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Ideal Gas Law with Mole Fractions and Multiple Gases (Example 2)
Example problem: A mixture of only O\(_2\) and N\(_2\) has a density of 1.185 g/L at 25\(^OC\) and a pressure of 101.3 kPa. Using the ideal gas law, find the mole fraction of O\(_2\) in the two-species mixture.
101.3 kPa is the same as 1 atm.
The way to solve this problem is to assume a quantity of gas, we usually call this a “basis”. It’s best to assume a simple number so that our calculations are slightly easier.
Assume there is 1 L of the gas mixture.
The density given was 1.185 g/L, meaning that the total mass is 1.185 g if we assume there is 1 L of it.
So,
\[{m_{total}} = 1.185g = {m_{{O_2}}} + {m_{{N_2}}}\]
We can work with this equation in terms of moles instead of masses.
To get the amount of moles in 1 L, use PV = nRT, and we will use n to represent quantity of moles.
As always, keep all decimals in your calculator, we will round here for simplicity.
\[{m_{total}} = 1.185g = {m_{{O_2}}} + {m_{{N_2}}}\]
\[{n_{total}} = \frac{{PV}}{{RT}}\]
\[{n_{total}} = \frac{{(1atm)(1L)}}{{\left( {0.08206\frac{{Latm}}{{molK}}} \right)(298K)}}\]
\[{n_{total}} = 0.041mol\]
And remember that moles is just the mass divided by the molar mass, so we found mass total = mass oxygen + mass nitrogen before, so divide everything by molar mass.
BUT, the trick here is that the mixture is binary, it only has two parts.
So if the total mass is 1.185g, then a part of that will be the mass of N\(_2\), and another part of that will be 1.185 – mass of N\(_2\), which is the remainder.
Because remember, both of the species must add up to the total mass. So we’re really just rearranging the equation: total mass = mass oxygen + mass nitrogen.
That way we will only have one variable to solve for!
\[{n_{total}} = 0.041mol = \frac{{{m_{{N_2}}}}}{{28.01g/mol}} + \frac{{(1.185 – {m_{{N_2}}})}}{{32g/mol}}\]
You can do it the other way around as well, but here we let the mass of oxygen be represented by the mass of nitrogen.
And by dividing throughout by molar masses we can use moles instead.
Now you just need to solve the equation above for the mass of nitrogen, it will be a little messy.
Solving you get,
\[{m_{{N_2}}} = 0.862g\]
And we let
\[{m_{{O_2}}} = 1.185 – {m_{{N_2}}}\]
So substituting our mass of nitrogen result in we obtain:
\[{m_{{O_2}}} = 0.323g\]
Now, to get the mole fraction of a species in a mixture, it needs to be the amount of moles of that species divided by the total moles in the mixture.
We usually let the variable x be the mole fraction.
So divide the mass of O\(_2\) found by 32g/mol to get the amount of O\(_2\) moles.
\[{n_{{O_2}}} = \frac{{0.323g}}{{32g/mol}} = 0.0101mol\]
And just do that over the total moles in the mixture, which was found previously.
\[{x_{{O_2}}} = \frac{{0.0101mol}}{{0.041mol}}\]
\[{x_{{O_2}}} = 0.247\]
And the mole fraction is unitless.
Try another Ideal Gas Law problem involving mixing two gases into a single vessel!
Thanks.