The percentage composition of iron(II) sulfate hexahydrate (FeSO₄·6H₂O) can be calculated by determining the molar mass of the compound and the mass contributions of each element. The molar mass is approximately 278.01 g/mol, with iron (Fe) contributing about 55.85 g/mol, sulfur (S) about 32.07 g/mol, oxygen (O) about 64.00 g/mol, and water (H₂O) contributing about 108.12 g/mol. The percentage of each component can then be calculated as follows: Fe (20.1%), S (11.5%), O (23.0%), and H₂O (38.9%).
Fe is 2+ and SO4 is 2- So... Iron (II) Sulfate would be FeSO4
No, it is not
FeI2
Fe(ClO4)2
Formula: (CH3COO)2Fe
Formula: Fe(ClO4)2
FeBr2 is Iron II Bromide
Formula: (NH4)2Fe(SO4)2
Formula: Fe(OH)2
1 : 1
Ferrous or iron(II) compounds can easily be oxidised to ferric or iron(III) compounds, when exposed to air. So iron(II) sulphate which is stored for some time will have iron(III) sulphate present.
The formula for iron(II) hypophosphite is Fe(H2PO2)2.