Answer :
Answer:
Fe₂O₃
Explanation:
When Fe is burned it can form iron (II) oxide or iron (III) oxide. The reactions are:
a) Fe + 1/2 O₂ → FeO
b) 4 Fe + 3 O₂ → 2 Fe₂O₃
In the reaction "a", the mass ratio of FeO to Fe is 71.84 g FeO:55.85 g Fe =
1.29 g FeO: 1 g Fe
In the reaction "b", the mass ratio of Fe₂O₃ to Fe is 319.3 g Fe₂O₃: 223.4g Fe =
1.43 g Fe₂O₃ : 1 g Fe
According to the experimental data, the mass ratio of the product to Fe is 165.5g compound: 115.0 g Fe =
1.43 g compound: 1 g Fe
Since this ratio is equal to that of reaction "b", the empirical formula of the solid is Fe₂O₃.
Answer:
The empirical formula is Fe2O3
Explanation:
Step 1: Data given
Mass of the iron sample = 115.0 grams
Mass of the pure solid = 164.5 grams
Molar mass of Fe = 55.845 g/mol
Molar mass of O = 16 g/mol
Step 2: Calculate mass of oxygen
mass of oxygen = mass of solid - mass of iron
Mass of oxygen = 164.5 grams - 115.0 grams
MAss of oxygen = 49.5 grams
Step 3: Calculate moles of oxygen
Moles oxygen = mass oxygen / molar mass oxygen
Moles oxygen = 49.5 grams / 16.0 g/mol
Moles oxygen = 3.09 moles
Step 4: Calculate moles of Fe
Moles Fe = mass Fe / molar mass Fe
Moles Fe = 115.0 grams / 55.845 g/mol
Moles Fe = 2.06 moles
Step 5: Calculate mol ratio
We divide by the smallest amount of moles
O: 3.09/2.06 = 1.5
Fe: 2.06 / 2.06= 1
This means for 1 mol Fe we have 1.5 mol O OR for each 2 moles Fe we have 3 moles O.
The empirical formula is Fe2O3