Answer :
Answer:
The empirical formula is = [tex]CCl_3[/tex]
The molecular formula = [tex]C_2Cl_6[/tex]
Explanation:
[tex]Moles =\frac {Given\ mass}{Molar\ mass}[/tex]
% of C = 10.13
Molar mass of C = 12.0107 g/mol
% moles of C = 10.13 / 12.0107 = 0.8434
% of Cl = 89.87
Molar mass of Cl = 35.453 g/mol
% moles of Cl = 89.87 / 35.453 = 2.5349
Taking the simplest ratio for C and Cl as:
0.8434 : 2.5349
= 1 : 3
The empirical formula is = [tex]CCl_3[/tex]
Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.
Thus,
Molecular mass = n × Empirical mass
Where, n is any positive number from 1, 2, 3...
Mass from the Empirical formula = 12*1 + 3*35.5 = 118.5 g/mol
Molar mass = 237 g/mol
So,
Molecular mass = n × Empirical mass
237 = n × 118.5
⇒ n ≅ 2
The molecular formula = [tex]C_2Cl_6[/tex]
1. The empirical formula of the compound containing 10.13% C and 89.87% Cl is CCl₃
2. The molecular formula of the compound is C₂Cl₆
1. How to determine the empirical formula
- C = 10.13%
- Cl = 89.87%
- Empirical formula =?
Divide by their molar mass
C = 10.13 / 12 = 0.844
Cl = 89.87 / 35.5 = 2.532
Divide by the smallest
C = 0.844 / 0.844 = 1
Cl = 2.532 / 0.844 = 3
Thus, the empirical formula of the compound is CCl₃
2. How to determine the molecular formula
- Molar mass of compound = 237 g/mol.
- Empirical formula = CCl₃
- Molecular formula =?
Molecular formula = n × empirical = molar mass
[CCl₃]n = 237
[12 + (3×35.5)]n = 237
118.5n = 237
Divide both side by 118.5
n = 237 / 118.5
n = 2
Molecular formula = [CCl₃]n
Molecular formula = [CCl₃]₂
Molecular formula = C₂Cl₆
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