Answer
Empirical formula = $C_{2}H_{5}Cl$
Work Step by Step
Percentage of Chlorine = 55.0
$6.02\times10^{23} $ H atoms = 1 g
Then, $4.19\times10^{23} $ H atoms = $(4.19\times10^{23})\div(6.02\times10^{23})$ = 0.70 g
9.00 g sample contains 0.70 g H
Therefore percentage of H in the sample = $(0.70\times100)\div9.00$ = 7.8
Hence percentage of Carbon = 100 –(55.0 + 7.8) = 37.2
Relative no.of atoms =Pecentage/Atomic mass
C $37.2\div12$ =3.1
H $7.8\div1$ = 7.8
Cl $55.0\div35.5$ = 1.5
Divide by smallest relative no.of atoms (here 1.5)
$3.1\div1.5$ = 2.0
$7.8\div1.5$ = 5.2
$1.5\div1.5$ = 1
Convert to the whole number
2
5
1
Empirical formula = $C_{2}H_{5}Cl$
