The empirical formula of the oxide containing 60 g of M and 24 g of oxygen is MO₃
Divide by their molar mass
M = 60 / 120 = 0.5
O = 24 / 16 = 1.5
Divide by the smallest
M = 0.5 /0.5 = 1
O = 1.5 / 0.5 = 3
Thus, the empirical formula of the oxide is MO₃
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