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A gas $\ce{X2O5}$ has the density of $\pu{5 g L-1}$ at STP. What is the atomic mass of $\ce{X}$ given $A_\mathrm{r}(\ce{O}) = 16?$

Note: The problem is my homework for school and has been translated from another language. It is probably either missing something or it contains wrong data. Just leaving it as a cautionary tale to anyone who might also have it.

I solved, or at least tried to, by expressing the given density $\rho = \pu{5 g L^-1}$ as

$$\rho = \frac{m}{V}.\tag{1}$$

Also,

$$m = Mn\tag{2}$$

$$V = V_\mathrm{m}n,\tag{3}$$

where $n$ is the amount of substance and $V_\mathrm{m} = \pu{22.4 L mol^-1}.$ So:

$$\rho = \frac{m}{V} = \frac{Mn}{V_\mathrm{m}n} = \frac{M}{V_\mathrm{m}}\tag{4}$$

$$\pu{5 g L^-1} = \frac{M}{\pu{22.4 L mol^-1}} \quad\Rightarrow\quad M = \pu{112 g mol^-1}\tag{5}$$

Then obviously by subtracting the atomic mass of oxygen: $$M_\mathrm{r} = 5A\mathrm{r}(\ce{O}) + 2A_\mathrm{r}(\ce{X}) \quad\Rightarrow\quad A_\mathrm{r}(\ce{X}) = 16\tag{6}$$

But that is not possible, as $16$ is the atomic mass of oxygen and there is no such thing as $\ce{O2O5}$ as far as I know. What am I doing wrong? Or is the problem wrong altogether?

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  • $\begingroup$ At STP, $V_\mathrm m$ is not $22.4\ \mathrm{l\ mol^{-1}}$; or maybe your school book is forty years old. $\endgroup$
    – Loong
    Mar 30, 2021 at 17:25
  • $\begingroup$ @Loong Then what is it? I mean that's what the textbook had and the professor told. Is there a more accurate measurement now? $\endgroup$
    – Gaurav Mall
    Mar 30, 2021 at 17:27
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    $\begingroup$ @Guarav At STP $V_\mathrm{m} = \pu{22.711 L mol^-1}.$ See Wikipedia. Still, that would result in molar mass of about $\pu{16.75 g mol^-1},$ so it would be nice if you could double-check the data; could the density be relative? $\endgroup$
    – andselisk ♦
    Mar 30, 2021 at 19:30
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    $\begingroup$ @andselisk Hey, thanks! Yeah, I actually researched it, since I didn't know anything about it and I found that since 1982, it's based on 1 bar and the value is $22.7$ approximately. Anyways, I talked with the teacher, and actually, the problem was wrong, posting a question about this might have been dumb, but thanks anyway! $\endgroup$
    – Gaurav Mall
    Mar 31, 2021 at 13:21

1 Answer 1

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Assuming it is given that $\ce{X2O5}$ behaves like an ideal gas, the best approach to do this problem is manipulating $pv = nRT$ to get the equation, which fits the given data. Suppose $m$ is the mass of the gas, $M$ is the molar mass of the gas, and $\rho$ is the density of the gas. Then $n = \frac{m}{M}$:

$$ pv = nRT = \frac{m}{M}RT \ \Rightarrow \ M = \frac{m}{v} \ \cdot \frac{RT}{p} = \rho \frac{RT}{p} \tag1$$

Now, you can apply $\rho = \pu{5 g L-1}$ at STP to the equation $(1)$ to find $M$. Then apply $2M_\ce{X} + 16 \times 5 = M$ to find $M_\ce{X}$, which is the atomic mass of $\ce{X}$.

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  • $\begingroup$ That seems like a really cool solution. Although unfortunately, we haven't done the ideal gas law yet. So, it really seems like there should be another solution. I'll look into it, if there isn't any other (cause maybe the problem was a bit unclear about something), I'll mark your answer as the accepted one. Thanks! $\endgroup$
    – Gaurav Mall
    Mar 30, 2021 at 17:18
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    $\begingroup$ Actually just to report: Using this equation and entering in the right quantities, it's the same answer I found with the way described above. So, I think it actually has to do with the fact that the problem was wrong. Thanks for your help, though :) $\endgroup$
    – Gaurav Mall
    Mar 30, 2021 at 17:33
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    $\begingroup$ Just FYI: $V_m = \pu{22.7 L mol-1}$ since 1982. Here standard pressure is considered $\pu{1 bar}$ (not $\pu{1 atm}$). $\endgroup$
    – Mathew Mahindaratne
    Mar 30, 2021 at 17:47
  • $\begingroup$ Oh okay, thanks! Guess my textbook is very old. Although I don't know, I'm still in the first grade of high school. Dunno it's weird. Thanks anyways! $\endgroup$
    – Gaurav Mall
    Mar 30, 2021 at 17:53
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    $\begingroup$ @Gaurav Mall: I agree with you about the question been wrong. Ask your teacher whether formula is really $\ce{XO5}$ or not. If it is it make sense, but still $\ce{SO5}$ is not possible. $\endgroup$
    – Mathew Mahindaratne
    Mar 30, 2021 at 17:57

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