Calculating atomic mass of unknown element in given compound (gas) at STP

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?

• At STP, $V_\mathrm m$ is not $22.4\ \mathrm{l\ mol^{-1}}$; or maybe your school book is forty years old. Mar 30 at 17:25
• @Loong Then what is it? I mean that's what the textbook had and the professor told. Is there a more accurate measurement now? Mar 30 at 17:27
• @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? Mar 30 at 19:30
• @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! Mar 31 at 13:21

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 $$\ce{X}$$.
• Just FYI: $V_m = \pu{22.7 L mol-1}$ since 1982. Here standard pressure is considered $\pu{1 bar}$ (not $\pu{1 atm}$). Mar 30 at 17:47
• @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. Mar 30 at 17:57