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Question
Solve the following problem:
Assuming the atomic weight of a metal M to be 56, find the empirical formula of its oxide containing 70.0% of M.
Solution
Given: Atomic mass of M = 56, Percentage of M = 70.0%
To find: The empirical formula of the compound
Calculation: %M = 70.0%
Hence, %O = 30.0%, Atomic mass of O = 16 u
Moles of M = `"% of M"/"Atomic mass of M"=70.0/56` = 1.25 mol
Moles of O = `"% of O"/"Atomic mass of O"=30.0/16` = 1.875 mol
Hence the ratio of number of moles of M : O is
`1.25/1.25` = 1 and `1.875/1.25` = 1.5
Convert the ratio into a whole number by multiplying by the suitable coefficient, i.e., 2.
Therefore, the ratio of the number of moles of M : O is 2 : 3.
Hence, the empirical formula is M2O3.
Empirical formula of the compound = M2O3
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