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17 April, 08:05

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?

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Answers (1)
  1. 17 April, 08:21
    0
    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
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