Answer:
The empirical formula is: Fe(CO)₅
Explanation:
According the global reaction:
Feₓ(CO)y + O₂ → Fe₂O₃ + CO₂
You should calculate Fe₂O₃ and CO₂ moles, thus:
0,799 Fe₂O₃ grams × [tex]\frac{1 mole}{159.69 Fe2O3 g}[/tex] = 5,00×10⁻³ Fe₂O₃ moles
2,200 CO₂ grams × [tex]\frac{1 mole}{44,01 CO2 g}[/tex] = 5,00×10⁻²CO₂ moles
The ratio between Fe₂O₃ moles and CO₂ moles is 1:10. Thus ratio between x and y must be 1:5 because Fe₂O₃ has 2 irons but CO₂ has just one carbon.
Assuming the formula is Fe₁(CO)₅ the molecular weight is 195,9 g/mol. Thus:
1,959 Fe(CO)₅ grams × [tex]\frac{1 mole}{195,9 Fe(CO)5 g}[/tex] = 1,00×10⁻² Fe(CO)₅ moles
Thus, assuming 1,00×10⁻² moles as basis for calculation, the global reaction is:
1 Fe(CO)₅ + ¹³/₂O₂ → ¹/₂ Fe₂O₃ + 5 CO₂
With this balanced equation the moles produced have sense, thus, the empirical formula is: Fe(CO)₅
I hope it helps!
