Empirical formula from mass percentages

Steps:

• Mass of Ru = 5.79 g - 1.39 g = 4.40 g
• Moles of Ru = 4.40 / 101 = 0.0436 mol (Ru atomic mass = 101 g/mol)
• Moles of O = 1.39 / 16 = 0.0869 mol (O atomic mass = 16 g/mol)
• Mole ratio Ru:O = 0.0436 : 0.0869 = 1:2, so empirical formula is RuO₂

Why C is correct:

• Empirical formula represents the simplest whole-number ratio of atoms in a compound, here 1 Ru to 2 O from mole calculation.

Why the others are wrong:

• A: Requires 2:1 Ru:O ratio, but moles show 1:2
• B: Requires 1:1 Ru:O ratio, but O moles are twice Ru moles
• D: Requires 1:4 Ru:O ratio, but O moles are only twice Ru moles

Final answer: C