Question

To find the formula of a compound composed of iron and carbon monoxide, Fex(CO)y, the compound is burned in pure oxygen, an reaction that proceeds according to the following unbalanced equation.
Fex(CO)y + O2 --> Fe2O3 + CO2

If you burn 1.959 g. of Fex(CO)y and obtain 0.799 g. of Fe2O3 and 2.200 g. of CO2, what is the empirical formula of Fex(CO)y?

Answer 1

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  × $\frac{1 mole}{159.69 Fe2O3 g}$ = 5,00×10⁻³ Fe₂O₃ moles

2,200 CO₂ grams  × $\frac{1 mole}{44,01 CO2 g}$ = 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  × $\frac{1 mole}{195,9 Fe(CO)5 g}$ = 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!