It would depend on your definition of percentage.
For example, if you were looking for percentage of atom count, you would count the atoms in the compound Fe2(SO4)3. You have 12 Oxygen, 3 Sulfur, and 2 Iron, leaving you with 3 sulfur out of 17 total atoms, or 3/17 or 0.176470588 (about 17.5%).
If, on the other hand, you are talking about percentage of mass, then you must calculate the mass of each element in the compound.
Iron's Atomic Mass is 55.847.
Sulfur's atomic mass is 32.066.
Oxygen's atomic mass is 15.999.
Multiply these by their respective ratios. Iron (55.847*2=111.694) Sulfur (32.066*3=96.198) Oxygen (15.999*12=191.988) Add them together to get (111.694+96.198+191.988=399.88)
Now, take the overall mass of the sulfur part (96.198) and divide it into the total mass (399.88). The answer is 96.198/399.88 or 0.24056717 or about 24%.
iron(II) sulfate, FeSO4 iron(III) sulfate Fe2 (SO4)3
Fe2(SO4)3 is Iron (III) Sulphate.
The formula unit for iron (III) sulfate is Fe2(SO4)3. This formula shows that there are two iron atoms, three sulfur atoms, and twelve oxygen atoms, totalling 17 atoms altogether.
Iron (III) Sulfide Ferrous Sulfide
There are two kinds of iron sulfate: iron (II) sulfate has the formula FeSO4 and iron (III) sulfate has the formula Fe2(SO4)3.
iron(II) sulfate, FeSO4 iron(III) sulfate Fe2 (SO4)3
iron(II) sulfate, FeSO4 iron(III) sulfate Fe2 (SO4)3
The chemical formula for iron(II) sulphate is FeSO4, for iron(III) sulfate is Fe2(SO4)3 The formula for iron(II) sulfide is FeS, iron(III) sulfide Fe2S3
The residue obtained from heating crystals of FeSO4 (iron(II) sulfate) is Fe2O3 (iron(III) oxide) and SO2 (sulfur dioxide) gases. This is because the heating process causes the iron(II) sulfate to decompose, releasing water vapor, sulfur dioxide gas, and leaving iron(III) oxide as a solid residue.
Fe2(SO4)3 is Iron (III) Sulphate.
Iron(III) Sulfate = Fe2(SO4)3
Fe(SO4)3 is Iron(III) sulfate Wrong: Fe2(SO4)3 is iron(III) sulfate.
The formula unit for iron (III) sulfate is Fe2(SO4)3. This formula shows that there are two iron atoms, three sulfur atoms, and twelve oxygen atoms, totalling 17 atoms altogether.
Iron(III) Sulfate = Fe2(SO4)3
The chemical formular for Iron(III) sulfate dihydrate is Fe2(SO4)3.2H2O.
The mass in grams of 1,40 mol of anhydrous iron(III) sulfate is 559,832.
Fe(SO4)3 is Iron(III) sulfate Wrong: Fe2(SO4)3 is iron(III) sulfate.