# Derivation of formula used in calculating molecular mass from a mixture

In a solution to a problem, to calculate Molecular mass of unknown gas from a mixture containing $\ce{O2}$ in $80~\%$, the author used the formula :

$\frac{1}{\sqrt{M_\text{mix}}}=\frac{X_{\ce{O2}}}{\sqrt{M_{\ce{O2}}}}+\frac{X_\text{gas}}{\sqrt{M_\text{gas}}}$.

The exact problem is:

Pure $\ce{O_2}$ diffuses through an aperture in $224\mathrm{s}$, where as mixture of $\ce{O_2}$ and another gas containing $80~\%\ \ce{O_2}$ diffuses from the same in $234\mathrm{s}$. The molecular mass of gas will be?

Solution by author:

\begin{align}\frac{t_\text{mix}}{t_{\ce{O2}}} &= \frac{r_{\ce{O2}}}{r_\text{mix}} = \sqrt{\frac{M_\text{mix}}{32}}\\[1em] \frac{234}{224} &= \sqrt{\frac{M_\text{mix}}{32}}\\[0.5em] M_\text{mix} &= 34.92\\[1em] \Longrightarrow \qquad \frac{1}{\sqrt{M_\text{mix}}} &= \frac{X_\text{gas}}{\sqrt{M_\text{gas}}} + \frac{X_{\ce{O2}}}{\sqrt{M_{\ce{O2}}}}\\[1em] \Longrightarrow \qquad \frac{1}{\sqrt{34.92}} &= \frac{0.2}{\sqrt{M_\text{gas}}} + \frac{0.8}{\sqrt{32}}\\[0.5em] M_\text{gas} &= 51.5\end{align}

I wish to know why this works here?

Hint: $$v_{rms}=\sqrt{\frac{3RT}M}$$

And weighted average of speeds for mixture.

EDIT:

$$v_{mix} ^{rms}= x_1 v_1^{rms}+ x_2 v_2^{rms}$$

This will complete the derivation of the formula you gave.

• This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review Commented Aug 29, 2017 at 6:07
• Actually there is just one step to the derivation after you write down both the equations.. I guess OP can solve that on his own. I think spoon feeding complete derivation won't do any good.
– ABC
Commented Aug 29, 2017 at 6:30
• Hint answers aren't really well-received here. If you want to give a hint, it's better to do it as a comment. Commented Aug 29, 2017 at 6:31