How to convert mol/m^2 to total mass ( e.g gram, kg etc )?

Question

I want to calculate the total $\rm{NO_2}$ amount in a year using satellite Sentinel-5p NO2 dataset. But the problem is sentinel satellite data stored in $\rm\frac{mol}{m^2}$ unit. But I have to compare it to lb or ton or any other unit to compare $\rm{NO_2}$ amount with ground-level $\rm{NO_2}$ measurement.

Example: In our satellite dataset, we have tropospheric $\rm{NO_2}$ column number density which unit is $\rm\frac{mol}{m^2}$. We want to calculate the total yearly $\rm{NO_2}$ of California. So we have summed all the year data of California. Then we get a total yearly $\rm{NO_2}$ 4200 $\rm\frac{mol}{m^2}$. I want to convert this to gram or ton or lb. For example, the total yearly $\rm{NO_2}$ of California is 35000 ton.

Also, if I convert that $\rm\frac{mol}{m^2}$ to $\rm\frac{molec}{cm^2}$, is it then possible to convert it to total gram or ton?

Research: I have asked that question in here. After getting answers, I thought I have found the solutions. But after some more research, it turns out not easy. Because in satellite data what we get is tropospheric vertical column number density and convert them to total mass like total $\rm{NO_2}$ in gram or ton is not straightforward. These forum posts (#1, #2, #3) also somewhat similar to mine, but I didn't get quite a solution to my problem.

EDIT: I know that we can easily multiply by area in $\rm m^2$ to cancel out $\rm m^2$ in the unit and multiple by $\rm{NO_2}$ molar mass, but this is not a ground-based problem. We are talking about satellite imagery dataset especially tropospheric vertical column density. #1, #2, #3 These posts shows why it is different for converting satellite imagery units to ground-based units.

Answer 1

$\rm \frac{mol}{m^2}$ shows the amount of $\rm{NO_2}$ in the atmosphere over a square meter of surface area - in mols.

The molar mass of the $\rm{NO_2}$ is $14+2\cdot 16=46$. It means, the mass of 1 mol of $\rm{NO_2}$ is $\rm{46g}$.

The surface area of the Earth is 510million $\rm{km^2}$. Thus, 1 $\rm \frac{mol}{m^2} \rm{NO_2}$ translates to $\rm{46 \frac{g}{m^2} \cdot 5.1 \cdot 10^8 km^2}$, what means a total mass of $23.46\cdot 10^9 \rm t$ in the whole atmosphere.

Here we assumed a constant distribution of the $\rm {NO_2}$ in the atmosphere, which is typically not the case in earth science-related problems. Also your research question is about the visualization of a highly non-level distribution. Here you need to integrate (or, considering your digital evironment, sum).