Mole Fraction of O2 to Mass of O2

[sirchicken]

Hi Everyone,

I am trying to calculate the mass of oxygen and keep running into dead ends and was wondering if anyone could offer any insight.

I am trying to find the resulting mass of oxygen in a 6m x 6m x 6m enclosure at standard temperature and pressure when the mole fraction of oxygen is reduced from 0.21 to 0.158. Can anybody offer any advice?

Thanks!

 

[Redbelly98]

What is your specific application, i.e. what is it you are trying to accomplish?

Anyway, I would start by calculating the mass as if the mole fraction were 100%, and use the ideal gas law to calculate how many moles of oxygen would be in the container.

p.s. welcome to PF.

 

[sirchicken]

I am trying to find out what the mass of the oxygen in the enclosure will be if the mole fraction of oxygen from 0.21 to 0.158.

I am assuming that at stp the mole fraction of oxygen is equal to 0.21 and the mole fraction of nitrogen is 0.79. Then a fire is introduced and reduces the mole fraction of oxygen to 75% of its ambient value. I am trying to calculate the resulting mass of oxygen in the enclosure so that I can determine how long the fire can burn.

Thanks!

 

[Redbelly98]

Okay. Not sure why you want to test this at 32 °F / 0 °C (that's what standard temperature is), rather than at room temperature, but you know more about what you are trying to do than I.

At standard temperature and pressure, 1 mole of an ideal gas occupies 22.4 liters.
You can use google to figure out how many 22.4-liter volumes would be contained in a 6x6x6 m^3 space; just go to google and enter

(6*6*6 m^3) / (22.4 liters)​

That will tell you how many moles of gas molecules (of all kinds) are in the 6x6x6 m³ space.

Multiply the number of moles of gas by the fraction that is oxygen, i.e. 0.21 or 0.158, to get the number of moles of oxygen.

Multiply the number of moles of oxygen by the grams-per-mole for oxygen -- that would be 32 -- and you'll have the mass in grams of oxygen that is in the container.

 

[LudusRex]

I would suggest using the ideal gas law to calculate the mass of oxygen in the given enclosure. The ideal gas law, PV = nRT, relates the pressure (P), volume (V), number of moles (n), gas constant (R), and temperature (T) of a gas. Since we are dealing with a fixed volume and temperature, the equation can be simplified to n = (PV)/RT.

To use this equation, we need to know the pressure and temperature of the enclosure. Standard temperature and pressure (STP) is defined at 0 degrees Celsius and 1 atmosphere of pressure. Therefore, we can use these values in the equation to calculate the number of moles of oxygen present in the enclosure at 0.21 mole fraction: n = (1 atm * 6m * 6m * 6m)/(0.0821 L*atm/mol*K * 273.15 K) = 0.585 moles of oxygen.

To calculate the mass of oxygen, we can use the molar mass of oxygen, which is 32 g/mol. Therefore, the mass of oxygen in the enclosure at 0.21 mole fraction would be 0.585 moles * 32 g/mol = 18.72 g.

To find the resulting mass of oxygen when the mole fraction is reduced to 0.158, we can use the same equation with the new mole fraction and the previously calculated number of moles: n = (1 atm * 6m * 6m * 6m * 0.158)/(0.0821 L*atm/mol*K * 273.15 K) = 0.463 moles of oxygen. The resulting mass of oxygen would then be 0.463 moles * 32 g/mol = 14.816 g.

I would also recommend double checking your calculations and ensuring that the units are consistent throughout the equation. I hope this helps!