Take all the CO2 in the atmosphere and put it into a layer on the surface of earth, how thick would the layer be? I teach a Middle School (year 8) Geometry course and have a student attempting to answer this question. My assumptions:


earth's radius is $3959 \text{ mi}$
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 $\dfrac{1562 \text{ kg}}{\text{cubic meter}}$
Conversion between cu meters and cu ft are $1:35.3147$. And cu ft to cu mi is $5280^3:1$


So, first, with assumption (3 and 4) we can compute the volume of CO2 in the atmosphere to be $\approx 1.92 \cdot 10^{12} \text{ m}^3$ or about $369 \text{ cu mi}$ of CO2.

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

$\begin{align*}
369 &amp;= \dfrac{4 \pi \cdot (3959 + x)^3}{3} - \dfrac{4 \pi (3959)^3}{3} \\
\end{align*}$

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

$\begin{align*}
369 \cdot \dfrac{3}{4 \pi} &amp;= (3959+x)^3 - (3959)^3 \\[.5pc]
369 \cdot \dfrac{3}{4 \pi}+3959^3 &amp;= (3959+x)^3 \\[.5pc]
\sqrt[3]{369 \cdot \dfrac{3}{4 \pi}+3959^3} &amp;= 3959 + x \\[.5pc]
3959.0000018746 &amp;\approx 3959 + x \\[.5pc]
.00000018746 &amp;\approx x
\end{align*}$

So, the layer of CO2 on earth's surface would be $\approx 1.87 \cdot 10^{-6} \text{ mi}$ or $.1187 \text{ in}$ in thickness. Which is less than half the thickness of an iPhone 6.

My Question: This result seems incredibly small. Where did I go wrong? Or, is this reasonable. I have no intuition and would be interested in a second opinion!
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...