Comprehensive Notes on Isotopes, Relative Atomic Mass Calculations, and Noble Gas Stability

Fundamental Concepts of Isotopes

• Definition of Isotopes: Isotopes are different atoms of the same element that possess the same number of protons but a different number of neutrons.
• Subatomic Particle Rules:
  • Across all isotopes of a given element, the number of protons remains identical.
  • Because neutral atoms have an equal number of protons and electrons, the number of electrons is also identical across all isotopes of an element.
  • The number of neutrons varies among different isotopes of the same element.

Isotopic Data of Common Elements

• Most chemical elements exist naturally as a mixture of more than one isotope.
• Hydrogen Isotopes:
  • ${}^{1}_{1}\text{H}$ (Protium): Contains $1$ proton, $1$ electron, and $0$ neutrons. Mass number = $1$ ($1p + 0n$). Natural percentage abundance = $99.984\%$.
  • ${}^{2}_{1}\text{H}$ (Deuterium): Contains $1$ proton, $1$ electron, and $1$ neutron. Mass number = $2$ ($1p + 1n$). Natural percentage abundance = $0.016\%$.
  • ${}^{3}_{1}\text{H}$ (Tritium): Contains $1$ proton, $1$ electron, and $2$ neutrons. Mass number = $3$ ($1p + 2n$). Natural percentage abundance = Small trace percentage.
• Carbon Isotopes:
  • ${}^{12}_{6}\text{C}$ (Carbon-12): Atomic number = $6$, Contains $6$ protons and $6$ neutrons. Mass number = $12$ ($6p + 6n$). Natural percentage abundance = $98.892\%$.
  • ${}^{13}_{6}\text{C}$ (Carbon-13): Atomic number = $6$, Contains $6$ protons and $7$ neutrons. Mass number = $13$ ($6p + 7n$). Natural percentage abundance = $1.108\%$.
  • ${}^{14}_{6}\text{C}$ (Carbon-14): Atomic number = $6$, Contains $6$ protons and $8$ neutrons. Mass number = $14$ ($6p + 8n$). Natural percentage abundance = Small trace percentage.
• Sodium Isotopes:
  • ${}^{23}_{11}\text{Na}$: Atomic number = $11$, Contains $11$ protons and $12$ neutrons. Mass number = $23$ ($11p + 12n$). Natural percentage abundance = $100\%$.
• Elements Without Natural Isotopes:
  • Not all chemical elements have multiple naturally occurring isotopes. Elements such as Fluorine ($\text{F}$), Sodium ($\text{Na}$), and Aluminium ($\text{Al}$) do not have multiple natural isotopes.

Physical and Chemical Properties of Isotopes

• Chemical Properties:
  • Isotopes of the same element exhibit identical chemical properties.
  • Chemical behavior is determined by the number of protons and outer-shell electrons. Since all isotopes of an element have the exact same number of protons and electrons, their chemical reactions and properties remain unchanged.
• Physical Properties:
  • Isotopes of the same element exhibit different physical properties.
  • The differing number of neutrons leads to different atomic masses.
  • Physical constants such as melting points and boiling points vary slightly with the mass of the isotope (i.e., physical properties change with isotopic mass).

Particle Identification Practice

• Particle Data Analysis:
  • Particle $\text{W}$: Protons = $6$, Neutrons = $6$, Electrons = $6$
  • Particle $\text{X}$: Protons = $8$, Neutrons = $8$, Electrons = $8$
  • Particle $\text{Y}$: Protons = $6$, Neutrons = $8$, Electrons = $6$
  • Particle $\text{Z}$: Protons = $7$, Neutrons = $7$, Electrons = $7$
• Identification:
  • Particles $\text{W}$ and $\text{Y}$ are isotopes of the same element (Carbon).
  • Reason: Both particles possess the exact same number of protons ($6$), identifying them as the same element, but different numbers of neutrons ($6$ vs $8$).

Relative Isotopic Mass and Relative Atomic Mass

• Necessity of Relative Mass:
  • The actual masses of individual atoms are extremely light. Expressing atomic masses using standard SI mass units such as grams ($\text{g}$) or kilograms ($\text{kg}$) is impractical for calculations.
• The Reference Standard: Carbon-12 Isotope:
  • The Carbon-12 isotope (${}^{12}\text{C}$) is universally designated as the reference standard atom for comparing atomic masses.
  • Key Reasons for Selecting Carbon-12:
  1. Carbon-12 does not undergo radioactive decay; its mass never changes over time.
  2. It is highly abundant in nature ($98.892\%$ natural abundance).
  3. It is assigned an exact fixed relative isotopic mass of $12.000\,\text{amu}$ (atomic mass units).
• Relative Isotopic Mass (RIM):
  • Definition: The mass of a specific isotope compared with $\frac{1}{12}$th of the mass of a Carbon-12 atom on the $\text{C-12}$ scale ($12.000\,\text{amu}$).
  • Relative Isotopic Mass is a relative ratio and has no units.
  • Mass Comparison Example: One Magnesium atom (${}^{24}\text{Mg}$) has a mass equal to two Carbon-12 atoms ($12 \times 2 = 24$). Consequently, the Relative Isotopic Mass of ${}^{24}\text{Mg}$ is exactly $24$ (dimensionless).
  • Approximation: The Relative Isotopic Mass of an isotope is approximately equal to its mass number:     $\text{RIM} \approx \text{Mass Number}$

Relative Atomic Mass (RAM) Calculations

• Need for Weighted Average:
  • Because an element can consist of multiple isotopes with different isotopic masses (for example, Hydrogen has three isotopes with masses $1$, $2$, and $3$), a single element-wide average mass value must be established.
• Definition of Relative Atomic Mass (RAM):
  • Relative Atomic Mass (RAM) is defined as the weighted average of the relative isotopic masses of all naturally occurring isotopes of an element, based on the Carbon-12 scale ($12.000\,\text{amu}$).
• Mathematical Formula for RAM:
  • To compute the average relative mass, the RIM of each isotope must be weighted by its percentage abundance in nature:     $\text{RAM} = (A\% \times \text{RIM}_A) + (B\% \times \text{RIM}_B) + (C\% \times \text{RIM}_C)$
  • Expressed as a single formula over $100$:     $\text{RAM} = \frac{(\text{RIM}_A \times x) + (\text{RIM}_B \times y)}{100}$     where $x$ and $y$ represent the percentage abundances of isotopes $\text{A}$ and $\text{B}$ respectively ($x + y = 100\%$), and $\text{RIM}_{A/B}$ represents the Relative Isotopic Mass of each respective isotope.

Practice Questions and Step-by-Step Worked Solutions

• Question 1: Name the isotope used by scientists to compare mass with other atoms.

  • Answer: Carbon-12 isotope (${}^{12}\text{C}$).

• Question 2: Find the Relative Isotopic Mass (RIM) of these atoms:

  • (a) ${}^{35}_{17}\text{Cl}$:
  • Answer: $\text{RIM} \approx 35$
  • (b) ${}^{37}_{17}\text{Cl}$:
  • Answer: $\text{RIM} \approx 37$
  • (c) ${}^{16}_{8}\text{O}$:
  • Answer: $\text{RIM} \approx 16$

• Question 3: Bromine ($\text{Br}$) has 2 isotopes, ${}^{79}\text{Br}$ and ${}^{81}\text{Br}$, with percentage abundances of $55\%$ and $45\%$ respectively. Calculate the Relative Atomic Mass (RAM).

  • Calculation:     $\text{RAM} = (79 \times 55\%) + (81 \times 45\%)$$\text{RAM} = (79 \times 0.55) + (81 \times 0.45)$$\text{RAM} = 43.45 + 36.45 = 79.9$
  • Answer: $79.9$

• Question 4: Natural chlorine has a Relative Atomic Mass (RAM) of $35.5$ and consists of two isotopes, ${}^{35}\text{Cl}$ and ${}^{37}\text{Cl}$. Calculate the percentage abundance of ${}^{35}\text{Cl}$ and ${}^{37}\text{Cl}$.

  • Step-by-Step Solution:     Let $x$ be the percentage abundance of ${}^{35}\text{Cl}$, and $y$ be the percentage abundance of ${}^{37}\text{Cl}$.     Since total percentage abundance equals $100\%$:     $x + y = 100 \implies y = 100 - x \quad \text{--- Equation (1)}$     Setting up the weighted average RAM equation:     $\frac{35x + 37y}{100} = 35.5 \quad \text{--- Equation (2)}$     Multiply both sides by $100$:     $35x + 37y = 3550$     Substitute Equation (1) into Equation (2):     $35x + 37(100 - x) = 3550$$35x + 3700 - 37x = 3550$$-2x = 3550 - 3700$$-2x = -150$$x = 75$     Substitute $x = 75$ back into Equation (1):     $75 + y = 100 \implies y = 25$
  • Answer: Percentage abundance of ${}^{35}\text{Cl} = 75\%$, Percentage abundance of {}^{37}\text{Cl} = 25\%$.\n\n# DSE Exam Checklist: Isotope Definition Requirements\n\n- When defining **isotopes** in Hong Kong DSE examination answers, responses **MUST** include the following four phrases:\n 1. **different atoms**\n 2. **same element**\n 3. **different amount of neutrons**\n 4. **same number of protons**\n\n# Electronic Arrangement and Stability of Noble Gases\n\n- **Noble Gas Stability**: Noble gases display exceptional chemical stability due to their electronic arrangement.\n- **Electronic Arrangements**:\n - Helium (\text{He}$, atomic number$2$):$2\n - Neon (\text{Ne}$, atomic number$10$):$2, 8\n - Argon (\text{Ar}$, atomic number$18$):$2, 8, 8\n - Krypton (\text{Kr}$, atomic number$36$):$2, 8, 18, 8\n - Xenon (\text{Xe}$, atomic number$54$):$2, 8, 18, 18, 8\n - Radon (\text{Rn}$, atomic number$86$):$2, 8, 18, 32, 18, 8\n- **Core Stability Rules**:\n - All noble gases except Helium possess 8 outermost shell electrons.\n - **Octet Rule**: A particle achieves great stability if it has an octet of electrons (8 outermost shell electrons).\n - **Duplet Rule**: A particle achieves great stability if it has a duplet of electrons (2 electrons in **ONLY one occupied electron shell**).\n- **Explanation of Single Occupied Shell Requirement for Duplet Stability**:\n - Helium is stable because 2$electrons fill the total capacity of its single electron shell ($n=1).\n - Other elements with 2 atoms/electrons in higher shells still have available space in their outer shell and are therefore reactive.\n\n# Applications and Practical Uses of Noble Gases\n\n- **Helium (\text{He})**:\n - Application: Used to fill airships and weather balloons.\n - Reasons:\n 1. It is much lighter than air.\n 2. It is chemically unreactive, making it non-explosive and completely safe.\n- **Argon (\text{Ar})**:\n - Application: Used to fill tungsten light bulbs.\n - Reasons:\n 1. It is chemically unreactive.\n 2. It will not react with hot tungsten filaments, giving the light bulb a longer lifetime.\n\n# Electronic Shell Rules and Common Misconceptions\n\n- **General Electronic Shell Rules**:\n - Occupying a full electron shell provides stability.\n - Outermost shell electrons cap at either 2$or$8\n - The Duet rule applies strictly to the 1\text{st} shell.\n - The Octet rule applies to all other occupied shells.\n- **Common Exam Trap & Misconception**:\n - **Question**: Is the statement *"Electrons move around the atom at high speed"* true or false?\n - **Answer**: The statement is **WRONG**.\n - **Explanation**: An atom consists of protons, neutrons, and electrons (p + n + e$). Electrons do not orbit the *atom*; instead, electrons orbit the **nucleus** ($p + n$$).