Chem Week 2 Lecture 4: Atomic Number vs Mass and Isotopes

Fundamental Subatomic Particles and Atomic Identifiers

• Proton Identity and Element Definition:

  • The number of protons in an atom's nucleus is the primary identity marker of an element. The number of protons strictly defines the element.

  • The atomic number represents the total number of protons in an atom's nucleus.

  • Standard notation utilizes the symbol $Z$ for the atomic number. In specific context-dependent textbook or exam applications, the letter $c$ may serve as a variable stand-in for the atomic number.

• Electrical Neutrality and Electrons:

  • Each proton carries an electrical charge of $+1$.

  • In a neutral atom, the overall charge is $0$. To balance the positive nuclear charge generated by $Z$ protons, a neutral atom must possess an equal number of electrons, each carrying a charge of $-1$.

  • $\text{Number of protons } (p^+) = \text{Number of electrons } (e^-) = Z \text{ (in a neutral atom)}$

  • If an atom contains $24$ protons, the total positive charge in the nucleus is $+24$. To achieve an overall neutral charge of $0$, there must be $24$ electrons present to provide a charge of $-24$.

  • Electrons can be gained or lost during chemical processes, forming charged atoms known as ions. Electron transfers dictate chemical reactivity.

• Mass Number and Neutrons:

  • The mass number is represented by the symbol $A$.

  • The mass number is defined as the total sum of protons and neutrons within the nucleus of an atom:         $A = \text{number of protons } (p^+) + \text{number of neutrons } (n)$

  • The number of neutrons in a given atom is derived by subtracting the atomic number ($Z$) from the mass number ($A$):         $\text{Number of neutrons } (n) = A - Z$

Symbolic Representation of Atoms and Isotopes

• Standard Isotopic Symbol Notation:

  • An isotope is represented symbolically using the format:         $\mathbf{^{A}_{Z}X}$         where:

    • $X$ is the chemical symbol of the element.

    • $A$ is the mass number, written as a upper-left superscript.

    • $Z$ is the atomic number, written as a lower-left subscript.

• Hyphenated Isotope Notation:

  • Isotopes are frequently represented by writing the chemical symbol or full element name followed by a dash and the mass number ($A$).

  • Examples include Carbon-13 (or $\text{C-13}$), Carbon-14 (or $\text{C-14}$), and Chlorine-35 (or $\text{Cl-35}$).

• Ion Charge Placement:

  • When an atom carries a net charge, the numerical charge is written in the upper-right superscript position relative to the chemical symbol ($X$).

Isotopes and Natural Abundance

• Definition of Isotopes:

  • Isotopes are atoms of the same element that share the same atomic number ($Z$, same number of protons) but possess different numbers of neutrons ($n$), resulting in differing mass numbers ($A$) and distinct physical masses.

• Natural Abundance:

  • The relative percentage of each naturally occurring isotope in a standardized sample of an element is constant. This fixed percentage distribution is termed the natural abundance.

• Isotopic Variations of Neon ($\text{Ne}$, $Z = 10$):

  • Neon exists as three distinct natural isotopes, all containing $10$ protons:

    • Neon-20 ($\mathbf{^{20}_{10}Ne}$): Contains $10$ protons and $10$ neutrons ($20 - 10 = 10$). Natural abundance = $90.48\%$. Out of $100$ naturally occurring neon atoms, $90.48$ will contain $10$ neutrons.

    • Neon-21 ($\mathbf{^{21}_{10}Ne}$): Contains $10$ protons and $11$ neutrons ($21 - 10 = 11$). Natural abundance = $0.27\%$.

    • Neon-22 ($\mathbf{^{22}_{10}Ne}$): Contains $10$ protons and $12$ neutrons ($22 - 10 = 12$). Natural abundance = $9.25\%$.

• Isotopic Variations of Carbon ($\text{C}$, $Z = 6$):

  • All carbon isotopes possess $6$ protons ($Z = 6$):

    • Carbon-12 ($\mathbf{^{12}_{6}C}$): Contains $6$ protons and $6$ neutrons ($12 - 6 = 6$).

    • Carbon-13 ($\mathbf{^{13}_{6}C}$): Contains $6$ protons and $7$ neutrons ($13 - 6 = 7$).

    • Carbon-14 ($\mathbf{^{14}_{6}C}$): Contains $6$ protons and $8$ neutrons ($14 - 6 = 8$).

Comprehensive Practice Problems and Solutions

• Problem 1: Chlorine Isotope with 18 Neutrons

  • Task: Determine the atomic number ($Z$), mass number ($A$), and isotopic symbol for a chlorine atom containing $18$ neutrons.

  • Step 1: Identify atomic number $Z$ from the periodic table for chlorine ($\text{Cl}$): $Z = 17$ ($17$ protons).

  • Step 2: Calculate mass number $A$:         $A = Z + n = 17 + 18 = 35$

  • Step 3: Construct isotopic symbol:         $\mathbf{^{35}_{17}Cl}$

• Problem 2: Neutral Chromium-52 Atom ($\mathbf{^{52}_{24}Cr}$)

  • Task: Determine the total number of protons, electrons, and neutrons in $\mathbf{^{52}_{24}Cr}$.

  • Step 1: Atomic number $Z = 24$, therefore Protons = $24$.

  • Step 2: Because it is a neutral atom, Electrons = Protons = $24$.

  • Step 3: Calculate neutrons:         $\text{Neutrons } (n) = A - Z = 52 - 24 = 28$

• Problem 3: Carbon Isotope with 7 Neutrons

  • Task: Determine the atomic number ($Z$), mass number ($A$), and symbol for the carbon isotope containing $7$ neutrons.

  • Step 1: Atomic number $Z$ for carbon ($\text{C}$) = $6$.

  • Step 2: Calculate mass number $A$:         $A = Z + n = 6 + 7 = 13$

  • Step 3: Construct symbol:         $\mathbf{^{13}_{6}C}$

• Problem 4: Neutral Potassium-39 Atom ($\mathbf{^{39}_{19}K}$)

  • Task: Determine the number of protons, neutrons, and electrons present in an atom of $\mathbf{^{39}_{19}K}$.

  • Step 1: Atomic number $Z = 19$, therefore Protons = $19$.

  • Step 2: Calculate neutrons:         $\text{Neutrons } (n) = A - Z = 39 - 19 = 20$

  • Step 3: Neutral atom, therefore Electrons = Protons = $19$.

Atomic Mass and Weighted Average Calculations

• Concept of Atomic Mass:

  • Atomic mass (also called atomic weight or standard atomic weight) is defined as the average mass of the isotopes that compose an element.

  • It appears directly beneath the chemical symbol on the periodic table (e.g., Phosphorus has an atomic mass of $30.97\,\text{amu}$).

  • Atomic mass is expressed in atomic mass units ($\text{amu}$).

  • It is a weighted average value based on the natural percentage abundance of each naturally occurring isotope.

• General Atomic Mass Equation:     $\text{Atomic Mass} = \sum_{n} \left(\text{Fraction of Isotope } n \times \text{Mass of Isotope } n\right)$

  • Where $\text{Fraction of Isotope } n = \frac{\text{Percent Abundance}}{100}$.

• Step-by-Step Calculation for Natural Chlorine:

  • Given Data: Chlorine consists of two stable isotopes:

    1. Chlorine-35: Natural abundance = $75.77\%$, Isotopic mass = $34.97\,\text{amu}$.

    2. Chlorine-37: Natural abundance = $24.23\%$, Isotopic mass = $36.97\,\text{amu}$.

  • Step 1: Convert percent abundances to decimal fractions:         $\text{Fraction of Cl-35} = \frac{75.77}{100} = 0.7577$         $\text{Fraction of Cl-37} = \frac{24.23}{100} = 0.2423$

  • Step 2: Calculate the mass contribution of each isotope:         $\text{Contribution from Cl-35} = 0.7577 \times 34.97\,\text{amu} = 26.4968\,\text{amu}$         $\text{Contribution from Cl-37} = 0.2423 \times 36.97\,\text{amu} = 8.9578\,\text{amu}$

  • Step 3: Sum the individual mass contributions to get total atomic mass:         $\text{Atomic Mass of Cl} = 26.4968\,\text{amu} + 8.9578\,\text{amu} = 35.4546\,\text{amu} \approx 35.45\,\text{amu}$

Conceptual Analogy: Charizard Card Grading Metaphor

• To understand weighted averages intuitively, consider evaluating the average grade of collectible Charizard cards graded on a numerical scale (6 to 10).

• Example 1: Sample Population of 10 Cards

  • Distribution Data:

    • $0$ cards grade as a $10$

    • $1$ card grades as a $9$

    • $3$ cards grade as an $8$

    • $3$ cards grade as a $7$

    • $3$ cards grade as a $6$

  • Method 1: Direct Summation Divided by Total Count         $\text{Average Grade} = \frac{9 + 8 + 8 + 8 + 7 + 7 + 7 + 6 + 6 + 6}{10}$

  • Method 2: Grouped Multiplication over Sample Size         $\text{Average Grade} = \frac{(1 \times 9) + (3 \times 8) + (3 \times 7) + (3 \times 6)}{10}$

  • Method 3: Fractional Expansion         $\text{Average Grade} = \left(1 \times \frac{9}{10}\right) + \left(3 \times \frac{8}{10}\right) + \left(3 \times \frac{7}{10}\right) + \left(3 \times \frac{6}{10}\right)$

  • Method 4: Decimal Fraction Conversion         $\text{Fraction of 9s} = \frac{1}{10} = 0.1 \quad (10\%)$         $\text{Fraction of 8s} = \frac{3}{10} = 0.3 \quad (30\%)$         $\text{Fraction of 7s} = \frac{3}{10} = 0.3 \quad (30\%)$         $\text{Fraction of 6s} = \frac{3}{10} = 0.3 \quad (30\%)$         $\text{Average Grade} = (0.1 \times 9) + (0.3 \times 8) + (0.3 \times 7) + (0.3 \times 6)$         $\text{Average Grade} = 0.9 + 2.4 + 2.1 + 1.8 = 7.2$

• Example 2: Sample Population of 100 Cards

  • Distribution Data:

    • $0.8$ cards grade as a $10$

    • $14.9$ cards grade as a $9$

    • $27.2$ cards grade as an $8$

    • $28$ cards grade as a $7$

    • $29$ cards grade as a $6$

  • Decimal Conversion and Calculation:         $\text{Fraction of 10s} = \frac{0.8}{100} = 0.008$         $\text{Fraction of 9s} = \frac{14.9}{100} = 0.149$         $\text{Fraction of 8s} = \frac{27.2}{100} = 0.272$         $\text{Fraction of 7s} = \frac{28}{100} = 0.280$         $\text{Fraction of 6s} = \frac{29}{100} = 0.290$         $\text{Average Grade} = (0.008 \times 10) + (0.149 \times 9) + (0.272 \times 8) + (0.280 \times 7) + (0.290 \times 6)$         $\text{Average Grade} = 0.08 + 1.341 + 2.176 + 1.96 + 1.74 = 7.297 \approx 7.3$

• Mathematical Implications of the Metaphor:

  • Dividing individual counts by total population size isolates relative percentages ($27.2$ out of $100 = 27.2\%$ or $0.272$).

  • Provided percentage abundance distributions remain fixed, the average calculation yields an equivalent value regardless of whether population size is $100$, $5,000$, or an arbitrary total number of atoms.

Questions & Discussion

• Question: Why are there fractional numbers of cards (e.g., $0.8$ cards graded as a $10$) in a sample size of $100$?

  • Response: Fractional quantities represent normalized statistical proportions derived from larger population distributions (analogous to sampling fragment percentages from large atomic populations).

• Question: Must percentages always be converted into decimal form when performing weighted average equations?

  • Response: Yes, percentages must always be converted into decimal form (by dividing by $100$) prior to executing any mathematical calculation involving percentage abundances.

• Question: Are atomic mass calculation problems always formatted to solve for average atomic mass, or can the setup vary?

  • Response: Problem setups can vary depending on which variables are given. For example, if the overall average atomic mass of chlorine ($35.45\,\text{amu}$) and the isotopic mass and abundance of Chlorine-35 ($75.77\%$, $34.97\,\text{amu}$) are provided, algebraic manipulation can be used to set up an equation with variables to solve for the unknown mass or abundance of Chlorine-37.