# Finding percentage of the gas in a binary gaseous mixture of the given density

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.

• 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). – andselisk Jan 25 '19 at 12:39
• This is correct if I reference the above. This is also another way that I will be analyze do to the time saving nature. – Kai Jan 25 '19 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%.

• 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. – andselisk Jan 25 '19 at 12:44
• @andselisk Yeah that quadratic solution didn't make much sense. Hopefully I ironed this out now. – Buck Thorn Jan 25 '19 at 14:20
• 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. – Kai Jan 25 '19 at 17:23
• @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. – Buck Thorn Jan 25 '19 at 17:32