Thank you for visiting 1 A chemical analysis of a gaseous compound shows its composition to be 36 4 carbon 57 5 fluorine and 6 1 hydrogen A sample. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
Answer:
1. Empirical formula = CFHâ‚‚
Molecular formula = C₉F₉Hâ‚₈
2. Empirical formula is CHâ‚‚Cl
Molecular formula = Câ‚‚Hâ‚„Clâ‚‚
3. Empirical formula = CFHâ‚‚
Molecular formula = C₉F₉Hâ‚₈
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 CFHâ‚‚
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 ( CFHâ‚‚)
Molecular formula = C₉F₉Hâ‚₈
Check the answer by calculating the molecular mass, it should be almost 296 g.
Molecular formula = C₉F₉Hâ‚₈
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 CHâ‚‚Cl
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 ( CHâ‚‚Cl)
Molecular formula = Câ‚‚Hâ‚„Clâ‚‚
Check the answer by calculating the molecular mass, it should be almost 99.8 g.
Molecular formula = Câ‚‚Hâ‚„Clâ‚‚
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 CFHâ‚‚
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 ( CFHâ‚‚)
Molecular formula = C₉F₉Hâ‚₈
Check the answer by calculating the molecular mass, it should be almost 296 g.
Molecular formula = C₉F₉Hâ‚₈
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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Rewritten by : Jeany
