Molecules, Compounds, Nomenclature, and Reaction Stoichiometry Study Guide

Fundamental Concepts: Atoms, Ions, Mixtures, and Compounds

  • Mixtures versus Compounds:

    • Mixtures: Formed when two or more different substances are physically combined. Mixtures possess variable composition (can exist in any proportion) and may be homogeneous or heterogeneous. Components retain their individual chemical identities and can be separated using physical or mechanical separation processes.

    • Compounds: Formed when two or more elements are chemically bonded together to produce a pure substance. Compounds possess a fixed, definite composition in accordance with the Law of Constant Composition. They cannot be separated by physical means and require chemical reactions (such as chemical decomposition) to break down into constituent elements.

  • Atoms versus Ions:

    • An atom is electrically neutral, containing equal numbers of protons and electrons.

    • An ion is an atom or group of atoms that carries a net positive or negative charge due to gaining or losing valence electrons.

    • Ionization: The process by which neutral atoms gain or lose electrons. Ions never form through the gain or loss of protons or neutrons, as altering nuclear protons changes the elemental identity.

    • Cations: Positively charged ions formed when neutral atoms lose electrons. Main-group metals typically lose electrons to form cations.

    • Anions: Negatively charged ions formed when neutral atoms gain electrons. Main-group nonmetals typically gain electrons to form anions.

    • Polyatomic Ions: Charged species consisting of two or more covalently bonded atoms that act as a single ionic unit.

Chemical Bonding: Ionic versus Covalent

  • Chemical Bonds:

    • Chemical compounds are held together by chemical bonds, which originate from electrostatic forces of attraction between charged subatomic particles (positively charged nuclei/protons and negatively charged electrons).

Types of Bonding

  • Ionic Bonds:

    • Mechanism: Result from the complete transfer of one or more valence electrons from one atom (typically a metal) to another (typically a nonmetal).

    • The metal atom loses electrons to become a cation (Mn+\text{M}^{n+}), while the nonmetal atom gains electrons to become an anion (Xn−\text{X}^{n-}).

    • The electrostatic attraction between these oppositely charged ions forms the ionic bond.

    • Solid State Structure: Ionic compounds do not form discrete individual molecules. Instead, they organize into an extended, three-dimensional crystalline lattice consisting of alternating, repeating cations and anions.

Formation of an Ionic Compound

  • Properties of Ionic Compounds:

    • High melting points and high boiling points due to strong inter-ionic electrostatic attractions across the lattice.

    • High heats of fusion and vaporization.

    • Electrical insulators in the solid state because ions are fixed in rigid lattice positions.

    • Electrical conductors when molten (liquid phase) or dissolved in water (aq\text{aq}), as ions become mobile charge carriers.

    • Hard and brittle solids; mechanical stress forces ions of like charge adjacent to one another, causing repulsion and sharp cleavage along crystal planes.

  • Covalent Bonds:

    • Mechanism: Result from the sharing of electron pairs between two or more nonmetal atoms.

    • No electron transfer occurs and discrete ions are not generated. Electrostatic attractions occur between the positively charged nuclei of both atoms and the negatively charged shared electron density localized between them.

    • Shared electrons occupy regions of lower potential energy compared to isolated, unbonded atoms, stabilizing the molecule.

    • Compounds bonded purely through covalent interactions are referred to as molecular compounds.

Potential Energy in Covalent Bonding

  • Properties of Covalent (Molecular) Compounds:

    • Relatively low melting and boiling points.

    • Low heats of fusion and vaporization.

    • Poor conductors of heat and electricity in all states of matter.

    • Exist as soft solids, liquids, or gases at room temperature.

    • Phase Change Behavior: When a molecular substance undergoes melting or boiling, covalent bonds within individual molecules remain intact. Thermal energy overcomes only the relatively weak intermolecular non-bonding attractive forces holding separate molecules together.

Intermolecular vs Intramolecular Forces

Representing Compounds: Chemical Formulas and Molecular Models

  • Types of Chemical Formulas:

    • Molecular (Chemical) Formula: Explicitly states the actual number and type of each element present in a single discrete molecule of a compound (e.g., Water = H2O\text{H}_2\text{O}, Carbon Dioxide = CO2\text{CO}_2, Glucose = C6H12O6\text{C}_6\text{H}_{12}\text{O}_6).

    • Empirical Formula: Expresses the simplest, lowest whole-number molar ratio of elements present in a compound (e.g., Glucose molecular formula C6H12O6\text{C}_6\text{H}_{12}\text{O}_6 reduces to the empirical formula CH2O\text{CH}_2\text{O}). Ionic compounds are exclusively represented by empirical formulas.

    • Structural Formula: Employs structural lines to represent covalent bonds between atoms, illustrating exact atomic connectivity and spatial bonding arrangements in two dimensions.

  • Three-Dimensional Molecular Models:

    • Perspective Drawing: Uses specific line conventions to depict 3D orientation on a 2D page: solid lines represent bonds in the plane of the page, dashed wedges represent bonds pointing away/behind the page, and solid wedges represent bonds pointing forward/out of the page.

    • Ball-and-Stick Model: Represents atoms as color-coded spheres and chemical bonds as rigid rods, accurately displaying 3D bond angles and geometric arrangements.

    • Space-Filling Model: Represents atoms as overlapping spheres whose radii are proportional to van der Waals radii, offering an estimate of overall molecular shape and filled volume scaled to visible size.

Molecular Representations of Methane

  • Examples of Compound Formulas and Models:

Table 3.1 Benzene Acetylene Glucose Ammonia
*   *Benzene*: Empirical Formula = CH\text{CH}, Molecular Formula = C6H6\text{C}_6\text{H}_6
*   *Acetylene*: Empirical Formula = CH\text{CH}, Molecular Formula = C2H2\text{C}_2\text{H}_2
*   *Glucose*: Empirical Formula = CH2O\text{CH}_2\text{O}, Molecular Formula = C6H12O6\text{C}_6\text{H}_{12}\text{O}_6
*   *Ammonia*: Empirical Formula = NH3\text{NH}_3, Molecular Formula = NH3\text{NH}_3
*   *Formaldehyde*: Empirical Formula = CH2O\text{CH}_2\text{O}, Molecular Formula = CH2O\text{CH}_2\text{O}
*   *Acetic Acid*: Empirical Formula = CH2O\text{CH}_2\text{O}, Molecular Formula = C2H4O2\text{C}_2\text{H}_4\text{O}_2
*   *Lactic Acid*: Empirical Formula = CH2O\text{CH}_2\text{O}, Molecular Formula = C3H6O3\text{C}_3\text{H}_6\text{O}_3
*   *Ribose*: Empirical Formula = CH2O\text{CH}_2\text{O}, Molecular Formula = C5H10O5\text{C}_5\text{H}_{10}\text{O}_5

Classification of Pure Substances and Molecular Units

  • Hierarchy of Pure Substances:

    • Elements:

      • Atomic Elements: Exist in nature with single unbonded atoms as basic structural units (e.g., Ne\text{Ne}, Na\text{Na}, He\text{He}).

      • Molecular Elements: Exist in nature as multi-atom molecules composed of two or more identical atoms covalently bound together (e.g., O2\text{O}_2, N2\text{N}_2, S8\text{S}_8).

    • Compounds:

      • Molecular Compounds: Composed of discrete covalently bound molecules consisting of two or more different nonmetals (e.g., H2O\text{H}_2\text{O}, CO2\text{CO}_2).

      • Ionic Compounds: Composed of extended ionic lattices containing cations and anions; basic repeating unit is defined as a formula unit (e.g., NaCl\text{NaCl}).

Classification of Pure Substances

  • The Seven Natural Diatomic Elements:

    • Seven elements exist naturally as neutral homonuclear diatomic molecules:

      1. Hydrogen (H2\text{H}_2)

      2. Nitrogen (N2\text{N}_2)

      3. Oxygen (O2\text{O}_2)

      4. Fluorine (F2\text{F}_2)

      5. Chlorine (Cl2\text{Cl}_2)

      6. Bromine (Br2\text{Br}_2)

      7. Iodine (I2\text{I}_2)

    • Mnemonic device: "Have No Fear Of Ice Cold Beer".

Diatomic Elements

Chemical Nomenclature and Ion Reference Tables

  • Predictable Monatomic Ions:

    • 1+ Cations: Group 1A Alkali Metals (Li+\text{Li}^+, Na+\text{Na}^+, K+\text{K}^+, Rb+\text{Rb}^+, Cs+\text{Cs}^+), Hydrogen (H+\text{H}^+), and Silver (Ag+\text{Ag}^+).

    • 2+ Cations: Group 2A Alkaline Earth Metals (Mg2+\text{Mg}^{2+}, Ca2+\text{Ca}^{2+}, Sr2+\text{Sr}^{2+}, Ba2+\text{Ba}^{2+}), Cadmium (Cd2+\text{Cd}^{2+}), and Zinc (Zn2+\text{Zn}^{2+}).

    • 3+ Cations: Group 3A Metal Aluminum (Al3+\text{Al}^{3+}).

    • 1- Anions: Group 7A Halogens (F−\text{F}^- fluoride, Cl−\text{Cl}^- chloride, Br−\text{Br}^- bromide, I−\text{I}^- iodide) and Hydride (H−\text{H}^-).

    • 2- Anions: Group 6A Chalcogens (O2−\text{O}^{2-} oxide, S2−\text{S}^{2-} sulfide).

    • 3- Anions: Group 5A (N3−\text{N}^{3-} nitride, P3−\text{P}^{3-} phosphide).

    • Note: Nonmetal elements only display these fixed negative charges when present as anions inside ionic compounds.

  • Polyatomic Ion Reference Table:

    • Cations: Ammonium (NH4+\text{NH}_4^+), Mercury(I) (Hg22+\text{Hg}_2^{2+}).

    • 1- Anions: Nitrate (NO3−\text{NO}_3^-), Nitrite (NO2−\text{NO}_2^-), Bicarbonate/Hydrogen Carbonate (HCO3−\text{HCO}_3^-), Perchlorate (ClO4−\text{ClO}_4^-), Chlorate (ClO3−\text{ClO}_3^-), Chlorite (ClO2−\text{ClO}_2^-), Hypochlorite (ClO−\text{ClO}^-), Perbromate (BrO4−\text{BrO}_4^-), Bromate (BrO3−\text{BrO}_3^-), Bromite (BrO2−\text{BrO}_2^-), Hypobromite (BrO−\text{BrO}^-), Periodate (IO4−\text{IO}_4^-), Iodate (IO3−\text{IO}_3^-), Iodite (IO2−IO_2^-), Hypoiodite (IO−\text{IO}^-), Acetate (CH3COO−\text{CH}_3\text{COO}^- or C2H3O2−\text{C}_2\text{H}_3\text{O}_2^-), Cyanide (CN−\text{CN}^-), Thiocyanate (SCN−\text{SCN}^-), Hydroxide (OH−\text{OH}^-), Permanganate (MnO4−\text{MnO}_4^-), Superoxide (O2−\text{O}_2^-).

    • 2- Anions: Chromate (CrO42−\text{CrO}_4^{2-}), Dichromate (Cr2O72−\text{Cr}_2\text{O}_7^{2-}), Sulfate (SO42−\text{SO}_4^{2-}), Sulfite (SO32−\text{SO}_3^{2-}), Thiosulfate (S2O32−\text{S}_2\text{O}_3^{2-}), Carbonate (CO32−\text{CO}_3^{2-}), Peroxide (O22−\text{O}_2^{2-}), Oxalate (C2O42−\text{C}_2\text{O}_4^{2-}).

    • 3- Anions: Arsenate (AsO43−\text{AsO}_4^{3-}), Phosphate (PO43−\text{PO}_4^{3-}), Phosphite (PO33−\text{PO}_3^{3-}).

  • Transition Metals with Variable Cation Charges:

Metal

Ion Formula

Systematic IUPAC Name

Older / Classical Name

Chromium

Cr2+\text{Cr}^{2+}

Chromium(II)

Chromous


Cr3+\text{Cr}^{3+}

Chromium(III)

Chromic

Iron

Fe2+\text{Fe}^{2+}

Iron(II)

Ferrous


Fe3+\text{Fe}^{3+}

Iron(III)

Ferric

Cobalt

Co2+\text{Co}^{2+}

Cobalt(II)

Cobaltous


Co3+\text{Co}^{3+}

Cobalt(III)

Cobaltic

Copper

Cu+\text{Cu}^+

Copper(I)

Cuprous


Cu2+\text{Cu}^{2+}

Copper(II)

Cupric

Tin

Sn2+\text{Sn}^{2+}

Tin(II)

Stannous


Sn4+\text{Sn}^{4+}

Tin(IV)

Stannic

Mercury

Hg22+\text{Hg}_2^{2+}

Mercury(I)

Mercurous


Hg2+\text{Hg}^{2+}

Mercury(II)

Mercuric

Lead

Pb2+\text{Pb}^{2+}

Lead(II)

Plumbous


Pb4+\text{Pb}^{4+}

Lead(IV)

Plumbic

  • Rules for Writing Formulas for Ionic Compounds:

    1. Write the formula and charge for the cation first, followed by the anion.

    2. Combine ions in the simplest whole-number ratio that balances positive and negative charges to yield net charge = 0.

    3. Swap and Drop Method: The magnitude of the charge of the cation becomes the subscript of the anion, and the magnitude of the charge of the anion becomes the subscript of the cation. Reduce subscripts to lowest integer terms.

    4. Enclose polyatomic ions in parentheses if a subscript greater than 1 is required.

  • Rules for Naming Binary Molecular Compounds:

    1. Name the element farthest left or lowest in the periodic table first, using its full elemental name.

    2. Name the second element using its root name and appending the suffix -ide.

    3. Use Greek numerical prefixes to represent the number of atoms of each element:

      • 1 = mono- (never used on the first element)

      • 2 = di-

      • 3 = tri-

      • 4 = tetra-

      • 5 = penta-

      • 6 = hexa-

      • 7 = hepta-

      • 8 = octa-

      • 9 = nona-

      • 10 = deca-

    4. Drop the final 'a' or 'o' of a prefix when preceding an element name beginning with a vowel (e.g., tetroxide instead of tetraoxide, monoxide instead of monooxide).

  • Common Retained Names for Binary Molecular Compounds:

    • H2O\text{H}_2\text{O}: Water (Dihydrogen monoxide)

    • NH3\text{NH}_3: Ammonia (Nitrogen trihydride)

    • H2O2\text{H}_2\text{O}_2: Hydrogen peroxide (Dihydrogen dioxide)

    • NaClO\text{NaClO}: Bleach (Sodium hypochlorite)

    • CH3COOH\text{CH}_3\text{COOH}: Vinegar (Acetic acid)

    • N2O\text{N}_2\text{O}: Nitrous oxide (Dinitrogen monoxide)

  • Naming Inorganic Acids:

    • Binary Acids (contains H+\text{H}^+ and a monatomic anion ending in -ide):

      • Prefix hydro- + root nonmetal name + -ic acid (e.g., HCl\text{HCl} = hydrochloric acid, HBr\text{HBr} = hydrobromic acid, HI\text{HI} = hydroiodic acid).

    • Oxyacids (contains H+\text{H}^+ and a polyatomic oxyanion):

      • If oxyanion ends in -ate, change suffix to -ic acid (e.g., HClO3\text{HClO}_3 = chloric acid, HClO4\text{HClO}_4 = perchloric acid, H2SO4\text{H}_2\text{SO}_4 = sulfuric acid, HNO3\text{HNO}_3 = nitric acid).

      • If oxyanion ends in -ite, change suffix to -ous acid (e.g., HClO2\text{HClO}_2 = chlorous acid, HClO\text{HClO} = hypochlorous acid, HNO2\text{HNO}_2 = nitrous acid).

  • Strong and Weak Acids/Bases Table:

    • Seven Strong Acids: HCl\text{HCl} (hydrochloric acid), HBr\text{HBr} (hydrobromic acid), HI\text{HI} (hydroiodic acid), H2SO4\text{H}_2\text{SO}_4 (sulfuric acid), HNO3\text{HNO}_3 (nitric acid), HClO4\text{HClO}_4 (perchloric acid), HClO3\text{HClO}_3 (chloric acid).

    • Selected Weak Acids: CH3COOH\text{CH}_3\text{COOH} (acetic acid), H3PO4\text{H}_3\text{PO}_4 (phosphoric acid), H2S\text{H}_2\text{S} (hydrosulfuric acid), HCN\text{HCN} (hydrocyanic acid), HCOOH\text{HCOOH} (formic acid), HF\text{HF} (hydrofluoric acid), HNO2\text{HNO}_2 (nitrous acid).

    • Strong Bases: Soluble alkali metal and heavy alkaline earth metal hydroxides: LiOH\text{LiOH}, NaOH\text{NaOH}, KOH\text{KOH}, RbOH\text{RbOH}, CsOH\text{CsOH}, Ca(OH)2\text{Ca(OH)}_2, Sr(OH)2\text{Sr(OH)}_2, Ba(OH)2\text{Ba(OH)}_2.

    • Selected Weak Bases: NH3\text{NH}_3 (ammonia), C5H5N\text{C}_5\text{H}_5\text{N} (pyridine), CH3NH2\text{CH}_3\text{NH}_2 (methylamine).

Formula Mass, Molecular Mass, and the Mole Concept

  • Mass Definitions:

    • Atomic Mass: Mass of a single atom, expressed in atomic mass units (amu\text{amu}).

    • Formula Mass (Formula Weight): Mass of one formula unit of an ionic compound. Calculated by summing atomic masses of all atoms appearing in the chemical formula (e.g., Formula Mass of NaCl=22.99+35.45=58.44 amu\text{NaCl} = 22.99 + 35.45 = 58.44\text{ amu}).

    • Molecular Mass (Molecular Weight): Mass of one individual molecule of a covalent compound (e.g., Molecular Mass of ethanol C2H5OH=2(12.011)+6(1.008)+1(16.00)=46.07 amu\text{C}_2\text{H}_5\text{OH} = 2(12.011) + 6(1.008) + 1(16.00) = 46.07\text{ amu}).

    • Molar Mass (MM): The mass in grams of one mole (6.022×10236.022 \times 10^{23} particles) of a substance, expressed in units of g mol−1\text{g}\,\text{mol}^{-1}. The numerical value of molar mass in g mol−1\text{g}\,\text{mol}^{-1} is equal to formula or molecular mass in amu\text{amu}.

  • Avogadro's Number and Mole Conversions:

    • Avogadro’s Number=6.02214×1023 particles mol−1\text{Avogadro's Number} = 6.02214 \times 10^{23}\,\text{particles}\,\text{mol}^{-1}.

    • Acts as the quantitative conversion bridge between macroscopic mass measurements and microscopic particle counts.

Molar Mass Conversion Visual

  • Mole Ratios Within Chemical Formulas:

    • Subscripts define mole-to-mole ratios of constituent elements.

    • For example, 1 mole1\,\text{mole} of C2H6\text{C}_2\text{H}_6 molecules contains:

      • 6.022×1023 C2H6 molecules6.022 \times 10^{23}\,\text{C}_2\text{H}_6\,\text{molecules}.

      • 2×(6.022×1023)=1.204×1024 atoms of C2 \times (6.022 \times 10^{23}) = 1.204 \times 10^{24}\,\text{atoms of C}.

      • 6×(6.022×1023)=3.613×1024 atoms of H6 \times (6.022 \times 10^{23}) = 3.613 \times 10^{24}\,\text{atoms of H}.

Percent Composition and Empirical/Molecular Formula Determination

  • Percent Composition Formula:

    • The mass percentage of each element in a compound is calculated as:         % Element=(number of atoms of element)×(atomic mass of element)molar mass of compound×100%\%\,\text{Element} = \frac{(\text{number of atoms of element}) \times (\text{atomic mass of element})}{\text{molar mass of compound}} \times 100\%

  • Percent Composition Example (MgCl2\text{MgCl}_2):

    • Formula Weight of MgCl2=24.31+2(35.45)=95.21 amu\text{MgCl}_2 = 24.31 + 2(35.45) = 95.21\,\text{amu}.

    • % Cl=2(35.45 amu)95.21 amu×100%=74.47% Cl\%\,\text{Cl} = \frac{2(35.45\,\text{amu})}{95.21\,\text{amu}} \times 100\% = 74.47\%\,\text{Cl}

    • % Mg=100%−74.47%=25.53% Mg\%\,\text{Mg} = 100\% - 74.47\% = 25.53\%\,\text{Mg}

  • Determining Empirical Formula from Analytical Mass Percent Data:

    1. Assume a total sample mass of exactly 100 g100\,\text{g}, converting mass percentages directly to grams.

    2. Convert mass in grams of each element to moles using elemental molar masses (g mol−1\text{g}\,\text{mol}^{-1}).

    3. Divide all calculated mole values by the smallest mole quantity obtained to establish a preliminary mole ratio.

    4. If fractional ratios remain, multiply all stoichiometric numbers by small integers to derive whole-number subscripts.

Empirical Formula Flowchart

  • Determining Molecular Formula from Empirical Formula and Molecular Mass:

    1. Calculate the formula mass of the empirical formula unit (EFM\text{EFM}).

    2. Calculate the whole-number multiple (nn):         n=Molar Mass or Molecular WeightEmpirical Formula Massn = \frac{\text{Molar Mass or Molecular Weight}}{\text{Empirical Formula Mass}}

    3. Multiply all subscripts in the empirical formula by nn to obtain the molecular formula.

Writing, Representation, and Balancing of Chemical Equations

  • Chemical Reaction Definition and Evidence:

    • A chemical reaction is a process where one or more substances (reactants) undergo chemical changes breaking and forming chemical bonds to generate new substances (products) with distinct chemical properties.

    • Macroscopic indicators of a chemical change include: gas evolution, color change, light emission, and heat exchange (exothermic heat release or endothermic absorption). Chemical reactions can also occur without visible changes.

Evidence of Chemical Reaction

  • Chemical Equations Structure:

    • Reactants are placed on the left side of the equation arrow (→\rightarrow), and products are placed on the right side. Plus signs (++) separate distinct chemical species.

    • Physical states of matter must be specified in parentheses after each chemical symbol:

      • (s)(s) = solid

      • (l)(l) = liquid

      • (g)(g) = gas

      • (aq)(aq) = aqueous solution (dissolved in water)

  • Law of Conservation of Mass and Balancing Guidelines:

    • Atoms cannot be created or destroyed during a chemical reaction. An equation must be balanced so that the total number of atoms of each element on the reactant side equals the total number of atoms on the product side.

    • Rule: Modify stoichiometric coefficients placed before chemical formulas. NEVER alter chemical formula subscripts, as changing subscripts alters chemical identity and properties.

    • Guideline 1: Balance elements appearing in the most complex compounds first.

    • Guideline 2: Balance pure uncombined elements or elemental homonuclear diatomics (like O2\text{O}_2 or N2\text{N}_2) last.

    • Guideline 3: Polyatomic ions existing intact on both sides of the equation should be balanced as a single unit.

    • Guideline 4: If fractional coefficients result during balancing, eliminate fractions by multiplying all coefficients by the common denominator.

Reaction Stoichiometry and Yield Calculations

  • Stoichiometry Roadmap:

    • Stoichiometry quantifies relationships between reactants and products using balanced chemical equations.

    • Balanced equation coefficients supply mole-to-mole ratios between species.

    • General conversion pathway for converting Mass of A to Mass of B:

Mass A (g)→Molar Mass AMoles A→Mole Ratio B:AMoles B→Molar Mass BMass B (g)\text{Mass A (g)} \xrightarrow{\text{Molar Mass A}} \text{Moles A} \xrightarrow{\text{Mole Ratio } \text{B}:\text{A}} \text{Moles B} \xrightarrow{\text{Molar Mass B}} \text{Mass B (g)}

Stoichiometry Roadmap

  • Limiting Reactants and Yield Definitions:

    • Limiting Reactant (Limiting Reagent): The reactant that is completely consumed first in a chemical reaction. It limits the maximum quantity of product formed.

    • Excess Reactant: The reactant present in a quantity greater than required to react completely with the limiting reactant.

    • Theoretical Yield: The maximum calculated quantity of product formed if all limiting reactant reacts completely.

    • Actual Yield: The mass of product measured experimentally upon completing a reaction.

    • Percent Yield:         % Yield=Actual YieldTheoretical Yield×100%\%\,\text{Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%

  • Methods for Identifying the Limiting Reactant:

    • Method 1 (Product Comparison): Convert the given starting amount of every reactant into moles (or grams) of a single common product. The reactant that yields the absolute smallest quantity of product is the limiting reactant.

    • Method 2 (Reactant Requirement Comparison): Calculate the exact moles of Reactant B required to completely react with the actual moles available of Reactant A. Compare the required moles of B against actual moles of B present. If actual B < required B, B is limiting; if actual B > required B, A is limiting.

Chemical Equilibrium and Reversible Reactions

  • Dynamic Equilibrium:

    • Reversible reactions proceed in both forward (→\rightarrow) and reverse (←\leftarrow) directions, represented using double equilibrium arrows (⇌\rightleftharpoons).

    • Dynamic Equilibrium is established when the rate of the forward reaction equals the rate of the reverse reaction.

    • At equilibrium, the concentrations of reactants and products remain constant over time. Dynamic equilibrium does NOT mean that reactant and product concentrations are equal.

  • Equilibrium Constant (KK):

    • For a generic reversible reaction:         aA+bB⇌cC+dDa\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}

    • The Law of Mass Action defines the equilibrium constant (KK):         K=[C]c[D]d[A]a[B]bK = \frac{[\text{C}]^c [\text{D}]^d}{[\text{A}]^a [\text{B}]^b}

Law of Mass Action

  • Meaning of the Magnitude of KK:

    • K≫1K \gg 1 (Large KK): Products are strongly favored at equilibrium. The equilibrium lies far to the right.

    • K≪1K \ll 1 (Small KK): Reactants are strongly favored at equilibrium. The equilibrium lies far to the left.

    • K≈1K \approx 1: Significant quantities of both reactants and products exist at equilibrium.

  • Disturbing Equilibrium:

    • When a system at dynamic equilibrium is disturbed (e.g., adding or removing reactants/products), the forward or reverse reaction rate temporarily shifts to relieve the disturbance until a new equilibrium is established.

    • Adding excess product shifts the reaction to the left (toward reactants) to consume added product.

Le Chatelier NO2 Shift

Concept Checks, Quiz Questions, and Worked Problems

  • Concept Check: Anion Properties

    • Question: Which of the following statements about anions is true?

      • A. An anion is the result of the loss of electrons.

      • B. An anion is typically a nonmetal element.

      • C. An anion is positively charged.

      • D. An anion generally appears first in a chemical formula.

    • Correct Answer: B (Anions gain electrons, carry negative charge, consist of nonmetal elements, and appear second in ionic chemical formulas).

  • Concept Check: Identifying Gaseous Molecular Compounds

    • Question: A diagram shows separate diatomic molecules consisting of one red sphere and one blue sphere moving freely in space. What does this depict?

      • A. A liquid molecular compound

      • B. A gaseous ionic compound

      • C. A gaseous molecular compound

      • D. A liquid mixture

      • E. A gaseous element

    • Correct Answer: C (A gaseous molecular compound).

    • Follow-up Question: What would the diagram represent if both bonded spheres in each molecule were blue?

    • Answer: A gaseous element (a homonuclear diatomic gas like N2\text{N}_2 or O2\text{O}_2).

  • Concept Check: Sodium Bicarbonate Bonding

    • Question: Sodium bicarbonate (baking soda, NaHCO3\text{NaHCO}_3) is used to deodorize refrigerators. Which best describes this compound?

      • A. A molecular compound containing only covalent bonds

      • B. An ionic compound containing only ionic bonds

      • C. An ionic compound containing both ionic and covalent bonds

      • D. A molecular compound containing both covalent and ionic bonds

    • Correct Answer: C (Ionic bond exists between Na+\text{Na}^+ and HCO3−\text{HCO}_3^-; covalent bonds exist inside HCO3−\text{HCO}_3^-).

  • Concept Check: Oxidation Number of Iron in Fe3(PO4)2\text{Fe}_3(\text{PO}_4)_2

    • Question: What is the charge on iron (Fe\text{Fe}) in Fe3(PO4)2\text{Fe}_3(\text{PO}_4)_2?

    • Solution: Phosphate (PO43−\text{PO}_4^{3-}) carries a −3-3 charge. Two phosphate ions total 2×(−3)=−62 \times (-3) = -6. Three iron ions must balance this with +6+6. Therefore, each iron ion has a charge of +63=+2\frac{+6}{3} = +2.

    • Correct Answer: +2.

  • Concept Check: Ammonium Sulfide Formula

    • Question: What is the correct formula for an ionic compound composed of ammonium and sulfur?

      • A. NH4S2\text{NH}_4\text{S}_2

      • B. NH4S\text{NH}_4\text{S}

      • C. (NH4)4S2(\text{NH}_4)_4\text{S}_2

      • D. (NH4)2S(\text{NH}_4)_2\text{S}

      • E. NH8S\text{NH}_8\text{S}

    • Correct Answer: D (NH4+\text{NH}_4^+ and S2−\text{S}^{2-} swap and drop to form (NH4)2S(\text{NH}_4)_2\text{S}).

  • Concept Check: Molecular Mass of Thioacetone

    • Question: What is the molecular mass (in amu\text{amu}) of thioacetone, CH3CH3CS\text{CH}_3\text{CH}_3\text{CS} (C3H6S\text{C}_3\text{H}_6\text{S})?

    • Solution:

      • C:3×12.011 amu=36.033 amu\text{C}: 3 \times 12.011\,\text{amu} = 36.033\,\text{amu}

      • H:6×1.008 amu=6.048 amu\text{H}: 6 \times 1.008\,\text{amu} = 6.048\,\text{amu}

      • S:1×32.06 amu=32.060 amu\text{S}: 1 \times 32.06\,\text{amu} = 32.060\,\text{amu}

      • Total Molecular Mass=36.033+6.048+32.060=74.14 amu\text{Total Molecular Mass} = 36.033 + 6.048 + 32.060 = 74.14\,\text{amu}

    • Correct Answer: 74.14.

  • Concept Check: Formula Mass of Calcium Gluconate

    • Question: What is the formula mass (in amu\text{amu}) of calcium gluconate, Ca(C6H11O7)2\text{Ca}(\text{C}_6\text{H}_{11}\text{O}_7)_2?

    • Solution:

      • Ca:1×40.078 amu=40.078 amu\text{Ca}: 1 \times 40.078\,\text{amu} = 40.078\,\text{amu}

      • C:12×12.011 amu=144.132 amu\text{C}: 12 \times 12.011\,\text{amu} = 144.132\,\text{amu}

      • H:22×1.008 amu=22.176 amu\text{H}: 22 \times 1.008\,\text{amu} = 22.176\,\text{amu}

      • O:14×15.999 amu=223.986 amu\text{O}: 14 \times 15.999\,\text{amu} = 223.986\,\text{amu}

      • Total Formula Mass=40.078+144.132+22.176+223.986=430.37 amu\text{Total Formula Mass} = 40.078 + 144.132 + 22.176 + 223.986 = 430.37\,\text{amu}

    • Correct Answer: 430.37.

  • Concept Check: Quantities of Carbon Atoms

    • Question: Which of the following has the largest number of carbon atoms?

      • A. 12 g12\,\text{g} of Carbon-12

      • B. 1 mole1\,\text{mole} of Acetylene (C2H2\text{C}_2\text{H}_2)

      • C. 9×1023 molecules9 \times 10^{23}\,\text{molecules} of CO2\text{CO}_2

    • Solution:

      • Option A: 12 g 12C=1.0 mol C atoms=6.022×1023 C atoms12\,\text{g } ^{12}\text{C} = 1.0\,\text{mol C atoms} = 6.022 \times 10^{23}\text{ C atoms}.

      • Option B: 1.0 mol C2H2=2.0 mol C atoms=1.204×1024 C atoms1.0\,\text{mol } \text{C}_2\text{H}_2 = 2.0\,\text{mol C atoms} = 1.204 \times 10^{24}\text{ C atoms}.

      • Option C: 9×1023 molecules CO2=9.0×1023 C atoms9 \times 10^{23}\text{ molecules } \text{CO}_2 = 9.0 \times 10^{23}\text{ C atoms}.

    • Correct Answer: B (1 mole1\,\text{mole} of acetylene).

  • Concept Check: Ba Atoms in Barium Carbonate

    • Question: How many Ba\text{Ba} atoms are in 3.50×103 mg3.50 \times 10^3\,\text{mg} of barium carbonate (BaCO3\text{BaCO}_3)?

    • Solution:

      • Mass=3.50 g BaCO3\text{Mass} = 3.50\,\text{g } \text{BaCO}_3

      • Molar Mass BaCO3=137.33+12.011+3(15.999)=197.34 g mol−1\text{Molar Mass } \text{BaCO}_3 = 137.33 + 12.011 + 3(15.999) = 197.34\,\text{g}\,\text{mol}^{-1}

      • Moles BaCO3=3.50 g197.34 g mol−1=0.0177355 mol\text{Moles } \text{BaCO}_3 = \frac{3.50\,\text{g}}{197.34\,\text{g}\,\text{mol}^{-1}} = 0.0177355\,\text{mol}

      • Atoms Ba=0.0177355 mol×6.022×1023 atoms mol−1=1.07×1022 atoms\text{Atoms } \text{Ba} = 0.0177355\,\text{mol} \times 6.022 \times 10^{23}\,\text{atoms}\,\text{mol}^{-1} = 1.07 \times 10^{22}\text{ atoms}

    • Correct Answer: 1.07×10221.07 \times 10^{22}.

  • Concept Check: Percent Oxygen Comparison

    • Question: Which compound has the greatest percent oxygen by mass? (MgO\text{MgO}, CaO\text{CaO}, K2O\text{K}_2\text{O}, Na2O\text{Na}_2\text{O})

    • Solution:

      • MgO:16.0040.31×100%=39.69%\text{MgO}: \frac{16.00}{40.31} \times 100\% = 39.69\%

      • CaO:16.0056.08×100%=28.53%\text{CaO}: \frac{16.00}{56.08} \times 100\% = 28.53\%

      • K2O:16.0094.20×100%=16.99%\text{K}_2\text{O}: \frac{16.00}{94.20} \times 100\% = 16.99\%

      • Na2O:16.0061.98×100%=25.81%\text{Na}_2\text{O}: \frac{16.00}{61.98} \times 100\% = 25.81\%

    • Correct Answer: A (MgO\text{MgO}).

  • Concept Check: Empirical Formula from Mass Analysis

    • Question: What is the empirical formula of a compound containing 27.0% S27.0\%\,\text{S}, 13.4% O13.4\%\,\text{O}, and 59.6% Cl59.6\%\,\text{Cl} by mass?

    • Solution:

      • S:27.0 g32.06 g mol−1=0.84217 mol\text{S}: \frac{27.0\,\text{g}}{32.06\,\text{g}\,\text{mol}^{-1}} = 0.84217\,\text{mol}

      • O:13.4 g16.00 g mol−1=0.83750 mol\text{O}: \frac{13.4\,\text{g}}{16.00\,\text{g}\,\text{mol}^{-1}} = 0.83750\,\text{mol}

      • Cl:59.6 g35.45 g mol−1=1.68124 mol\text{Cl}: \frac{59.6\,\text{g}}{35.45\,\text{g}\,\text{mol}^{-1}} = 1.68124\,\text{mol}

      • Divide by 0.837500.83750: S=1\text{S} = 1, O=1\text{O} = 1, Cl=2\text{Cl} = 2.

    • Correct Answer: B (SOCl2\text{SOCl}_2).

  • Concept Check: Balancing Combustion Reaction

    • Question: Balance C7H16(g)+‾O2(g)→‾H2O(g)+‾CO2(g)\text{C}_7\text{H}_{16}(g) + \underline{\quad}\text{O}_2(g) \rightarrow \underline{\quad}\text{H}_2\text{O}(g) + \underline{\quad}\text{CO}_2(g). What is the coefficient for oxygen?

    • Solution: Balance C\text{C} first (7 CO27\,\text{CO}_2), balance H\text{H} second (8 H2O8\,\text{H}_2\text{O}). Product oxygen total = 7(2)+8(1)=22 O atoms7(2) + 8(1) = 22\,\text{O}\text{ atoms}. Therefore, 11 O211\,\text{O}_2 molecules are required on the reactant side.

    • Correct Answer: B (1111).

  • Concept Check: Lead(II) Carbonate Decomposition

    • Question: PbCO3(s)→PbO(s)+CO2(g)\text{PbCO}_3(s) \rightarrow \text{PbO}(s) + \text{CO}_2(g). What mass of PbO\text{PbO} (in grams) is produced by decomposition of 2.50 g2.50\,\text{g} PbCO3\text{PbCO}_3?

    • Solution:

      • Molar Mass PbCO3=267.21 g mol−1\text{Molar Mass } \text{PbCO}_3 = 267.21\,\text{g}\,\text{mol}^{-1}

      • Molar Mass PbO=223.20 g mol−1\text{Molar Mass } \text{PbO} = 223.20\,\text{g}\,\text{mol}^{-1}

      • Mass PbO=2.50 g×1 mol PbCO3267.21 g×1 mol PbO1 mol PbCO3×223.20 g PbO1 mol PbO=2.09 g\text{Mass } \text{PbO} = 2.50\,\text{g} \times \frac{1\,\text{mol } \text{PbCO}_3}{267.21\,\text{g}} \times \frac{1\,\text{mol } \text{PbO}}{1\,\text{mol } \text{PbCO}_3} \times \frac{223.20\,\text{g } \text{PbO}}{1\,\text{mol } \text{PbO}} = 2.09\,\text{g}

    • Correct Answer: 2.09.

  • Concept Check: Limiting Reactant for Water Production

    • Question: 2 H2(g)+O2(g)→2 H2extO(g)2\,\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\,\text{H}_2 ext{O}(g). If 5.0 g5.0\,\text{g} H2\text{H}_2 reacts with 10.0 g10.0\,\text{g} O2\text{O}_2, what mass of water (in grams) is produced?

    • Solution:

      • Moles H2=5.0 g2.016 g mol−1=2.48 mol→2.48 mol H2O\text{Moles } \text{H}_2 = \frac{5.0\,\text{g}}{2.016\,\text{g}\,\text{mol}^{-1}} = 2.48\,\text{mol} \rightarrow 2.48\,\text{mol } \text{H}_2\text{O}

      • Moles O2=10.0 g32.00 g mol−1=0.3125 mol→2×0.3125=0.625 mol H2O\text{Moles } \text{O}_2 = \frac{10.0\,\text{g}}{32.00\,\text{g}\,\text{mol}^{-1}} = 0.3125\,\text{mol} \rightarrow 2 \times 0.3125 = 0.625\,\text{mol } \text{H}_2\text{O}

      • O2\text{O}_2 is limiting. Mass H2O=0.625 mol×18.02 g mol−1=11.3 g\text{Mass } \text{H}_2\text{O} = 0.625\,\text{mol} \times 18.02\,\text{g}\,\text{mol}^{-1} = 11.3\,\text{g}.

    • Correct Answer: 11.3.

  • Concept Check: Titanium Chloride Percent Yield

    • Question: Ti(s)+2 Cl2(g)→TiCl4(s)\text{Ti}(s) + 2\,\text{Cl}_2(g) \rightarrow \text{TiCl}_4(s). 3.00 g3.00\,\text{g} Ti\text{Ti} reacts with 6.00 g6.00\,\text{g} Cl2\text{Cl}_2 to form 7.7 g7.7\,\text{g} TiCl4\text{TiCl}_4. True or False: The percent yield is 96%96\%?

    • Solution:

      • Moles Ti=3.0047.87=0.06267 mol\text{Moles } \text{Ti} = \frac{3.00}{47.87} = 0.06267\,\text{mol}

      • Moles Cl2=6.0070.90=0.08463 mol\text{Moles } \text{Cl}_2 = \frac{6.00}{70.90} = 0.08463\,\text{mol}

      • Required Cl2=2×0.06267=0.1253 mol\text{Cl}_2 = 2 \times 0.06267 = 0.1253\,\text{mol}. Cl2\text{Cl}_2 is limiting.

      • Theoretical Yield TiCl4=0.084632×189.67 g mol−1=8.025 g\text{Theoretical Yield } \text{TiCl}_4 = \frac{0.08463}{2} \times 189.67\,\text{g}\,\text{mol}^{-1} = 8.025\,\text{g}

      • % Yield=7.7 g8.025 g×100%=95.95%≈96%\%\,\text{Yield} = \frac{7.7\,\text{g}}{8.025\,\text{g}} \times 100\% = 95.95\% \approx 96\%

    • Correct Answer: A (True).