What is the percent of O in
Cr203?
(Cr = 52.00 amu, O = 16.00 amu)
[?]%

Answer:
31.6%
Explanation:
find the molecular formular
(52×2)+(16×3)=152
since the %of oxygen is required
MF of o in Cr2O3 =16×3=48
%O= 48/152×100= 31.6%