A 1.00 L flask holds a mixture of methane, CH₄ (16.04 g/mol), and ethane, C₂H₆ (30.07 g/mol)…

Step 1. The total amount. $n = \dfrac{PV}{RT} = \dfrac{0.98 \times 1.00}{0.08314 \times 300} = 0.03929\ \text{mol}$.

Step 2. The average molar mass. $\bar{M} = \dfrac{0.851\ \text{g}}{0.03929\ \text{mol}} = 21.66\ \text{g/mol}$.

Step 3. The mole fraction. We want $x$, the mole fraction of methane. From $\bar{M} = 16.04x + 30.07(1 - x)$ we get $x = \dfrac{30.07 - 21.66}{30.07 - 16.04} = 0.600$.

Step 4. The partial pressures. $P_{\text{CH}_4} = 0.600 \times 0.98 = 0.588\ \text{bar}$, and $P_{\text{C}_2\text{H}_6} = 0.98 - 0.588 = 0.392\ \text{bar}$.

Step 5. The mass percent. Methane contributes $0.600 \times 16.04 = 9.62\ \text{g}$ of every $21.66\ \text{g}$ of a mole of mixture, which is $44.4\%$.