What is the emperical formula for a compound containing 68.3% lead, 10.6% sulfur, and the remainder oxygen? a. Pb2SO4 b. PbSO3 c. PbSO4 d. PbSO2 e. PbS2O3

Molecular formula → PbSO₄ → Lead sulfate

Option c.

Explanation:

The % percent composition indicates that in 100 g of compound we have:

68.3 g of Pb, 10.6 g of S and  (100 - 68.3 - 10.6) = 21.1 g of O

We divide each element by the molar mass:

68.3 g Pb / 207.2 g/mol = 0.329 moles Pb

10.6 g S / 32.06 g/mol = 0.331 moles S

21.1 g O / 16 g/mol = 1.32 moles O

We divide each mol by the lowest value to determine, the molecular formula

0.329 / 0.329 = 1 Pb

0.331 / 0.329 = 1 S

1.32 / 0.329 = 4 O

Molecular formula → PbSO₄ → Lead sulfate

The empirical formula is PbSO4 (option C)

Explanation:

Step 1: Data given

Suppose the mass of a compound is 100.0 grams

The compound contains:

68.3 % lead = 68.3 grams

10.6 % sulfur = 10.6 grams

Rest = oxygen

Molar mass Pb = 207.2 g/mol

Molar mass S = 32.065 g/mol

Molar mass O = 16.0 g/mol

Step 2: Calculate mass O

MAss O = 100 - 68.3 - 10.6 = 21.1 grams

Step 3: Calculate moles

Moles = mass / molar mass

Moles Pb = 68.3 grams / 207.2 g/mol

Moles Pb = 0.3296 moles

Moles S = 10.6 grams / 32.065 g/mol

Moles S = 0.3306 moles

Moles O = 21.1 grams / 16.0 g/mol

Moles O = 1.319 moles

Step 4: Calculate mol ratio

We divide by the smallest amount of moles

Pb: 0.3296/0.3296 = 1

S: 0.3306/0.3296 = 1

O: 1.319/0.3296 = 4

The empirical formula is PbSO4 (option C)