Chapter 3, Problem 55Q

(a)

Interpretation Introduction

Interpretation:

The chemical formula for CFC-12 has to be determined.

Concept-introduction:

Prediction of chemical formula for CFC:

Given: CFC-11

The code number is added by 90 to get a three digit number.

$90+11=101$

The first digit ‘1’ indicates the number of Carbon atoms,

The second digit ‘0’ indicates the number of Hydrogen atoms,

The third digit ‘1’ indicates the number of Fluorine atoms.

Then the remaining bonds are assumed to be chlorine.

Explanation of Solution

Given code: CFC-12

Adding the code with 90 gives $90+12=102$

The first digit ‘1’ indicates the number of Carbon atoms that is $\left(\text{1}\text{C}\right)$,

The second digit ‘0’ indicates the number of Hydrogen atoms $\left(\text{0}\text{H's}\right)$,

The third digit ‘2’ indicates the number of Fluorine atoms $\left(2\text{F's}\right)$.

Carbon forms four covalent bonds, consist of $\left({\text{CF}}_{\text{2}}\right)$ which lacks two more bonds and that is bonded to two chlorine atoms.

Therefore, the chemical formula is $\left(CC{l}_2{F}_2\right)$

(b)

Interpretation Introduction

Interpretation:

The code for the chemical formula $\left({\text{CCl}}_{\text{4}}\right)$ has to be determined.

Concept-introduction:

Prediction of chemical formula for CFC:

Given: CFC-11

The code number is added by 90 to get a three digit number.

$90+11=101$

The first digit ‘1’ indicates the number of Carbon atoms,

The second digit ‘0’ indicates the number of Hydrogen atoms,

The third digit ‘1’ indicates the number of Fluorine atoms.

Then the remaining bonds are assumed to be chlorine.

Explanation of Solution

Given chemical formula: $\left({\text{CCl}}_{\text{4}}\right)$

The formula consists of One Carbon atom which the first digit of chemical formula is ‘1’; and has no hydrogen and fluorine atoms become ‘0’ for the second and third digit.

Therefore, the chemical formula of $\left({\text{CCl}}_{\text{4}}\right)$ has the three digit value 100

By subtracting the value with 90 gives,

$\begin{array}{c}\text{90}\text{+}\text{x}\text{=}\text{100}\\ \\ \text{x=}\text{100}\text{-}\text{90}\text{=}\text{10}\end{array}$

Therefore, the chemical formula is $\left({\text{CCl}}_{\text{4}}\right)$ coded as $CFC-10$

(c)

Interpretation Introduction

Interpretation:

Using ‘90’ method work HCFC-22 has to be explained.

Concept-introduction:

Prediction of chemical formula for CFC:

Given: CFC-11

The code number is added by 90 to get a three digit number.

$90+11=101$

The first digit ‘1’ indicates the number of Carbon atoms,

The second digit ‘0’ indicates the number of Hydrogen atoms,

The third digit ‘1’ indicates the number of Fluorine atoms.

Then the remaining bonds are assumed to be chlorine.

Explanation of Solution

Given code: HCFC-22

Yes, ‘90’ method work can be done as follows,

Adding the code with 90 gives $90+22=112$

The first digit ‘1’ indicates the number of Carbon atoms that is $\left(\text{1}\text{C}\right)$,

The second digit ‘1’ indicates the number of Hydrogen atoms $\left(\text{1}\text{H's}\right)$,

The third digit ‘2’ indicates the number of Fluorine atoms $\left(2\text{F's}\right)$.

Carbon forms four covalent bonds, consist of $\left({\text{CHF}}_{\text{2}}\right)$ which lacks one more bond and that is bonded to one chlorine atom.

Therefore, the chemical formula is $\left(CH{F}_{2}Cl\right)$

(d)

Interpretation Introduction

Interpretation:

Whether Halon-1301 can be explained using ‘90’ method work or not has to be interpreted.

Concept-introduction:

Prediction of chemical formula for CFC:

Given: CFC-11

The code number is added by 90 to get a three digit number.

$90+11=101$

The first digit ‘1’ indicates the number of Carbon atoms,

The second digit ‘0’ indicates the number of Hydrogen atoms,

The third digit ‘1’ indicates the number of Fluorine atoms.

Then the remaining bonds are assumed to be chlorine.

Explanation of Solution

The ‘90’ method work can be done and useful only for Fluorine and Chlorine Halogen. There are no guidelines for handling bromine.

Hence, this method does not work for Halon-1301 $\left({\text{CF}}_{\text{3}}\text{Br}\right)$.