Atoms, Ions, and Elements: Foundations of Atomic Theory and the Periodic Table

Chapter 2: Atoms, Ions, and Elements

Introduction

Atoms are the fundamental building blocks of matter, and chemistry is essentially the study of matter and its transformations. Understanding the structure and behavior of different atoms provides a deeper grasp of the macroscopic world.

• Macroscopic: Observable properties of matter (e.g., gold bar, water in a beaker).

• Microscopic: Atomic and molecular level (e.g., arrangement of atoms in a solid or liquid).

2.2 Early Atomic Theory

Fundamental Chemical Laws

• Law of Conservation of Mass (Matter): Matter can neither be created nor destroyed. The total mass of substances remains unchanged during a chemical reaction.

Example:

• Calcium oxide + carbon dioxide → calcium carbonate

• $\text{CaO} + \text{CO}_2 \rightarrow \text{CaCO}_3$

• 56.08 g + 44.00 g = 100.08 g

• Law of Multiple Proportions: When elements combine to form more than one compound, the ratios of the masses of one element that combine with a fixed mass of the other are small whole numbers.

Example:

• Water ($\text{H}_2\text{O}$): 2.02 g H + 16.00 g O

• Hydrogen peroxide ($\text{H}_2\text{O}_2$): 2.02 g H + 32.00 g O

• Ratio of O in $\text{H}_2\text{O}_2$ to $\text{H}_2\text{O}$: $\frac{32.00}{16.00} = 2:1$

• Law of Constant Proportion (Definite Composition): A given compound always contains the same proportion of elements by mass, regardless of its source.

Sample of Water	Oxygen (g)	Hydrogen (g)	Mass Ratio O/H
A	16.00	2.00	8.00/1.00
B	32.00	4.00	8.00/1.00

Water always has a ratio of 8.00 g O to 1.00 g H.

Dalton's Atomic Theory

• All matter consists of atoms.

• Atoms of an element are identical in mass and properties, and differ from those of other elements.

• Compounds are formed by the chemical combination of atoms in specific ratios.

• Atoms are not created or destroyed in chemical reactions; only their arrangements change.

Modern View: Atoms are not indestructible (can be split in nuclear reactions), and isotopes exist (atoms of the same element with different masses).

2.3 The Electron and the Atom

Discovery of the Electron (J. J. Thomson)

• Used cathode-ray tubes to discover the electron.

• Observations: Rays bent in magnetic/electric fields, identical for all cathodes.

• Conclusions: Rays consist of negatively charged particles (electrons) found in all matter.

• Charge-to-mass ratio: $\frac{e}{m} = -1.76 \times 10^8 \ \text{C/g}$

• Proposed the Plum Pudding Model: Electrons embedded in a sphere of positive charge.

Millikan's Oil Drop Experiment

• Measured the charge of a single electron: $-1.6 \times 10^{-19} \ \text{C}$

• Confirmed the quantization of electric charge.

2.4 Rutherford's Gold Foil Experiment and the Nuclear Model of the Atom

Discovery of the Nucleus (Ernest Rutherford)

• Gold foil experiment: Most alpha particles passed through, but some were deflected at large angles.

• Conclusion: Atom is mostly empty space with a small, dense, positively charged nucleus.

• Electrons move around the nucleus at relatively large distances.

• Protons discovered by Rutherford (1919); Neutrons by Chadwick (1932).

Particle	Charge (C)	Relative Charge	Mass (amu)	Relative Mass	Location
Proton (p+)	+1.6022 × 10−19	+1	1.0073	1836	Nucleus
Neutron (n0)	0	0	1.0087	1839	Nucleus
Electron (e−)	−1.6022 × 10−19	−1	0.00055	1	Outside nucleus

• Magnitude of charge on protons and electrons is equal and opposite.

• Protons and neutrons have similar masses; electrons have negligible mass.

• Atomic mass unit (amu): $1 \ \text{amu} = 1.66 \times 10^{-24} \ \text{g}$

2.5 Atomic Number

Atomic Number, Symbol, and Element Name

• The atomic number (Z) is the number of protons in the nucleus and defines the element.

• All atoms of a given element have the same atomic number.

• Periodic table is arranged by increasing atomic number.

Example:

• Potassium (K): Atomic number = 19, Number of protons = 19, Number of electrons (neutral atom) = 19

2.6 Ions

Formation and Types of Ions

• Ions are atoms or groups of atoms with a net positive or negative charge due to loss or gain of electrons.

• Cation: Positively charged ion (loss of electrons). Example: $\text{Na} \rightarrow \text{Na}^+ + e^-$

• Anion: Negatively charged ion (gain of electrons). Example: $\text{Cl} + e^- \rightarrow \text{Cl}^-$

• Some elements can form more than one ion (e.g., Cu+, Cu2+).

Atomic Number	Protons	Electrons	Ion Charge	Ion Symbol
16	16	18	−2	S2−
13	13	10	+3	Al3+

Ion charge = number of protons − number of electrons

2.7 Atomic Mass and Isotopes

Isotopes

• Isotopes are atoms of the same element with different numbers of neutrons (and thus different masses).

• Identified by their mass number (A): $A = Z + n$

• Isotopes have nearly identical chemical properties because they have the same number of protons and electrons.

Example notation: $^{12}_6\text{C}$ or C-12 (6 protons, 6 neutrons)

Protons	Neutrons	Electrons	Atomic Number	Mass Number	Atomic Symbol
6	7	6	6	13	$^{13}_6\text{C}$
42	54	42	42	96	$^{96}_{42}\text{Mo}$

• Atomic Mass Unit (amu): Standard unit for atomic and molecular masses.

• 1 amu = $1.66 \times 10^{-24}$ g

Average Atomic Mass

• The atomic mass of an element is the weighted average of the masses of its naturally occurring isotopes.

Formula:

$\text{Atomic Mass} = \sum (\text{mass of isotope}_i \times \text{fractional abundance of isotope}_i)$

Example (Gallium):

• Ga-69: 68.9256 amu, 60.11%

• Ga-71: 70.9247 amu, 39.89%

• $\text{Atomic mass} = (68.9256 \times 0.6011) + (70.9247 \times 0.3989) = 41.431 + 28.292 = 69.72$ amu

Common mistake: Do not simply average the isotope masses; use weighted average.

Sample Calculation: Chlorine Isotopes

Isotope	Mass Value (amu)	% Natural Abundance
Cl-35	34.968853	75.77%
Cl-37	?	?

Given the average atomic mass (35.45 amu), you can solve for the unknowns using the weighted average formula.

Sample Calculation: Americium

• Given: 6 atoms of Am, 1 Am atom = 243 u

• Mass = $6 \times 243 = 1458$ u

2.8 Atoms in Review

• Atoms: Identity determined by number of protons; chemistry determined by electrons.

• Ions: Change in number of electrons.

• Isotopes: Change in number of neutrons.

Number of electrons, protons, and neutrons in neutral atoms:

• Number of protons = atomic number

• Number of electrons = number of protons (neutral atom)

• Number of neutrons = mass number − atomic number

2.9 Introduction to the Periodic Table

Structure of the Periodic Table

• Elements with similar chemical properties are found in vertical columns called groups or families (18 groups).

• Horizontal rows are called periods (7 periods).

• Main group elements: "A" groups; Transition elements: "B" groups; Inner transition elements: lanthanides and actinides.

Classification of Elements

Type	Properties
Metals	Malleable, ductile, lustrous, good conductors, high density/melting point, tend to form cations, few colors
Nonmetals	Brittle, dull, poor conductors, low density/melting point, tend to form anions, various colors
Metalloids	Intermediate properties, semiconductors

• Group IA: Alkali metals (soft, react violently with water)

• Group IIA: Alkaline earth metals (harder, less reactive than IA)

• Group VIIA: Halogens (very reactive, diatomic)

• Group VIIIA: Noble gases (colorless, unreactive, monatomic)

Periodic Table and Ionic Charge

• Elements in the same group have similar preferred ionic charges.

• Metals tend to lose electrons (form cations); nonmetals tend to gain electrons (form anions).

• Ion charge = number of protons − number of electrons

2.10 Counting Atoms, Ions, and Molecules: The Concept of the Mole

Avogadro's Number and the Mole

• 1 mole (mol) = $6.022 \times 10^{23}$ individual units (Avogadro's number).

• Units can be atoms, ions, molecules, electrons, etc.

Examples:

• 1 mol Cl2 = $6.022 \times 10^{23}$ molecules of Cl2

• 1 mol Ne = $6.022 \times 10^{23}$ atoms of Ne

• 1 mol NaCl = $6.022 \times 10^{23}$ formula units of NaCl

Conversions Using the Mole

• To convert moles to number of particles: $\text{number of particles} = \text{moles} \times 6.022 \times 10^{23}$

• To convert number of particles to moles: $\text{moles} = \frac{\text{number of particles}}{6.022 \times 10^{23}}$

Example: 3.5 moles He × $6.022 \times 10^{23}$ atoms/mol = $2.108 \times 10^{24}$ atoms He

Atomic Mass (amu/atom) vs. Molar Mass (g/mol)

• Atomic mass: Mass of 1 atom in amu

• Molar mass: Mass of 1 mole of atoms in grams (numerically equal to atomic mass in amu)

Element	Atomic Mass (amu/atom)	Molar Mass (g/mol)	Number of Atoms in 1 mol
H	1.0079	1.0079	$6.022 \times 10^{23}$
C	12.011	12.011	$6.022 \times 10^{23}$
U	238.029	238.029	$6.022 \times 10^{23}$

To convert between mass and moles: $\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

Practice Problems

• How many moles of Ca contain $3.01 \times 10^{23}$ atoms? $\text{moles} = \frac{3.01 \times 10^{23}}{6.022 \times 10^{23}} = 0.500$ mol

• How many copper atoms are in a penny weighing 3.10 g? $\text{moles Cu} = \frac{3.10}{63.55} = 0.0488$ mol; $0.0488 \times 6.022 \times 10^{23} = 2.94 \times 10^{22}$ atoms

Additional info: These notes cover the foundational concepts of atomic structure, isotopes, ions, and the periodic table, as well as the quantitative concept of the mole, which are essential for understanding chemical reactions and stoichiometry in general chemistry.