Answer:
20.11% of oxygen
Explanation:
The reaction is:
2CuO ⇒ 2Cu + O2
If you have an initial amoung of a sample, thi may be have impurities, so you have to start from products to know what you had at the begining.
to make a relation between reagents and products you need to change the mass to mol, then you can relate them with the molar ratio.
[tex]1,098gCu x \frac{1mol Cu}{63.546g Cu} x \frac{2mol CuO}{2molCu} x \frac{1mol O}{1molCuO} x \frac{16gO}{1molO} = 0,276gO[/tex]
Now you have the initial amount of oxygen, to convert to percent you must divide the mass of oxygen between the mass of the sample and multiply by 100
[tex]percent composition of O = \frac{0.276gO}{1.375g sample} *100 = 20.11[/tex]
