Equal masses of methane and oxygen are mixed in an empty container at $25^{\circ }\text{C}$. The fraction of the total pressure exerted by oxygen is:

Let the mass of each gas be $w$ grams.Molar mass of ${\text{CH}}_4=16\text{g/mol}$.Molar mass of ${\text{O}}_2=32\text{g/mol}$.Number of moles:
$$n_{{\text{CH}}_4}=\frac{w}{16}$$
$$n_{{\text{O}}_2}=\frac{w}{32}$$
Mole fraction of ${\text{O}}_2$ ($X_{{\text{O}}_2}$):
$$X_{{\text{O}}_2}=\frac{n_{{\text{O}}_2}}{n_{{\text{O}}_2}+n_{{\text{CH}}_4}}=\frac{\frac{w}{32}}{\frac{w}{32}+\frac{w}{16}}$$
Divide by $w$:
$$X_{{\text{O}}_2}=\frac{1/32}{1/32+2/32}=\frac{1/32}{3/32}=\frac{1}{3}$$
According to Dalton's law, the partial pressure of a gas is equal to its mole fraction times the total pressure ($P_i=X_i{P}_{total}$).The fraction of total pressure exerted by ${\text{O}}_2$ is $1/3$.