Take all the CO2 in the atmosphere and put it into a layer on the surface of earth, how thick would the layer be?
Ok, based on the comments above, this is where I am at... 


earth's radius is $6371 \text{ km}$
earth is a sphere
There is approximately $3 \cdot 10^{15}$ kg of CO2 in earth's atmosphere
The density of CO2 at 1atm is $\dfrac{1.977 \text{ kg}}{\text{ m}^3}$


So, first, with assumption (3 and 4) we can compute the volume of CO2 in the atmosphere to be $\approx 1.52 \cdot 10^{15} \text{ m}^3$ or $1.52 \cdot 10^6 \text{ km}^3$.

So, now we can find the thickness of the layer by solving the equation

$\begin{align*}
1.52 \cdot 10^6 &amp;= \dfrac{4 \pi \cdot (6371 + x)^3}{3} - \dfrac{4 \pi (6371)^3}{3} \\
\end{align*}$

Where $x$ is the thickness of the layer of CO2 on earth's surface in km.

$\begin{align*}
1.52 \cdot 10^6 \text{ km}^3 &amp;= \dfrac{4 \pi \cdot (6371 + x)^3}{3} - \dfrac{4 \pi (6371)^3}{3} \\[.5pc]
\dfrac{3}{4\pi} \cdot 1.52 \cdot 10^6 \text{ km}^3 &amp;= (6371+x)^3 - 6371^3 \\
\dfrac{1.14}{\pi} \cdot 10^6 \text{ km}^3&amp;= (6371+x)^3 - 6371^3 \\
3.63 \cdot 10^5 +6371^3 \text{ km}^3 &amp;= (6371 + x)^3 \\
\sqrt[3]{3.63 \cdot 10^5 +6371^3 \text{ km}^3} &amp;= 6371 + x \\
6371.0029811 \text{ km} &amp;\approx 6371 + x \\
.0029811 \text{ km} &amp;\approx x
\end{align*}$

So, the layer of CO2 on earth's surface would be 2.98 meters thick...