Chapter 4, Problem 25Q

(a)

Interpretation Introduction

Interpretation:

The mass percent of chlorine in ${\text{CCl}}_{\text{3}}\text{F}\left(\text{Freon}\text{-}\text{11}\right)$ molecule has to be calculated.

Concept Introduction:

Molar mass: It is obtained by dividing the mass of substance with the amount of substance and the S.I. unit of molar mass is $\text{kg/mol}$. It is numerically equal to the molecular weight since molecular weight is sum of all individual atom weight present in molecule which is represented in $amu$.

  $\text{Molar}\text{mass}\left(\text{g/mol}\right)=\frac{\text{Given}\text{mass}\text{of}\text{substance}\left(\text{g}\right)}{\text{Moles}\text{of}\text{substance}\left(mol\right)}$

Molar mass of $H_{2}O$ can be determined as follows,

$\text{Molar}\text{mass}\text{of}{\text{H}}_{\text{2}}\text{O}=\underset{MassoftwoHydrogen}{\left(\text{2}\text{×}\text{1}\text{g}\right)}\text{+}\underset{massofoneOxygen}{\left(\text{1×}\text{16}\text{g}\right)}=18g/mol$

Mass ratio of a compound with formula AB can be determined as follows,

$\boxed{\text{Mass}\text{ratio}\text{of}\text{A}=\text{100}\text{.0}\text{g}\text{AB}\text{×}\frac{\text{Mass}\text{of}\text{A}}{\text{Mass}\text{of}\text{AB}}}$

Mass percentage will give an idea about how many parts exist in every 100 parts of the whole substance.

Mass percentage of a compound with formula AB can be determined as follows,

$\boxed{\text{Mass}\text{percent}\text{of}\text{A}=\frac{\text{Mass}\text{of}\text{A}}{\text{Mass}\text{of}\text{AB}}\text{×}\text{100}}$

Explanation of Solution

Given molecule is ${\text{CCl}}_{\text{3}}\text{F}\left(\text{Freon}\text{-}\text{11}\right)$ and its molar mass is 137.5 g/mole and this value can be taken as the total mass of the compound to find the mass percentage of any element in it.

Molar mass of chlorine is 35.5 g/mol and there are three chlorine atoms in the given Freon-11 molecule.

The mass percent of chlorine in ${\text{CCl}}_{\text{3}}\text{F}\left(\text{Freon}\text{-}\text{11}\right)$ can be calculated as follows,

$\begin{array}{c}\text{Mass}\text{percent}\text{of}\text{Cl}\text{=}\frac{\text{3}\text{×}\left(\text{35}\text{.5}\text{g/mol}\right)}{\text{12}\text{.0}\text{g/mol}\text{+}\text{3×}\left(\text{35}\text{.5}\text{g/mol}\right)\text{+}\text{19}\text{.0}\text{g/mol}}\text{×}\text{100}\\ \text{=}77.5\%\end{array}$

(b)

Interpretation Introduction

Interpretation:

Interpretation:

The mass percent of chlorine in ${\text{CCl}}_{\text{3}}{\text{F}}_2\left(\text{Freon}\text{-}\text{12}\right)$ molecule has to be calculated.

Concept Introduction:

Molar mass: It is obtained by dividing the mass of substance with the amount of substance and the S.I. unit of molar mass is $\text{kg/mol}$. It is numerically equal to the molecular weight since molecular weight is sum of all individual atom weight present in molecule which is represented in $amu$.

  $\text{Molar}\text{mass}\left(\text{g/mol}\right)=\frac{\text{Given}\text{mass}\text{of}\text{substance}\left(\text{g}\right)}{\text{Moles}\text{of}\text{substance}\left(mol\right)}$

Molar mass of $H_{2}O$ can be determined as follows,

$\text{Molar}\text{mass}\text{of}{\text{H}}_{\text{2}}\text{O}=\underset{MassoftwoHydrogen}{\left(\text{2}\text{×}\text{1}\text{g}\right)}\text{+}\underset{massofoneOxygen}{\left(\text{1×}\text{16}\text{g}\right)}=18g/mol$

Mass ratio of a compound with formula AB can be determined as follows,

$\boxed{\text{Mass}\text{ratio}\text{of}\text{A}=\text{100}\text{.0}\text{g}\text{AB}\text{×}\frac{\text{Mass}\text{of}\text{A}}{\text{Mass}\text{of}\text{AB}}}$

Mass percentage will give an idea about how many parts exist in every 100 parts of the whole substance.

Mass percentage of a compound with formula AB can be determined as follows,

$\boxed{\text{Mass}\text{percent}\text{of}\text{A}=\frac{\text{Mass}\text{of}\text{A}}{\text{Mass}\text{of}\text{AB}}\text{×}\text{100}}$

Explanation of Solution

Given molecule is ${\text{CCl}}_{\text{3}}{\text{F}}_2\left(\text{Freon}\text{-}\text{12}\right)$ and its molar mass is 156.5 g/mole and this value can be taken as the total mass of the compound to find the mass percentage of any element in it.

Molar mass of chlorine is 35.5 g/mol and there are three chlorine atoms in the given Freon-12 molecule.

The mass percent of chlorine in ${\text{CCl}}_{\text{3}}{\text{F}}_2\left(\text{Freon}\text{-}\text{12}\right)$ can be calculated as follows,

$\begin{array}{c}\text{Mass}\text{percent}\text{of}\text{Cl}\text{=}\frac{\text{3}\text{×}\left(\text{35}\text{.5}\text{g/mol}\right)}{\text{12}\text{.0}\text{g/mol}\text{+}\text{3×}\left(\text{35}\text{.5}\text{g/mol}\right)\text{+}2\times \text{19}\text{.0}\text{g/mol}}\text{×}\text{100}\\ \text{=}58.7\%\end{array}$

(c)

Interpretation Introduction

Interpretation:

The maximum mass of chlorine that could be released in the stratosphere by 100 g of Freon-11 and Freon -12 has to be calculated.

Concept Introduction:

Molar mass: It is obtained by dividing the mass of substance with the amount of substance and the S.I. unit of molar mass is $\text{kg/mol}$. It is numerically equal to the molecular weight since molecular weight is sum of all individual atom weight present in molecule which is represented in $amu$.

  $\text{Molar}\text{mass}\left(\text{g/mol}\right)=\frac{\text{Given}\text{mass}\text{of}\text{substance}\left(\text{g}\right)}{\text{Moles}\text{of}\text{substance}\left(mol\right)}$

Molar mass of $H_{2}O$ can be determined as follows,

$\text{Molar}\text{mass}\text{of}{\text{H}}_{\text{2}}\text{O}=\underset{MassoftwoHydrogen}{\left(\text{2}\text{×}\text{1}\text{g}\right)}\text{+}\underset{massofoneOxygen}{\left(\text{1×}\text{16}\text{g}\right)}=18g/mol$

Mass ratio of a compound with formula AB can be determined as follows,

$\boxed{\text{Mass}\text{ratio}\text{of}\text{A}=\text{100}\text{.0}\text{g}\text{AB}\text{×}\frac{\text{Mass}\text{of}\text{A}}{\text{Mass}\text{of}\text{AB}}}$

Mass percentage will give an idea about how many parts exist in every 100 parts of the whole substance.

Mass percentage of a compound with formula AB can be determined as follows,

$\boxed{\text{Mass}\text{percent}\text{of}\text{A}=\frac{\text{Mass}\text{of}\text{A}}{\text{Mass}\text{of}\text{AB}}\text{×}\text{100}}$

Explanation of Solution

The mass percentage of chlorine in Freon-11 and Freon -12 are $77.5\%and58.7\%$ respectively.

And so, the maximum mass of chorine that could be released in the stratospher by 100 g of each compound will be 77.5 g and 58.7 g respectively.

(d)

Interpretation Introduction

Interpretation:

The number of atoms in 77.5 g and 58.7 g of chlorine has to be determined.

Concept Introduction:

Molar mass: It is obtained by dividing the mass of substance with the amount of substance and the S.I. unit of molar mass is $\text{kg/mol}$. It is numerically equal to the molecular weight since molecular weight is sum of all individual atom weight present in molecule which is represented in $amu$.

  $\text{Molar}\text{mass}\left(\text{g/mol}\right)=\frac{\text{Given}\text{mass}\text{of}\text{substance}\left(\text{g}\right)}{\text{Moles}\text{of}\text{substance}\left(mol\right)}$

The relation between the number of moles and mass of the substance is,

$Numberofmole=\frac{Massingram}{Molarmass}$

     $Massingramofthesubs\tan ce=Numberofmole\times Molarmass$

$\begin{array}{c}6.02\times {10}^{23}atoms=1mol\\ Moles\times 6.02\times {10}^{23}=Atoms\end{array}$

Explanation of Solution

The mass percentage of chlorine in Freon-11 and Freon -12 are $77.5\%and58.7\%$ respectively.

And so, the maximum mass of chorine that could be released in the stratospher by 100 g of each compound will be 77.5 g and 58.7 g respectively.

The number of chlorine atoms in these masses can be calculated as follows,

In Freon-11

$\frac{77.5gCl}{35.5g/molCl}\times 6.02\times {10}^{23}atoms=1.3\times 10^{24}Clatoms$

In Freon-12

$\frac{58.7gCl}{35.5g/molCl}\times 6.02\times {10}^{23}atoms=1.0\times 10^{24}Clatoms$