Atoms, Molecules, and Ions: Foundations of General Chemistry

Dalton’s Atomic Theory

Foundations of Atomic Theory

Dalton’s Atomic Theory, proposed in the early 19th century, laid the groundwork for modern chemistry by describing the nature of matter at the atomic level.

• All matter is composed of indivisible particles called atoms.

• Atoms of a given element are identical in mass and properties.

• Atoms of different elements have different masses and properties.

• Compounds are formed by the combination of atoms of different elements in fixed, simple ratios.

• Chemical reactions involve the rearrangement of atoms; atoms are not created or destroyed in chemical reactions.

Example: Water (H2O) always contains two hydrogen atoms and one oxygen atom, illustrating the fixed ratio of atoms in a compound.

Atomic Structure

Subatomic Particles and Atomic Models

Atoms consist of a dense nucleus containing protons and neutrons, surrounded by a cloud of electrons. The properties of these subatomic particles are fundamental to understanding chemical behavior.

• Proton: Positively charged particle (+1), mass ≈ 1.673 × 10−24 g.

• Neutron: Neutral particle (0 charge), mass ≈ 1.675 × 10−24 g.

• Electron: Negatively charged particle (−1), mass ≈ 9.11 × 10−28 g.

• Atomic mass unit (amu): Defined as 1/12 the mass of a carbon-12 atom ($1\ \text{amu} = 1.6605 \times 10^{-24}\ \text{g}$).

• Elementary charge (e): The magnitude of the charge of a proton or electron ($e = 1.602 \times 10^{-19}\ \text{C}$).

	Electron	Proton	Neutron
Mass (g)	9.11 × 10−28	1.673 × 10−24	1.675 × 10−24
Mass (amu)	5.49 × 10−4	1.007	1.009
Charge (C)	−1.60 × 10−19	+1.60 × 10−19	0
Charge unit	−1	+1	0
Mass (Å)	<0.001	0.88	0.88

Example: The helium atom (He) consists of 2 protons, 2 neutrons, and 2 electrons.

Atomic Symbol and Notation

Representing Atoms and Ions

Atoms are represented using atomic symbols that convey the number of protons, neutrons, and the charge of the species.

• Atomic number (Z): Number of protons in the nucleus; defines the element.

• Mass number (A): Total number of protons and neutrons.

• Ion: A charged atom or molecule.

• Cation: Formed when an atom loses one or more electrons; positively charged.

• Anion: Formed when an atom gains one or more electrons; negatively charged.

Example: An atom with 13 protons and 10 electrons is the Al3+ ion (aluminum ion).

Example: The ion with 34 protons and 36 electrons is Se2− (selenide ion).

Isotopes

Atoms of the Same Element with Different Masses

Isotopes are atoms of the same element (same Z) with different numbers of neutrons (different A), resulting in different masses.

Symbol	Atomic Number	Number of Protons	Number of Neutrons	Mass (amu)	% Natural Abundance
$^1_1\text{H}$ (protium)	1	1	0	1.0078	99.9885
$^2_1\text{H}$ (deuterium)	1	1	1	2.0141	0.0115
$^3_1\text{H}$ (tritium)	1	1	2	3.01605	~0
$^3_2\text{He}$	2	2	1	3.01603	1.34
$^4_2\text{He}$	2	2	2	4.0026	98.66

Example: Hydrogen has three isotopes: protium, deuterium, and tritium.

Average Atomic Mass

Weighted Average of Isotopic Masses

The atomic mass of an element on the periodic table is a weighted average of the masses of all naturally occurring isotopes.

• Formula:

$\text{Average mass} = \sum_i (\text{fractional abundance}_i \times \text{isotopic mass}_i)$

Or, for two isotopes:

$\text{Average mass} = \left(\frac{\%\ \text{Isotope A}}{100\%}\right) \times \text{mass of Isotope A} + \left(\frac{\%\ \text{Isotope B}}{100\%}\right) \times \text{mass of Isotope B}$

Example: Boron has two isotopes: 10B (10.0129 amu, 19.9%) and 11B (11.0093 amu, 80.1%).

$\text{Average mass} = (0.199 \times 10.0129) + (0.801 \times 11.0093) = 1.99 + 8.82 = 10.81\ \text{amu}$

Calculation of Average Atomic Masses

To calculate the average atomic mass, multiply the atomic mass of each isotope by its fractional abundance and sum the results.

Isotope	Atomic Mass (amu)	% Abundance
24Mg	23.98504	78.99% / 100% = 0.7899
25Mg	24.9858	10.00% / 100% = 0.1000
26Mg	25.9829	11.01% / 100% = 0.1101

Example: The average atomic mass of magnesium is calculated as:

$\text{Average mass} = (0.7899 \times 23.98504) + (0.1000 \times 24.9858) + (0.1101 \times 25.9829) = 24.31\ \text{amu}$

Percent Abundance Calculation

Given the average atomic mass and the masses of isotopes, you can calculate the percent abundance of each isotope using algebraic methods.

Example: Chlorine has two isotopes, 35Cl and 37Cl, with an average atomic mass of 35.45 amu. The percent composition can be determined by setting up equations based on the average mass formula.

*Additional info: For more complex calculations, see worked examples in standard chemistry textbooks such as OpenStax Chemistry 2e.*