Question:
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)
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)
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Empirical formula steps:
- Convert to moles
- Divide by smallest
- Adjust to nearest whole number ratio
Updated On: Apr 30, 2026
- $\mathrm{M_3O_2}$
- $\mathrm{MO_2}$
- $\mathrm{M_2O_3}$
- $\mathrm{M_2O_5}$
- $\mathrm{M_2O}$
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The Correct Option is C
Solution and Explanation
Concept:
Empirical formula is obtained from simplest whole number ratio of moles.
Step 1: Convert percentages to moles.
\[ \text{Moles of M} = \frac{54}{27} = 2 \] \[ \text{Moles of O} = \frac{46}{16} = 2.875 \]
Step 2: Find simplest ratio.
\[ 2 : 2.875 \] Divide by $0.125$: \[ 16 : 23 \Rightarrow \text{approx} \; 2 : 3 \]
Step 3: Write empirical formula.
\[ \mathrm{M_2O_3} \]
Step 1: Convert percentages to moles.
\[ \text{Moles of M} = \frac{54}{27} = 2 \] \[ \text{Moles of O} = \frac{46}{16} = 2.875 \]
Step 2: Find simplest ratio.
\[ 2 : 2.875 \] Divide by $0.125$: \[ 16 : 23 \Rightarrow \text{approx} \; 2 : 3 \]
Step 3: Write empirical formula.
\[ \mathrm{M_2O_3} \]
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