Atom and Atomic Theory: Foundations of Chemistry

Atom and Atomic Theory

Introduction

This section introduces the fundamental concepts of atomic theory, which form the basis of modern chemistry. Understanding atoms, their structure, and the laws governing their behavior is essential for studying chemical reactions and properties of matter.

Atom Kuramı (Atomic Theory)

Law of Conservation of Mass (Antoine Lavoisier, 1774)

The law of conservation of mass states that matter cannot be created or destroyed in a chemical reaction. The total mass of reactants equals the total mass of products.

• Definition: In any chemical reaction, the mass of the substances produced (products) is equal to the mass of the substances that react (reactants).

• Equation Example:

$2H_2 + O_2 \rightarrow 2H_2O$

• 4 g H2 + 32 g O2 = 36 g H2O

Law of Definite Proportions (Joseph Proust, 1799)

This law states that a chemical compound always contains the same elements in the same proportion by mass, regardless of the source or method of preparation.

• Definition: The ratio of the masses of the elements in a compound is always the same.

• Example: Water (H2O) always contains 11.19% hydrogen and 88.81% oxygen by mass.

Sample Calculations

• Example 1: 0.455 g magnesium is burned in 2.315 g oxygen. After the reaction, 2.015 g oxygen remains unreacted. The mass of magnesium oxide formed is:

Tepkime Öncesi Kütle = 0.455 g + 2.315 g = 2.770 g Tepkime Sonrası Kütle = 2.770 g Magnezyum Oksit = 2.770 g - 2.015 g = 0.755 g

• Example 2: 0.100 g magnesium reacts with oxygen to form 0.166 g magnesium oxide. How much magnesium oxide forms from 0.144 g magnesium?

$0.166 \text{ g MgO} / 0.100 \text{ g Mg} = 1.66$ $0.144 \text{ g Mg} \times (0.166 / 0.100) = 0.239 \text{ g MgO}$

Table: Mass Relationships in Compounds

The following table demonstrates the application of the laws of definite proportions and conservation of mass to sodium oxide compounds:

Compound	Na (sodium) (g)	O (oxygen) (g)	Na2O (sodium oxide) (g)	mNa/mO
1st Compound	4.6	1.6	6.2	2.88
2nd Compound	0.32	1.24	1.56	0.26
3rd Compound	3.68	1.28	4.96	2.88

Additional info: The mNa/mO column is calculated by dividing the mass of sodium by the mass of oxygen for each compound.

Dalton's Atomic Theory

• All matter is composed of extremely small particles called atoms.

• Atoms of a given element are identical in mass and properties; atoms of different elements differ in mass and properties.

• Atoms cannot be created, divided, or destroyed in chemical reactions.

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

Law of Multiple Proportions (John Dalton, 1803-1808)

If two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in ratios of small whole numbers.

• Example: CO and CO2 both contain carbon and oxygen. The ratio of oxygen masses that combine with a fixed mass of carbon is 2:1.

Compound	C (g)	O (g)	O(CO2)/O(CO)
CO	12	16	2/1
CO2	12	32

Example: Manganese Oxides

Three manganese oxides have the following compositions:

Compound	Mn %	O %	O (g)
1st	74.44	25.56	22.56
2nd	69.60	30.40	33.82
3rd	63.18	36.82	45.13

• Divide each O (g) value by the smallest (22.56): 1, 1.5, 2

• Multiply by 2 to get whole numbers: 2:3:4

Electrons and Discoveries in Atomic Physics

Forces Between Charged Bodies

• Like charges repel, opposite charges attract.

• Uncharged objects do not exert electrical forces on each other.

• Charged particles are deflected in magnetic fields: negative particles deflect one way, positive the opposite.

Discovery of the Electron

• Electrons were discovered using cathode ray tubes (CRT).

• Electrons are negatively charged, much lighter than atoms, and are a fundamental component of all atoms.

Rutherford Atomic Model

• Most of the atom's mass and all its positive charge are concentrated in a small nucleus.

• The nucleus contains protons (positive) and neutrons (neutral).

• Electrons orbit the nucleus in the surrounding space.

Properties of Subatomic Particles

Protons, Neutrons, and Electrons

• Proton: Positively charged, mass ≈ 1 atomic mass unit (amu).

• Neutron: No charge, mass ≈ 1 amu.

• Electron: Negatively charged, mass ≈ 1/1836 of a proton.

Atomic Number and Mass Number

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

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

• Isotopes: Atoms of the same element (same Z) with different numbers of neutrons (different A).

$A = Z + N$ $N = A - Z$

Ions

• Cation: Atom that loses electrons (positive charge).

• Anion: Atom that gains electrons (negative charge).

• Example: Cl- has 17 protons and 18 electrons.

Atomic Masses

Atomic Mass Unit (amu)

• 1 amu is defined as 1/12 the mass of a carbon-12 atom.

• Atomic masses are weighted averages based on isotopic abundance.

$\text{Average atomic mass} = (\text{isotope 1 mass} \times \text{fraction}) + (\text{isotope 2 mass} \times \text{fraction}) + \ldots$

• Example: Carbon: 98.9% $^{12}$C (12.0000 amu), 1.1% $^{13}$C (13.00335 amu)

$\text{Average mass} = (0.9890 \times 12.0000) + (0.0110 \times 13.00335) = 12.01$ amu

The Mole Concept and Avogadro's Number

Definition of the Mole

• 1 mole (mol) is the amount of substance containing as many entities (atoms, molecules, ions) as there are atoms in 12 g of carbon-12.

• Avogadro's number ($N_A$): $6.022 \times 10^{23}$ entities per mole.

Calculations with Moles

• Number of moles: $n = \frac{N}{N_A}$

• Mass-mole relationship: $n = \frac{m}{M}$

• Example: 1 mol Mg = $6.022 \times 10^{23}$ atoms = 24.31 g

Sample Problems

• How many atoms in 2.35 mol Fe?

$2.35 \text{ mol Fe} \times 6.022 \times 10^{23} \text{ atoms/mol} = 1.42 \times 10^{24} \text{ atoms}$

• How many moles in $1.00 \times 10^{22}$ Mg atoms?

$\frac{1.00 \times 10^{22}}{6.022 \times 10^{23}} = 0.0166 \text{ mol}$

• What is the mass of 0.0166 mol Mg?

$0.0166 \text{ mol} \times 24.31 \text{ g/mol} = 0.404 \text{ g}$

Isotopic Abundance and Average Atomic Mass

• Average atomic mass is calculated using the masses and natural abundances of isotopes.

• Example: Magnesium has three isotopes: $^{24}$Mg (78.99%), $^{25}$Mg (10.00%), $^{26}$Mg (11.01%).

$\text{Average mass} = (0.7899 \times 23.985) + (0.1000 \times 24.986) + (0.1101 \times 25.983) = 24.31$ amu

Chemical Formulas

Empirical and Molecular Formulas

• Empirical formula: The simplest whole-number ratio of elements in a compound.

• Molecular formula: The actual number of atoms of each element in a molecule.

• Empirical formulas can be determined from percent composition data.

Additional info: Calculations for empirical and molecular formulas involve converting mass percentages to moles, finding the simplest ratio, and comparing to the molar mass.