17. (Ch1) A 50 g sample of a compound is analyzed and found to contain 21.85 g of phosphorus and 28.15 g of oxygen.The empirical formula for this compound is a. PO_(3) b. P_(2)O C. P_(2)O_(5) d. P_(3)O_(2) e. none of these

To find the empirical formula, we need to determine the mole ratio of phosphorus to oxygen in the compound.First, let's calculate the moles of phosphorus and oxygen in the sample:Moles of phosphorus = mass of phosphorus / molar mass of phosphorusMoles of phosphorus = 21.85 g / 30.97 g/mol (molar mass of phosphorus)Moles of phosphorus ≈ 0.705 molesMoles of oxygen = mass of oxygen / molar mass of oxygenMoles of oxygen = 28.15 g / 16.00 g/mol (molar mass of oxygen)Moles of oxygen ≈ 1.765 molesNext, we need to find the mole ratio of phosphorus to oxygen:Mole ratio of phosphorus to oxygen = moles of phosphorus / moles of oxygenMole ratio of phosphorus to oxygen ≈ 0.705 / 1.765Mole ratio of phosphorus to oxygen ≈ 0.4To simplify the mole ratio, we can divide both sides by 0.4:Mole ratio of phosphorus to oxygen ≈ 0.4 / 0.4Mole ratio of phosphorus to oxygen ≈ 1Therefore, the empirical formula for this compound is .So, the correct answer is e. none of these.