Let x be the wight of the pure Ag2O in the 1.60 sample.
The reaction is:
2Ag2O => 4Ag + O2
x/232 moles
moles of O2 would be => x/(232*2)
Weight of O2 => (x/464)*32
Hence, given is the mass of O2 => 0.0689x = 0.104 => x = 1.509 gms
Hence the % by mass of silver oxide is:
=> (1.509/1.60)*100
=> 94.33 %
This is the percentage of pure Ag2O in impure sample.