Chemistry, asked by tejarorasimran5445, 4 months ago

An organic compound has the percentage composition by mass
C = 36.36% H = 6.06% F = 57.58%
Show that the empirical formula of the compound is CH2F

Answers

Answered by Rameshjangid

Answer: The empirical formula of the compound: $CH_2F$

GIVEN:-

An organic compound is given in which

Amount of carbon = 36.36 g

Amount of hydrogen = 6.06 g

Amount of fluorine = 57.58 g

TO FIND:-

The empirical formula of the given compound.

SOLUTION:-

We know,

The molar mass of carbon = 12 g

The molar mass of hydrogen = 1 g

The molar mass of fluorine = 57.58g

The number of Moles of carbon, hydrogen, and fluorine,

No. of mole of carbon = 36.36/12 = 3.03

No. of mole of hydrogen = 6.06/1 = 6.06

No. of mole of fluorine = 57.58/19= 3.03

Simplest ratio,

$3.03:6.06:3.03\\\frac{3.03}{3.03}:\frac{6.06}{3.03}:\frac{3.03}{3.03} \\1:2:1$

The empirical formula of the compound is as follows

$CH_2F$