# What is the partial pressure of oxygen in this mixture of oxygen and argon?

A $19.3\:\mathrm{g}$ mixture of oxygen and argon is found to occupy a volume of $16.2\:\mathrm{L}$ when measured at $675.9\:\mathrm{mmHg}$ and $43.4\:\mathrm{^\circ{}C}$. What is the partial pressure of oxygen in this mixture?

The answer is $436\:\mathrm{mmHg}$, but I didn't manage to get to it.

I know that it is found by this formula. $P_i=X_i\cdot P_\text{total}$, where $P_i$ is the partial pressure and $X_i$ is the mole fraction.

$P_\text{total}$ is given to find $X_i$. I should divide the mole of $\ce{O2}$ by the total number of mole of mixture $n_{\ce{O2}}/n_\text{total}$.

$n_\text{total}$ is easy to find with $n=PV/RT$, but my problem is, that I can't find the number of moles of $\ce{O2}$. The mass in the problem is given for the mixture, so how to extract only the mass of $\ce{O2}$, or in other words, how to get the number of moles for $\ce{O2}$?

How is the mixture composed? Find expressions for mass, volume and moles.

$V_\text{total} = V_{\ce{O2}} + V_{\ce{Ar}}$
$n_\text{total} = n_{\ce{O2}} + n_{\ce{Ar}}$
$m_\text{total} = m_{\ce{O2}} + m_{\ce{Ar}}$

Transform $V_\text{total}$ with ideal gas and solve for $n_{\ce{Ar}}$

$V_\text{total} = \frac{\mathcal{R}T}{p_\text{total}}(n_{\ce{O2}} + n_{\ce{Ar}})$
$n_{\ce{Ar}} = \frac{p_\text{total}V_\text{total}}{\mathcal{R}T} - n_{\ce{O2}}$

Involve total mass and plug in previous equation and transform with molar mass to moles equation

$m_\text{total} = m_{\ce{O2}} + m_{\ce{Ar}}$
$m_\text{total} = n_{\ce{O2}}M_{\ce{O2}} + n_{\ce{Ar}}M_{\ce{Ar}}$
$m_\text{total} = n_{\ce{O2}}M_{\ce{O2}} + \left(\frac{p_\text{total}V_\text{total}}{\mathcal{R}T} - n_{\ce{O2}}\right)M_{\ce{Ar}}$

Now there is only one variable left, which you do not know.

$m_\text{total} = n_{\ce{O2}}\left(M_{\ce{O2}} - M_{\ce{Ar}}\right) + \frac{p_\text{total}V_\text{total}}{\mathcal{R}T}\cdot M_{\ce{Ar}}$
$n_{\ce{O2}} = \left(m_\text{total} - \frac{p_\text{total}V_\text{total}}{\mathcal{R}T} \cdot M_{\ce{Ar}}\right)\cdot \left(M_{\ce{O2}} - M_{\ce{Ar}}\right)^{-1}$

You have found the total number of moles, $n_{\rm{total}}$, and you want the number of moles of oxygen, $n_{\ce{O_2}}$. Write and solve a first-degree equation relating $n_{\ce{O_2}}$ to the total mass, 19.3$\,$g. Remember that $n_{\ce{Ar}}=n_{\rm{total}}-n_{\ce{O_2}}$.

Another visualization:

Assume $19.3$ g of pure oxygen; use the Ideal Gas law to find the volume $V_1$

Assume $19.3$ g of pure argon; use the Ideal Gas law to find the volume $V_2$, smaller than $V_1$ since argon is heavier than oxygen

Hopefully, the actual volume will be somewhere in between $V_1$ and $V_2$.

How far along the gap from $V_2$ to $V_1$ is the actual volume?

That's how far you are from zero oxygen to 100% oxygen,

With this fraction and the total pressure, the partial pressure is simple,,,