Introductory Chemistry Study Notes: Chemical Composition, Reactions, Gases, Solutions, and Acids/Bases

Chapter 6: Chemical Composition – Moles and Molar Mass

Understanding Avogadro’s Number

Avogadro’s number is a fundamental constant in chemistry, linking the macroscopic scale of substances to the atomic and molecular scale.

• Definition: Avogadro’s number is $6.022 \times 10^{23}$ particles per mole. These particles can be atoms, molecules, formula units, or ions, depending on the substance.

• Example: 1.00 mol of carbon atoms contains $6.022 \times 10^{23}$ carbon atoms.

• Historical Context: Named after Amedeo Avogadro, who proposed that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

Converting Between Moles and Atoms

Conversions between moles and the number of particles are essential for quantitative chemical calculations.

• Conversion Process: Use Avogadro’s number and the chemical formula’s subscripts to relate moles to atoms or molecules.

• Example: To find the number of hydrogen atoms in 2.00 mol of H2O:

$2.00\ \text{mol H}_2\text{O} \times \frac{6.022 \times 10^{23}\ \text{molecules}}{1\ \text{mol}} \times \frac{2\ \text{H atoms}}{1\ \text{molecule}} = 2.41 \times 10^{24}\ \text{H atoms}$

• Importance: Understanding molecular formulas is crucial for stoichiometry.

Calculating Molar Mass

Molar mass connects the mass of a substance to the number of moles present.

• Definition: The molar mass of an element is its atomic mass (from the periodic table) expressed in grams per mole (g/mol).

• Example: Carbon’s atomic mass is approximately 12.01 amu, so its molar mass is 12.01 g/mol.

• Application: Used for converting between grams and moles in chemical calculations.

Molar Mass of Compounds

The molar mass of a compound is calculated by summing the molar masses of its constituent elements, each multiplied by its subscript in the formula.

• Calculation: For H2O: $2 \times 1.01\ \text{g/mol} + 1 \times 16.00\ \text{g/mol} = 18.02\ \text{g/mol}$

• Importance: Essential for quantitative analysis in chemistry.

Converting Moles to Grams and Vice Versa

Molar mass serves as a conversion factor between the amount of substance (in moles) and its mass (in grams).

• Conversion: Multiply by molar mass to convert moles to grams; divide by molar mass to convert grams to moles.

• Example: $3.00\ \text{mol CO}_2 \times 44.01\ \text{g/mol} = 132\ \text{g CO}_2$

Chapter 7: Chemical Reactions

Evidence of Chemical Reactions

Chemical reactions are identified by observable changes indicating the formation of new substances.

• Indicators: Color change, gas formation (bubbling), precipitate formation, temperature change, light production, odor change.

• Example: Mixing two clear solutions that form a cloudy solid (precipitate).

Law of Conservation of Mass

This law states that matter is neither created nor destroyed in a chemical reaction; atoms are simply rearranged.

• Principle: Total mass of reactants equals total mass of products in a closed system.

• Example: 10 g of reactants yield 10 g of products.

Chemical Equations

Chemical equations use formulas and symbols to represent chemical reactions, showing reactants and products.

• Format: Reactants on the left, products on the right, separated by an arrow.

• Example: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$

Balancing Chemical Equations

Balancing ensures the same number of each atom on both sides of the equation, upholding the law of conservation of mass.

• Technique: Adjust coefficients (not subscripts) to balance atoms.

• Example: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ is balanced.

Writing Balanced Chemical Equations

Translate word descriptions into chemical formulas and include states of matter.

• Example: Sodium chloride solution reacts with silver nitrate solution to form silver chloride solid and sodium nitrate solution:

• $\text{NaCl}(aq) + \text{AgNO}_3(aq) \rightarrow \text{AgCl}(s) + \text{NaNO}_3(aq)$

Solubility and Precipitation Reactions

Solubility rules help predict whether a precipitate will form in a reaction.

• Solubility: Maximum amount of solute that can dissolve in a solvent at a given temperature.

• Key Rules: All nitrates and Group 1 salts are soluble; most chlorides are soluble except AgCl, PbCl2, Hg2Cl2.

• Example: AgCl is insoluble and forms a precipitate.

Types of Chemical Reactions

• Precipitation: Two aqueous solutions form an insoluble solid (precipitate). Example: $\text{AgNO}_3(aq) + \text{NaCl}(aq) \rightarrow \text{AgCl}(s) + \text{NaNO}_3(aq)$

• Redox: Involves electron transfer (Oxidation Is Loss, Reduction Is Gain). Example: $\text{Zn} + \text{CuCl}_2 \rightarrow \text{ZnCl}_2 + \text{Cu}$

• Combination: Two or more reactants form one product. Example: $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$

• Decomposition: One reactant breaks into two or more products. Example: $2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$

• Exchange/Displacement: Ions or elements swap places. Example: $\text{AB} + \text{CD} \rightarrow \text{AD} + \text{CB}$

Single Displacement/Replacement Reactions

• Definition: One element replaces another in a compound. Example: $\text{Fe} + \text{CuSO}_4 \rightarrow \text{FeSO}_4 + \text{Cu}$

Double Displacement/Replacement Reactions

• Definition: Two compounds exchange ions. Example: $\text{NaCl} + \text{AgNO}_3 \rightarrow \text{AgCl} + \text{NaNO}_3$

Stoichiometry and Chemical Calculations

Stoichiometry uses balanced equations to calculate the relationships between reactants and products.

• Example: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ means 2 mol H2 react with 1 mol O2 to produce 2 mol H2O.

Mole-to-Mole Conversions

• Definition: Use coefficients from balanced equations to relate moles of substances.

• Example: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$, so $\frac{2\ \text{mol H}_2\text{O}}{2\ \text{mol H}_2}$ or $\frac{1\ \text{mol O}_2}{2\ \text{mol H}_2\text{O}}$

Theoretical Yield and Limiting Reactants

• Theoretical Yield: Maximum product possible from given reactants.

• Limiting Reactant: The reactant that is completely consumed first.

• Example: If 32.0 g O2 is used in $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$, theoretical yield is 36.0 g H2O.

Percent Yield Calculations

• Formula: $\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$

• Example: If actual yield is 8.5 g and theoretical yield is 10.0 g, percent yield is 85%.

Chapter 11: Gases

Key Properties of Gases

• No fixed shape or volume; expand to fill their container.

• Compressible due to large spaces between particles.

• Low density compared to solids and liquids.

• Mix easily with other gases (diffusion).

• Exert pressure due to collisions with container walls.

Molecular Characteristics of Gases

• Particles are far apart and in constant, random motion.

• Weak attractions between particles.

• Behavior modeled by kinetic molecular theory (KMT).

• Average kinetic energy is proportional to temperature (Kelvin).

Kinetic Molecular Theory (KMT) and Gas Behavior

• Gas particles are tiny and far apart (explains low density, high compressibility).

• Constant random motion causes pressure.

• Collisions are elastic (no net loss of kinetic energy).

• Average kinetic energy increases with temperature.

Gas Laws and Calculations

• Boyle’s Law: $P_1V_1 = P_2V_2$ (pressure and volume inversely related at constant T and n).

• Charles’s Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$ (volume and temperature directly related at constant P and n).

• Combined Gas Law: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ (relates P, V, and T at constant n).

Gas Pressure and Units

• Definition: Force exerted by gas particles colliding with container walls.

• Common Units: atm, mmHg, torr, kPa, psi.

Unit	Equality
1 atm	760 mmHg = 760 torr = 101.325 kPa

Ideal Gas Law

• Formula: $PV = nRT$

• R (gas constant): 0.0821 L·atm/(mol·K)

• Example: 1.00 mol gas at 1.00 atm and 273 K occupies 22.4 L.

Avogadro’s Law

• Formula: $\frac{V_1}{n_1} = \frac{V_2}{n_2}$

• Equal volumes of gases at same T and P contain equal numbers of molecules.

Dalton’s Law of Partial Pressures

• Formula: $P_{\text{total}} = P_1 + P_2 + P_3 + \ldots$

• Application: Used for gas mixtures (e.g., air composition).

Standard Temperature and Pressure (STP)

• STP: 0°C (273.15 K) and 1 atm.

• At STP: 1 mol ideal gas = 22.4 L.

Chapter 12: Liquids, Solids, and Intermolecular Forces

Characteristics of Solids and Liquids

• Solids: Definite shape and volume; particles vibrate in fixed positions.

• Liquids: Definite volume, take shape of container; particles can move past each other.

• Both are much less compressible than gases.

Surface Tension

• Definition: Resistance of a liquid’s surface to stretching or breaking due to intermolecular forces.

• Example: Water beads on a surface.

Viscosity

• Definition: Resistance to flow; higher viscosity means slower flow.

• Example: Honey is more viscous than water.

Intermolecular Forces

• Hydrogen Bonding: Strong dipole attraction when H is bonded to N, O, or F (e.g., H2O).

• Dipole-Dipole Forces: Attractions between polar molecules (e.g., HCl).

• Dispersion Forces: Weak attractions present in all molecules (e.g., noble gases).

• Ion-Dipole Forces: Attraction between an ion and a polar molecule (e.g., Na+ in water).

Type	Occurs In	Example
Hydrogen Bonding	H bonded to N, O, or F	H2O, NH3
Dipole-Dipole	Polar molecules	HCl
Dispersion	All molecules	I2, F2
Ion-Dipole	Ions in polar solvents	NaCl in water

Chapter 13: Solutions

Definition of Solutions

• Solution: Homogeneous mixture with uniform composition.

• Example: Salt water (NaCl dissolved in H2O).

Solute and Solvent

• Solute: Substance being dissolved (e.g., salt).

• Solvent: Substance doing the dissolving (e.g., water).

Types of Solutions

• Unsaturated: Less than maximum solute dissolved; more can dissolve.

• Saturated: Maximum solute dissolved; excess remains undissolved.

• Supersaturated: More solute dissolved than normally possible; unstable.

Electrolytes and Non-Electrolytes

• Electrolyte: Solution conducts electricity due to dissolved ions (e.g., NaCl in water).

• Non-Electrolyte: Solution does not conduct electricity; contains molecules, not ions (e.g., sugar in water).

Factors Affecting Solubility

• Temperature: Solubility of solids increases with temperature; solubility of gases decreases with temperature.

• Pressure: Gas solubility increases with pressure (Henry’s Law).

Concentration of Solutions

• Concentrated: Large amount of solute relative to solvent.

• Dilute: Small amount of solute relative to solvent.

Mass Percent Concentration

• Formula: $\text{Mass percent} = \frac{\text{mass solute}}{\text{mass solution}} \times 100$

• Example: 5.0 g salt in 100.0 g solution: $\frac{5.0}{100.0} \times 100 = 5.0\%$

Molarity (M)

• Definition: Moles of solute per liter of solution ($M = \frac{\text{mol}}{\text{L}}$)

• Example: 0.50 mol NaCl in 2.0 L solution: $0.25\ M$

Ion Concentration in Solutions

• Example: 0.20 M CaCl2 yields [Ca2+] = 0.20 M, [Cl-] = 0.40 M

Dilution Calculations

• Formula: $M_1V_1 = M_2V_2$

• Example: To make 100.0 mL of 0.50 M from 2.0 M stock: $V_1 = \frac{(0.50)(100.0)}{2.0} = 25.0\ \text{mL}$

Chapter 14: Acids and Bases

Properties of Acids

• Sour taste, react with metals, turn blue litmus red, pH < 7, increase [H3O+].

• Example: HCl (hydrochloric acid).

Properties of Bases

• Bitter taste, slippery feel, turn red litmus blue, pH > 7, increase [OH-].

• Example: NaOH (sodium hydroxide).

Arrhenius Theory of Acids and Bases

• Arrhenius Acid: Produces H+ (or H3O+) in water.

• Arrhenius Base: Produces OH- in water.

• Example: HCl is an acid, NaOH is a base.

Bronsted-Lowry Theory of Acids and Bases

• Bronsted-Lowry Acid: Proton (H+) donor.

• Bronsted-Lowry Base: Proton (H+) acceptor.

• Example: NH3 accepts H+ to become NH4+.

Strong and Weak Acids/Bases

• Strong Acid: Completely ionizes in water (e.g., HCl).

• Weak Acid: Partially ionizes (e.g., CH3COOH).

• Strong Base: Completely dissociates (e.g., NaOH).

• Weak Base: Partially reacts with water (e.g., NH3).

Water and pH

• Water autoionizes: $2\text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-$

• At 25°C: $[\text{H}_3\text{O}^+] = [\text{OH}^-] = 1.0 \times 10^{-7}\ \text{M}$

• pH = 7 is neutral.

Ion-Product Constant for Water (Kw)

• Formula: $K_w = [\text{H}_3\text{O}^+][\text{OH}^-]$

• At 25°C: $K_w = 1.0 \times 10^{-14}$

Understanding pH and pOH

• pH: $pH = -\log[\text{H}_3\text{O}^+]$

• pOH: $pOH = -\log[\text{OH}^-]$

• Relationship: $pH + pOH = 14$ (at 25°C)

• To find [H3O+]: $[\text{H}_3\text{O}^+] = 10^{-pH}$

• To find [OH-]: $[\text{OH}^-] = 10^{-pOH}$

Concept	Formula	Description	Example Calculation
pH	$pH = -\log[\text{H}_3\text{O}^+]$	Acidity based on [H3O+]	$pH = -\log(1.0 \times 10^{-4}) = 4.00$
pOH	$pOH = -\log[\text{OH}^-]$	Basicity based on [OH-]	$pOH = -\log(1.0 \times 10^{-3}) = 3.00$
Relationship	$pH + pOH = 14$	Sum at 25°C is constant	$pH = 3.25 \rightarrow pOH = 10.75$
[H3O+] from pH	$[\text{H}_3\text{O}^+] = 10^{-pH}$	Find [H3O+] from pH	$[\text{H}_3\text{O}^+] = 10^{-4.50} = 3.2 \times 10^{-5}\ \text{M}$
[OH-] from pOH	$[\text{OH}^-] = 10^{-pOH}$	Find [OH-] from pOH	$[\text{OH}^-] = 10^{-5.00} = 1.0 \times 10^{-5}\ \text{M}$

Acidic, Basic, and Neutral Solutions

• Acidic: pH < 7, $[\text{H}_3\text{O}^+] > [\text{OH}^-]$

• Basic: pH > 7, $[\text{OH}^-] > [\text{H}_3\text{O}^+]$

• Neutral: pH = 7, $[\text{H}_3\text{O}^+] = [\text{OH}^-]$