Question

A compound contains 4.07% hydrogen,24.27%carbon &71.65% chlorine .Its molar mass is 98.96g .What are its empirical and molecular formula

Answer 1

Explanation:

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

therefore the empirical formula is

CH2Cl

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 C2H4Cl2.

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Answer 2

Explanation:

Explanation:

Given :

Percentage composition of Hydrogen in the compound is 4.07 %

Percentage composition of Carbon in the compound is 24.47 %

Percentage composition of Chlorine in the compound is 71.65 %

Molar mass = 99 g/mol

Tofind:

Molecular formula of the compound

Solution :

Step 1 : Calculating relative number of atoms

a. Hydrogen

Percentage composition = 4.07 %

Atomic weight of hydrogen = 1

Relative number of atoms = 4.07 / 1 = 4.07

b. Carbon

Percentage comppsition = 24.47 %

Aomic weight of carbon = 12

Relative number of atoms = 24.47 /12 = 2.02

c. Chlorine

Percentage comppsition = 71.65%

Atomic weight of chlorine = 35.5

Relative number of atoms = 71.65 / 35.5 = 2.01

Step 2 : Calculating Simplest ratio for each element

Simplest ratio is calculted by dividing all the values we got by the smallest value we got.

The smallest value we got is 2.01

a. Hydrogen

Simplest ratio = 2.01 / 4.07 ≈ 2

b. Carbon

Simplest ratio = 2.01 / 2.02 ≈ 1

c. Chlorine

Simplest ratio = 2.01 / 2.01 = 1

The empirical formula becomes CH₂Cl

Molecular weight of Empirical formula = 12 + 2(1) + 35.5 = 49.5

We are already given that molecular weight as 99.

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Relation between Empirical formula and Molecular formula is given by ,

Molecular Formula=Empirical formula×n

n is some integer which is given by ,

n = Empirical formula weight/Molecular Formula weight

We know the values for calculating the value of n . So , by substituting we get ;

:⟹ n= 99/49.5

:⟹ n=2

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Applying the first formulae , (i.e , relation between empirical formulae and Molecular formulae) we get :

⟹MolecularFormula=Empiricalformula×n

⟹MolecularFormula=CH₂CI × 2

⟹MolecularFormula=C₂H₄Cl

Hence , The molecular formula of the given compound is C₂H₄Cl