Chapter 5, Problem 5.78QP

Interpretation Introduction

Interpretation:

The given interhalogen compounds have to be arranged in increasing order of percentage of fluorine by mass.

Concept introduction:

Percent composition by mass of each element present in a compound is known as the compound’s percent composition by mass.

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{an}\text{element}\text{=}\frac{\text{n}\text{×atomic}\text{mass}\text{of}\text{element}}{\text{molecular}\text{or}\text{formula}\text{mass}\text{of}\text{compound}}\text{×100\%}\\ \text{Where}\text{n}\text{is}\text{the}\text{number}\text{of}\text{atoms}\text{of}\text{the}\text{element}\text{in}\text{a}\text{molecule}\text{or}\text{formula}\text{unit}\text{of}\text{the}\text{compound}\end{array}$

Answer to Problem 5.78QP

The percentage fluorine by mass of interhalogen compounds is given below:

$\text{percentage}\text{mass}\text{of}\text{\%}\text{F}\text{in}{\text{IF}}_{\text{3}}\text{=}\text{30}\text{.9925}\text{\%}$

$\text{percentage}\text{mass}\text{of}\text{\%}\text{F}\text{in}\text{ClF}\text{=}\text{34}\text{.8906}\text{\%}$

$\text{percentage}\text{mass}\text{of}\text{\%}\text{F}\text{in}{\text{BrF}}_{\text{3}}\text{=}\text{41}\text{.6330}\text{\%}$

$\text{percentage}\text{mass}\text{of}\text{\%}\text{F}\text{in}{\text{IF}}_{\text{7}}\text{=}\text{51}\text{.1705}\text{\%}$

The percentage of fluorine by mass of given interhalogen compounds was arranged in the increasing order as,

${\text{IF}}_{\text{3}}{\text{<ClF<BrF}}_{\text{3}}{\text{<IF}}_{\text{7}}$

Explanation of Solution

Percent composition by mass: Percent composition by mass of each element present in a compound is known as the compound’s percent composition by mass.  The formula is expressed mathematically as follows:

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{an}\text{element}\text{=}\frac{\text{n}\text{×atomic}\text{mass}\text{of}\text{element}}{\text{molecular}\text{or}\text{formula}\text{mass}\text{of}\text{compound}}\text{×100\%}\\ \text{Where}\text{n}\text{is}\text{the}\text{number}\text{of}\text{atoms}\text{of}\text{the}\text{element}\text{in}\text{a}\text{molecule}\text{or}\text{formula}\text{unit}\text{of}\text{the}\text{compound}\end{array}$ By using the above formula, percentage fluorine by mass of interhalogen compounds is calculated as follows:

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{\%}\text{F}\text{in}{\text{BrF}}_{\text{3}}\text{=}\frac{\text{3×atomic}\text{mass}\text{F}\text{in}\text{}\text{amu}}{\text{molecular}\text{mass}\text{of}{\text{BrF}}_{\text{3}}\text{in}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{3×18}\text{.9984}\text{amu}}{\text{136}\text{.8992}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{56}\text{.9952}\text{amu}}{\text{136}\text{.8992}\text{amu}}\text{×}\text{100\%}\\ \text{=}\text{41}\text{.6330}\text{\%}\end{array}$

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{\%}\text{F}\text{in}{\text{IF}}_{\text{7}}\text{=}\frac{\text{7×atomic}\text{mass}\text{F}\text{in}\text{}\text{amu}}{\text{molecular}\text{mass}\text{of}{\text{IF}}_{\text{7}}\text{in}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{7×18}\text{.9984}\text{amu}}{\text{259}\text{.8933}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{132}\text{.9888}\text{amu}}{\text{259}\text{.8933}\text{amu}}\text{×}\text{100\%}\\ \text{=}\text{51}\text{.1705}\text{\%}\end{array}$

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{\%}\text{F}\text{in}{\text{IF}}_{\text{3}}\text{=}\frac{\text{3×atomic}\text{mass}\text{F}\text{in}\text{}\text{amu}}{\text{molecular}\text{mass}\text{of}{\text{IF}}_{\text{3}}\text{in}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{3×18}\text{.9984}\text{amu}}{\text{183}\text{.8997}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{56}\text{.9952}\text{amu}}{\text{183}\text{.8997}\text{amu}}\text{×}\text{100\%}\\ \text{=}\text{30}\text{.9925}\text{\%}\end{array}$

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{\%}\text{F}\text{in}\text{ClF}\text{=}\frac{\text{1×atomic}\text{mass}\text{F}\text{in}\text{}\text{amu}}{\text{molecular}\text{mass}\text{of}\text{ClF}\text{in}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{18}\text{.9984}\text{amu}}{\text{54}\text{.4514}\text{amu}}\text{×}\text{100\%}\\ \text{=}\text{34}\text{.8906}\text{\%}\end{array}$

The percentage fluorine by mass of interhalogen compounds is given below:

$\text{percentage}\text{mass}\text{of}\text{\%}\text{F}\text{in}{\text{IF}}_{\text{3}}\text{=}\text{30}\text{.9925}\text{\%}$

$\text{percentage}\text{mass}\text{of}\text{\%}\text{F}\text{in}\text{ClF}\text{=}\text{34}\text{.8906}\text{\%}$

$\text{percentage}\text{mass}\text{of}\text{\%}\text{F}\text{in}{\text{BrF}}_{\text{3}}\text{=}\text{41}\text{.6330}\text{\%}$

$\text{percentage}\text{mass}\text{of}\text{\%}\text{F}\text{in}{\text{IF}}_{\text{7}}\text{=}\text{51}\text{.1705}\text{\%}$

Conclusion

The percentage fluorine by mass of interhalogen compounds was arranged in the below order:

${\text{IF}}_{\text{3}}{\text{<ClF<BrF}}_{\text{3}}{\text{<IF}}_{\text{7}}$