Determine the empirical formula for a 100.00-g sample of a compound having the following percent composition:

80.68% mercury, 12.87% oxygen, and 6.45% sulfur

Empirical formula: Hg₂SO₄

Explanation:

80.68 g of Hg, 12.87g of O and 6.45 g of S are contained in 100 g of compount. That's what the percent composition indicates.

Let's convert the mass to moles:

80.68 g / 200.59 g /mol = 0.402 moles of Hg

12.87 g / 16 g/mol = 0.804 moles of O

6.45 g / 32.06 g/mol = 0.201 moles of S

We divide the value by the lowest number obtained (0.201)

0.402 moles of Hg / 0.201 = 2 Hg

0.804 moles of O / 0.201 = 4 O

0.201 moles of S / 0.201 = 1 S

Empirical formula: Hg₂SO₄

im not exactly sure but i think its b!

the volume increases