The empirical formula of a compound containing 0.067mol Cu and 0.066mol O is [Cu = 63.5, O = 16]
- \(Cu_{2}O\)
-
\(CuO\)
- \(CuO_{2}\)
- \(CuO_{4}\)
The correct answer is: B
Explanation
Cu : O
0.067 : 0.066
mole ratio \(\frac{0.067}{0.066}\) : \(\frac{0.066}{0.066}\)
= 1 : 1
= CuO