Discussion
The empirical formula of a metal oxide which has 54 % metal (M) and 46 % oxygen (O) is (Atomic mass of M = 27 amu and O= 16 amu)
Answer: C
The empirical formula is M2O3.
Divide each percentage by the atomic mass to get the mole ratio:
\[ \begin{aligned} \mathrm{M} &: \frac{54}{27} = 2 \\ \mathrm{O} &: \frac{46}{16} = 2.875 \end{aligned} \]
The ratio M : O is 2 : 2.875, or 1 : 1.44, which is closest to 1 : 1.5, that is 2 : 3. (A metal of atomic mass 27 is aluminium, and Al2O3 contains about 53 % metal, matching the data.)
M2O and MO2 give ratios of 1 : 0.5 and 1 : 2, and M3O2 and M2O5 are further from 1 : 1.44.
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