An oxide of metal has 47% by mass of oxygen.Metal M has atomic mass of 24. What is the empirical formula of the oxide?
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The formulae of the oxide is :
M x O y
Let X be moles of M and y be moles of Oxygen.
Mass of oxygen is 0.47g
Mass of M is 0.53g
Moles of M = 0.53 / 24 = 0.02
Moles of Oxygen = 0.47 / 16 = 0.03
Mole ratio :
M : O = 0.02 : 0.03
= 2:3
The empirical formula of the oxide is thus :
M₂O₃
M x O y
Let X be moles of M and y be moles of Oxygen.
Mass of oxygen is 0.47g
Mass of M is 0.53g
Moles of M = 0.53 / 24 = 0.02
Moles of Oxygen = 0.47 / 16 = 0.03
Mole ratio :
M : O = 0.02 : 0.03
= 2:3
The empirical formula of the oxide is thus :
M₂O₃
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