Atoms, Elements, Compounds, and Chemical Composition: Study Guide

Chapter 4: Atoms and Elements

Dalton’s Atomic Theory

Dalton’s atomic theory laid the foundation for modern chemistry by describing the nature of atoms and their role in chemical reactions.

• Atoms: The smallest unit of matter that retains the properties of an element.

• Key Points:

  • All matter is composed of indivisible atoms.

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

  • Atoms cannot be created or destroyed in chemical reactions.

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

• Example: Water (H2O) is always composed of two hydrogen atoms and one oxygen atom.

Discovery of Electrons

The discovery of electrons revealed that atoms are divisible and contain smaller charged particles.

• Key Point: J.J. Thomson used cathode ray experiments to discover the electron, a negatively charged subatomic particle.

• Example: Electrons are responsible for electricity and chemical bonding.

Plum Pudding Model

The plum pudding model was an early attempt to describe atomic structure.

• Key Point: Proposed by J.J. Thomson, it suggested that electrons were embedded in a positively charged "pudding".

• Example: Like raisins in a cake, electrons are scattered throughout the atom.

Rutherford’s Experiment Results and Conclusions

Rutherford’s gold foil experiment changed the understanding of atomic structure.

• Key Point: Most alpha particles passed through the foil, but some were deflected, indicating a dense, positively charged nucleus.

• Conclusion: Atoms have a small, dense nucleus surrounded by electrons.

• Example: The nuclear model replaced the plum pudding model.

Subatomic Particles and Their Properties

Atoms are composed of three main subatomic particles.

• Proton: Positively charged, found in the nucleus, mass ≈ 1 amu.

• Neutron: Neutral, found in the nucleus, mass ≈ 1 amu.

• Electron: Negatively charged, found outside the nucleus, mass ≈ 0.0005 amu.

Particle	Charge	Location	Mass (amu)
Proton	+1	Nucleus	1
Neutron	0	Nucleus	1
Electron	-1	Outside nucleus	0.0005

The Periodic Table

The periodic table organizes elements based on their properties and atomic structure.

• Metals: Located on the left and center; shiny, conductive, malleable.

• Nonmetals: Located on the right; dull, poor conductors, brittle.

• Metalloids: Border between metals and nonmetals; properties intermediate.

• Main Group Elements: Groups 1, 2, and 13-18.

• Transition Elements: Groups 3-12.

Type	Location	Properties
Metals	Left/Center	Conductive, malleable, shiny
Nonmetals	Right	Poor conductors, brittle, dull
Metalloids	Border	Intermediate properties

Groups Numbering and Naming

Groups are columns in the periodic table, each with characteristic properties.

• Group 1: Alkali metals

• Group 2: Alkaline earth metals

• Group 17: Halogens

• Group 18: Noble gases

Cations and Anions – Predicting Ions for Main Group Elements

Main group elements form ions by gaining or losing electrons to achieve stable electron configurations.

• Cations: Positively charged ions formed by losing electrons (e.g., Na+).

• Anions: Negatively charged ions formed by gaining electrons (e.g., Cl-).

• Prediction: Group 1 forms +1 cations, Group 2 forms +2 cations, Group 17 forms -1 anions, Group 16 forms -2 anions.

Calculating Charge for Transition Elements

Transition elements can form multiple ions with different charges.

• Key Point: The charge is often indicated by Roman numerals in the compound name (e.g., Fe(III) for Fe3+).

• Example: Iron can form Fe2+ and Fe3+.

Isotopes and Calculating Weighted Average Mass

Isotopes are atoms of the same element with different numbers of neutrons.

• Key Point: The atomic mass is a weighted average of all naturally occurring isotopes.

• Formula:

$\text{Average atomic mass} = \sum (\text{isotope mass} \times \text{fractional abundance})$

• Example: Carbon has isotopes C-12 and C-13; the average atomic mass reflects their abundances.

Chapter 5: Molecules and Compounds

Law of Constant Composition

The law of constant composition states that a given compound always contains the same proportion of elements by mass.

• Key Point: Water always contains 11% hydrogen and 89% oxygen by mass.

Writing the Symbol of a Compound

Compounds are represented by chemical formulas indicating the elements and their ratios.

• Example: NaCl for sodium chloride, H2O for water.

Chemical Formulas

Chemical formulas show the types and numbers of atoms in a compound.

• Empirical Formula: Simplest whole-number ratio of elements.

• Molecular Formula: Actual number of atoms in a molecule.

• Example: Glucose: Empirical formula CH2O, molecular formula C6H12O6.

Polyatomic Ions and Representation in Compounds

Polyatomic ions are groups of atoms with a charge, acting as a unit in compounds.

• Example: Nitrate (NO3-), sulfate (SO42-).

• Representation: Compounds with polyatomic ions: NaNO3, CaSO4.

Classification of Elements and Compounds

Elements and compounds are classified based on their structure and bonding.

• Atomic Elements: Exist as single atoms (e.g., Ne, Ar).

• Molecular Elements: Exist as molecules (e.g., O2, N2).

• Ionic Compounds: Formed from cations and anions (e.g., NaCl).

• Molecular Compounds: Formed from nonmetals (e.g., CO2).

Writing Formulas for Ionic Compounds

Ionic compounds are written by balancing the charges of cations and anions.

• Key Point: The total positive and negative charges must equal zero.

• Example: Ca2+ and Cl- combine to form CaCl2.

Nomenclature

Nomenclature is the system for naming chemical compounds.

• Naming Binary Ionic Compounds: Name cation, then anion with "-ide" ending (e.g., sodium chloride).

• Naming Compounds with Metals with More Than One Ion: Use Roman numerals (e.g., iron(III) oxide).

• Naming Compounds with Polyatomic Ions: Name cation, then polyatomic ion (e.g., potassium nitrate).

• Naming Molecular Compounds: Use prefixes (mono-, di-, tri-) (e.g., carbon dioxide).

• Naming Binary Acids: "Hydro-" + root + "-ic acid" (e.g., hydrochloric acid).

• Naming Oxyacids: Based on polyatomic ion (e.g., sulfuric acid from sulfate).

Formula Mass of a Molecule and Formula Unit

The formula mass is the sum of the atomic masses of all atoms in a molecule or formula unit.

• Formula:

$\text{Formula mass} = \sum (\text{number of atoms} \times \text{atomic mass})$

• Example: Formula mass of NaCl: $1 \times 22.99 + 1 \times 35.45 = 58.44$ amu.

Chapter 6: Chemical Composition

The Mole and Counting Using the Mole

The mole is a counting unit used to relate the number of particles to mass.

• Key Point: 1 mole = $6.022 \times 10^{23}$ particles (Avogadro’s number).

• Example: 1 mole of H2O contains $6.022 \times 10^{23}$ molecules.

Molar Mass and Atomic Mass

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol).

• Key Point: Molar mass is numerically equal to atomic mass in amu, but in g/mol.

• Example: Molar mass of NaCl is 58.44 g/mol.

Conversion from Mass to Moles to Molecules

Conversions allow calculation between mass, moles, and number of particles.

• Formulas:

$\text{Moles} = \frac{\text{Mass (g)}}{\text{Molar mass (g/mol)}}$

$\text{Number of particles} = \text{Moles} \times 6.022 \times 10^{23}$

• Example: 18 g H2O = 1 mole = $6.022 \times 10^{23}$ molecules.

Conversion from Molecules to Moles to Mass

Reverse conversions are used to find mass from number of molecules.

• Formulas:

$\text{Moles} = \frac{\text{Number of particles}}{6.022 \times 10^{23}}$

$\text{Mass (g)} = \text{Moles} \times \text{Molar mass (g/mol)}$

Mass Percent

Mass percent expresses the proportion of an element in a compound by mass.

• Formula:

$\text{Mass percent} = \frac{\text{mass of element in 1 mole}}{\text{mass of 1 mole of compound}} \times 100\%$

• Example: In H2O, mass percent of H = $\frac{2 \times 1.01}{18.02} \times 100\% = 11.2\%$

Using Mass Percent as a Conversion Factor

Mass percent can be used to calculate the mass of an element in a given mass of compound.

• Example: 100 g H2O contains 11.2 g H.

Calculating Empirical Formula from Experimental Mass Data

Empirical formulas are determined from experimental data by finding the simplest ratio of elements.

• Steps:

  1. Convert mass of each element to moles.

  2. Divide by the smallest number of moles.

  3. Round or multiply to get whole numbers.

• Example: 40 g C and 6.7 g H yields empirical formula CH2.

Calculating Empirical Formula Using Mass Percent Composition

Mass percent composition can be used to determine empirical formulas.

• Key Point: Assume 100 g sample, convert mass percent to grams, then to moles.

Calculating Molecular Formulas Using Empirical Formulas and Molar Masses

The molecular formula is determined by comparing the empirical formula mass to the molar mass.

• Formula:

$\text{n} = \frac{\text{Molar mass}}{\text{Empirical formula mass}}$

$\text{Molecular formula} = (\text{Empirical formula}) \times n$

• Example: Empirical formula CH2, molar mass 28 g/mol, empirical mass 14 g/mol, n = 2, molecular formula C2H4.