How can atmospheric $CO_2$ absorption of infrared be 100% when its atmospheric concentration is 0.04%? An absorption spectrum from high in the atmosphere of infrared radiation  emitted from the earth, shows that for the 15µm wavelength there is almost complete absorption.  This is attributed to absorption by CO2 in the atmosphere at this wavelength.
However, the concentration of CO2 in the atmosphere is generally about 0.04%, or 400 parts per million.  This says to me that for any cubic metre of volume, there would be a large space not occupied by CO2 molecules through which such radiation would pass uninhibited, and therefore ultimately simply pass out to space.
This seems contradictory to me.  Could someone perhaps elaborate on what is happening here.
 A: Let's say a 15µm photon has a nonzero probability of interacting with any $CO_2$ molecule that it passes within a distance of 1 wavelength.  How many $CO_2$ molecules are there in a cylinder with radius 15µm and extending from the earth's surface out to space? More than $10^{16}$.
I think what bothers you is the idea that $CO_2$ can be so tremendously much more effective at absorbing this wavelength than $O_2$ or $N_2$.  Else why would the ratio be relevant?  It's a bit like wondering how someone can die from arsenic poisoning when only 0.04% of their meal was arsenic. It's a small fraction of the total, but that doesn't mean it's not important.
A: I thought I might answer my own question.  
The absorption of radiation by matter follows a decreasing exponential relationship.
I = Iº exp(−α n X)  where
Iº is radiation flux at origin, I is radiation flux at distance X from origin, n is number of molecules per m³, α is absorption cross section m²/molecule.
This equation is derived here http://astrowww.phys.uvic.ca/~tatum/stellatm/atm5.pdf
It is a version of the Beer-Lambert law.
The international standard atmosphere at sea-level is T = 288K, pressure = 101325 Pa.  
For a relative volume of CO2 of 0.04%, this gives a partial pressure of CO2 = (101325 x 0.0004) Pa
Solving the ideal gas law pV = nRT using the partial pressure of CO2, and applying Avogadro’s number as the number of molecules per mole, gives 10.2 x 10^21 molecules of CO2 in 1 m³.
Now we can determine the distance to absorb a given proportion of incident radiation, by using the equation above, and solving for X.
We need to know the absorption cross section of CO2.
This is found here http://vpl.astro.washington.edu/spectra/co2.htm from the PNNL link cm2/molecule vs. wavenumbers.
I am interested in the 15µm wavelength (667 per cm), which looks to be about 4.5 x 10^-18  cm2/molecule (or 4.5 x 10^-22 m²/molecule)
So now, for 99% absorption, or I/Iº = 0.01, we can solve for X.
The result is almost complete absorption in 1 metre!  
The result is sensitive to the order of magnitude of the absorption cross section.  It compares to the sometimes quoted experimental finding of Heinz Hug of complete absorption at 10 metres.
A: Even at this low concentration, a cubic millimeter of air still contains about $10^{13}$ $CO_2$ molecules. Plenty of opportunities to absorb photons. 
A: The atmosphere weighs as much as 10 meters of water. Taking 0.04 % of that means 4 millimeter. A glass pane of 4 mm absorbs all IR.
A: At 0.04% I get that the spacing between the CO2 molecules is about 10 nm. So to get through, the electromagnetic waves would have to get past a lot of these molecules. Each one does a bit of absorbing. It's not so surprising that the net effect is that they absorb pretty much all the radiation at some wavelengths.
A: In the nineteenth century it was already discovered that concentration times pathlength is an experimental invariant. So to compare lab results with atmosphere you have to multiply the concentration with the atmospheric path length.
For the atmosphere I use a scale height of 8043 m, this gives the following absorption spectrum with a central saturated band:

