Determine the empirical formulas for the compounds with the following percent composition: 28.6% oxygen 57.6% strontium, 13.8% phosphorus, and 28.6% Sr_(3)P_(4)O_(5) Sr_(2)P_(2)O_(8) Sr_(3)P_(2)O_(8) Sr_(3)P_(3)O_(6) Multiple Choice 1 point

To determine the empirical formula for a compound with the given percent composition, we need to convert the percentages to grams and then to moles.Given:- 28.6% oxygen- 57.6% strontium- 13.8% phosphorusAssuming we have 100 grams of the compound, we can calculate the mass of each element:- Oxygen: 28.6 grams- Strontium: 57.6 grams- Phosphorus: 13.8 gramsNext, we convert the masses to moles using the molar masses of each element:- Oxygen (O): 16.00 g/mol- Strontium (Sr): 87.62 g/mol- Phosphorus (P): 30.97 g/molMoles of each element:- Oxygen: 28.6 g / 16.00 g/mol = 1.79 mol- Strontium: 57.6 g / 87.62 g/mol = 0.66 mol- Phosphorus: 13.8 g / 30.97 g/mol = 0.44 molNow, we need to find the simplest whole number ratio of moles of each element. We do this by dividing each mole value by the smallest mole value among them:- Oxygen: 1.79 mol / 0.44 mol = 4.07- Strontium: 0.66 mol / 0.44 mol = 1.50- Phosphorus: 0.44 mol / 0.44 mol = 1To get whole numbers, we can multiply all ratios by a common factor to get the smallest whole number ratio. In this case, multiplying by 2 gives us:- Oxygen: 4.07 * 2 = 8.14- Strontium: 1.50 * 2 = 3- Phosphorus: 1 * 2 = 2The empirical formula is therefore Sr3P2O8.So, the correct answer is: