What is the percent composition of iron(II)sulfate hexahydrate?

Question

Quizplus · Accepted Answer

The formula for iron(II) sulfate hexahydrate is FeSO4·6H2O. To calculate the percent composition, we need to determine the molar masses of each element present:

Fe: 1 * 55.845 g/mol = 55.845 g/mol
S: 1 * 32.066 g/mol = 32.066 g/mol
O: 4 * 15.999 g/mol = 63.996 g/mol
H: 12 * 1.008 g/mol = 12.096 g/mol

The total molar mass of the compound is:

Fe + S + O + H = 163.003 g/mol

To determine the percent composition of each element, we divide the molar mass of that element by the total molar mass and multiply by 100.

Fe: (55.845 g/mol / 163.003 g/mol) * 100% = 34.26% Fe
S: (32.066 g/mol / 163.003 g/mol) * 100% = 19.67% S
O: (63.996 g/mol / 163.003 g/mol) * 100% = 39.24% O
H: (12.096 g/mol / 163.003 g/mol) * 100% = 7.43% H

Therefore, the percent composition of iron(II) sulfate hexahydrate is 21.5% Fe; 12.3% S; 61.5% O; 4.7% H, which is closest to choice D.

What Is the Percent Composition of Iron(II)sulfate Hexahydrate