Given a mixture of $\ce{He}$ and $\ce{O2}$ with density $\pu{0.47 g/L}$, temperature $\pu{298.15 K}$ and pressure $\pu{0.948 atm}$. What is the weight % of $\ce{He}$?
I get stuck in the formula
$$\frac{m(\ce{He})}{m_\mathrm{tot}} = \frac{p(\ce{He})}{p_\mathrm{tot}} \cdot \frac{4}{32}$$
I don't seem to have enough information to calculate $p(\ce{He})$.
There is no lack of data. Keep in mind that usually you are allowed to approximate a given gas mixture using ideal gas law:
$$pV = nRT = \frac{mRT}{M} \tag{1}\label{eq:1}$$
On the other hand, density by definition is
$$\rho = \frac{m}{V} \tag{2}\label{eq:2}$$
Combining \eqref{eq:1} and \eqref{eq:2}, and applying the equation to a binary mixture of gases ($\bar{\rho}$ and $\bar{M}$ are average density and average molecular weight, respectively):
$$\bar{\rho} = \frac{m_\mathrm{tot}}{V_\mathrm{tot}} = \frac{p_\mathrm{tot} \bar{M}}{RT} \tag{3}\label{eq:3}$$
From \eqref{eq:3} one can determine average molecular weight of the mixture. Remember to use SI units: $\bar{\rho} = \pu{0.47 g L-1} = \pu{0.47 kg m-3}$; $p_\mathrm{tot} = \pu{0.948 atm} = \pu{96056 Pa}$:
\begin{align} \bar{M} &= \frac{\bar{\rho} RT}{p_\mathrm{tot}} \\ &= \frac{\pu{0.74 kg m-3} \cdot \pu{8.314 J mol-1 K-1} \cdot \pu{298.15 K}}{\pu{96056 Pa}} \\ &= \pu{1.213e-2 kg mol-1} \\ &= \pu{12.13 g mol-1} \tag{4}\label{eq:4} \end{align}
At the same time average molecular weight of a binary mixture is
$$\bar{M} = \omega_1 M_1 + (1 - \omega_1)M_2 \tag{5}\label{eq:5}$$
Equating \eqref{eq:4} and \eqref{eq:5}, and solving the obtained equation for $\omega_1$ gives the answer ($M_1 = M(\ce{He}) = \pu{4.00 g mol-1}$; $M_2 = M(\ce{O2}) = \pu{31.99 g mol-1}$).
$$4.00\omega_1 + 31.99 - 31.99\omega_1 = 12.13 \implies \omega_1 = 0.71\, \text{or}\,71\%$$
$$a=b\tag{a}\label{eq:a}$$ [...] $\eqref{eq:a}$ Only automatic enumeration does not work. $$a=b\tag{a}\label{eq:a}$$ See equation $\eqref{eq:a}$.
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