Answer:
Concentration of Flourine = 24.756%
Step-by-step explanation:
Given that :
High-density polyethylene may be flourinated by inducing the random substitution of Flourine atoms for hydrogen.
the objective is to determine he concentration of Flourine (in wt%) that must be added if this substitution occurs for 12% of all the original hydrogen atoms.
At standard conditions , the atomic weight of the these compounds are as follows:
Carbon = 12.01 g/mol
Chlorine = 35.45 g/mol
Fluorine = 19.00 g/mol
Hydrogen = 1.008 g/mol
Oxygen = 16.00 g/mol
The chemical formula for polyethylene = (CH₂ - CH₂)ₙ
Therefore, for two carbons, there will be 4 hydrogens;
i.e
(CH₂ - CH₂)₂
( C₂H₄ - C₂H₄ )
Suppose the number of original hydrogen = 4moles
number of moles of Flourine F = 12% of 4
= 0.12 × 4
= 0.48 mol
∴ the number of remaining moles of Hydrogen is:
= 4 - 0.48
= 3.52 moles
number of moles of Carbon = 2 moles
∴ the mass of flourine F = number of moles of F × molar mass of F
= 0.48 × 19
= 9.12
The total mass of the compound now is = (0.48 × 19 ) + (3.52 × 1) + (2× 12)
= 9.12 + 3.52 + 24
= 36.64
Concentration of Flourine = (mass of flourine/total mass) × 100
Concentration of Flourine = (9.12/36.84 ) × 100
Concentration of Flourine = 0.24756 × 100
Concentration of Flourine = 24.756%