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21. What is the empirical formula of a compound which is \( 57.19 \% \) sulfur and \( 42.81 \% \) oxygen?

Ask by Chavez Peterson. in the United States
Feb 25,2025

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The empirical formula of the compound is S₂O₃.

Solución

To determine the empirical formula, assume a 100 g sample of the compound. Then: • Mass of sulfur (S) = 57.19 g • Mass of oxygen (O) = 42.81 g Next, calculate moles for each element. For sulfur (atomic mass ≈ 32.06 g/mol): Moles of S = 57.19 g / 32.06 g/mol ≈ 1.78 moles For oxygen (atomic mass = 16.00 g/mol): Moles of O = 42.81 g / 16.00 g/mol ≈ 2.67 moles Now, determine the simplest mole ratio by dividing each value by the smallest number of moles (1.78 moles): • Ratio for S = 1.78 / 1.78 = 1 • Ratio for O = 2.67 / 1.78 ≈ 1.5 Since the ratio for oxygen is 1.5 (which is not a whole number), multiply both ratios by 2 to get whole numbers: • Sulfur: 1 × 2 = 2 • Oxygen: 1.5 × 2 = 3 Thus, the simplest (empirical) formula is S₂O₃.

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Beyond the Answer

To determine the empirical formula from the percentage composition, first convert the percentages to grams (assuming 100 grams of the compound gives us 57.19 g of sulfur and 42.81 g of oxygen). Next, convert grams to moles using the molar masses (sulfur = 32.07 g/mol, oxygen = 16.00 g/mol): - Moles of sulfur: \( \frac{57.19 \, \text{g}}{32.07 \, \text{g/mol}} \approx 1.78 \, \text{mol} \) - Moles of oxygen: \( \frac{42.81 \, \text{g}}{16.00 \, \text{g/mol}} \approx 2.68 \, \text{mol} \) Now, divide both by the smaller number of moles (1.78) to find the simplest ratio: - Sulfur: \( \frac{1.78}{1.78} = 1 \) - Oxygen: \( \frac{2.68}{1.78} \approx 1.50 \) To convert this ratio to whole numbers, multiply by 2 to get: - Sulfur: \( 1 \times 2 = 2 \) - Oxygen: \( 1.50 \times 2 = 3 \) Thus, the empirical formula is \( \text{S}_2\text{O}_3 \).

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