If 60 g of M combines with 24 g of oxygen, what would the empirical formula of the oxide be?
( molar mass of M = 120, O = 16)
- MO
- MO\(_{2}\)
-
MO\(_{3}\)
- MO\(_{4}\)
The correct answer is: C
Explanation
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