Two oxides of a metal contain `27.6%` and `30.0%` of Oxygen, respecttively. If the formula of the first be `M_(3) O_(4)`. Find that of the second.

Explanation

`{: ("Frist oxide", "Second oxide"), ("Oxygen" = 27.6%, "Oxygen" = 30.0%),("Metal" = 72.4%, "Metal" = 70.0%), ("Formula" = M_(3) O_(4), "Formula" = ?):}`
Let atomic weight the metal be `x`.
Therefore, percentage by weight of the metal in the compound `M_(3) O_(4)`
`= (3 xx x xx 100)/(3 x + 64) = 7.24` (given)
Hence, `x = 55.97 ~~ 56`.
Thus having known the atomic weights of the metal and oxygen, empirical formula of the second oxide can be worked out as under"
`{: ("Element", %, {:("Atomic"),("ratio"):}, {:("Least"),("ratio"):}, {:("Whole"),("number"),("ratio"):}),("Oxygen", 30.00, (30.00)/(16) = 1.88, (1.88)/(1.25) = 1.5, 3), ("Metal", 70.00, (70.00)/(56) = 1.25, (1.25)/(1.25) = 1, 2):}`
Thus, the formula of the second oxide is `M_(2) O_(3)`.