Answer :
Answer:
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
Answer:
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)