What is the chemical formula for moist air at 20°C?

[davidwinth]

I am trying to find the chemical formula for moist air. By using an online psychrometric calculator, I find that moist air at 20°C has a mol fraction of water=0.0235 and mol fraction dry air=0.9765.

Does this mean like this?

0.9765(O₂ + 3.76N₂) + 0.0235H₂O ----> 41.55(O₂ + 3.76N₂) + H₂0

My confusion comes when I think about 1 mol of air. Does 1 mol of air have 1 mol of oxygen and 3.76 mols of N2 or does 1 mol of air contain .21 mols of oxygen and .79 mols of N2? If the latter, then I think the formula above is wrong and it should be:

0.9765(0.21O₂ + 0.79N₂) + 0.0235H₂O
----> 0.205O₂ + 0.771N₂ + 0.0235H₂O
----> O₂+ 3.76N₂ + 0.115H₂O.

Which one is right, or is neither one right?

Thanks!

 

[BvU]

Does this mean like this?

No.
If O₂ and N₂ are present in a mol ratio of 1 : 3.76 , then 1 mol of air is 1/(1+3.76) mol of O₂ and 3.76/(1+3.76) mol of N₂, so

0.9765(O₂ + 3.76N₂)/4.76 + 0.0235H₂O​

adding up to 1 mol/mol

http://staff.www.ltu.se/~lassew/ene/moist.pdf

 

[DaveC426913]

I may not be correctly understanding what you're after but, as far as I am aware,moist air does not have a "formula". It's not in solution - it is simply a bunch of chemicals (some gas, one liquid) mixed together, so I'm not sure the equations will generate meaningful results. They can combine in almost any quantity, and they aren't producing byproducts because they're not interacting chemically.

But I could be wrong.

 

[davidwinth]

There should be a formula for the given mol fractions. We should at least be able to write one down. I hope so anyway, because I need to use this formula in the next step!

 

[davidwinth]

No.
If O₂ and N₂ are present in a mol ratio of 1 : 3.76 , then 1 mol of air is 1/(1+3.76) mol of O₂ and 3.76/(1+3.76) mol of N₂, so

0.9765(O₂ + 3.76N₂)/4.76 + 0.0235H₂O​

adding up to 1 mol/mol

http://staff.www.ltu.se/~lassew/ene/moist.pdf

O.k, this is the same as my second answer. Thank you!

 

[Borek]

You can express the composition of a mixture with a "formula" - while technically mixture is not a compound it is often convenient to use this approach.

Exact numbers don't matter much - what matters is their ratio, to some extent O₂+3.76N₂ and 0.21O₂+0.79N₂ mean the same thing. Sometimes it is more convenient to choose one of the numbers (typically the lowest) and make it 1, sometimes it is more convenient to make the numbers sum to 1 (so that each expresses actual molar fraction of a mixture component). There is no convention that I am aware of that would make one approach more correct than the other (could be some trades do have such conventions).

By 1 mole of air I would understand such an amount of air that contains exactly 1 mole of molecules of all types, it will behave as 1 mole of gas when it comes to volume it occupies (that is, as long as the ideal gas approximation holds reasonably well). Then if numbers in the "formula" sum to 1 each refers to actual number of moles of a component in "1 mole".

 

[Chestermiller]

For the mole fractions given, one mole of moist air contains (0.9765)(0.21) moles of oxygen, (0.9765)(0.79) moles of nitrogen, and 0.0235 moles of water vapor. There is no chemical reaction formula after this, because there is no reaction.

 

[davidwinth]

For the mole fractions given, one mole of moist air contains (0.9765)(0.21) moles of oxygen, (0.9765)(0.79) moles of nitrogen, and 0.0235 moles of water vapor. There is no chemical reaction formula after this, because there is no reaction.

I just wanted the "formula". The reaction happens in the next step.

 

[Chestermiller]

I just wanted the "formula". The reaction happens in the next step.

There is no formula because no reaction takes place (yet).

 

[davidwinth]

There is no formula because no reaction takes place (yet).

But there is... we found it.

How to make moist air? Follow this formula...

O₂ + 3.76N₂ + 0.115H₂O

 

[Chestermiller]

But there is... we found it.

How to make moist air? Follow this formula...

O₂ + 3.76N₂ + 0.115H₂O

That would be 4.875 moles of moist air.

 

[davidwinth]

That would be 4.875 moles of moist air.

Yes! Thanks.

I thought the next part would be easy but I am having trouble there too. I am sending dry air to mix with propane in stoichiometric ratio. So, for every volume of propane that I send to the burner, I useC₃H₈ + 5(O₂+3.76N₂) -> 3CO₂ + 4H₂O + 18.8N₂

So that is 5+18.8 = 23.8 volumes of propane. So far so good. But the question arose: what happens if I saturate those 23.8 volumes of air on the way to the burner, such that it ends up saturated at the same temperature and pressure? It should have a larger volume due to the addition of water vapor, but one person says it will no longer be stoichiometric while another says it is still stoichiometric because the oxygen and fuel are in the same relative ratio even if water is added to the air along the way.

We end up with:

C₃H₈ + 5(O₂+ 3.76N₂ + 0.115H₂O)

My thinking is that for every 23.8 volumes of air I send, even though it gets water added to it, all this does is dilute the final fuel-wetair mixture but it is still stoichiometric. I am just not sure how to prove myself wrong or right. Do I need to use mass instead of volumes here? Thanks for any input.

 

[Chestermiller]

Yes! Thanks.

I thought the next part would be easy but I am having trouble there too. I am sending dry air to mix with propane in stoichiometric ratio. So, for every volume of propane that I send to the burner, I useC₃H₈ + 5(O₂+3.76N₂) -> 3CO₂ + 4H₂O + 18.8N₂

So that is 5+18.8 = 23.8 volumes of propane. So far so good. But the question arose: what happens if I saturate those 23.8 volumes of air on the way to the burner, such that it ends up saturated at the same temperature and pressure? It should have a larger volume due to the addition of water vapor, but one person says it will no longer be stoichiometric while another says it is still stoichiometric because the oxygen and fuel are in the same relative ratio even if water is added to the air along the way.

We end up with:

C₃H₈ + 5(O₂+ 3.76N₂ + 0.115H₂O)

My thinking is that for every 23.8 volumes of air I send, even though it gets water added to it, all this does is dilute the final fuel-wetair mixture but it is still stoichiometric. I am just not sure how to prove myself wrong or right. Do I need to use mass instead of volumes here? Thanks for any input.

The important thing for a stoichiometric ratio is the ratio of moles oxygen to moles pentane. The relationship you gave is correct. For using moist air, you need (23.8/0.9765) moles of moist air to get the same ratio of oxygen to pentane.

 

[davidwinth]

Thank you ChesterMiller.

I think I may be missing something. It sounds like you are saying two opposite things. Am I correct in thinking that the final mixture is stoichiometric regardless of whether the 23.8 volumes of dry air gets saturated along the way to the burner? You say the relationship I gave is correct, but then you say I need 23.8/0.9765 mols of moist air. (I thank you for your help.)

 

[Chestermiller]

The 23.8 moles of dry air have the required stoichiometric amount of oxygen. If this air gets saturated with moisture along the way, you will have 24.4 moles of moist air, but it will still contain the correct stoichiometric amount of oxygen. I hope this explains it satisfactorily.

 

[davidwinth]

Thank you!