Q3.150CP - Chemistry: Molecular Nature Of Matter And Change

Question

When 1.5173 g of an organic iron compound containing Fe, C, H, and O was burned in O₂, 2.838 g of CO₂ and 0.8122 g of H₂O were produced. In a separate experiment to determine the mass % of iron, 0.3355 g of the compound yielded 0.0758 g of Fe2O₃. What is the empirical formula of the compound?

Step-by-Step Solution

Answer

The empirical formula is C₁₅H₂₁O₆Fe.

Step 1: Forming chemical equation

 On translating the statement,

$Compound+O_2\left(g\right)\to C{O}_2\left(g\right)+H_{2}O\left(g\right)$

Step 2: Finding the moles and masses of carbon and hydrogen

On calculating the moles and masses of carbon and hydrogen,

$$\begin{gathered} Moles of C=2.838 g C{O}_2\times \frac{1 mol C{O}_2}{18.02 g C{O}_2}\times \frac{1 mol C}{1 mol C{O}_2}=0.09014 mol H \\ Mass of C=0.06449 molC\times \frac{12.01 g C}{1 mol C}=0.7745 g C \\ Moles of H=0.8122 g H_{2}O\times \frac{1 mol H_{2}O}{44.01 g H_{2}O}\times \frac{1 mol C}{1 mol C{O}_2}=0.06449 mol C \\ Mass of H=0.09014 molH\times \frac{1.01 g H}{1 mol H}=0.0910 g H \end{gathered}$$

Step 3: Finding the moles and masses of Fe

On finding moles, mass and Mass % can be,

$$\begin{gathered} Moles of Fe=0.0758 g F{e}_2{O}_3\times \frac{1 mol F{e}_2{O}_3}{159.69 g F{e}_2{O}_3}\times \frac{2mol Fe}{1 mol F{e}_2{O}_3}=9.493\times {10}^{-4}mol Fe \\ Mass of Fe=9.493\times {10}^{4}mol Fe\times \frac{55.85 g Fe}{1 mol Fe}=0.05302 g Fe \end{gathered}$$

On converting the mass % of Fe,

$Mass \% Fe=\frac{0.05302 g Fe}{0.3355 g}\times 100=15.8\% Fe$

The mass and moles of Fe can be,

$$\begin{gathered} Mass of Fe=0.158\times 1.5173 g compound=0.2397 g Fe \\ Moles of Fe=0.2397 g Fe\times \frac{1 mol Fe}{55.85 g Fe}=4.292\times {10}^{-3}mol Fe \end{gathered}$$

Step 4: Finding the moles and masses of O

On finding the mass and moles of O,

$$\begin{gathered} Mass of O=1.5173 g compound-0.2397 g Fe-0.7745 g C-0.0910 g H \\ Mass of O=0.4121 g O \\ Moles of O=0.4121 g O\times \frac{1 mol O}{16 g O}=0.02576 mol O \end{gathered}$$

The empirical formula can be as follows,

${\mathrm{C}}_{\frac{\text{0.06449}}{{\text{4.292×10}}^{\text{-3}}}}{\mathrm{H}}_{\frac{\text{0.09014}}{{\text{4.292×10}}^{\text{-3}}} }{\text{O}}_{\frac{\text{0.02576}}{{\text{4.292×10}}^{\text{-3}}}}{\text{Fe}}_{\frac{{\text{4.292×10}}^{\text{-3}}}{{\text{4.292×10}}^{\text{-3}}}}\to {\text{C}}_{\text{15}}{\text{H}}_{\text{21}}{\text{O}}_{\text{6}}\text{Fe}$