Dieldrin, an insecticide, contains $C,H,Cl$ and $O.$ Combustion of 29.72mg of Dieldrin gave $41.21mgC{O}_2$ and $5.63mgof{H}_{2}O.$ In a separate analysis 25.31 mg of Dieldrin was converted into $\text{57}.\text{13 mg AgCl.}$ What is the empirical formula of Dieldrin?

$$\begin{gathered} C,H,Cl,O_2\to C{O}_2+H_{2}O \\ 41.21 5.63 \\ mg mg \\ \text{Millimoles of\hspace{0.17em}}C{O}_2=\frac{41.21}{44}=0.937\text{\hspace{0.17em}Millimoles.} \\ \text{millimoles carbon = 1}\times {\text{\hspace{0.17em}millimole of CO}}_2\text{=0.937} \\ {\text{millimoles of\hspace{0.17em}H}}_{2}O=\frac{5.63}{18}=0.313\text{\hspace{0.17em}mmoles= 2}\times \text{0.313 = 0.626 mmoles.} \end{gathered}$$

25.31mg of dieldrin was converted to 57.13 mg AgCl

$$\begin{gathered} \therefore 29.72mg\text{\hspace{0.17em}of dieldrin will be converted to x mg AgCl } \\ x=\frac{29.72\times 57.13}{25.31}=67.084mg \\ \text{Millimoles of\hspace{0.17em}}AgCl=\frac{67.084}{143.5}=\text{\hspace{0.17em}0.467 mmoles} \\ \text{Millimoles of Cl =1\hspace{0.17em}}\times \text{\hspace{0.17em}Millimoles of AgCl = 1 × 0.467= 0.467 millimoles.} \\ \text{Mass of Cl = 0.467 × 35.5 = 16.596 mg } \\ \therefore \text{\hspace{0.17em}\hspace{0.17em}Mass of O = 29.72 - }\left(\text{0.937 × 12 + 0.62 × 12 + 16.596}\right)=1.260mg \\ \text{Millimoles of O = }\frac{1.260}{16}\text{=0.079 m moles} \\ \text{Millimoles of }C=0.937, \\ \text{Millimoles of }H=0.626, \\ \text{Millimoles of }Cl=0.467, \\ \text{Millimoles of }O=0.79, \\ \therefore C:H:Cl:O=0.937:0.626:0.467:0.079=12:8:6:1. \\ \therefore \text{\hspace{0.17em}\hspace{0.17em}The empirical formula\hspace{0.17em}}=C_{12}{H}_{8}OC{l}_6. \end{gathered}$$