A compound of sodium oxide has 74.2% of sodium and 25.8% of Oxygen. The empirical formula of the compound is?
Correct answer: C. Na2O
- A. Nao
- B. NaO2
- C. Na2O
- D. Na2O2
Explanation
Let's find the empirical formula step-by-step 👇1️⃣ Assume 100 g of the compound → Na = 74.2 g, O = 25.8 g2️⃣ Convert to moles: Na = 74.2 ÷ 23 = 3.23 mol O = 25.8 ÷ 16 = 1.61 mol3️⃣ Find the simplest ratio: Na : O = 3.23 ÷ 1.61 : 1.61 ÷ 1.61 = 2 : 1✅ Empirical formula = Na₂O
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Mole calculations connect mass, number of particles and Avogadro's number, while balanced equations provide the mole ratios used in stoichiometry. Questions cover limiting and excess reactants, theoretical yield and percentage yield, including identifying which reactant is consumed first and comparing the actual product with the maximum possible product.
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