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A compound has the following percentages by mass: barium, 58.8%; sulfur, 13.74%; oxygen, 27.43%. Determine the empirical formula.​

User Gleeb
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1 Answer

4 votes

Answer:

BaSO₄

Step-by-step explanation:

Let's assume we have 100 g of the compound. If that's the case we would have:

  • 58.8 g of Ba
  • 13.74 g of S
  • 27.43 g of O

Now we convert the masses of each element into moles, using their respective molar masses:

  • 58.8 g Ba ÷ 137.327 g/mol = 0.428 mol Ba
  • 13.74 g S ÷ 32 g/mol = 0.429 mol S
  • 27.43 g O ÷ 16 g/mol = 1.71 mol O

We divide those moles by the lowest number among them:

  • 0.428 mol Ba / 0.428 = 1
  • 0.429 mol S / 0.428 ≈ 1
  • 1.71 mol O / 0.428 = 4

We can express those results as Ba₁S₁O₄, meaning the empirical formula is thus BaSO₄.

User AbhayBohra
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