Chapter 10, Problem 100QRT

Interpretation Introduction

Interpretation:

The mass of chloroform and $\text{HF}$ required to make starting material ${\text{CHClF}}_{\text{2}}$ has to be stated.

Concept Introduction:

Polymers are the high molecular mass compounds.  The small monomers units joined together to form large molecule, this reaction is known as polymerization reaction.  Addition polymerization is the polymerization in which monomer units are linked together with no side product formation.  Addition polymerization is given by the molecules which contains carbon-carbon double bond.

Explanation of Solution

The polymer Teflon is formed by the addition reaction of tetrafluoroethylene.

Consider the given reaction as follows.

${\text{F}}_{\text{2}}{\text{C=CF}}_{\text{2}}\stackrel{\text{peroxide}\text{catalyst}}{\to }\text{Teflon}$

From the above reaction, one mole of tetrafluoroethylene gives one mole of Teflon.  The molar mass of tetrafluoroethylene and teflon is same.  Therefore, $1\text{kg}$ of Teflon is obtained by $1\text{kg}$ tetrafluoroethylene.

Conversion of kilograms to grams is as follows.

$1\text{kg}=1000\text{g}$

The molar mass of tetrafluoroethylene is $100.02\text{g}{\text{mol}}^{-1}$.

Use the following expression for moles of a compound.

$\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Substitute $1000\text{g}$ for given mass and $100.02\text{g}{\text{mol}}^{-1}$ for molar mass.

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{tetrafluoroethylene}=\frac{1000\text{g}}{100.02\text{g}{\text{mol}}^{-1}}\\ =9.99\text{mol}\end{array}$

Consider the reaction for the formation of tetrafluoroethylene.

$2{\text{ CHClF}}_2\stackrel{highT}{\to }{\text{F}}_{\text{2}}\text{C}={\text{CF}}_2+2\text{ HCl}$

From the given reaction, $2\text{mol}$ of ${\text{CHClF}}_2$ gives $1\text{mol}$ of ${\text{F}}_{\text{2}}\text{C}={\text{CF}}_2$.  Therefore, $9.99\text{mol}$ of ${\text{F}}_{\text{2}}\text{C}={\text{CF}}_2$ is formed from $19.98\text{mol}$ of ${\text{CHClF}}_2$.

Consider the given reaction as follows.

${\text{CHCl}}_3+2\text{ HF}\to {\text{CHClF}}_2+2\text{ HCl}$

One mole of ${\text{CHClF}}_2$ is formed by the one mole of ${\text{CHCl}}_{\text{3}}$.  Therefore, $19.98\text{mol}$ of ${\text{CHClF}}_2$ is formed by $19.98\text{mol}$ of ${\text{CHCl}}_{\text{3}}$.

The molar mass of ${\text{CHCl}}_{\text{3}}$ is $119.37\text{g}{\text{mol}}^{-1}$.

Use the following expression to calculate the amount of ${\text{CHCl}}_{\text{3}}$ required.

$\text{Amount}\text{of}{\text{CHCl}}_{\text{3}}=\text{Number}\text{of}\text{moles}\times \text{Molar}\text{mass}$

Substitute $119.37\text{g}{\text{mol}}^{-1}$ for molar mass and $19.98\text{mol}$ for moles in the above equation.

$\begin{array}{c}\text{Amount}\text{of}\text{CHCl}{}_{\text{3}}=19.98\text{mol}\times 119.37\text{g}{\text{mol}}^{-1}\\ =\underset{\_}{2.38\times 10^{3}g}\end{array}$

One mole of ${\text{CHClF}}_2$ is formed by the two moles of $\text{HF}$.  Therefore, $19.98\text{mol}$ of ${\text{CHClF}}_2$ is formed by $39.96\text{mol}$ of $\text{HF}$.

The molar mass of $\text{HF}$ is $20.01\text{g}{\text{mol}}^{-1}$.

Use the following expression to calculate the amount of $\text{HF}$ required.

$\text{Amount}\text{of HF}=\text{Number}\text{of}\text{moles}\times \text{Molar}\text{mass}$

Substitute $20.01\text{g}{\text{mol}}^{-1}$ for molar mass and $39.96\text{mol}$ for moles in above equation.

$\begin{array}{c}\text{Amount}\text{of}\text{HF}=39.96\text{mol}\times 20.01\text{g}{\text{mol}}^{-1}\\ =\underset{\_}{7.94\times 10^{2}g}\end{array}$

Therefore, for preparing $1\text{kg}$ Teflon, $\underset{\_}{2.38\times 10^{3}g}$ of chloroform and $\underset{\_}{7.94\times 10^{2}g}$ of $\text{HF}$ is required.