<p>It is easier to calculate:</p>
<ol>
<li>The pressure of air on the ocean level is <span class="math-container">$\approx 100 \rm {kPa}$</span>.</li>
<li>Thus, its weight is <span class="math-container">$\approx$</span> 100000N over <span class="math-container">$1 \rm m^2$</span>.</li>
<li>Considering that the overwhelming majority of the atmosphere is below some km over the surface of the Earth, we can ignore the decrease of the gravity with height.</li>
</ol>
<p>Now calculate the mass of the air above a square meter (ten tons). And multiple it with the surface of the Earth (in square meters). Then, divide it with the mean molar mass of air and multiple with the partial pressure of <span class="math-container">$\ce{O2}$</span> in it (<span class="math-container">$\approx$</span> 29%).</p>

CC BY-SA 4.0
2020-11-25T23:04:38.907Z
Change "above" to "below".
Spencer
CC BY-SA 4.0
2020-11-26T07:29:05.360Z
@Spencer Thanks!
peterh
CC BY-SA 4.0
2020-12-28T08:42:13.960Z
“neglige” isn’t a word, but it should definitely be 
Gimelist
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2020-12-28T08:46:35.357Z
@Gimelist Yeah :-) But I fixed my English extension. :-) Thanks!
peterh
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2020-12-28T13:05:23.597Z
The only problem is that this assumes the same molecular composition throughout the air column.
Spencer
CC BY-SA 4.0
2020-12-28T13:17:55.667Z
@Spencer Not if we see the partial pressure (I extended it into the post).
peterh
CC BY-SA 4.0
2020-06-26T16:17:37.640Z
2020-12-28T13:17:04.847Z
2020-12-28T13:17:04.847Z
peterh
<p><a href="https://www.newscientist.com/article/mg15020308-500-the-last-word/" rel="nofollow noreferrer">https://www.newscientist.com/article/mg15020308-500-the-last-word/</a> says the atmosphere weighs 5 × 10^21 grams. The molar weight of oxygen gas is 32 gram per mole. Based on that estimate for the weight of atmosphere, the total quantity of O2 in atmosphere is around 5*10^21/32 = 1.5625e+20 mole.</p>

CC BY-SA 4.0
2020-11-24T03:55:31.597Z
you forgot to account for the fact oxygen only makes up ~23% of the atmosphere by mass.
John
CC BY-SA 4.0
2020-06-26T16:03:31.327Z
2020-06-26T16:03:31.327Z
ElnorCat
<p>At standard conditions, one liter of air, 0.21 L of oxygen gas, contains 0.0094 moles of <span class="math-container">$\rm O_2$</span>. What is the total quantity of oxygen gas molecules in the atmosphere?</p>

CC BY-SA 4.0
2020-06-26T16:04:31.147Z
more about accuracy. I have provided an answer to the best accuracy I had, but, likely possible to make much more educated guesses.
ElnorCat
CC BY-SA 4.0
2020-11-23T23:02:20.063Z
Need to be careful here... I don't think your question is really asking about how much gas is in the entire world's atmosphere (which the others answer), but the amount in the atmosphere you are looking at in the 1st sentence (it sounds like a homework question, yes?). If so, what they're really wanting you to do is convert the moles to molecules. You should have learned a conversion ratio value for that... may have a "special" name for a scientist.
JeopardyTempest
CC BY-SA 4.0
2020-11-23T23:05:43.083Z
If it is a hw question, and they are really asking about the entire world's atmosphere, there should be another value for the world's atmosphere given in the assignment that you'll then use.  If this is what you're working on, we can try to guide you more... [chat](https://chat.stackexchange.com/?tab=site&host=earthscience.stackexchange.com) may be a good place to get more input.  We don't generally just tell you the answer, but want to help you work through it by guiding.
JeopardyTempest
<atmosphere><oxygen>
How many mole of oxygen gas is there in the atmosphere?
ElnorCat
# How many mole of oxygen gas is there in the atmosphere?

(Asked by: username_0 on 2020-06-26)

At standard conditions, one liter of air, 0.21 L of oxygen gas, contains 0.0094 moles of $\rm O\_2$. What is the total quantity of oxygen gas molecules in the atmosphere?

**Tags:** atmosphere, oxygen

> **Comments:**
> **username_0**: more about accuracy. I have provided an answer to the best accuracy I had, but, likely possible to make much more educated guesses.  
> **username_1**: Need to be careful here... I don't think your question is really asking about how much gas is in the entire world's atmosphere (which the others answer), but the amount in the atmosphere you are looking at in the 1st sentence (it sounds like a homework question, yes?). If so, what they're really wanting you to do is convert the moles to molecules. You should have learned a conversion ratio value for that... may have a "special" name for a scientist.  
> **username_1**: If it is a hw question, and they are really asking about the entire world's atmosphere, there should be another value for the world's atmosphere given in the assignment that you'll then use.  If this is what you're working on, we can try to guide you more... [chat](https://chat.stackexchange.com/?tab=site&host=earthscience.stackexchange.com) may be a good place to get more input.  We don't generally just tell you the answer, but want to help you work through it by guiding.  

---

## 2 answers

### Answer by username_2 on 2020-06-26. Score: 2

It is easier to calculate:

1. The pressure of air on the ocean level is $\approx 100 \rm {kPa}$.
2. Thus, its weight is $\approx$ 100000N over $1 \rm m^2$.
3. Considering that the overwhelming majority of the atmosphere is below some km over the surface of the Earth, we can ignore the decrease of the gravity with height.

Now calculate the mass of the air above a square meter (ten tons). And multiple it with the surface of the Earth (in square meters). Then, divide it with the mean molar mass of air and multiple with the partial pressure of $\ce{O2}$ in it ($\approx$ 29%).

> **Comments:**
> **username_3**: Change "above" to "below".  
> **username_2**: @username_3 Thanks!  
> **username_4**: “neglige” isn’t a word, but it should definitely be   
> **username_2**: @username_4 Yeah :-) But I fixed my English extension. :-) Thanks!  
> **username_3**: The only problem is that this assumes the same molecular composition throughout the air column.  
> **username_2**: @username_3 Not if we see the partial pressure (I extended it into the post).  

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### Answer by username_0 on 2020-06-26. Score: 0

<https://www.newscientist.com/article/mg15020308-500-the-last-word/> says the atmosphere weighs 5 × 10^21 grams. The molar weight of oxygen gas is 32 gram per mole. Based on that estimate for the weight of atmosphere, the total quantity of O2 in atmosphere is around 5\*10^21/32 = 1.5625e+20 mole.

> **Comments:**
> **username_5**: you forgot to account for the fact oxygen only makes up ~23% of the atmosphere by mass.  

---

