The distance that CO2 molecules travel in the air before they collide with each other
You need to calculate the probability of one (CO2) molecule being the nearest neighbour of another.

Let $w(r)$ be the probability that the nearest neighbour occurs between distance  $r$ and $r+dr$ (on a spherical surface). This must be the probability than no molecules exist closer to the victim molecule,  which occurs at a distance $r$, and that this molecule exists in the shell $r$ to $r+dr$. Thus, 

$$ w(r)=\left(  1-\int_0^r w(r)dr \right)4\pi r^2 n$$

where $n$ is the average number of molecules / unit volume. (1 mol/m3 $\equiv 10^3N_A/10^{27} = 0.6023 $ molecules / nm$^3$).

Solving this gives

$$ w(r)=4\pi r^2n\exp(-4\pi r^3n/3)$$

This function is zero at zero goes through a maximum and is  small at large. distances, see figure. 

If you want a single number then the average distance between molecules is 

$$\langle r \rangle =\int_0^\infty rw(r)dr =\Gamma (4/3)/(4\pi n/3)^{1/3} =0.55396n^{-1/3}$$

where $n$ is in number of molecules/volume then $n^{-1/3}$ is in units of length.

( $\Gamma(4/3)$ is the gamma function, sort of equivalent to a factorial but for a non-integer number)

The figure below shows $w(r)$ the probability of a near neighbour at distance $r$ for some concentrations.



If what you want is the total path taken accounting for all collisions then you would have to start with one the of the $w(r)$ distributions corresponding to your concentration and do a Monte Carlo computer calculation to work out the total path, not easy and v long calculation probably. There will be a statistical mechanics way of calculating this I'm sure but I'm not aware of it at present.