Chapter 2, Problem 2.28QP

(a)

Interpretation Introduction

Interpretation: The number of neutrons in given isotopes of platinum is to be represented.

The average atomic mass of platinum and the comparison of average atomic mass with the value of average atomic mass, given in the periodic table are to be represented.

Concept introduction: The number of neutrons present in an isotope is calculated by the formula,

$N=A-Z$

The average atomic mass of an isotope is calculated from the formula,

$\text{Average}\text{atomic}\text{mass}=m_1\times x_1+m_2\times x_2+m_3\times x_3+\mathrm{...}+m_n{x}_n$

To determine: The number of neutrons in given isotopes of platinum.

Answer to Problem 2.28QP

Solution

The number of neutrons in given isotopes of platinum is calculated in explanation.

Explanation of Solution

Explanation

The isotopes are the group of atoms which have same number of protons and electrons but differ in number of neutrons. The neutrons are uncharged, sub atomic particles. The number of neutrons present in an isotope is calculated by the formula,

$N=A-Z$ $\left(1\right)$

Where,

• $N$ is number of neutrons.
• $A$ is atomic mass.
• $Z$ is atomic number.

There are six isotopes of platinum ${}^{190}\text{Pt,}{}^{192}\text{Pt,}{}^{\text{194}}\text{Pt,}{}^{\text{195}}\text{Pt,}{}^{\text{196}}\text{Pt}\text{and}{}^{\text{198}}\text{Pt}$. The number of neutrons present in platinum is calculated below as (I), (II), (III), (IV), (V) and (VI).

(I)

This isotope of symbol ${}^{190}\text{Pt}$ indicates that the mass number $\left(A\right)$ of platinum is $190$.

From periodic table, the atomic number $\left(Z\right)$ of platinum is $78$.

Substitute the value of atomic number $\left(Z\right)$ and the mass number $\left(A\right)$ in equation $\left(1\right)$.

$\begin{array}{c}N=A-Z\\ =190-78\\ =112\end{array}$

The number of neutrons is $\underset{\_}{112}$ in ${}^{190}\text{Pt}$.

(II)

This isotope of symbol ${}^{192}\text{Pt}$ indicates that the mass number $\left(A\right)$ of platinum is $192$.

From periodic table, the atomic number $\left(Z\right)$ of platinum is $78$.

Substitute the value of atomic number $\left(Z\right)$ and the mass number $\left(A\right)$ in equation $\left(1\right)$.

$\begin{array}{c}N=A-Z\\ =192-78\\ =114\end{array}$

The number of neutrons is $\underset{\_}{114}$ in ${}^{192}\text{Pt}$.

(III)

This isotope of symbol ${}^{194}\text{Pt}$ indicates that the mass number $\left(A\right)$ of platinum is $194$.

From periodic table, the atomic number $\left(Z\right)$ of platinum is $78$.

Substitute the value of atomic number $\left(Z\right)$ and the mass number $\left(A\right)$ in equation $\left(1\right)$.

$\begin{array}{c}N=A-Z\\ =194-78\\ =116\end{array}$

The number of neutrons is $\underset{\_}{116}$ in ${}^{194}\text{Pt}$.

(IV)

This isotope of symbol ${}^{195}\text{Pt}$ indicates that the mass number $\left(A\right)$ of platinum is $195$.

From periodic table, the atomic number $\left(Z\right)$ of platinum is $78$.

Substitute the value of atomic number $\left(Z\right)$ and the mass number $\left(A\right)$ in equation $\left(1\right)$.

$\begin{array}{c}N=A-Z\\ =195-78\\ =117\end{array}$

The number of neutrons is $\underset{\_}{117}$ in ${}^{195}\text{Pt}$.

(V)

This isotope of symbol ${}^{196}\text{Pt}$ indicates that the mass number $\left(A\right)$ of platinum is $196$.

From periodic table, the atomic number $\left(Z\right)$ of platinum is $78$.

Substitute the value of atomic number $\left(Z\right)$ and the mass number $\left(A\right)$ in equation $\left(1\right)$.

$\begin{array}{c}N=A-Z\\ =196-78\\ =118\end{array}$

The number of neutrons is $\underset{\_}{118}$ in ${}^{196}\text{Pt}$.

(VI)

This isotope of symbol ${}^{198}\text{Pt}$ indicates that the mass number $\left(A\right)$ of platinum is $198$.

From periodic table, the atomic number $\left(Z\right)$ of platinum is $78$.

Substitute the value of atomic number $\left(Z\right)$ and the mass number $\left(A\right)$ in equation $\left(1\right)$.

$\begin{array}{c}N=A-Z\\ =198-78\\ =120\end{array}$

The number of neutrons is $\underset{\_}{120}$ in ${}^{198}\text{Pt}$.

(b)

Interpretation Introduction

To determine: The average atomic mass of platinum and the comparison of average atomic mass with the value of average atomic mass, given in the periodic table.

Answer to Problem 2.28QP

Solution

The average atomic mass of platinum is $\underset{\_}{195.081amu}$.

Explanation of Solution

Explanation

The average atomic mass of an isotope is calculated from the formula,

$\text{Average}\text{atomic}\text{mass}=m_1\times x_1+m_2\times x_2+m_3\times x_3+\mathrm{...}+m_n{x}_n$ $\left(2\right)$

Where,

• $m$ is mass of an isotope.
• $x$ is natural abundance expressed in decimals.

The atomic masses of isotopes $\left({}^{190}\text{Pt,}{}^{192}\text{Pt,}{}^{\text{194}}\text{Pt,}{}^{\text{195}}\text{Pt,}{}^{\text{196}}\text{Pt}\text{and}{}^{\text{198}}\text{Pt}\right)$ are given as,

$\begin{array}{l}{m}_1=189.96\text{amu}\\ m_2=191.96\text{amu}\\ m_3=193.96\text{amu}\\ m_4=194.97\text{amu}\end{array}$

$\begin{array}{l}{m}_5=195.97\text{amu}\\ m_6=197.97\text{amu}\end{array}$

Where,

• $m_1$ is the atomic mass of ${}^{190}\text{Pt}$.
• $m_2$ is the atomic mass of ${}^{192}\text{Pt}$.
• $m_3$ is the atomic mass of ${}^{194}\text{Pt}$.
• $m_4$ is the atomic mass of ${}^{195}\text{Pt}$.
• $m_5$ is the atomic mass of ${}^{196}\text{Pt}$.
• $m_6$ is the atomic mass of ${}^{198}\text{Pt}$.

The natural abundance of isotopes $\left({}^{190}\text{Pt,}{}^{192}\text{Pt,}{}^{\text{194}}\text{Pt,}{}^{\text{195}}\text{Pt,}{}^{\text{196}}\text{Pt}\text{and}{}^{\text{198}}\text{Pt}\right)$ are given $0.014,0.782,32.967,33.832,25.242\text{and}7.163$ in percentage respectively.

The natural abundance of isotopes $\left({}^{190}\text{Pt,}{}^{192}\text{Pt,}{}^{\text{194}}\text{Pt,}{}^{\text{195}}\text{Pt,}{}^{\text{196}}\text{Pt}\text{and}{}^{\text{198}}\text{Pt}\right)$ in decimals are given as,

$\begin{array}{l}{x}_1=0.00014\\ x_2=0.00782\\ x_3=0.32967\\ x_4=0.33832\end{array}$

$\begin{array}{l}{x}_5=0.2542\\ x_6=0.07163\end{array}$

Where,

• $x_1$ is the natural abundance of isotope ${}^{190}\text{Pt}$ in decimals.
• $x_2$ is the natural abundance of isotope ${}^{192}\text{Pt}$ in decimals.
• $x_3$ is the natural abundance of isotope ${}^{194}\text{Pt}$ in decimals.
• $x_4$ is the natural abundance of isotope ${}^{195}\text{Pt}$ in decimals.
• $x_5$ is the natural abundance of isotope ${}^{196}\text{Pt}$ in decimals.
• $x_6$ is the natural abundance of isotope ${}^{198}\text{Pt}$ in decimals.

Hence,

• The value of $m_1\times x_1$ is $0.027$.
• The value of $m_2\times x_2$ is $1.501$.
• The value of $m_3\times x_3$ is $63.943$.
• The value of $m_4\times x_4$ is $65.962$.
• The value of $m_5\times x_5$ is $49.467$.
• The value of $m_6\times x_6$ is $14.181$.

Substitute the all values of $m$ and $x$ in equation $\left(2\right)$.

$\begin{array}{c}\text{Average}\text{atomic}\text{mass}=m_1\times x_1+m_2\times x_2+m_3\times x_3+m_4\times x_4+m_5\times x_5+m_6\times x_6\\ =0.027+1.501+63.943+65.962+49.467+14.181\\ \text{Average}\text{atomic}\text{mass}=195.081\text{amu}\end{array}$

Therefore, the average atomic mass of platinum is $\underset{\_}{195.081}$. The value of average atomic mass of platinum in periodic table is $195.084\text{amu}$. This information indicates that the calculated result is very close to actual value given in periodic table.

Conclusion

1. a. The number of neutrons in given isotopes of platinum is calculated in explanation.
2. b. The average atomic mass of platinum is $\underset{\_}{195.081amu}$.