compound contains 4.07 percentage of hydrogen and 24. 27 percentage of carbon and 71.65 percentage of chlorine and its molar mass is 98.96 grams of molar mass .find its empirical formula and molecular formula
Answers
Explanation:
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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