What is the molecular formula of a compound with a molar mass of 219.9 g/mol and empirical formula of P2O3?
The Correct Answer and Explanation is:
Correct Answer: P4O6
To determine the molecular formula of a compound, you need both the empirical formula and the molar mass of the compound. The empirical formula represents the simplest whole-number ratio of atoms in a compound. In this case, the empirical formula is P₂O₃, which means there are two phosphorus atoms and three oxygen atoms in the simplest ratio.
Step 1: Calculate the molar mass of the empirical formula (P₂O₃)
- Atomic mass of phosphorus (P) ≈ 30.97 g/mol
- Atomic mass of oxygen (O) ≈ 16.00 g/mol
Now calculate the total molar mass of P₂O₃:
- (2 × 30.97) + (3 × 16.00)
= 61.94 + 48.00
= 109.94 g/mol
Step 2: Divide the compound’s actual molar mass by the empirical formula mass
- Actual molar mass = 219.90 g/mol
- Empirical formula molar mass = 109.94 g/mol
Now divide:
- 219.90 ÷ 109.94 ≈ 2.00
This result tells us that the molecular formula contains 2 times the number of atoms in the empirical formula.
Step 3: Multiply the empirical formula by this factor
- Empirical formula: P₂O₃
- Multiply each subscript by 2:
(P₂O₃) × 2 = P₄O₆
So, the molecular formula of the compound is P₄O₆.
Conclusion
This approach confirms that the molecular formula is a simple multiple of the empirical formula. The molecular formula P₄O₆ maintains the same atom ratio as P₂O₃ but reflects the true number of atoms present in a molecule with a mass of 219.9 g/mol. This method is standard in chemistry for finding molecular formulas when given empirical formulas and molar masses.
