Concept:To find the empirical formula from mass percentages, convert the percentages to grams (assuming a 100g sample), divide by the respective atomic masses to find moles, and establish the simplest whole-number ratio.
Formula:$$ \text{Moles} = \frac{\text{Percentage mass}}{\text{Atomic mass}} $$
Solution:- For Sodium (Na): Mass = 74.2g. Atomic mass = 23 g/mol.
- Moles of Na = \( 74.2 / 23 = 3.22 \text{ moles} \).
- For Oxygen (O): Mass = 25.8g. Atomic mass = 16 g/mol.
- Moles of O = \( 25.8 / 16 = 1.61 \text{ moles} \).
- Divide both mole values by the smallest mole value (1.61) to find the ratio.
- Na ratio = \( 3.22 / 1.61 = 2 \).
- O ratio = \( 1.61 / 1.61 = 1 \).
- The simplest whole-number ratio is 2 Na for every 1 O, yielding the empirical formula \( \text{Na}_2\text{O} \).
Why other options are incorrect:- NaO (1:1), \( \text{NaO}_2 \) (1:2), \( \text{Na}_2\text{O}_2 \) (1:1): None of these match the calculated mathematically simplified 2:1 stoichiometric ratio.
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