Question

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?

Answer 1

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