This is my first post, I read quite a bit of available answers in this forum (and other sources) but I cant find my mistakes nor the answer to the problem. I am not experienced in chemistry, thus I put this under "undergraduate" and hope that someone of you can give me some hints on how this can be solved. Thank you for your time!

I - The main problem/question:

A combustion calculator gives me the quantity of oxygen in a volume (say 1m3) at various temperatures in, "mol frac", starting at 20.49 [mol frac] at normal condition (T=20°C and p0=101300Pa), decreasing to e.g 5.00 [mol frac] at 500°C.

I have difficulties to understand the unit [mol frac] especially in the conext of the change of the density. The user manual didnt help; the only hint I found was that *"assumes that the molecular weight of the gaseous mixture throughout the domain is approximately that of air, 29 g/mol."* which doesnt help me at all.

Thus, I started to calculate "something" to see some pattern but created more questions...

**II - The problems I met on the way:**

Using the following equation (Eq. 1) should help

M = m / n (Eq. 1)

At normal conditions (101300 Pa, 20°C) the air has a density of about 1.2 kg/m3.

Knowing that oxygen as molecule (O2) has a molar mass M=2x16u (u as the atomic mass unit) and

that the share at normal conditions is 23.135 [mass-%] I can derive

the amount of substance n [mol] by using Eq. (1) as following:

n(O2)= (1200g *23.135%) / 32 g/mol =8.68 mol oxygen per m3 air

the mass m would be

8.68 mol x 32g/mol =277.8g oxygen per m3 air

Thus, in one m3 air, there are 8.68 mol oxygen equivalent to 277.8g oxygen (a check delivered: 289/1200=23.15% of 1.2kg)

**Q1:** Is this correct, isn't it?

However:

The Gay Lussac's Law of definite proportions says, 1 mol of a gas requires a volume of 22.4 litres.

1m3 has 1000 litres, thus 1000/22.4 =44.64 ["mol-spaces"]

**Q2:** Is there a better wording for "mol-spaces"?

I will use the vol-% of oxygen, given to 20.942%:

and that 1 mol oxygen has m=32g:

44.64 x 20.942% gives 9.37 ["mol-spaces"] for oxygen

32g/mol x =299.9 [g] oxgen per m3 air

**Q3:** Where is this inconsistency coming from; or to be honest: what is wrong in my calculations?

When it comes to the original question, the mol-fraction at a certain temperature I am lost as the density of air changes.

At 500°C I calculate the density of air with the changing air pressure using Eq. (2):

ρ_gas,L = p_0/[287.1∙ (T+273.15)] (2)

Using Eq. (2) for 500°C, this results in a density of 0.49kg/m3 while at 20°C it would be 1.20 kg/m3.

Thus, for "no change of the oxygen concentration", I would expect still less mol per m3 for the higher temperatures. I expect a reduction to 0.49/1.20=0.40 for all moles as the "mol-density" would decrease

From the upper section I learned that at normal condition, there were 8.68 mol oxygen in one m3, thus, at the higher temperature, that would mean 8.68 mol x 0.40 = 3.54mol - which would be still 21% (as the other fractions would also decrease)

**Q4: **Is this reduction factor determination reasonable?

Phew, that became quite a long question, I hope it is not too confusing.

Thank you so much for reading, considering and eventually answering to my "stupid" questions.

Jo