Assuming the atomic weight of a metal
M to be 56, find the empirical formula
of its oxide containing 70.0% of M.
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Atomic weight of M = 56
Percentage of M = 70.0 %
Since, the oxide contains 70.0 % of M, the amount of oxygen will be 30.0%
The atomic weight of Oxygen = 16
1) Moles of M
= % of M / Atomic mass of M
= 70.0 / 56
= 1.25
2) Moles of O
= % of O / Atomic mass of O
= 30.0 / 16
= 1.875
Hence, the ratio of the number of moles of the metal with oxygen would be,
• For M = 1.25 / 1.25 = 1
• For O = 1.875 / 1.25 = 1.5
i.e., 2 : 3
Hence, the empirical formula of the compound would be M₂O₃.
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