi \approx 4$, we get $V\approx 4r^3$. The atmosphere is the outer shell between $r=6400$km and $r=6410$km, so

$V_{\text{atmosphere}}\approx 4((6400+10{)}^3-{6400}^3){\text{km}}^3\approx 4.9\times {10}^9{\text{km}}^3\approx 5\times {10}^9{\text{km}}^3=5\times {10}^{18}{\text{m}}^3.$

Volume of a breath
If you inflate a balloon with a normal breath, it probably has a radius of ≈ 5 cm. Thus

$V_{\text{breath}}=\frac{4}{3}\pi r^3\approx 4\times 5^3\approx 500{\text{cm}}^3=5\times {10}^{-4}{\text{m}}^3.$

Fraction of the atmosphere per breath

${\displaystyle \frac{5\times {10}^{-4}}{5\times {10}^{18}}=\frac{1}{{10}^{22}}}$ of the atmosphere.

Molecules per breath
The mass of a breath is

$5\times {10}^{-4}{\text{m}}^3\times 1{\text{kg/m}}^3=5\times {10}^{-4}\text{kg}=5\times {10}^{-1}\text{g}.$

Then

$6\times {10}^{23}\frac{\text{molecules}}{\text{mol}}\times \frac{1\text{mol}}{30\text{g}}\times 0.5\text{g}={10}^{22}\frac{\text{molecules}}{\text{breath}}.$

Putting it all together

${10}^{22}\times \frac{1}{{10}^{22}}=1$

So we inhale about one molecule of Caesar’s last breath with each breath we take. Obviously, many simplifying liberties were taken. But, we can be confident we’re on the right order of magnitude!

If you want more, here’s a fun site where you can practice. If you’re into software engineering like me and want more practical experience, check out this repo and its corresponding tech talk. And, finally, here’s another good article about Fermi estimation.