Given a sample containing 3.72 g of phosphorus (P) and 21.28 g of chlorine (Cl), what is the empirical formula of the compound formed?

• A. PCl2
• B. P2Cl5
• C. PCl5
• D. PCl3

Answer: (C)

Verified step by step guidance

1. Step 1: Determine the number of moles of each element by dividing the given mass by the atomic mass. Use the atomic masses: phosphorus (P) = 30.97 g/mol, chlorine (Cl) = 35.45 g/mol. Calculate moles of P as \(\frac{3.72}{30.97}\) and moles of Cl as \(\frac{21.28}{35.45}\).
2. Step 2: Find the mole ratio of phosphorus to chlorine by dividing both mole values by the smaller number of moles obtained in Step 1. This will give a simplified ratio of atoms in the compound.
3. Step 3: If the mole ratio is not a whole number, multiply both ratios by the smallest integer that converts them into whole numbers. This step ensures the empirical formula has whole number subscripts.
4. Step 4: Write the empirical formula using the whole number mole ratios as subscripts for each element. For example, if the ratio is 1:5, the formula is \(\mathrm{PCl_5}\).
5. Step 5: Verify the empirical formula by checking that the mole ratio corresponds to the given masses and that the formula matches one of the provided options.