Dieldrin, an insecticide, contains $C,H,Cl$  and O. Combustion of 29.72 mg of Dieldrin gave 41.21 mg $C{O}_2$ and 5.63 mg $H_{2}O$. In a separate analysis 25.31 mg of Dieldrin was converted  into 57.13mg AgCl. What is the empirical formula of Dieldrin ?

$C_x{H}_{y}C{l}_z{O}_w+O_2\to C{O}_2+H_{2}O$

Millimoles of  $C=\frac{41.21}{44}$
Millimoles of  $H=\frac{5.63}{18}\times 2$
29.72 mg of dieldrin contains $\frac{57.13}{25.31}\times 29.72mg$ of AgCl
Millimoles of  $Cl=\frac{57.13\times 29.72}{25.31\times 143.5}$
Weight of  $O=29.72-[\frac{41.21}{44}\times 12+\frac{5.63}{18}\times 2\times 1+\frac{57.13\times 29.72\times 35.5}{25.31\times 143.5}]$
Weight of  $O=29.72-[11.239+0.6255+16.59]$
Weight of oxygen =1.265 mg
$$\begin{gathered} n_C=\frac{41.21}{44} \Rightarrow \frac{41.21}{44}\times \frac{16}{1.265}=1184\approx 12 \\ {n}_H=\frac{5.63}{9} \Rightarrow \frac{5.63}{9}\times \frac{16}{1.265}=7.9\approx 8 \\ {n}_O=\frac{1.265}{16} \Rightarrow \frac{1.265}{16}\times \frac{16}{1.265}=1 \\ {n}_{Cl}=\frac{57.13\times 29.72}{25.31\times 143.5} \Rightarrow \frac{57.13\times 29.72}{25.31\times 143.5}\times =5.95\approx 6 \end{gathered}$$