<p><span class="math-container">$\rm \frac{mol}{m^2}$</span> shows the amount of <span class="math-container">$\rm{NO_2}$</span> in the atmosphere over a square meter of surface area - in mols.</p>

<p>The molar mass of the <span class="math-container">$\rm{NO_2}$</span> is <span class="math-container">$14+2\cdot 16=46$</span>. It means, the mass of 1 mol of <span class="math-container">$\rm{NO_2}$</span> is <span class="math-container">$\rm{46g}$</span>.</p>

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

<p>Here we assumed a constant distribution of the <span class="math-container">$\rm {NO_2}$</span> 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).</p>

CC BY-SA 4.0
2020-03-12T16:05:47.937Z
Thanks! I have added an update. Would you please like to see!
Niyamat Ullah
CC BY-SA 4.0
2020-03-12T15:33:36.977Z
2020-03-12T15:33:36.977Z
peterh
<p>I want to calculate the total <span class="math-container">$\rm{NO_2}$</span> amount in a year using satellite <a href="https://developers.google.com/earth-engine/datasets/catalog/COPERNICUS_S5P_OFFL_L3_NO2" rel="nofollow noreferrer">Sentinel-5p NO2 dataset</a>. But the problem is sentinel satellite data stored in <span class="math-container">$\rm\frac{mol}{m^2}$</span> unit. But I have to compare it to <code>lb</code> or <code>ton</code> or any other unit to compare <span class="math-container">$\rm{NO_2}$</span> amount with ground-level <span class="math-container">$\rm{NO_2}$</span> measurement. </p>

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

<p>Also, if I convert that <span class="math-container">$\rm\frac{mol}{m^2}$</span> to <span class="math-container">$\rm\frac{molec}{cm^2}$</span>, is it then possible to convert it to total <code>gram</code> or <code>ton</code>? </p>

<p><strong>Research</strong>: I have asked that question in <a href="https://gis.stackexchange.com/questions/353368/correlating-mol-m2-with-ton-and-lb">here</a>. 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 <span class="math-container">$\rm{NO_2}$</span> in gram or ton is not straightforward. These forum posts (<a href="https://earthscience.stackexchange.com/questions/8860/convert-omi-no2-vertical-column-density-molecules-cm2-into-mixing-ratio-ppm">#1</a>, <a href="https://earthscience.stackexchange.com/questions/19391/convert-no2-concentration-in-sentinel-5p-data-from-mol-m2-to-%CE%BCg-m3-on-the-ground">#2</a>, <a href="https://www.researchgate.net/post/How_can_I_convert_the_unit_from_molecules_cm2_to_ppm" rel="nofollow noreferrer">#3</a>) also somewhat similar to mine, but I didn't get quite a solution to my problem. </p>

<p><strong>EDIT</strong>: I know that we can easily multiply by area in <span class="math-container">$\rm m^2$</span> to cancel out <span class="math-container">$\rm m^2$</span> in the unit and multiple by <span class="math-container">$\rm{NO_2}$</span> molar mass, but this is not a ground-based problem. We are talking about satellite imagery dataset especially <strong>tropospheric vertical column density</strong>. <a href="https://earthscience.stackexchange.com/questions/8860/convert-omi-no2-vertical-column-density-molecules-cm2-into-mixing-ratio-ppm">#1</a>, <a href="https://earthscience.stackexchange.com/questions/19391/convert-no2-concentration-in-sentinel-5p-data-from-mol-m2-to-%CE%BCg-m3-on-the-ground">#2</a>, <a href="https://www.researchgate.net/post/How_can_I_convert_the_unit_from_molecules_cm2_to_ppm" rel="nofollow noreferrer">#3</a> These posts shows why it is different for converting satellite imagery units to ground-based units. </p>

CC BY-SA 4.0
2020-03-12T15:35:19.587Z
Your question reduces to relative easy unit conversions/multiplications, but I am not sure, if I detailed enough well. If something is not clear, feel free to ask it and I will improve my answer.
peterh
CC BY-SA 4.0
2020-03-12T15:49:43.817Z
"So we have summed all the year data of California": Not sure if you can do that. NO$_2$ might have a longer (or shorter) lifetime than the temporal resolution of your satellite data, so you might end up counting the same molecule multiple times (or missing some molecules).
Jean-Marie Prival
CC BY-SA 4.0
2020-03-12T15:58:50.037Z
@Jean-MariePrival What I did is use reduceRegion function of google earth engine to sum no2 of an entire year. I know this has a problem. But that is different. But I want to know how to convert my yearly total no2 (mol/m^2 ) to other units ( ton, gram, lb )
Niyamat Ullah
CC BY-SA 4.0
2020-03-12T23:42:01.983Z
The question is utterly unclear. If you have a uniform mass/area number and want to know the mass, you multiply by the area. If the distribution is not uniform, you must integrate. If you are, by contrast, not sure what your numbers mean you should *ask about that.*
Peter - Reinstate Monica
CC BY-SA 4.0
2020-03-13T00:00:03.257Z
By the way, I concur with @Jean-MariePrival. What if the concentration was measured every minute? What if once a year? What if the measurements are downwind from the actual producer in Nevada? Ok, they aren't, probably, but NOx produced in California may go out to sea or *into* Nevada so won't show up in the atmosphere over CA. What's needed are densities of the neighboring areas, and a model for the NOx gas kinetics and transportation (i.e. where does it go, how and how fast does it react with other atmospheric gases, precipitation, and the ground including plants).
Peter - Reinstate Monica
CC BY-SA 4.0
2020-03-13T04:27:56.130Z
https://en.wikipedia.org/wiki/Area_density#Column_density This is relevant
Gimelist
<geophysics><geochemistry><remote-sensing><atmospheric-chemistry><satellites>
How to convert mol/m^2 to total mass ( e.g gram, kg etc )?
Niyamat Ullah
# How to convert mol/m^2 to total mass ( e.g gram, kg etc )?

(Asked by: username_0 on 2020-03-12)

I want to calculate the total $\rm{NO\_2}$ amount in a year using satellite [Sentinel-5p NO2 dataset](https://developers.google.com/earth-engine/datasets/catalog/COPERNICUS_S5P_OFFL_L3_NO2). 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](https://gis.stackexchange.com/questions/353368/correlating-mol-m2-with-ton-and-lb). 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](https://earthscience.stackexchange.com/questions/8860/convert-omi-no2-vertical-column-density-molecules-cm2-into-mixing-ratio-ppm), [#2](https://earthscience.stackexchange.com/questions/19391/convert-no2-concentration-in-sentinel-5p-data-from-mol-m2-to-%CE%BCg-m3-on-the-ground), [#3](https://www.researchgate.net/post/How_can_I_convert_the_unit_from_molecules_cm2_to_ppm)) 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](https://earthscience.stackexchange.com/questions/8860/convert-omi-no2-vertical-column-density-molecules-cm2-into-mixing-ratio-ppm), [#2](https://earthscience.stackexchange.com/questions/19391/convert-no2-concentration-in-sentinel-5p-data-from-mol-m2-to-%CE%BCg-m3-on-the-ground), [#3](https://www.researchgate.net/post/How_can_I_convert_the_unit_from_molecules_cm2_to_ppm) These posts shows why it is different for converting satellite imagery units to ground-based units.

**Tags:** geophysics, geochemistry, remote-sensing, atmospheric-chemistry, satellites

> **Comments:**
> **username_1**: Your question reduces to relative easy unit conversions/multiplications, but I am not sure, if I detailed enough well. If something is not clear, feel free to ask it and I will improve my answer.  
> **username_2**: "So we have summed all the year data of California": Not sure if you can do that. NO$_2$ might have a longer (or shorter) lifetime than the temporal resolution of your satellite data, so you might end up counting the same molecule multiple times (or missing some molecules).  
> **username_0**: @username_2 What I did is use reduceRegion function of google earth engine to sum no2 of an entire year. I know this has a problem. But that is different. But I want to know how to convert my yearly total no2 (mol/m^2 ) to other units ( ton, gram, lb )  
> **username_3**: The question is utterly unclear. If you have a uniform mass/area number and want to know the mass, you multiply by the area. If the distribution is not uniform, you must integrate. If you are, by contrast, not sure what your numbers mean you should *ask about that.*  
> **username_3**: By the way, I concur with @username_2. What if the concentration was measured every minute? What if once a year? What if the measurements are downwind from the actual producer in Nevada? Ok, they aren't, probably, but NOx produced in California may go out to sea or *into* Nevada so won't show up in the atmosphere over CA. What's needed are densities of the neighboring areas, and a model for the NOx gas kinetics and transportation (i.e. where does it go, how and how fast does it react with other atmospheric gases, precipitation, and the ground including plants).  
> **username_4**: https://en.wikipedia.org/wiki/Area_density#Column_density This is relevant  

---

## 1 answers

### Answer by username_1 on 2020-03-12. Score: 7

$\rm \frac{mol}{m^2}$ 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).

> **Comments:**
> **username_0**: Thanks! I have added an update. Would you please like to see!  

---

