Question

a compound contans 4.07% hydrogen,24.27% corbon 71.65% chlorine it's motas mas is 9896 gmo-1 what are ils empirical and molecular formula​

Answer 1

Answer:

step 1

Divide the given percentages of atoms with their molecular masses

H---4.07/1 =4.07

C---24.27/12 =2.02

Cl--71.65/35.5=2.01

step 2

divide all values with the lowest value obtained.

H---4.07/2.01=2

C---2.02/2.01=1

Cl---2.01/2.01=1

$\large\rm { \therefore }$ the empirical formula is

$\large\rm { CH_{2} Cl}$

WEIGHT OF EMPIRICAL FORMULA=12+2+35.5=49.5

Given molecular weight =98.96 which is double of empirical weight .therefore molecular formula is $\large\rm { C_{2} H_{4} Cl_{2}}$