Question

a compound contains 4.07 % H 24.27 c and 71.65 % cl its molecular formula is 98.97 what are empirical and molecular formula​

Answer 1

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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

therefore the empirical formula is

$</p><p>CH2Cl</p><p></p><p>$

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

$C </p><p>2</p><p> </p><p> H </p><p>4</p><p> </p><p> Cl </p><p>2</p><p> </p><p> .$