Comprehensive Notes on Atoms: History, Structure, and Measurement

The History of Atoms

• • Leucippus and Democritus: Originators of the idea of "atomos" (indivisible particles). Democritus famously stated, "Nothing exists except atoms and empty space; everything else is opinion" (460-370 B.C.).
  • Aristotle and Plato: Disagreed with the atomos theory. Their popularity and the lack of scientific proof at the time caused the theory to disappear for centuries.

Antoine Lavoisier (1789)

• Law of Conservation of Mass: States that matter is conserved in chemical reactions.
  • The overall mass of reactants equals the overall mass of products.
  • Example: For the reaction $2 ext{ Na(s)} + ext{Cl}<em>2 ext{(g)} o 2 ext{ NaCl(s)}$. If $7.7 ext{ g}$ of $ext{Na(s)}$ and $11.9 ext{ g}$ of $ext{Cl}</em>2 ext{(g)}$ are mixed, $19.6 ext{ g}$ of $ext{NaCl(s)}$ is formed. This demonstrates that $7.7 ext{ g} + 11.9 ext{ g} = 19.6 ext{ g}$.

John Dalton (1797)

• Law of Multiple Proportions: If element B can combine with $1 ext{ g}$ of element A in multiple proportions to form different compounds, the different masses of B that combine per gram of A can be expressed as a whole number ratio.
  • Example: Consider reactions $ext{A} + ext{B} o ext{AB}$ and $ext{A} + 2 ext{B} o ext{AB}_2$.
    • If $1 ext{ g}$ of A mixes with $1.5 ext{ g}$ of B to make AB, then $1 ext{ g}$ of A should mix with $3 ext{ g}$ of B to make $ext{AB}_2$. The ratio of B masses is $3:1.5 = 2:1$, a whole number ratio.

Dalton's Atomic Theory (1808)

1. Elements are composed of tiny, indestructible particles called atoms.
2. All atoms of one element have the same mass and other properties that distinguish them from the atoms of other elements.
3. Atoms combine in simple, whole-number ratios to form molecules.
4. Atoms of one element cannot change into atoms of another element.

J.J. Thomson (late 1800s)

• Cathode Ray Experiments: Explored the properties of cathode rays, streams of particles originating from the negative terminus (cathode) in a vacuum tube.
  • Experimental Setup: Used electrically charged plates and magnetic fields to deflect cathode rays.
  • Observations/Properties of Cathode Rays:
    1. Travel in straight lines.
    2. Are negatively charged (deflected towards a positive plate/away from a negative plate).
    3. Have a charge/mass ratio of $-1.76 imes 10^8 ext{ coulombs/gram (C/g)}$.
    4. This specific charge/mass ratio was obtained for any materials used, suggesting universality.
  • Discovery: These results indicated that "cathode rays" (later named electrons) could be found in every element, implying they were fundamental, subatomic particles. The mass of a cathode ray particle (electron) was approximately $9.1094 imes 10^{-31} ext{ g}$, almost $2000$ times less than hydrogen.
  • New Model: Proposed the "plum-pudding model," where negatively charged electrons were embedded in a sphere of diffuse positive charge.

Ernest Rutherford (1909)

• Gold Foil Experiment: Aimed to confirm Thomson's plum-pudding model.
  • Experimental Setup: Alpha ($\alpha$) particles (positively charged, relatively massive particles) were directed at a thin sheet of gold foil. A detector surrounding the foil observed the deflection of the alpha particles.
  • Observations:
    • Most $\alpha$ particles passed straight through the gold foil or with very little deflection.
    • A small fraction of $\alpha$ particles were deflected through large angles.
    • A very few $\alpha$ particles were deflected backward.
  • Conclusion: These observations contradicted the plum-pudding model, which predicted only minor deflections. Rutherford concluded that a new model was needed.
• Nuclear Theory of the Atom (Rutherford's Model):
  1. Most of the atom's mass and all of its positive charge are concentrated in a tiny, dense region at the center called the nucleus.
  2. The immense volume surrounding the nucleus consists mostly of empty space and is occupied by dispersed negatively charged electrons (e$^-$).
  3. Since atoms are electrically neutral, there are always an equal number of positively charged particles (protons) in the nucleus and negatively charged electrons ($e^-$) orbiting it. (While he proposed positive particles, the term 'proton' was formalized later).

James Chadwick (1932)

• Discovery of the Neutron: Identified the missing link in Rutherford's model.
  • Discovered the neutron, a third subatomic particle located in the nucleus alongside the proton.
  • Neutrons weigh roughly the same as protons but carry no electrical charge.
  • This discovery was crucial for understanding atomic stability and eventually led to work on the fission of Uranium-235 and atomic bombs.

Atomic Structure

• Every element is characterized by differing amounts of protons, neutrons, and electrons.
• Atomic Notation: Represented as $_{Z}^{A}\text{X}$
  • X: Chemical symbol of the element.
  • Z: Atomic Number – Represents the number of protons in the nucleus. It uniquely identifies an element.
  • A: Mass Number – Represents the sum of protons and neutrons in the nucleus.

Key Definitions

• Atomic Number (Z): Number of protons in the nucleus. For a neutral atom, it also equals the number of electrons.
  • Examples:
    • $ext{Z} = 1$ corresponds to Hydrogen (H).
    • $ext{Z} = 6$ corresponds to Carbon (C).
    • $ext{Z} = 11$ corresponds to Sodium (Na).
    • $ext{Z} = 82$ corresponds to Lead (Pb – from "Plumbum").
• Mass Number (A): Sum of protons and neutrons.
  • Number of neutrons = $ext{A} - ext{Z}$.
• Charge of an Atom (or ion): Indicates the relative number of electrons compared to protons.
  • $ext{Charge} = ( ext{number of protons}) - ( ext{number of electrons})$.

Examples: Calculating Subatomic Particles in Neutral Atoms

• Example: Indicate how many protons, neutrons, and electrons are present in $_{24}^{52}\text{Cr}$.
  1. $ext{Z} = 24$, therefore $24$ protons.
  2. $ext{A} = 52$ (total protons and neutrons). Number of neutrons = $ext{A} - ext{Z} = 52 - 24 = 28$ neutrons. Therefore, $28$ neutrons.
  3. The atom has no explicit charge (neutral), so the number of electrons equals the number of protons. Therefore, $24$ electrons.

Ions

• Definition: Elements that have either gained or lost electrons, resulting in a net electrical charge.
  • Cations: Positively charged ions (when an atom loses electrons), denoted as $_{Z}^{A}\text{X}^{m+}$.
  • Anions: Negatively charged ions (when an atom gains electrons), denoted as $_{Z}^{A}\text{X}^{m-}$.
• Important Note: When an element forms an ion, its atomic number (Z) and mass number (A) do not change. Only the number of electrons changes.
• Charge Calculation: $ext{Charge} = ( ext{number of protons}) - ( ext{number of electrons})$.
  • Rearranged for electrons: $ext{Number of electrons} = ( ext{number of protons}) - ( ext{charge})$.

Example: Calculating Subatomic Particles in Ions

• Example: Indicate how many protons, neutrons, and electrons are present in $_{8}^{16}\text{O}^{2-}$.
  1. $ext{Z} = 8$, therefore $8$ protons.
  2. $ext{A} = 16$. Number of neutrons = $ext{A} - ext{Z} = 16 - 8 = 8$ neutrons. Therefore, $8$ neutrons.
  3. Charge = $-2$. Number of electrons = $( ext{number of protons}) - ( ext{charge}) = 8 - (-2) = 10$ electrons. Therefore, $10$ electrons.

Isotopes

• Definition: Elements that have gained or lost neutrons. They are atoms of the same element (same Z) but with different mass numbers (A).
• Notation: Expressed in two ways:
  1. Symbolic: $<em>{Z}^{A}\text{X}$, e.g., $</em>{10}^{20}\text{Ne}$, $<em>{10}^{21}\text{Ne}$, $</em>{10}^{19}\text{Ne}$.
  2. Name-Mass Number: neon-20, Ne-20, neon-21, Ne-21, neon-19, Ne-19.
• Important Note: By adding or removing neutrons, the mass number (A) changes, but the atomic number (Z) (and thus the element identity) remains the same.

Example: Calculating Subatomic Particles in Isotopes

• Example: Indicate how many protons, neutrons, and electrons are present in $<em>{15}^{32}\text{P}$ and $</em>{15}^{30}\text{P}$.
  • For $_{15}^{32}\text{P}$:
    • Atomic number (Z) = $15$
    • Mass number (A) = $32$
    • Number of protons = $15$
    • Number of neutrons = $ext{A} - ext{Z} = 32 - 15 = 17$
    • Charge = $0$
    • Number of electrons = $15$
  • For $_{15}^{30}\text{P}$:
    • Atomic number (Z) = $15$
    • Mass number (A) = $30$
    • Number of protons = $15$
    • Number of neutrons = $ext{A} - ext{Z} = 30 - 15 = 15$
    • Charge = $0$
    • Number of electrons = $15$

Atomic Mass Units (amu)

• Definition: A standard unit used to express the mass of atoms and other subatomic particles.
• $1 ext{ amu} = 1.66053873 imes 10^{-24} ext{ g}$.

Measuring Atoms

Ways of Expressing Quantities of an Element

1. Mass: typically in grams ($ext{g}$).
2. Atomic Mass Unit: typically in amu.
3. Amount of a substance: in moles ($ext{mol}$).
4. Number of Particles: in atoms or molecules,

Mass Spectrometry

• Technique: Determines the mass of atoms by measuring the mass/charge ratio of positive ions (originally called anode or "canal" rays).
• Principle: Ions are produced, accelerated, and then passed through a magnetic field. The extent of deflection in the magnetic field depends on the ion's mass-to-charge ratio. Lighter ions or ions with higher charge are deflected more.
• Application: Used to separate and quantify different isotopes of an element based on their mass differences.

Discrepancy in Theoretical vs. Experimental Atomic Mass

• Theoretical Mass Calculation (Example for Cl-35):
  • $m<em>{ ext{Cl-35, theo}} = (17 imes m</em>{ ext{proton}}) + (18 imes m<em>{ ext{neutron}}) + (17 imes m</em>{ ext{electron}})$
  • $= (17 imes 1.6726 imes 10^{-24} ext{g}) + (18 imes 1.6749 imes 10^{-24} ext{g}) + (17 imes 9.1094 imes 10^{-28} ext{g})$
  • $= 5.8598 imes 10^{-23} ext{g}$.
• Experimental Mass (for Cl-35): $m_{ ext{Cl-35, experimental}} = 5.8069 imes 10^{-23} ext{g}$.
• Mass Difference: The experimental mass is lower than the predicted theoretical mass.
• Reason (Nuclear Binding Energy):
  • This mass difference is due to the nuclear binding energy that is released when protons and neutrons combine to form the nucleus. Some mass is converted into energy (E = mc$^2$).
  • Therefore, it is not suitable to predict the true masses of atoms simply by summing the standard molecular weights of their constituent subatomic particles.

Atomic Mass Units (amu) – Revisited

• Purpose: Scientists needed a scale to compare the masses of elements based on their mass numbers (A).
• Standard: The amu unit is defined based on the mass of the Carbon-12 atom ($^{12}{~6} ext{C}$). The mass of a $^{12}</em>{~6} ext{C}$ atom is exactly $12 ext{ amu}$.
  • The mass of $^{12}_{~6} ext{C}$ is $1.99265 imes 10^{-23} ext{ g}$, containing $6$ protons and $6$ neutrons.
  • Thus, $1 ext{ amu} = \frac{1.99265 imes 10^{-23} ext{ g}}{12} = 1.66054 imes 10^{-24} ext{ g}$.
• Reliability: amu values are reliable because they are relative to the mass of the C-12 isotope. All other elements' atomic masses are determined by their mass ratio to C-12.
• Example: Mass Ratio of Cl-35 to C-12:
  • $\frac{ ext{Mass of Cl-35}}{ ext{Mass of C-12}} = \frac{5.8069 imes 10^{-23} ext{ g}}{1.99265 imes 10^{-23} ext{ g}} = 2.91415$. So, Cl-35 is $2.91415$ times the mass of a C-12 atom.
  • $m<em>{ ext{Cl-35}} = m</em>{ ext{C-12}} imes ext{mass ratio}_{ ext{Cl-35/C-12}} = 12.00 ext{ amu} imes 2.91415 = 34.9698 ext{ amu} ag{rounded to } 34.97 ext{ amu}$.

Atomic Mass (Weighted Average)

• Significance of Isotopes: Most elements have multiple naturally occurring isotopes.
• Atomic Mass on Periodic Table: The value shown on the periodic table is a weighted average of the masses of all naturally occurring isotopes of that element.
• Formula: $ext{Atomic Mass} = \sum [( ext{fraction of isotope n}) imes ( ext{mass of isotope n})]$

Example: Calculating Atomic Mass of Chlorine

• Given Data:
  • Cl-35: $34.97 ext{ amu}$, 75.77 ext{%} abundant.
  • Cl-37: $36.97 ext{ amu}$, 24.23 ext{%} abundant.
• Step 1: Check isotopic fractions sum to $1$ (or 100 ext{%}):
  • $(75.77/100) + (24.23/100) = 0.7577 + 0.2423 = 1$. (Confirms all isotopes accounted for).
• Step 2: Calculate Atomic Mass:
  • $ext{Atomic Mass} = (0.7577 imes 34.97 ext{ amu}) + (0.2423 imes 36.97 ext{ amu})$
  • $= 26.490769 ext{ amu} + 8.956621 ext{ amu} = 35.44739 ext{ amu} ag{rounded to } 35.45 ext{ amu}$.

Example: Calculating Atomic Mass of Magnesium (with abundance calculation)

• Given Data:
  • Mg-24: $23.99 ext{ amu}$, 78.99 ext{%} abundant.
  • Mg-25: $24.99 ext{ amu}$, 10.00 ext{%} abundant.
  • Mg-26: $25.98 ext{ amu}$. (Abundance needs to be determined).
• Step 1: Determine the natural abundance of Mg-26:
  • $ext{Fraction of Mg-26} = 1 - ( ext{fraction of Mg-24}) - ( ext{fraction of Mg-25})$
  • = 1 - (78.99/100) - (10.00/100) = 1 - 0.7899 - 0.1000 = 0.1101 ag{or } 11.01 ext{%}.
• Step 2: Determine the atomic mass of Mg:
  • $ext{Atomic Mass} = (23.99 ext{ amu})(0.7899) + (24.99 ext{ amu})(0.1000) + (25.98 ext{ amu})(0.1101)$
  • $= 18.950101 ext{ amu} + 2.499 ext{ amu} + 2.859698 ext{ amu} = 24.308799 ext{ amu} ag{rounded to } 24.31 ext{ amu}$.
• Step 3: Convert units from amu to grams:
  • $m_{ ext{Mg}} = 24.31 ext{ amu} imes (1.66053873 imes 10^{-24} ext{ g/amu})$
  • $= 4.035 imes 10^{-23} ext{ g}$.

Avogadro's Number ($N_A$)

• Definition: A precise number used by chemists to express an amount of a substance.
  • $N_A = 6.022 imes 10^{23} ext{ particles/mol}$.
  • This means that $6.022 imes 10^{23}$ particles (atoms, molecules, ions, etc.) equal $1 ext{ mol}$ of those particles.
• Determination: Avogadro's number was determined using the relationship:
  • $1 ext{ mol C-12} = 12 ext{ grams of C-12}$.
  • Knowing the mass of a single C-12 atom (which is $12 ext{ amu} imes 1.6605 imes 10^{-24} ext{ g/amu} = 1.9926 imes 10^{-23} ext{ g C-12}$), the number of atoms in $12 ext{ g}$ of C-12 can be calculated:
    • $\frac{12 ext{ g C-12}}{1.9926 imes 10^{-23} ext{ g C-12/atom}} = 6.022 imes 10^{23} ext{ atoms of C-12}$.
• Universality: While determined relative to C-12, Avogadro's number is a constant for all elements and substances.

Calculations Involving Avogadro's Number

• Number of objects: $ext{Number of objects of X} = ( ext{mol of X}) imes N_A$
• Moles from objects: $ext{Mol of X} = \frac{ ext{Number of objects of X}}{N_A}$

Example: Moles to Atoms Conversion

• Problem: A student was told to measure exactly $1.31 ext{ mol}$ of elemental copper. How many copper atoms does this value correspond to?
• Answer:
  • $ext{Number of Cu atoms} = 1.31 ext{ mol Cu} imes (6.022 imes 10^{23} ext{ atoms Cu/mol Cu})$
  • $= 7.88882 imes 10^{23} ext{ atoms Cu} ag{rounded to } 7.89 imes 10^{23} ext{ atoms Cu}$

Molar Mass (M)

• Definition: The mass in grams of one mole of a substance. Its units are usually grams per mole ($ext{g/mol}$).
  • For C-12, $1 ext{ mol C-12} = 12.00 ext{ grams of C-12}$.
  • Thus, $M_{ ext{C-12}} = 12.00 ext{ g/mol C-12}$.
• Relationship to Atomic Mass: An element's molar mass is numerically equal to its atomic mass in amu (e.g., $1 ext{ mol of C} = 12.01 ext{ g}$, so $M_{ ext{C}} = 12.01 ext{ g/mol}$).

Calculations Involving Molar Mass

• Moles from mass: $n = \frac{m}{M}$
• Mass from moles: $m = n imes M$
  • Where: $n = ext{number of moles}$, $m = ext{mass (g)}$, $M = ext{molar mass (g/mol)}$.

Example: Comparing Number of Atoms in Given Mass

• Problem: Are there more atoms present in $1.00 ext{ g}$ of Mn or $1.00 ext{ g}$ of Fe?
• Step 1: Determine the number of moles (n) for both Mn and Fe:
  • For Mn: $n_{ ext{Mn}} = \frac{1.00 ext{ g}}{54.94 ext{ g/mol}} = 0.01820 ext{ mol Mn}$.
  • For Fe: $n_{ ext{Fe}} = \frac{1.00 ext{ g}}{55.85 ext{ g/mol}} = 0.01791 ext{ mol Fe}$.
• Step 2: Determine the number of atoms for both Mn and Fe using ext{# atoms} = n imes N_A:
  • For Mn: ext{# atoms Mn} = 0.01820 ext{ mol Mn} imes (6.022 imes 10^{23} ext{ atoms Mn/mol Mn}) = 1.096 imes 10^{22} ext{ atoms Mn}.
  • For Fe: ext{# atoms Fe} = 0.01791 ext{ mol Fe} imes (6.022 imes 10^{23} ext{ atoms Fe/mol Fe}) = 1.078 imes 10^{22} ext{ atoms Fe}.
• Conclusion: There are more atoms in $1.00 ext{ g}$ of Manganese than in $1.00 ext{ g}$ of Iron.

Converting Between Known Quantities

• Diagram of Interconversions:
  • $ext{Mass} \xleftrightarrow{\text{Molar Mass}} \text{Moles} \xleftrightarrow{\text{Avogadro's Number}} \text{Particles}$
  • $ext{Volume} \xleftrightarrow{\text{Density}} \text{Mass}$