Chapter 2, Problem 2.29QP

Interpretation Introduction

Interpretation: The exact mass of ${}^{48}\text{Ti}$ isotope is to be calculated from the given table.

Concept introduction: The exact 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 exact mass of ${}^{48}\text{Ti}$ isotope from the given table.

Answer to Problem 2.29QP

Solution

The exact mass of ${}^{48}\text{Ti}$ isotope is $\underset{\_}{47.95amu}$.

Explanation of Solution

Explanation

The exact mass of an isotope is calculated from the formula,

$\text{Average}\text{atomic}\text{mass}=m_1\times x_1+m_2\times x_2+m_3\times x_3+\mathrm{...}+m_n{x}_n$ $\left(1\right)$

Where,

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

The given values and natural abundance in decimals are summarized in Table $1$.

Symbol	Exact mass $\left(m\right)$ in amu	Natural abundance in percentage	Natural abundance in decimals $\left(x\right)$
${}^{46}\text{Ti}$	$45.95263=m_1$	$8.25\%$	$0.0825=x_1$
${}^{47}\text{Ti}$	$46.9518=m_2$	$7.44\%$	$0.0744=x_2$
${}^{48}\text{Ti}$	$m_3$	$73.72\%$	$0.7372=x_3$
${}^{49}\text{Ti}$	$48.94787=m_4$	$5.41\%$	$0.0541=x_4$
${}^{50}\text{Ti}$	$49.9448=m_5$	$5.18\%$	$0.0518=x_5$
Average	$47.87$

Table $1$

In Table $1$,

• $m_1$ is the exact  mass of ${}^{46}\text{Ti}$ isotope.
• $m_2$ is the exact mass of ${}^{47}\text{Ti}$ isotope.
• $m_3$ is the exact mass of ${}^{48}\text{Ti}$ isotope
• $m_4$ is the exact mass of ${}^{49}\text{Ti}$ isotope.
• $m_5$ is the exact mass of ${}^{50}\text{Ti}$ isotope.
• $x_1$ is the natural abundance in decimals for ${}^{46}\text{Ti}$ isotope.
• $x_2$ is the natural abundance in decimals for ${}^{47}\text{Ti}$ isotope.
• $x_3$ is the natural abundance in decimals for ${}^{48}\text{Ti}$ isotope.
• $x_4$ is the natural abundance in decimals for ${}^{49}\text{Ti}$ isotope.
• $x_5$ is the natural abundance in decimals for ${}^{50}\text{Ti}$ isotope.

The $m_3$, exact mass of ${}^{48}\text{Ti}$ is to be calculated.

Substitute the all values of $\text{m,}\text{x}$ and average atomic mass in equation $\left(1\right)$.

$\begin{array}{c}\text{Average}\text{atomic}\text{mass}=m_1\times x_1+m_2\times x_2+m_3\times x_3+m_4\times x_4+m_5\times x_5\\ 47.87=\left(\begin{array}{c}\left(45.95263\times 0.0825\right)+\left(46.9518\times 0.0744\right)+\left(m_3\times 0.7372\right)\\ +\left(48.94787\times 0.0541\right)+\left(49.9448\times 0.0518\right)\end{array}\right)\\ =3.7911+3.4932+0.7372m_3+2.6481+2.5871\end{array}$

Simplifying,

$\begin{array}{c}47.87=12.5195+0.7372m_3\\ 0.7372m_3=47.87-12.5195\\ 0.7372m_3=35.3505\\ m_3=\frac{35.3505}{0.7372}\end{array}$

So,

$m_3=47.95\text{amu}$

Hence, the exact mass of ${}^{48}\text{Ti}$ isotope is $\underset{\_}{47.95amu}$.

Conclusion

The exact mass of ${}^{48}\text{Ti}$ isotope is $\underset{\_}{47.95amu}$