91Ó°ÊÓ

On translating the statement,

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

On calculating the moles and masses of carbon and hydrogen,

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

On finding moles, mass and Mass % can be,

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

On converting the mass % of Fe,

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

The mass and moles of Fe can be,

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

On finding the mass and moles of O,

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

The empirical formula can be as follows,

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