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

Solution:

The molecular formula is PbSO₄, indicating lead sulfate

Option c.

Explanation:

The percentage makeup shows that in 100 g of this compound, there are:

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

To find the moles of each element, we divide by their molar masses:

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

Next, we find the mole ratio by dividing each by the smallest number of moles:

0.329 / 0.329 = 1 Pb

0.331 / 0.329 = 1 S

1.32 / 0.329 = 4 O

Thus, the molecular formula is PbSO₄, representing lead sulfate.

Solution:

The empirical formula is PbSO4 (option C)

Justification:

Step 1: Given data

Assuming the compound's mass is 100.0 grams

The breakdown of the compound includes:

68.3% lead = 68.3 grams

10.6% sulfur = 10.6 grams

Remaining mass = oxygen

Molar mass of Pb = 207.2 g/mol

Molar mass of S = 32.065 g/mol

Molar mass of O = 16.0 g/mol

Step 2: Determine the mass of O

Mass of O = 100 - 68.3 - 10.6 = 21.1 grams

Step 3: Calculate moles

Moles = mass / molar mass

Moles of Pb = 68.3 grams / 207.2 g/mol = 0.3296 moles

Moles of S = 10.6 grams / 32.065 g/mol = 0.3306 moles

Moles of O = 21.1 grams / 16.0 g/mol = 1.319 moles

Step 4: Find mole ratio

Moles are divided by the smallest mole count

Pb: 0.3296/0.3296 = 1

S: 0.3306/0.3296 = 1

O: 1.319/0.3296 = 4

Thus, the empirical formula is PbSO4 (option C)