Chapter 4, Problem 64A

Interpretation Introduction

Interpretation : The approximate atomic mass of lead is to be calculated.

Concept Introduction : The weighted average mass of the atoms in a naturally occurring sample of an element is the atomic mass. An element's atomic mass is calculated as the weighted average of all of its isotopes' masses. The formula of atomic mass is given as:

  $\text{Atomic mass}\text{=}\text{Isotope}\text{mass number}\times \text{Natural percent abundance}$

Answer to Problem 64A

The approximate atomic mass of lead is $207.24\text{amu}$ .

Explanation of Solution

Given information :

There are four isotopes of lead.

The isotope of lead one has 82 protons and 122 neutrons with 1.4% abundance.

The isotope of lead two has 82 protons and 125 neutrons with 22.1% abundance.

The isotope of lead three has 82 protons and 124 neutrons with 24.1% abundance.

The isotope of lead four has 82 protons and 126 neutrons with 52.4% abundance.

The weighted average mass of the atoms in a naturally occurring sample of an element is the atomic mass. An element's atomic mass is calculated as the weighted average of all of its isotopes' masses. The formula of atomic mass is given as:

  $\text{Atomic mass}\text{=}\text{Isotope}\text{mass number}\times \text{Natural percent abundance}$

To calculate the atomic mass, first, calculate the mass number of given isotopes of lead.

The formula of mass number is given as:

  $\text{mass}\text{number}\text{=}\text{number}\text{of}\text{protons +}\text{number}\text{of}\text{neutrons}$

The mass number of lead one is given as:

  $\begin{array}{c}\text{mass}\text{number}\text{=}\text{number}\text{of}\text{protons +}\text{number}\text{of}\text{neutrons}\\ \text{=}\text{82+122}\\ \text{=}\text{204}\end{array}$

The mass number of lead two is given as:

  $\begin{array}{c}\text{mass}\text{number}\text{=}\text{number}\text{of}\text{protons +}\text{number}\text{of}\text{neutrons}\\ \text{=}\text{82+125}\\ \text{=}\text{207}\end{array}$

The mass number of lead three is given as:

  $\begin{array}{c}\text{mass}\text{number}\text{=}\text{number}\text{of}\text{protons +}\text{number}\text{of}\text{neutrons}\\ \text{=}\text{82+124}\\ \text{=}\text{206}\end{array}$

The mass number of lead four is given as:

  $\begin{array}{c}\text{mass}\text{number}\text{=}\text{number}\text{of}\text{protons +}\text{number}\text{of}\text{neutrons}\\ \text{=}\text{82+126}\\ \text{=}\text{208}\end{array}$

The given elements are rewritten as Lead-204, Lead-207, Lead-206 and Lead-208.

The formula to calculate the atomic mass is also rewritten as:

  $\text{Atomic mass}\text{=}\left(\text{Pb-204×1}\text{.4\%}\right)\text{+}\left(\text{Pb-207×22}\text{.1\%}\right)\text{+}\left(\text{Pb-206×24}\text{.1\%}\right)\text{+}\left(\text{Pb-208×52}\text{.4\%}\right)$

To calculate the atomic mass, substitute the values in the above formula:

  $\begin{array}{c}\text{Atomic mass}\text{=}\left(\text{Pb-204×1}\text{.4\%}\right)\text{+}\left(\text{Pb-207×22}\text{.1\%}\right)\text{+}\left(\text{Pb-206×24}\text{.1\%}\right)\text{+}\left(\text{Pb-208×52}\text{.4\%}\right)\\ =\left(204\times 0.014\right)+\left(207\times 0.221\right)+\left(206\times 0.241\right)+\left(208\times 0.524\right)\\ =2.856+45.747+49.646+108.992\\ =207.24\text{amu}\end{array}$

The atomic mass of lead is $207.24\text{amu}$ .