How to find the molecular weight of a given gas based on its density?

Question

The density of a gas $\ce{X}$ is $10$ times that of hydrogen. In that case, what is the molecular weight of gas $\ce{X}$?

Well what I've done up to now is this:

The molecular mass of hydrogen is $M(\ce{H2})=2$, the density of hydrogen is $\rho(\ce{H2})= \pu{0.089 kg/m3}$. Assuming that equal volumes $V$ of both gases are taken, the mass of hydrogen $m(\ce{H2})= (0.089 \cdot v) \pu{kg}$ and the mass of the unknown gas $m(\ce{X}) = (0.89 \cdot v) \pu{kg}$.

Although this hasn't really gotten me anywhere. How will it be done?

Answer 1

Indeed you can use Avogadro's law here, but I guess it is useful to give a little bit more insight in Avogadro's Law. What you can do is apply the ideal gas law:

$$p V = n R T $$

Here $n$ is the amount of substance of gas, which you can also write as: $$ n=\frac{m}{M},$$ where $m$ is the total mass of the gas and $M$ is the molecular mass. You can rewrite the ideal gas law now to: $$p M = \frac{m}{V} R T,$$ where you should recognize that $\frac{m}{V}=\rho$.

The gas constant $R$ is (as the name shows) a constant so you can write \begin{align} R &=\frac{p M}{\rho T} & \rightarrow && \frac{p_1 M_1}{\rho_1 T_1}&=\frac{p_2 M_2}{\rho_2 T_2} \end{align}

And therefore, for equal pressure and temperature you will find that $$\frac{M_1}{\rho_1}=\frac{M_2}{\rho_2},$$ which means that if the density of gas 2 (the unknown gas $\ce{X}$) is $10$ times as high, the molecular mass should be as well.

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What is interesting to add is that this assumes the ideal gas law to be applicable. If, for example, your gases would be at high pressure (e.g. $\pu{200 bar}$) this is typically not a good assumption so the molecular mass that you would find for a $10$ times as dense gas will not be equal to $10$ times the molecular mass of hydrogen.

Answer 2

Assuming equal volume $V$, equal temperature $T$, and equal pressure conditions $p$, we can apply Avogadro's Law: An equal volume of two gasses contain an equal number of molecules at the same temperature and pressure.

Now, you can proceed:

\begin{align} \rho &= \frac{\text{Mass}}{\text{Volume}}\\ &= \frac{n\times M (\text{molar mass})}{\text{Volume}}\\ \implies\frac{\rho_1}{\rho_2}&=\frac{M_1}{M_2}\\ \implies M_{\ce{X}}&=10\times M_{\ce{H2}} \end{align}