1) chemical analysis of a gaseous compound show its composition to be 36.4% carbon, 57.5% fluorine, and 6.1%
hydrogen. a sample of 1.00 l of this gas has a mass of 2,96 g. what molecular formula do these data suggest for
this compound?

2)analysis of a compound shows that it consists of 24.3% carbon, 4.1% hydrogen, and 71.6% chlorine. the molecular
mass of the compound is determined to be 99.8 g/mol. what molecular formula corresponds to these data?

3)chemical analysis of a gaseous compound show its composition to be 36.4% carbon, 57.5% fluorine, and 6.1%
hydrogen. a sample of 1.00 l of this gas has a mass of 2,96 g. what molecular formula do these data suggest for
this compound?

1.  Empirical formula =  CFH₂

   Molecular formula = C₉F₉H₁₈  

2. Empirical formula is  CH₂Cl

   Molecular formula = C₂H₄Cl₂

3. Empirical formula =  CFH₂

   Molecular formula = C₉F₉H₁₈

Explanation:

1) Given data

  Percentage of carbon= 36.4%

  Percentage of fluorine  = 57.5%

  Percentage of hydrogen = 6.1%

  mass = 296 g

  Empirical formula = ?

  Molecular formula  = ?

Solution:

Number of gram atoms of C = 36.4/12 = 3.03

Number of gram atoms of F = 57.5/19 = 3.03

Number of gram atoms of H = 6.1/ 1 = 6.1

Atomic ratio:

    C              :            F             :    H

3.03/3.03      :       3.03/3.03    :   6.1/3.03

    1                :            1              :     2 

      C : F : H = 1 : 1 : 2

Empirical formula is  CFH₂

Now we calculate Molecular formula

   Molecular formula = n (empirical formula)

   n = molar mass of compound / empirical formula mass

   n = 296  / 32

   n = 9.25 we take it as 9

    Molecular formula = n (empirical formula)

    Molecular formula = 9 ( CFH₂)

  Molecular formula = C₉F₉H₁₈

Check the answer by calculating the molecular mass, it should be almost 296 g.

   Molecular formula = C₉F₉H₁₈

   molecular mass = 12 (9) + 18.998 (9) + 1 (18)

   molecular mass = 108 +  170.98 + 18

   molecular mass = 296.98 g/mol

2) Given data

   Percentage of carbon= 24.3%

   Percentage of hydrogen  = 4.1%

   Percentage of chlorine = 71.6%

   molecular mass = 99.8 g/mol

   Molecular formula  = ?

Solution:

Number of gram atoms of C = 24.3/12 = 2

Number of gram atoms of H = 4.1/1 = 4.1

Number of gram atoms of Cl = 71.6/35.5 = 2

Atomic ratio:

    C         :      H        :    Cl

    2/2      :       4/2    :   2/2

    1          :      2        :     1 

      C : H : Cl = 1 : 2 : 1

Empirical formula is  CH₂Cl

Now we calculate Molecular formula

   Molecular formula = n (empirical formula)

   n = molar mass of compound / empirical formula mass

   n = 99.8  / 49.5

   n = 2.02 we take it as 2

   Molecular formula = n (empirical formula)

    Molecular formula = 2 ( CH₂Cl)

  Molecular formula = C₂H₄Cl₂

Check the answer by calculating the molecular mass, it should be almost 99.8 g.

   Molecular formula = C₂H₄Cl₂

   molecular mass = 12 (2) + 1 (4) + 2 (35.5)

   molecular mass = 24 +  4 + 71

   molecular mass = 99 g/mol

3) Given data

  Percentage of carbon= 36.4%

  Percentage of fluorine  = 57.5%

  Percentage of hydrogen = 6.1%

  mass = 296 g

  Empirical formula = ?

  Molecular formula  = ?

Solution:

Number of gram atoms of C = 36.4/12 = 3.03

Number of gram atoms of F = 57.5/19 = 3.03

Number of gram atoms of H = 6.1/ 1 = 6.1

Atomic ratio:

    C              :            F             :    H

3.03/3.03      :       3.03/3.03    :   6.1/3.03

    1                :            1              :     2 

      C : F : H = 1 : 1 : 2

Empirical formula is  CFH₂

Now we calculate Molecular formula

   Molecular formula = n (empirical formula)

   n = molar mass of compound / empirical formula mass

   n = 296  / 32

   n = 9.25 we take it as 9

    Molecular formula = n (empirical formula)

    Molecular formula = 9 ( CFH₂)

  Molecular formula = C₉F₉H₁₈

Check the answer by calculating the molecular mass, it should be almost 296 g.

   Molecular formula = C₉F₉H₁₈

   molecular mass = 12 (9) + 18.998 (9) + 1 (18)

   molecular mass = 108 +  170.98 + 18

   molecular mass = 296.98 g/mol

a