A sure metal, M forms two oxides M2O and MO.?

If the percent by mass of M in M2O is 70.0 % what is the percent by mass of M in MO ?
Answers:
If in M2O, M is 70% then by difference O = 30%

So clearly %M surrounded by MO = 35/(35+30) = 53.8%
2*M=70/100
M=35/100
O=30/100

M/MO=35/(35+30)=53%
The percentage of Oxygen is 30%, use this...
Oxygen (O) is roughly 16 g/mol.
16/(x+16)=0.3
4.8+0.3X=16, 11.2=.3x, x=37.33
x=2M, M=37.33/2=18.67

So Percent mass of M within MO is 18.67/(18.67+ 16)=18.67/34.67=0.5385=53.85%

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