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