Fundamental Concepts in Chemistry

Atomic Structure

Atoms are composed of three fundamental particles: Protons, Neutrons, and Electrons.

Protons

Positively charged particles (+1 charge) located in the nucleus.

Neutrons

Neutral particles (no charge) located in the nucleus.

Electrons

Negatively charged particles (-1 charge) orbiting the nucleus.

States of Matter

The physical forms in which substances can exist, including solids, liquids, and gases.

Properties of Solids

Solids have a definite shape, fixed positions of particles, compact particles, relatively high densities, and are difficult to compress.

Physical Properties

Characteristics that can be observed or measured without changing the substance's chemical identity (e.g., color, density, melting point).

Chemical Properties

Describe how a substance changes or reacts to form new substances (e.g., flammability, corrosiveness).

Zinc

A bluish-white metal that has physical properties such as a density of 7.14 g/cm3 and a melting point of 419°C, and a chemical property of corroding upon prolonged contact with moist air.

Substance

A material with a constant composition and properties.

Element

A pure substance that cannot be broken down into simpler substances by chemical means.

Compound

A substance formed when two or more elements are chemically bonded, which can be decomposed into simpler substances by chemical means.

Mixture

A combination of two or more substances that are physically combined, not chemically bonded.

Homogeneous Mixture

A mixture with uniform composition throughout (e.g., ethyl alcohol and water).

Heterogeneous Mixture

A mixture with non-uniform composition.

SI System

The International System of Units that uses prefixes to denote multiples or submultiples of base units.

SI Prefixes

Prefixes used in the SI system to express very large or very small quantities, such as mega- (10^6), kilo- (10^3), centi- (10^-2), and micro- (10^-6).

Dimensional Analysis

A method that involves using conversion factors to change units.

Conversion Example

To convert 16.0 lb to grams using the factor 2.2 lbs = 1 kg results in approximately 7260 g.

Key Concept of Chemical Properties

Chemical properties involve a change in the substance's composition.

Key Concept of Physical Properties

Physical properties can be observed without changing the substance's identity.

Important Note on Mixtures

Mixtures can have varying compositions, while pure substances have fixed compositions.

Common Mistake in SI Prefixes

The abbreviation for micro- is μ (mu), not m (which is milli-).

Important Note on Calculations

Always include units in your calculations and ensure they cancel out correctly.

Density

Density is defined as mass per unit volume.

Density Calculation

Density = Mass / Volume

Example of Density Calculation

A metal cube with a mass of 112 g is dropped into a graduated cylinder containing 30.00 mL of water, causing the water level to rise to 39.50 mL. Volume of cube = 39.50 mL - 30.00 mL = 9.50 mL. Density = 112 g / 9.50 mL = 11.8 g/mL.

Volume Calculation

Volume = Mass / Density

Example of Volume Calculation

A gold ring has a mass of 15.37 g. If the ring is pure gold (density = 16.1 g/mL): Volume = 15.37 g / 16.1 g/mL = 0.955 mL.

Real-world Application of Density

Density helps identify substances and determine purity.

Summary of Density

Density is a fundamental property relating mass and volume, and it can be used to calculate either mass or volume if the other is known.

Naming Binary Compounds

Naming binary compounds involves using prefixes to indicate the number of atoms of each element, following specific rules based on the type of compound.

Nomenclature Rules for Binary Molecular Compounds

Use prefixes to indicate the number of atoms of each element (e.g., dichlorine heptoxide for Cl2O7).

Key Prefixes for Binary Compounds

mono- (1), di- (2), tri- (3), tetra- (4), penta- (5), hexa- (6), hepta- (7)

Chemical Formulas

Chemical formulas represent the composition of compounds, and acids are named following specific conventions based on their composition.

Acid Nomenclature

Oxyacids: Acids containing oxygen. Named based on the polyatomic ion. For example, HNO3 (aq) is nitric acid.

Common Acids

Hydrochloric acid (HCl), Sulfuric acid (H2SO4), Nitric acid (HNO3).

Ionic Compounds

Compounds formed by the electrostatic attraction between ions.

Naming Ionic Compounds

The cation (positive ion) is named first, followed by the anion (negative ion).

Examples of Ionic Compounds

calcium sulfate / CaSO4, ammonium sulfide / (NH4)2S, magnesium hydroxide / Mg(OH)2, iron (II) carbonate / FeCO3, aluminum nitrite / Al(NO2)3.

Polyatomic Ions

Memorize common polyatomic ions and their charges.

Importance of Ion Charges

It is important to know the charges of ions when writing formulas.

Mole Calculation

A mole is a unit of amount (approximately 6.022 × 10^23 entities).

Moles Calculation Formula

Moles = Number of molecules / Avogadro's number

Example of Moles Calculation

How many moles of CO2 are present in 2.2 × 10^9 CO2 molecules? Moles of CO2 = 2.2 × 10^9 / 6.022 × 10^23 = 3.7 × 10^−15 moles.

Avogadro's Number

6.022 × 10^23 is the number of entities in one mole.

Percent Composition Calculation

Percent by mass = (Mass of element / Mass of compound) × 100%

Example of Percent Composition Calculation

What is the percent by mass of sulfur in Na2S? Molar mass of Na2S = (2×23) + 32.06 = 78.06 g/mol. Percent of S = (32.06 / 78.06) × 100% = 41%.

Application of Percent Composition

Percent composition is used to identify and characterize compounds.

Summary of Percent Composition

Percent composition indicates the proportion of each element's mass in a compound.

Molecular Formula Determination

Find the empirical formula from the percent composition. Determine the molecular weight. Calculate the ratio between the molecular weight and the empirical formula weight.

Molecular Formula

The molecular formula gives the actual number of atoms in a molecule.

Empirical Formula

The empirical formula is determined from the ratios of the elements in a compound.

Empirical Formula Weight

(2×12.01)+(4×1.008)+16.00=44.05 g/mol

Ratio

88 g/mol / 44.05 g/mol ≈ 2

Molecular Formula Calculation

C2×2 H4×2 O1×2 = C4H8O2

Balancing Chemical Equations

Balancing chemical equations involves adjusting coefficients to ensure the number of atoms of each element is the same on both sides of the equation.

Conservation of Mass

Balancing ensures mass is conserved in chemical reactions.

Stoichiometry

Stoichiometry involves using balanced equations to calculate the amounts of reactants and products in a chemical reaction.

Mole Ratios

Use coefficients from the balanced equation to determine mole ratios.

Limiting Reactant

The reactant that is completely consumed in a reaction, determining the maximum amount of product formed.

Percent Yield

The ratio of actual yield to theoretical yield, expressed as a percentage.

Theoretical Yield of ClF3

0.116 mol ClF3 × 92.45 g/mol = 10.72 g ClF3

Percent Yield Calculation

Percent Yield = (Actual Yield / Theoretical Yield) × 100% = (8.5 g / 10.72 g) × 100% ≈ 79%

Significance of Percent Yield

Percent yield indicates the efficiency of a chemical reaction.

Summary of Molecular Formula

The molecular formula is determined from the empirical formula and the molecular weight of the compound.

Summary of Balancing Equations

Balancing chemical equations ensures the conservation of mass by having an equal number of atoms for each element on both sides of the equation.

Summary of Stoichiometry

Stoichiometry uses balanced equations to quantitatively relate reactants and products in chemical reactions.

Summary of Limiting Reactant

The limiting reactant determines the maximum product yield, and the percent yield measures the reaction's efficiency.

Example of Empirical Formula Calculation

C: 54.53 g / 12.01 g/mol ≈ 4.54 mol; H: 9.15 g / 1.008 g/mol ≈ 9.08 mol; O: (100−54.53−9.15) g = 36.32 g / 16.00 g/mol ≈ 2.27 mol

Balanced Equation Example

P4 + 10 Cl2 → 4 PCl5

Moles of Reactants Calculation

Cl2: 5.6 g / 70.9 g/mol ≈ 0.079 mol; F2: 6.6 g / 38.0 g/mol ≈ 0.174 mol

Limiting Reactant Determination Example

For every 1 mol of Cl2, we need 3 mol of F2. 0.079 mol Cl2 × 3 = 0.237 mol F2 needed. Since we only have 0.174 mol of F2, F2 is the limiting reactant.

Molarity

Molarity (M) is the number of moles of solute per liter of solution.

Molarity Formula

Molarity = Moles of solute / Liters of solution

Moles of H2SO4

Moles of H2SO4 = 70.0 g / 98.08 g/mol = 0.714 mol

Volume of solution

Volume of solution = 280. mL = 0.280 L

Molarity Calculation Example

Molarity = 0.714 mol / 0.280 L = 2.55 M

Volume Calculation Formula

Volume = Moles of solute / Molarity

Moles of KOH

Moles of KOH = 6.31 g / 56.11 g/mol = 0.112 mol

Volume of KOH Solution

Volume = 0.112 mol / 0.250 M = 0.448 L = 448 mL ≈ 450 mL

Practical Application of Molarity

Molarity is essential for preparing solutions of specific concentrations.

Titration

Titration is a technique used to determine the concentration of a solution by reacting it with a solution of known concentration.

Titration Calculation Formula

M1V1 = M2V2

Molarity of KOH Example

M KOH = (0.1982 M × 25.84 mL) / 38.65 mL = 0.1325 M

Equivalence Point

The point in a titration where the acid and base have completely reacted.

Protons

Protons determine the element.

Neutrons

Neutrons affect the mass number (isotope).

Electrons

Electrons determine the charge.

41K Isotope Example

Potassium (K) has an atomic number of 19, so it has 19 protons. Neutrons = 41 - 19 = 22. There are 19 electrons.

Ion Symbol Example

The symbol for a species composed of 38 protons, 52 neutrons, and 36 electrons is 90Sr2+.

Isotopic Notation

Isotopic Notation: Z_A X, where A is the mass number, Z is the atomic number, and X is the element symbol.

Electromagnetic Radiation Relationship

c = λν, where c is the speed of light (3.00 × 10^8 m/s).

Wavelength Calculation Example

λ = c / ν = 3.00 × 10^8 m/s / 6.10 × 10^14 Hz = 4.92 × 10^-7 m

Convert Meters to Ångstroms

4.92 × 10^-7 m × 1 × 10^10 Å / 1 m = 4.92 × 10^3 Å

Planck's Equation

E = hν, where E is energy, h is Planck's constant, and ν is frequency.

Electromagnetic radiation

Wavelength and frequency are inversely proportional, and energy is directly proportional to frequency.

Electron Configuration

Describes the arrangement of electrons in an element's ground state.

Ground State Configuration

The electron configuration of an element in its lowest energy state.

Aufbau Principle

States that electrons fill the lowest energy levels first.