At STP, the density of a gas in a vessel is $0.9002$. If the gas is a mixture of argon and helium, what percentage of the gas is argon?

I am stuck on this. From what I can gather, the only influencing characters would be moles and grams. This idea is based of off of STP and given constants.

I have a couple equations written down but I can’t seem to the flow going.

R, T, P are all known along with acccording densities. I also want to say $22.4$ is also a known at $V_\mathrm{tot}$ and therefore $n_\mathrm{tot} = 1$.

$$n(\ce{Ar}) + n(\ce{He}) = 1$$

$$\frac{n(\ce{Ar})}{n(\ce{Ar}) + n(\ce{He})}d(\ce{Ar}) + \frac{n(\ce{He})}{n(\ce{Ar}) + n(\ce{He})}d(\ce{He}) = d_\mathrm{tot}$$

$$d = PM/RT$$


Well, I can see all the relations you require knowledge of in the question itself!

Calculate the effective molecular mass from

$$M = \frac{dRT}{P}$$

$R$ is known, $d$ given. $P$ and $T$ are available from the fact that it is at STP.

$M$ comes out to be ${20.176 u}$.

Effective molar mass is easily calculated below:

$$M_\mathrm{eq} = M_1x_1 + M_2x_2$$

where $x_i$ is the mole fraction of each gas. You can replace $x_2$ by $1-x_1$.

Plugging in the values and solving for $x_1$, argon mole fraction comes out to be $0.4493$.

I believe you can now calculate the demanded percentage, be it by mass or by moles.

  • $\begingroup$ This seems to be the correct answer. The only thing I'd probably add for clarity is that the density given is the relative density $d = ρ/ρ^\circ$, where $ρ^\circ = ρ(\ce{H2O},\pu{4 °C})$ (usually). $\endgroup$ – andselisk Jan 25 at 12:39
  • $\begingroup$ This is correct if I reference the above. This is also another way that I will be analyze do to the time saving nature. $\endgroup$ – Kai Jan 25 at 17:28

I'd start the solution with this equation:

$n_{Ar}d_{Ar}/(n_{Ar} + n_{He}) + n_{He}d_{He}/(n_{Ar} + n_{He}) = d_{avg}$

where I emphasize that the density of the gas is a molar average value, and rewrite it as

$\chi_{Ar} d_{Ar} + (1-\chi_{Ar})d_{He} = d_{avg}$

where $\chi_{Ar} = n_{Ar}/n$ is the mole fraction of argon in the gas mixture and the $d_{i}$ are densities computed assuming all of the gas corresponds to He or Ar, for instance

$d_{Ar} = M_{Ar}n/V = M_{Ar}/V_{m}$

where $V_{m}$ is the molar volume at STP (22.414 $m^3/kg mol$).

It follows that

$ d_{avg} = \chi_{Ar}M_{Ar}/V_{m} + (1-\chi_{Ar})M_{He}/V_{m} = (\chi_{Ar}M_{Ar} + (1-\chi_{Ar})M_{He})/V_{m} = M_{avg}/V_{m} $

which can be solved for $ \chi_{Ar} $:

$ \chi_{Ar} = ((d_{avg}V_{m})-M_{He})/(M_{Ar}-M_{He})$

Finally, the molar percentage argon is computed as $f_{Ar}=100\chi_{Ar}$.

For your particular problem I get $f_{Ar}$= 45.00%.

  • $\begingroup$ Please check your math; I suspect there shouldn't be any quadratic equations. Also, if you check the tabulated values of the gas densities at STP ($ρ(\ce{He}) = \pu{0.179 g L-1}$; $ρ(\ce{Ar}) = \pu{1.784 g L-1}$, it's obvious that they should be in a roughly 1:1 ratio in this mix. In fact, it's about 55% helium and 45% argon. $\endgroup$ – andselisk Jan 25 at 12:44
  • $\begingroup$ @andselisk Yeah that quadratic solution didn't make much sense. Hopefully I ironed this out now. $\endgroup$ – Night Writer Jan 25 at 14:20
  • $\begingroup$ Okay this makes sense. I had the right idea a few times but I got lost in the algebra. The key piece of information in your answer that would have allowed me to figure the problem out is the Vm. Additionally, I didn’t condense the ratio into terms of x. So I had nHe = 1 - nAr plugged into the original equation you started with. That is where I got lost in the algebra and as I had mentioned, the value of Vm that I didn’t realize. $\endgroup$ – Kai Jan 25 at 17:23
  • $\begingroup$ @William R. Ebenezer got the answer right before I did but I thought since I was already on the board to clean up mine and leave it. $\endgroup$ – Night Writer Jan 25 at 17:32

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