Chapter 2, Problem ISP

An isotope of an element contains 63 protons and 91 neutrons.

1. (a) Identify the element and give its symbol.
2. (b) Give the element’s atomic number.
3. (c) Give the mass number of the isotope.
4. (d) This element has two naturally occurring isotopes. Given the information in the table, calculate the atomic weight of the element.

1. (e) In which region of the periodic table is the element found? Explain your answer.
2. (f) Is the element a metal, metalloid, or nonmetal? Explain your answer.
3. (g) This element, used in compact fluorescent light bulbs and computer screens, has an atomic radius of 180 pm. Calculate how long the chain of atoms would be if all the atoms in a 1.25-mg sample of this element were put into a row.

Interpretation Introduction

Interpretation:

The isotope of an element contains 63 protons and 91 neutrons.  The element has to be identified and the symbol of the element has to be given.

Explanation of Solution

Number of protons of the element is 63.

Number of neutrons of the element is 91.

The atomic number is same as that of number of protons, 63.  Also the number of protons is same as that number of electrons in an uncharged atom. The element is Europium with a symbol of $\text{Eu}$.

Interpretation Introduction

Interpretation:

The atomic number of element with 63 protons and 91 neutrons has to be given.

Explanation of Solution

Number of protons of the element is 63.

The atomic number is same as that of number of protons, 63.

Atomic number $\text{=}\text{Z=63}$.

Interpretation Introduction

Interpretation:

The mass number of the isotope with 63 protons and 91 neutrons has to be given.

Explanation of Solution

Number of protons of the element is 63.

Number of neutrons of the element is 91.

Atomic number $\text{=}\text{Z=63}$.

Mass number is the sum of atomic number of the element and number of neutrons.

Isotope’s mass number $\text{=}$ $\text{A}\text{=}\text{Z}\text{+}\text{number}\text{of}\text{neutrons}\text{=}\text{63}\text{+91}\text{=}\text{151}$.

Interpretation Introduction

Interpretation:

The element has two naturally occurring isotopes.  From the table given the atomic weight of the element has to be calculated.

$\begin{array}{l}\text{Isotope}\text{Mass}\text{number}\text{percent}\text{abundance}\text{isotopic}\text{mass}\text{(u)}\\ \\ \text{1}\text{151}\text{47}\text{.80}\text{150}\text{.920}\\ \\ \text{2}\text{153}\text{52}\text{.20}\text{152}\text{.921}\end{array}$

Explanation of Solution

Every 10000 atoms of the element contain 4780 atoms of the ${}^{\text{151}}\text{Eu}$ isotope with an atomic mass of $\text{150}\text{.920}\text{u}$ and 5220 atoms of the ${}^{\text{153}}\text{Eu}$ isotope with an atomic mass of $\text{152}\text{.922}\text{u}$.

$\frac{\text{4780}\text{atoms}{}^{\text{151}}\text{Eu}}{\text{10000}\text{Eu}\text{atoms}}\text{×}\left(\frac{\text{150}\text{.920}\text{u}}{\text{1}\text{atom}{}^{\text{151}}\text{Eu}}\right)\text{+}\left(\frac{\text{5220}\text{atoms}{}^{\text{153}}\text{Eu}}{\text{10000}\text{Eu}\text{atoms}}\right)\text{×}\left(\frac{\text{152}\text{.922}\text{u}}{\text{1}\text{atom}{}^{\text{153}}\text{Eu}}\right)\text{=}\text{151}\text{.965u/Eu}\text{atoms}$

Interpretation Introduction

Interpretation:

The region in which the element is found in periodic table has to be explained.

Explanation of Solution

The element is found in the lanthanide series because it shows up in the row labeled “Lanthanides”.

Interpretation Introduction

Interpretation:

The given element is a metal, metalloid or non-metal has to be found and the answer has to be explained.

Explanation of Solution

The element Europium is a transition metal.

Interpretation Introduction

Interpretation:

The element given has an atomic radius of $\text{180}\text{pm}$.  The length of the chain of atoms would be if all the atoms in a $\text{1}\text{.25-mg}$ sample of this element were put into a row has to be calculated.

Explanation of Solution

$\text{1}\text{.25}\text{mg}\text{Eu}\text{×}\left(\frac{\text{1}\text{g}}{\text{1000}\text{mg}}\right)\text{×}\left(\frac{\text{1}\text{mol}\text{Eu}}{\text{151}\text{.965}\text{g}\text{Eu}}\right)\text{×}\left(\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{Eu}\text{atoms}}{\text{1}\text{mol}\text{Eu}\text{atoms}}\right)\text{=}\text{4}\text{.95}\text{×}{\text{10}}^{\text{18}}\text{Eu}\text{atoms}$.

The atomic radius is $\text{180}\text{pm}$.  The diameter is twice the radius and by multiplying the number of atoms by the atomic diameter and using the metric conversions to determine the number of meters.

    $\text{4}\text{.95}\text{×}{\text{10}}^{\text{18}}\text{Eu}\text{atoms}\text{×}\text{2}\text{×}\text{(180}\text{pm)}\text{×}\frac{\text{1}\text{×}{\text{10}}^{\text{-12}}\text{m}}{\text{1}\text{pm}}\text{=}\text{1}\text{.78}\text{×}{\text{10}}^{\text{9}}\text{m}$.