Atomic Theory and Isotopes – Study Notes

History of Atomic Theory

Early claim: matter is composed of small indestructible particles (atoms) proposed by Leucippus and his student Democritus (transcript mispronounced as 'Lucifus').
Counterview (more popular at the time): Plato and Aristotle argued matter was not made of tiny particles but reduced to four elements: fire, earth, air, and water. The idea that everything is made of small particles was ultimately proven correct, though long after.
John Dalton (English chemist) helped convince the public that matter is made of small particles and formulated Dalton's Atomic Theory of Matter.

Dalton's Atomic Theory of Matter (four postulates)

1) Each element is composed of tiny, indivisible particles called atoms.
2) All atoms of a given element have the same mass and other properties that distinguish them from atoms of other elements (e.g., oxygen vs hydrogen differ).
3) Atoms combine in simple whole-number ratios to form compounds (e.g., water is formed from two hydrogens and one oxygen).
4) Atoms of one element cannot change into atoms of another element in a chemical reaction; atoms merely rearrange the way they are bound together.

These postulates laid the groundwork for the modern atomic theory, which builds on three key laws: conservation of mass, definite proportions, and multiple proportions.

Modern Foundations: Three Core Laws

Law of Conservation of Mass: In a chemical reaction, matter is neither created nor destroyed. The mass of reactants equals the mass of products: $m{ ext{reactants}} = m{ ext{products}}.$ For example, in a generic reaction forming water, the total mass of H and O before the reaction equals the total mass of H2O after.
Law of Definite Proportions (also called the Law of Definite/Constant Proportions): All samples of a given compound have the same proportion of constituent elements regardless of source or preparation. Think of it as a fixed ratio of elements in a compound.
- Simplified interpretation: the ratio of elements in a compound is constant.
- Example (as described in lecture): decomposition of water yielding a fixed ratio of oxygen to hydrogen; described as an eight-to-one ratio in the talk (note: this specific numeric example in the transcript may reflect a teaching simplification; the key point is that the ratio is fixed).
Law of Multiple Proportions: When two elements (A and B) form two different compounds, the masses of element B that combine with a fixed mass of element A are in a simple small whole-number ratio. Example: for carbon and oxygen formed as CO and CO₂:
- One part carbon with one part oxygen forms CO.
- One part carbon with two parts oxygen forms CO₂.
- The ratio of the masses of oxygen that combine with a fixed mass of carbon is a simple ratio.

Subatomic Particles and the First Atomic Models

Discovery of the Electron

J.J. Thomson conducted the cathode ray experiment and discovered the electron.
Key findings:
- Electrons travel in straight lines.
- Electrons carry a negative charge.
Charge-to-mass ratio (q/m) for the electron was determined:
$\frac{q}{m} = -1.76 \times 10^{8}\ \,\text{C g}^{-1}.$
Separately, Robert Millikan’s oil-drop experiment measured the elementary charge magnitude:
$|e| = 1.6 \times 10^{-19}\ \,\text{C}.$
Combining q/m with |e| yields the electron mass:
$m_e = \frac{|e|}{|q/m|} \approx 9.1 \times 10^{-28}\ \,\text{g}.$ (transcript value)
Thomson proposed the plum pudding model: electrons embedded in a positively charged sphere (like blueberries in a muffin). The “plum pudding” idea depicted a diffuse electron cloud within a positively charged mass, but this model was later shown to be incorrect.

Rutherford and the Nuclear Model

Ernest Rutherford performed the gold foil experiment by directing a beam of positively charged particles at a very thin piece of gold foil.
Observations:
- Most alpha particles passed through with little deflection.
- A tiny fraction were deflected backward or at large angles.
Implications:
- Matter is not a uniform, continuous distribution; it contains small, dense regions—nuclei—surrounded by mostly empty space.
- The nucleus contains protons (positive charge) and neutrons (no charge) in close proximity.
Nuclear atom model (key points):
- Most of the atom's mass and all of its positive charge are in the nucleus at the center.
- A cloud or electron cloud surrounds the nucleus and contains negatively charged electrons.
- The number of electrons outside the nucleus equals the number of protons in the nucleus for a neutral atom.

Building Blocks of the Nucleus and Electron Cloud

Three main particles:
- Protons (p): positively charged; located in the nucleus; mass is about the same as neutrons.
- Neutrons (n): electrically neutral; located in the nucleus; mass similar to protons.
- Electrons (e⁻): negatively charged; located in the surrounding electron cloud; much smaller mass than protons/neutrons.
Key facts:
- Protons and neutrons have nearly identical masses; electrons are much lighter.
- Proton charge is +1; electron charge is −1; magnitudes are equal in a neutral atom.
Summary table (as described in the transcript):
- Masses: protons ≈ neutrons ≈ 1 (in atomic mass units, roughly), electrons ≪ protons.
- Charges: proton +; electron −; neutron 0.
Importance for identification: the number of protons uniquely identifies the element (the atomic number).

Atomic Number, Mass Number, and Isotopes

Atomic number (Z): the number of protons in the nucleus; symbol Z. This is the defining property of an element and is what the periodic table is arranged by.
Mass number (A): the total number of protons and neutrons in the nucleus:
$A = Z + N,$ where N is the number of neutrons.
Neutrons (N): $N = A - Z.$
Isotopes: atoms of the same element (same Z) that differ in the number of neutrons (A differs). They have the same chemical behavior but different masses.
Ions vs neutral atoms:
- Neutral atom: number of electrons equals the number of protons (e⁻ = Z).
- Ions: electrons differ from protons, giving a net charge.
- Cation: positively charged (electrons lost, e⁻ < Z).
- Anion: negatively charged (electrons gained, e⁻ > Z).
Isotopes vs ions summary:
- Isotopes differ in neutrons (A varies, Z fixed).
- Ions differ in electrons (e⁻ varies from Z).

Isotopic Notation and Examples

Isotopic notation basics:
- Nuclear notation: $^{A}_{Z}X,$ where X is the element symbol, A is the mass number, and Z is the atomic number.
- For a neutral atom: $A = Z + N.$ The mass number A is the sum of protons and neutrons; Z is the number of protons.
Example practice (as described in the transcript):
- Neon with mass number 20 and Z = 10: $^{20}_{10}Ne.$ Neutrons = $N = A - Z = 20 - 10 = 10.$ Electrons for a neutral Neon atom = 10.
- If a Neon species has mass number 22 (A = 22) and Z = 10: neutrons = 12; electrons depend on charge (neutral would be 10).
Ion example from the transcript:
- If a species has Z = 10 (Neon) and A = 22 with electrons = 10, it is neutral. If electrons ≠ Z, it is an ion. For instance, a Neon ion with 9 electrons would be Ne⁻? (in the lecture example an ion with a different electron count is discussed).
Practice sequence described:
- Atomic number (Z) identifies the element on the periodic table.
- Given a mass number A and Z, you can compute N and e⁻ (if neutral) or determine the charge if e⁻ ≠ Z.
- Example: Bromine with some given A and Z may yield a charged ion description such as Br^{4+} or Br^{4+} notation.

Atomic Mass and Isotopes

Atomic mass vs mass number:
- Mass number A is an integer (protons + neutrons).
- Atomic mass (often shown as atomic weight on the periodic table) is a decimal and represents a weighted average of all naturally occurring isotopes, based on their natural abundances.
Definitions:
- Atomic mass (also called atomic weight or standard atomic weight): weighted average mass of the isotopes of an element.
- It is typically given in atomic mass units (amu). 1 amu is defined as 1/12 the mass of a carbon-12 atom, and 1 amu is approximately equal to 1 g/mol when considering molar mass relationships.
Isotopes vs abundance:
- Isotopes have different numbers of neutrons; natural abundance describes what fraction of each isotope is found in nature.
- The weighted average mass (atomic mass) is calculated from the isotopic masses and their fractional abundances.

How to Calculate Atomic Mass (Example: Chlorine)

Chlorine has two main isotopes:
- Chlorine-35 (mass 34.97 amu) with abundance 75.77%.
- Chlorine-37 (mass 36.97 amu) with abundance 24.23%.
Steps:
- Convert percent abundances to decimals:
- f{35} = 0.7577, \quad f{37} = 0.2423.
- Multiply each isotope mass by its fractional abundance:
- $f{35} \times m{35} = 0.7577 \times 34.97 = 26.4978\ ext{amu},$
- $f{37} \times m{37} = 0.2423 \times 36.97 = 8.9578\ ext{amu}.$
- Sum to get the atomic mass (weighted average):
- $\text{Atomic Mass} = 26.4978 + 8.9578 \approx 35.4556\ ext{amu}.$
- With significant figures from the data (four sig figs in the inputs), the atomic mass is reported as $\approx 35.46\ ext{amu}.$
Notes:
- Atomic mass on the periodic table is the weighted average of isotopes, not a whole-number mass.
- The decimal value reflects both the masses of isotopes and their natural abundances.
- The term amu stands for atomic mass unit and is related to molar mass: 1 amu ≈ 1 g/mol.

Quick Worked Examples and Practice (from the transcript)

Isotope example 1:
- Given: isotope with A = 13, Z = 7, neutral atom.
- Neutrons: $N = A - Z = 13 - 7 = 6.$
- Electrons (neutral): $e^- = Z = 7.$
- If the third row has a charge of +1 (as discussed), then electrons would be 6, making the ion a cation (lost 1 electron).
Isotope example 2 (neutral Neon-like scenario):
- Given: A = 14, Z = 7; neutral → $N = 14 - 7 = 7,$ $e^- = 7.$
Isotope example 3 (charged Neon ion):
- Given: A = 14, Z = 7; electrons = 6 → charge +1 (since one fewer electron than protons).
Isotope identification practice (additional, from transcript):
- Atomic number Z, mass number A, and electrons for a neutral atom: e⁻ = Z.
- For an ion with e⁻ ≠ Z, determine the charge and label (e.g., Cr^{4+}, Cr^{4+} notation variations: Cr⁴⁺ or Cr4+).
- Example: Cr with A = 90 and Z = 40; electrons = 36; charge = +4 (since 40 protons but only 36 electrons).
- Notation flexibility: Cr^{4+} or Cr4+ (both widely used; the important part is indicating the charge).
Summary exercise (conceptual):
- Given A and Z, compute N and e⁻ (for neutral) or determine the charge if e⁻ differs from Z.

Key Takeaways and Connections

Atomic number Z is the defining property of an element; it equals the number of protons and determines the element's identity.
Mass number A is the total number of protons and neutrons; it can vary for isotopes of the same element.
Isotopes differ in neutron number; isotopic mass is expressed in amu and contributes to the atomic mass via natural abundance.
Ions differ from neutral atoms by the electron count; cations have fewer electrons than protons; anions have more.
The modern understanding of atomic structure is grounded in Dalton’s ideas but refined by Thomson, Millikan, Rutherford, and subsequent models, leading to the concept of a dense nucleus surrounded by an electron cloud.
The periodic table organizes elements by increasing atomic number; the mass of an element (as listed on the table) is a weighted average of isotopes, not a simple integer.

Terminology recap (quick reference)

Atomic number: Z = ext{# of protons}
Mass number: $A = Z + N$
Neutrons: $N = A - Z$
Isotopes: same Z, different A (different N)
Ions: atoms with unequal numbers of electrons and protons; charge indicated as X^{n+} or X^{n-}
Atomic mass unit: $1\ \text{amu} = 1/12 \times m(^{12}\text{C})$ ; 1 amu ≈ 1 g/mol

Practice prompts to try on your own:

Given an element with Z = 15 and A = 31, compute N and e⁻ for a neutral atom. Then determine the ion if e⁻ = 14.
Compute the atomic mass for chlorine given the isotopes and abundances used in the example, showing all steps with four significant figures.
Write the isotopic notation for an isotope with A = 22 and Z = 10.