One compound contains $4.07\%$ Hydrogen,$24.47\%$ Carbon and $71.65\%$ Chlorine. Its molar mass is $98.96\ g/mol$. Find the empirical formula and molecular formula.

(N/A) $1$. Calculate moles of each element in $100\ g$ of the compound:
$Moles \ of \ H = \frac{4.07}{1.008} \approx 4.04$
$Moles \ of \ C = \frac{24.47}{12.01} \approx 2.04$
$Moles \ of \ Cl = \frac{71.65}{35.45} \approx 2.02$
$2$. Divide by the smallest value $(2.02)$ to get the simplest ratio:
$C: 2.04/2.02 \approx 1$
$H: 4.04/2.02 \approx 2$
$Cl: 2.02/2.02 = 1$
Empirical formula is $CH_{2}Cl$.
$3$. Calculate empirical formula mass: $12.01 + 2(1.008) + 35.45 = 49.48\ g/mol$.
$4$. Find $n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{98.96}{49.48} \approx 2$.
$5$. Molecular formula = $n \times (CH_{2}Cl) = C_{2}H_{4}Cl_{2}$.