Answer:
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
<em> Molecular formula = n (empirical formula)</em>
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
<em> Molecular formula = n (empirical formula)</em>
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
<em> Molecular formula = n (empirical formula)</em>
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
