What is the percent sulfur in iron(III) sulfate?
A) 28%
B) 32%
C) 24%
D) 48%
E) 42%

Question

Quizplus · Accepted Answer

Each mole of iron(III) sulfate contains one mole of iron and three moles of sulfate. The percent sulfur by mass is calculated as the mass of sulfur divided by the mass of the compound, multiplied by 100%.

The molar mass of iron(III) sulfate (Fe2(SO4)3) is:

Fe: 2 x 55.85 g/mol = 111.7 g/mol
S: 3 x 32.06 g/mol = 96.18 g/mol
O: 12 x 15.99 g/mol = 191.88 g/mol

Total: 399.76 g/mol

The mass of sulfur in one mole of iron(III) sulfate is:

3 x 32.06 g/mol = 96.18 g/mol

The percent sulfur by mass is:

(96.18 g/mol / 399.76 g/mol) x 100% = 24%

Therefore, the answer is choice C.

What Is the Percent Sulfur in Iron(III) Sulfate