assume that atomic weight of a metal 'M' to be 56,find the empirical formula of its oxide containing 70% of M
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275
atomic weight of metal M = 56
in an oxide, metal is 70% by mass and so oxygen is 30% by mass.
let there be n atoms of oxygen per 1 atom of metal.
Then proportion of the masses of metal and oxygen :
56 : n * 16 = 70 : 30
16 n = 56 * 30 / 70
n = 24/16 = 1.5
There are two atoms of metal for 3 atoms of Oxygen.: M2 O3
in an oxide, metal is 70% by mass and so oxygen is 30% by mass.
let there be n atoms of oxygen per 1 atom of metal.
Then proportion of the masses of metal and oxygen :
56 : n * 16 = 70 : 30
16 n = 56 * 30 / 70
n = 24/16 = 1.5
There are two atoms of metal for 3 atoms of Oxygen.: M2 O3
Answered by
100
Ans: M2O3
Ratio of moles of metal and oxide:
M : O
70/56 : 30/16
1 : 1.5
2 : 3
Hence formula is M2O3
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