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A certain metal M forms two oxides M3O and M2O. If the percent by mass of M in M3O is 78.0 what is the percent by mass of M in M2O?
Wiki User
∙ 12y ago
There is a confusing part in this question: Do you mean M2O and M3O or MO2 and MO3? I presume you mean the latter (I.e. M Dioxide and M Trioxide). Because I can't think of any chemicals that form 3M to 1O.
A little algebra will help you.
- Let M be the mass of the metal.
- Let O be the mass of Oxygen.
therefore:
m+3O=total mass
since M is 78% of the total mass,
M/(M+3O)=0.78
Since the otomic mass of Oxygen is 16,
M/(M+3*16)=0.78
multiplying out the denominator,
M=0.78M+37.44
deducting .78M,
0.22M=37.44
So therefore,
M=37.44/0.22
M=170.181818
In terms of the uncertainty, this would make sense, as Thulium, with a mass of 169 forms a 4+ and 6+ oxidation state. Therefore in the Dioxide M2O
total mass=170.181818+2*16
=202.181818
then mass of M in this is equal to:
170.181818/202.181818
Which is equal to 0.84173 or 84%.
If it is M2O and M3O the same procedure can be used, but tou will have to re-work it.
Wiki User
∙ 12y ago
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