Q:

How to calculate the average molar mass of the atmosphere?

The task: Determine the average molar mass of the atmosphere - each for the moist air and the dry air. One example that I have calculated is this one: I have the gas Nitrogen and the percentage for the moist air(77,0%) and the percentage for the dry air(78,08%) aswell as the molar mass(28,014) in $$\left(\frac{g}{mol}\right)$$
My solution: $$\text{Nitrogen}=  0,78 \cdot 0,28+ 0,77 \cdot 0,28 = 43,4 \frac{g}{mol}$$ 
Is that correct? Kind regards, iloveoov.

A:

The basic idea is correct. For moist air you can calculate the molar mass given by the part of nitrogen with
$$M_{N_2,moist}=0.77\cdot 28.014\frac{g}{\text{mol}} = 21.57\frac{g}{\text{mol}}.$$
Note, that I multiplied with $28.014$ and not with 0.28. 
The next step would be to also do the same for all other gases (oxygen, argon, carbon dioxide, etc) and then sum that up:
$$M_{air,moist}=M_{N_2,moist}+M_{O_2,moist}+ …$$
That will give you the molar mass of moist air. 
Afterwards do the same steps for dry air. 
What you also did, is to sum both values of dry and moist air. That is not asked in your task.

