The empirical formula of the oxide-containing 60 g of M and 24 g of oxygen is MO₃.
A chemical formula showing the simplest ratio of elements in a compound rather than the total number of atoms in the molecule [tex]CH_2O[/tex] is the empirical formula for glucose.
Data obtained from the question
M = 60 g
O = 24 g
Empirical formula =?
How to determine the empirical formula
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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