chem實驗式(急~~)

There are 4 possibilities for the metal to form an oxide, depending on the valency of the metal.
(1)M2O, (valency of the metal is one, in Group 1)
2M2O = 4M + O2
(2)MO, (valency of the metal is two, in Group II)
2MO = 2M + O2
(3)M2O3, (valency of the metal is three, in Group III)
2M2O3 = 4M + 3O2
(4)MO2, (valency of the metal is four, in Group IV)
MO2 = M + O2

First, we consider case (1) where a metallic oxide M2O is formed (valency of M is 1)
2M2O = 4M + O2
Since the atomic mass of the metal is given = 107.9,
Atomic mass of oxygen is 16 (From periodic table of elements or any chem. textbook)
Molecular weight of M2O = (2 x 107.9 + 16) = 231.8
2M2O = 4M + O2
2(231.8) = 4(107.9) + 2 x (16)
463.6 = 431.6 + 32
463.6 g of M2O decomposes into 431.6 g of M and 32 g of oxygen
The product, oxygen, is a gas which escapes into the atmosphere, leaving 431.6 g of metal M
We try to find out percentage loss of oxygen with reference to the original metallic oxide

32
-------- x 100% = 6.9%
463.6

The answer is 6.9 % which agrees with the given percentage loss, so we know that empirical formula is M2O (two atoms of M combine with one atom of oxygen to form one molecule of M2O. )

If the answer is not 6.9% , you have to consider case 2, case 3 and case 4 in that order until the correct answer comes out.

CORRECT ANSWER to the problem is M2O

===================================================
Actually the metal is silver, Ag, with atomic mass = 107.9
2Ag2O = 4Ag + O2

I am sorry that the solution to your problem is in English as I cannot type in Chinese on the computer. I also cannot put the 2 into subscript as in M2O.