Answer to Question #249875 in General Chemistry for Marifi Pusing

Question #249875
Consider Teflon, the polymer made from tetrafluoroethylene. (a) Draw a portion of the Teflon molecule, (b) Calculate the molar mass of a Teflon molecule that contains 5.0x10^4 CF_2 units (c) What are the mass percents of C and F in Teflon? [NOTE: underscores (_) refer to subscripts, while (^) indicates superscripts].
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Expert's answer
2021-10-12T02:07:46-0400

The monomer unit is tetrafluoroethylene which has the following basic unit.



The polymeric chain of Tetrafluoroethylene that is Teflon can be represented as follows:



The number of molecules of Tetrafluoroethylene contains in a polymer.

5.0x104 molecules.

Moral mass of the polymer=Number of molecules of Tetrachloroethylene x Molar mass of Monomer


The molar mass of Monomer (Tetrafluoroethylene) = 100.02g/mol

Molar mass of the polymer

=5.0x104 x100.02

=5.0x106 g/mol

% Mass of C in the given polymer=

"\\frac{weight \\; of \\; carbon \\; in\\; monomer}{total \\; weight \\; of\\; the\\; monomer} \\times 100 \\\\\n\n= \\frac{2 \\times12.01}{100.2} \\times 100 \\\\"

=24.02%

% Mass of F in the given polymer = "\\frac{weight \\; of \\; fluorine \\; in\\; monomer}{total \\; weight \\; of\\; the\\; monomer} \\times 100 \\\\\n\n= \\frac{4 \\times 19.0}{100.2} \\times 100 \\\\"

=75.98%


Therefore, % Mass of C is 24.02% and F is 75.98%.


The monomer unit is tetrafluoroethylene which has the following basic unit.




The polymeric chain of Tetrafluoroethylene that is Teflon can be represented as follows:




The number of molecules of Tetrafluoroethylene contains in a polymer 5.0x104 molecules.


Molar mass of the polymer= Number of molecules of Tetrachloroethylene x Molar mass of Monomer

The molar mass of Monomer (Tetrafluoroethylene) = 100.02 g/mol

Molar mass of the polymer

=5.0x104 x100.02 g/mpl

=5.0x106 g/mol


% Mass of C in the given polymer =

"\\frac{weight \\; of \\; carbon \\; in\\; monomer}{total \\;weight \\; of \\; the\\; monomer} \\times 100 \\\\\n=\\frac{2 \\times 12.01}{100.2}\\times 100 \\\\"

=24.02%


% Mass of F in the given polymer ="\\frac{weight \\; of \\; fluorine \\; in\\; monomer}{total \\; weight \\; of\\; the\\; monomer} \\times 100 \\\\\n\n=\\frac {4 \\times 19.0}{100.02} \\times 100 \\\\"

=75.98%


Therefore, % Mass of C is 24.02% and F is 75.98%.



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