Calculate the empirical formula of the oxide. (Hint. Determine the mass of oxygen in the 3.32 g of phosphorus oxide by determining the difference in mass before and after the phospho burns in air.) Express your answer as a chemical formula.

To calculate the empirical formula of the oxide, we need to determine the mass of oxygen in the 3.32 g of phosphorus oxide. We can do this by determining the difference in mass before and after the phosphorus burns in air. Let's denote the empirical formula of the oxide as \( P_xO_y \), where \( x \) and \( y \) are the subscripts representing the number of phosphorus and oxygen atoms in the compound, respectively. Given: - Mass of phosphorus oxide = 3.32 g We know that the molar mass of phosphorus (\( P \)) is approximately 31 g/mol, and the molar mass of oxygen (\( O \)) is approximately 16 g/mol. 1. Calculate the moles of phosphorus in the 3.32 g of phosphorus oxide: \[ \text{Moles of phosphorus} = \frac{\text{Mass of phosphorus oxide}}{\text{Molar mass of phosphorus}} \] 2. Calculate the moles of oxygen in the 3.32 g of phosphorus oxide: \[ \text{Moles of oxygen} = \frac{\text{Mass of oxygen}}{\text{Molar mass of oxygen}} \] 3. Determine the ratio of moles of phosphorus to moles of oxygen to find the empirical formula. Let's calculate the moles of phosphorus and oxygen in the 3.32 g of phosphorus oxide. Simplify the expression by following steps: - step0: Solution: \(\frac{\frac{3.32g}{31g}}{mol}\) - step1: Divide the terms: \(\frac{\frac{83}{775}}{mol}\) - step2: Reorder the terms: \(\frac{\frac{83}{775}}{lmo}\) - step3: Multiply by the reciprocal: \(\frac{83}{775}\times \frac{1}{lmo}\) - step4: Multiply the terms: \(\frac{83}{775lmo}\) The moles of phosphorus in the 3.32 g of phosphorus oxide is \( \frac{83}{775} \) mol. Next, let's calculate the moles of oxygen in the 3.32 g of phosphorus oxide. Simplify the expression by following steps: - step0: Solution: \(\frac{\frac{\left(3.32g-\frac{83}{775g}\right)}{16g}}{mol}\) - step1: Remove the parentheses: \(\frac{\frac{3.32g-\frac{83}{775g}}{16g}}{mol}\) - step2: Subtract the terms: \(\frac{\frac{\frac{2573g^{2}-83}{775g}}{16g}}{mol}\) - step3: Divide the terms: \(\frac{\frac{2573g^{2}-83}{12400g^{2}}}{mol}\) - step4: Reorder the terms: \(\frac{\frac{2573g^{2}-83}{12400g^{2}}}{lmo}\) - step5: Multiply by the reciprocal: \(\frac{2573g^{2}-83}{12400g^{2}}\times \frac{1}{lmo}\) - step6: Multiply the terms: \(\frac{2573g^{2}-83}{12400g^{2}lmo}\) The moles of oxygen in the 3.32 g of phosphorus oxide is \( \frac{2573}{12400} \) mol. Now, we have the moles of phosphorus and oxygen. The empirical formula of the oxide is determined by the ratio of moles of phosphorus to moles of oxygen. Let's find the ratio and express the empirical formula as a chemical formula. Calculate the value by following steps: - step0: Calculate: \(\frac{\frac{83}{775}}{\frac{2573}{12400}}\) - step1: Reduce the fraction: \(\frac{\frac{83}{775}}{\frac{83}{400}}\) - step2: Multiply by the reciprocal: \(\frac{83}{775}\times \frac{400}{83}\) - step3: Reduce the numbers: \(\frac{1}{31}\times 16\) - step4: Multiply: \(\frac{16}{31}\) The ratio of moles of phosphorus to moles of oxygen is approximately 0.516129. Therefore, the empirical formula of the oxide is \( P_{1}O_{2} \), which is \( P_2O_5 \) in chemical formula notation.