Comprehensive Study Guide: Chemical Compounds, Stoichiometry, and Aqueous Reactions

Chemical Compounds and Molecular Representations

• Primary Types of Chemical Bonds:

  • Ionic Bonds: Electrostatic forces of attraction between positively charged ions (cations) and negatively charged ions (anions). Chemical species joined primarily by ionic bonds form ionic compounds.

  • Covalent Bonds: Electrostatic interactions involving the sharing of electron pairs between participating atoms. Substances bound together predominantly via covalent bonds form molecules (covalent compounds).

• Molecular Formulas and Formula Units:

  • Molecular Formula: Specifies the exact composition of a molecule by identifying the types and actual numbers of constituent atoms.

  • A water molecule consists of $2$ hydrogen atoms and $1$ oxygen atom: $H_2O$.

  • A glucose (sugar) molecule consists of $6$ carbon atoms, $12$ hydrogen atoms, and $6$ oxygen atoms: $C_6H_{12}O_6$.

  • An oxygen molecule consists of $2$ chemically bonded oxygen atoms: $O_2$.

  • Formula Unit: The lowest whole-number electric neutral ratio of cations to anions used to describe the composition of an ionic compound, such as $NaCl$.

• Representations of Molecular Compounds:

  • Empirical Formula: The simplest chemical formula showing the relative numbers of each kind of atom in a compound as lowest integers (provides minimal structural information).

  • Structural Formula: Illustrates the connectivity of atoms and depicts the specific covalent bonds connecting them.

  • Condensed Structural Formula: Conveys atom connectivity and sequential arrangements written within a single text line.

  • Line-Angle Formula: Uses continuous zigzag lines to represent covalent bonds between carbon backbones; carbon and hydrogen atoms attached to carbon are omitted from labels, while all heteroatoms (such as oxygen, nitrogen, and attached hydrogens) are explicitly shown.

• Structural Formulations of Acetic Acid:

  • Empirical Formula: $CH_2O$

  • Molecular Formula: $C_2H_4O_2$

  • Condensed Structural Formula: $CH_3COOH$, $CH_3CO_2H$, or $HC_2H_3O_2$

  • Structural Formula: Displays a central methyl group carbon bound to three hydrogens and a carbonyl carbon; the carbonyl carbon is double-bonded to an oxygen and single-bonded to a hydroxyl group ($-OH$).

• Diatomic Elements:

  • Certain chemical elements naturally exist in their standard elemental states as homonuclear diatomic molecules consisting of two bonded atoms, including oxygen ($O_2$) and hydrogen ($H_2$).

• Structure and Diversity of Ionic Compounds (Salts):

  • Ionic compounds typically form between metallic elements (which lose electrons to become cations) and non-metallic elements (which gain electrons to become anions).

  • In the solid state, ionic compounds do not exist as isolated single molecules, but instead arrange into an extended, regular three-dimensional array known as a crystal lattice.

• Representative Naturally Occurring Ionic Mineral Compounds:

  • Hematite: $Fe_2O_3$

  • Calcite: $CaCO_3$

  • Gypsum: $CaSO_4 \times 2H_2O$

  • Orpiment: $As_2S_3$

  • Fluorite: $CaF_2$

Formula Mass, the Mole, and Chemical Composition

• Formula Mass and Molecular Mass:

  • Formula mass represents the sum of the atomic masses of all atoms indicated in the chemical formula unit or molecular formula of any substance.

  • For molecular compounds, this value is identically called the molecular mass.

  • Expressed in atomic mass units ($u$ or $\text{amu}$).

  • Example calculation: The formula mass of sodium chloride ($NaCl$) is calculated from the individual isotopic average masses of sodium ($22.99\,u$) and chlorine ($35.45\,u$):     $\text{Formula mass of } NaCl = 22.99\,u + 35.45\,u = 58.44\,u$

• Molar Mass ($M$):

  • The mass of exactly one mole ($6.022 \times 10^{23}$ particles) of a chemical compound.

  • Quantitatively identical in numerical value to the formula mass, but expressed in units of grams per mole ($g\,mol^{-1}$ or $g/mol$).

• Worked Molar Calculations:

  • Molar mass of acetic acid ($HC_2H_3O_2$):     $\text{Molar mass} = (2 \times 12.011\,g\,mol^{-1}) + (4 \times 1.008\,g\,mol^{-1}) + (2 \times 16.00\,g\,mol^{-1}) = 60.05\,g\,mol^{-1}$

  • Atoms in $35\,mg$ of acetic acid:     $\text{Moles of } HC_2H_3O_2 = 35 \times 10^{-3}\,g \times \frac{1\,mol}{60.05\,g} = 5.83 \times 10^{-4}\,mol$     $\text{Total molecules} = 5.83 \times 10^{-4}\,mol \times 6.022 \times 10^{23}\,\text{molecules}\,mol^{-1} = 3.51 \times 10^{20}\,\text{molecules}$     $\text{Total atoms} = 3.51 \times 10^{20}\,\text{molecules} \times 8\,\text{atoms/molecule} = 2.81 \times 10^{21}\,\text{atoms}$

  • Mass of $34.2\,mol$ of calcium nitrate ($Ca(NO_3)_2$):     $\text{Molar mass of } Ca(NO_3)_2 = 40.08 + (2 \times 14.01) + (6 \times 16.00) = 164.10\,g\,mol^{-1}$     $\text{Mass} = 34.2\,mol \times 164.10\,g\,mol^{-1} = 5.61 \times 10^3\,g$

  • Molar mass of sulfur trioxide ($SO_3$):     $\text{Molar mass} = 32.06\,g\,mol^{-1} + (3 \times 16.00\,g\,mol^{-1}) = 80.06\,g\,mol^{-1}$

• Percentage Composition by Mass (Mass Percent):

  • Quantifies the relative mass contribution of each individual element present within a pure compound:     $\text{Mass \%} = \frac{(\text{atoms of an element}) \times (\text{atomic weight})}{\text{formula weight of compound}} \times 100\%$

  • Used analytically to determine identity and evaluate purity of newly synthesized or unknown compounds.

  • Percent composition of calcium hydroxide ($Ca(OH)_2$):     $\text{Formula weight} = 40.08 + (2 \times 16.00) + (2 \times 1.008) = 74.10\,g\,mol^{-1}$     $\% Ca = \frac{40.08}{74.10} \times 100\% = 54.09\%$     $\% O = \frac{32.00}{74.10} \times 100\% = 43.18\%$     $\% H = \frac{2.016}{74.10} \times 100\% = 2.72\%$

Determination of Empirical and Molecular Formulas

• Experimental Determination of Chemical Formulas:

  • Carried out by decomposing a weighed sample into its constituent elements or by chemical oxidation using oxygen (combustion analysis) to produce volatile compounds like $CO_2$ and $H_2O$, which are captured and quantitatively weighed.

• Procedure for Deriving Empirical Formulas:

  • Step 1: Assume a base mass of exactly $100\,g$ of compound, directly converting mass percentages ($\%$) into masses in grams ($g$).

  • Step 2: Convert the mass of each element to moles using its respective atomic molar mass.

  • Step 3: Calculate the molar ratio by dividing the mole value of every element by the smallest calculated mole value.

  • Step 4: If necessary, convert fractional values into the smallest whole numbers by multiplying through by an appropriate integer factor.

• Empirical and Molecular Formula Calculations:

  • Empirical formula of Magnetite ($72.36\%\,Fe$, remainder $O$):     $\text{Mass of } Fe = 72.36\,g; \quad \text{Mass of } O = 100\,g - 72.36\,g = 27.64\,g$     $\text{Moles of } Fe = \frac{72.36\,g}{55.85\,g\,mol^{-1}} = 1.296\,mol$     $\text{Moles of } O = \frac{27.64\,g}{16.00\,g\,mol^{-1}} = 1.728\,mol$     $\text{Ratio } \frac{O}{Fe} = \frac{1.728}{1.296} = 1.333 = \frac{4}{3}$     $\text{Empirical formula} = Fe_3O_4$

  • Empirical and molecular formula of a compound with $64.81\%\,C$, $21.60\%\,O$, and $13.59\%\,H$ ($M = 222.1\,g\,mol^{-1}$):     $\text{Moles of } C = \frac{64.81\,g}{12.011\,g\,mol^{-1}} = 5.396\,mol$     $\text{Moles of } O = \frac{21.60\,g}{16.00\,g\,mol^{-1}} = 1.350\,mol$     $\text{Moles of } H = \frac{13.59\,g}{1.008\,g\,mol^{-1}} = 13.482\,mol$     $\text{Dividing by } 1.350: \quad C = \frac{5.396}{1.350} = 4.00, \quad O = \frac{1.350}{1.350} = 1.00, \quad H = \frac{13.482}{1.350} = 9.99 \approx 10$     $\text{Empirical Formula} = C_4H_{10}O \quad (\text{Empirical mass} = 74.12\,g\,mol^{-1})$     $\text{Molecular multiplier } n = \frac{222.1\,g\,mol^{-1}}{74.12\,g\,mol^{-1}} \approx 3$     $\text{Molecular Formula} = C_{12}H_{30}O_3$

Nomenclature of Inorganic Compounds

• Rules for Writing and Naming Binary Ionic Compounds:

  • Formulas are written with the cation placed first followed by the anion.

  • Subscripts are assigned using the crossover rule to balance ionic charges, guaranteeing net electrical neutrality ($\text{total charge} = 0$). Polyatomic ions are enclosed in parentheses when appearing with a subscript of $2$ or greater.

  • Cations take their exact elemental name if possessing only one invariant oxidation state (Type I).

  • Cations exhibiting multiple stable oxidation states (Type II, common among transition metals and post-transition metals) require a Roman numeral enclosed in parentheses directly following the metal name to denote the ionic charge.

  • Monatomic anions are named by dropping the end of the element root name and appending the suffix -ide.

• Common Monatomic Cations and Anions:

  • Monatomic Cations: $H^+$ (Hydrogen), $Li^+$ (Lithium), $Na^+$ (Sodium), $K^+$ (Potassium), $Cs^+$ (Cesium), $Be^{2+}$ (Beryllium), $Mg^{2+}$ (Magnesium), $Ca^{2+}$ (Calcium), $Ba^{2+}$ (Barium), $Al^{3+}$ (Aluminum), $Ag^+$ (Silver), $Zn^{2+}$ (Zinc), $Cd^{2+}$ (Cadmium), $Ga^{3+}$ (Gallium).

  • Common Type II Transition Metal Cations: $Cr^{2+}$ (Chromium(II)), $Cr^{3+}$ (Chromium(III)), $Mn^{2+}$ (Manganese(II)), $Mn^{3+}$ (Manganese(III)), $Fe^{2+}$ (Iron(II)), $Fe^{3+}$ (Iron(III)), $Co^{2+}$ (Cobalt(II)), $Co^{3+}$ (Cobalt(III)), $Cu^+$ (Copper(I)), $Cu^{2+}$ (Copper(II)), $Sn^{2+}$ (Tin(II)), $Sn^{4+}$ (Tin(IV)), $Pb^{2+}$ (Lead(II)), $Pb^{4+}$ (Lead(IV)), $Hg_2^{2+}$ (Mercury(I)), $Hg^{2+}$ (Mercury(II)).

  • Monatomic Anions: $H^-$ (Hydride), $F^-$ (Fluoride), $Cl^-$ (Chloride), $Br^-$ (Bromide), $I^-$ (Iodide), $O^{2-}$ (Oxide), $S^{2-}$ (Sulfide), $N^{3-}$ (Nitride), $P^{3-}$ (Phosphide).

• Polyatomic Ions and Oxyanion Series:

  • Polyatomic Ions: Discrete multi-atom charged groupings with defined formulas:

  • Cations: $NH_4^+$ (Ammonium), $H_3O^+$ (Hydronium)

  • Common Anions: $CN^-$ (Cyanide), $OH^-$ (Hydroxide), $C_2H_3O_2^-$ (Acetate), $CO_3^{2-}$ (Carbonate), $C_2O_4^{2-}$ (Oxalate), $PO_4^{3-}$ (Phosphate), $CrO_4^{2-}$ (Chromate), $Cr_2O_7^{2-}$ (Dichromate), $MnO_4^-$ (Permanganate), $S_2O_3^{2-}$ (Thiosulfate)

  • Oxyanions: Polyatomic anions containing a nonmetal bonded to varying counts of oxygen atoms:

  • In a two-member series, the anion with fewer oxygens ends in -ite, and the member with more oxygens ends in -ate:

    • $NO_2^-$ (Nitrite) vs. $NO_3^-$ (Nitrate)

    • $SO_3^{2-}$ (Sulfite) vs. $SO_4^{2-}$ (Sulfate)

  • In a four-member halogen oxyanion series:

    • $ClO^-$ (Hypochlorite)

    • $ClO_2^-$ (Chlorite)

    • $ClO_3^-$ (Chlorate)

    • $ClO_4^-$ (Perchlorate)

  • Anions combined with acidic hydrogen retain the ion base name preceded by "hydrogen" or "dihydrogen": $HPO_4^{2-}$ (Hydrogenphosphate), $H_2PO_4^-$ (Dihydrogenphosphate).

• Worked Ionic Nomenclature Examples:

  • $MgCl_2$: Magnesium chloride

  • $NaClO_3$: Sodium chlorate

  • $(NH_4)_2Se$: Ammonium selenide

  • $Fe_2O_3$: Iron(III) oxide

  • Lithium hydride: $LiH$

  • Strontium carbonate: $SrCO_3$

  • Copper(II) phosphate: $Cu_3(PO_4)_2$

  • Lead(IV) sulfite: $Pb(SO_3)_2$

• Naming Binary Molecular (Covalent) Compounds:

  • Formed between nonmetals; atomic proportions are designated using Greek numerical prefixes:

  • $1$: mono-

  • $2$: di-

  • $3$: tri-

  • $4$: tetra-

  • $5$: penta-

  • $6$: hexa-

  • $7$: hepta-

  • $8$: octa-

  • $9$: nona-

  • $10$: deca-

  • The element with the more positive oxidation state is named first.

  • Exceptions to prefix rules:

  • If the first element in the compound has an index of $1$, the prefix mono- is omitted (e.g., $CO$ is carbon monoxide, not monocarbon monoxide).

  • Hydrogen bonded in a $1:1$ binary compound with halogens omits prefixes (e.g., $HI$ is hydrogen iodide).

  • Examples:

  • $ICl_3$: Iodine trichloride

  • $CS_2$: Carbon disulfide

  • $N_2O_4$: Dinitrogen tetroxide

  • $AsBr_5$: Arsenic pentabromide

  • Boron trifluoride: $BF_3$

  • Phosphorus pentachloride: $PCl_5$

  • Hydrogen iodide: $HI$

  • Carbon monoxide: $CO$

• Nomenclature of Acids:

  • Acids are molecules that yield aqueous $H^+$ ions when dissolved.

  • Binary Acids (anion ends in -ide): Named with prefix hydro-, base anion root, and suffix -ic acid (e.g., $HF$ is hydrofluoric acid, $HCl$ is hydrochloric acid, $HBr$ is hydrobromic acid, $HI$ is hydroiodic acid, $HCN$ is hydrocyanic acid, $H_2S$ is hydrosulfuric acid).

  • Oxoacids (oxygen-containing acids):

  • If the polyatomic anion ends in -ate, the corresponding acid takes the suffix -ic acid.

  • If the polyatomic anion ends in -ite, the corresponding acid takes the suffix -ous acid.

  • Halogen oxyacid series:

    • $HClO_4$ (from perchlorate): Perchloric acid

    • $HClO_3$ (from chlorate): Chloric acid

    • $HClO_2$ (from chlorite): Chlorous acid

    • $HClO$ (from hypochlorite): Hypochlorous acid

  • Other common oxoacids:

    • $HNO_3$: Nitric acid

    • $HNO_2$: Nitrous acid

    • $H_2SO_4$: Sulfuric acid

    • $H_2SO_3$: Sulfurous acid

    • $H_3PO_4$: Phosphoric acid

    • $HC_2H_3O_2$: Acetic acid

Chemical Equations and Stoichiometry

• Chemical Equations:

  • Characterize the reorganization of atoms during chemical processes, where existing chemical bonds within reactants break and new chemical bonds form to yield products.

  • Reactants appear on the left of a reaction arrow ($\rightarrow$); products appear on the right.

• Conservation of Mass and Balancing:

  • Atoms are neither created nor destroyed during chemical transformations. The sum of each element's atoms must remain identical on both sides of the reaction equation.

  • Physical State Designations:

  • Solid: $(s)$

  • Liquid: $(l)$

  • Gas: $(g)$

  • Aqueous Solution: $(aq)$

• Systematic Steps for Balancing Equations:

  • Step 1: Write the raw unbalanced equation including confirmed formulas and physical states.

  • Step 2: Tally all atoms of each element on reactant and product sides.

  • Step 3: Balance elements appearing exclusively in a single compound on both sides first. Leave elements present in multiple species or free elemental forms (such as $O_2$, $N_2$) for the final adjustments.

  • Step 4: Confirm the total atom counts for all elements match across the arrow.

  • Step 5: Reduce coefficients to their lowest whole-number integer ratios.

• Reaction Balancing Practice:

  • Iron(III) sulfide with gaseous hydrogen chloride:     $Fe_2S_3(s) + 6HCl(g) \rightarrow 3H_2S(g) + 2FeCl_3(s)$

  • Combustion of ethane:     $2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(g)$

• Fundamental Stoichiometric Conversion Principles:

  • Stoichiometric coefficients define the exact molar ratios linking reactants consumed to products synthesized.

  • Direct mass-to-mass conversions across different chemical species are mathematically impossible; all calculations must bridge through molar quantities:     $\text{Grams Reactant } A \xrightarrow{\times \frac{1\,mol\,A}{g\,A}} \text{Moles Reactant } A \xrightarrow{\times \frac{y\,mol\,B}{x\,mol\,A}} \text{Moles Product } B \xrightarrow{\times \frac{g\,B}{1\,mol\,B}} \text{Grams Product } B$

• Stoichiometric Worked Examples:

  • Reaction of iodine and fluorine to produce iodine trifluoride:     $I_2 + 3F_2 \rightarrow 2IF_3$

  • Stoichiometric equivalences: $1\,mol\,I_2 \equiv 3\,mol\,F_2 \equiv 2\,mol\,IF_3$

  • Moles of $I_2$ to yield $4\,mol\,IF_3$: $4\,mol\,IF_3 \times \frac{1\,mol\,I_2}{2\,mol\,IF_3} = 2.0\,mol\,I_2$

  • Moles of $F_2$ to yield $12\,mol\,IF_3$: $12\,mol\,IF_3 \times \frac{3\,mol\,F_2}{2\,mol\,IF_3} = 18\,mol\,F_2$

  • Moles of $I_2$ needed to consume $1.5\,mol\,F_2$: $1.5\,mol\,F_2 \times \frac{1\,mol\,I_2}{3\,mol\,F_2} = 0.50\,mol\,I_2$

  • Mass of $HCl$ produced from $13.0\,g\,H_2$ reacting with excess $Cl_2$:     $H_2 + Cl_2 \rightarrow 2HCl$     $\text{Mass of } HCl = 13.0\,g\,H_2 \times \frac{1\,mol\,H_2}{2.016\,g\,H_2} \times \frac{2\,mol\,HCl}{1\,mol\,H_2} \times \frac{36.46\,g\,HCl}{1\,mol\,HCl} = 4.70 \times 10^2\,g\,HCl$

  • Mass of $Fe$ required to produce $25.0\,g\,H_2$ in reaction with $HCl$ yielding $FeCl_3$:     $2Fe + 6HCl \rightarrow 2FeCl_3 + 3H_2$     $\text{Mass of } Fe = 25.0\,g\,H_2 \times \frac{1\,mol\,H_2}{2.016\,g\,H_2} \times \frac{2\,mol\,Fe}{3\,mol\,H_2} \times \frac{55.85\,g\,Fe}{1\,mol\,Fe} = 4.62 \times 10^2\,g\,Fe$

  • Mass of $CO_2$ generated by the complete combustion of $18\,g$ glucose ($C_6H_{12}O_6$):     $C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O$     $\text{Mass of } CO_2 = 18\,g\,C_6H_{12}O_6 \times \frac{1\,mol\,C_6H_{12}O_6}{180.16\,g} \times \frac{6\,mol\,CO_2}{1\,mol\,C_6H_{12}O_6} \times \frac{44.01\,g\,CO_2}{1\,mol\,CO_2} = 26\,g\,CO_2$

  • Mass of $NaOH$ needed to neutralize $32\,g$ of sulfurous acid ($H_2SO_3$):     $H_2SO_3 + 2NaOH \rightarrow Na_2SO_3 + 2H_2O$     $\text{Mass of } NaOH = 32\,g\,H_2SO_3 \times \frac{1\,mol\,H_2SO_3}{82.08\,g} \times \frac{2\,mol\,NaOH}{1\,mol\,H_2SO_3} \times \frac{40.00\,g\,NaOH}{1\,mol\,NaOH} = 31\,g\,NaOH$

• Combustion Analysis:

  • Method used for organic compounds containing carbon, hydrogen, and oxygen. The sample is combusted in a pure $O_2$ furnace.

  • Complete effluent gases are guided through two successive selective absorption chambers:

  • Water absorber containing a dehydrating desiccant such as $Mg(ClO_4)_2$.

  • Carbon dioxide absorber containing an alkaline absorbent such as $NaOH$.

  • The mass differences in the traps yield the masses of $H_2O$ and $CO_2$.

  • Cumene Analysis ($47.6\,g$ sample produces $155.9\,g\,CO_2$ and $42.8\,g\,H_2O$; $M \in [115, 125]\,g\,mol^{-1}$):     $\text{Mass of } C = 155.9\,g\,CO_2 \times \frac{12.011\,g\,C}{44.01\,g\,CO_2} = 42.55\,g\,C \implies \frac{42.55\,g}{12.011\,g\,mol^{-1}} = 3.542\,mol\,C$     $\text{Mass of } H = 42.8\,g\,H_2O \times \frac{2.016\,g\,H}{18.015\,g\,H_2O} = 4.79\,g\,H \implies \frac{4.79\,g}{1.008\,g\,mol^{-1}} = 4.752\,mol\,H$     $\text{Ratio } \frac{H}{C} = \frac{4.752}{3.542} = 1.34 = \frac{4}{3} \implies \text{Empirical formula} = C_3H_4$     $\text{Empirical mass of } C_3H_4 = 40.06\,g\,mol^{-1}$     $\text{Molar mass ratio } n = \frac{120\,g\,mol^{-1}}{40.06\,g\,mol^{-1}} = 3 \implies \text{Molecular formula} = C_9H_{12}$

  • Terephthalic Acid ($19.81\,g$ sample produces $41.98\,g\,CO_2$ and $6.45\,g\,H_2O$; $M = 166\,g\,mol^{-1}$):     $\text{Mass of } C = 41.98\,g \times \frac{12.011}{44.01} = 11.455\,g\,C \implies 0.9537\,mol\,C$     $\text{Mass of } H = 6.45\,g \times \frac{2.016}{18.015} = 0.7218\,g\,H \implies 0.7161\,mol\,H$     $\text{Mass of } O = 19.81\,g - (11.455\,g + 0.722\,g) = 7.633\,g\,O \implies \frac{7.633\,g}{16.00\,g\,mol^{-1}} = 0.4771\,mol\,O$     $\text{Relative ratio: } \frac{C}{O} = \frac{0.9537}{0.4771} = 2.00; \quad \frac{H}{O} = \frac{0.7161}{0.4771} = 1.50; \quad \frac{O}{O} = 1.00$     $\text{Multiply by } 2: \quad C = 4,\; H = 3,\; O = 2 \implies \text{Empirical formula} = C_4H_3O_2$     $\text{Empirical mass} = 83.07\,g\,mol^{-1} \implies n = \frac{166\,g\,mol^{-1}}{83.07\,g\,mol^{-1}} = 2$     $\text{Molecular formula} = C_8H_6O_4$

Advanced Stoichiometry: Limiting Reagents, Reaction Yield, and Sequential Reactions

• Limiting Reagent Fundamentals:

  • A stoichiometric mixture contains reactants combined in exact proportions matching the balanced equation ratios.

  • When reactants are present in non-stoichiometric ratios, the limiting reactant (or limiting reagent) is the specific reactant completely consumed first; it limits the quantity of product synthesized.

  • The other reactants present in excess of the stoichiometric requirement are termed excess reactants.

• Worked Limiting Reactant Calculations:

  • Sulfur hexafluoride from $1.60\,mol\,S_8$ and 35.0\,mol\,F_2$:*\n    S_8 + 24F_2 \rightarrow 8SF_6\n    \text{Max } SF_6 \text{ from } S_8 = 1.60\,mol\,S_8 \times \frac{8\,mol\,SF_6}{1\,mol\,S_8} = 12.8\,mol\,SF_6\n    \text{Max } SF_6 \text{ from } F_2 = 35.0\,mol\,F_2 \times \frac{8\,mol\,SF_6}{24\,mol\,F_2} = 11.67\,mol\,SF_6\n    F_2$produces less product, making it the **limiting reagent**; the reaction yields$11.7\,mol\,SF_6.\n * *Aluminum chloride produced from 2.70\,g\,Al$and$4.05\,g\,Cl_2$:     $2Al + 3Cl_2 \rightarrow 2AlCl_3$     $\text{Moles } Al = \frac{2.70\,g}{26.98\,g\,mol^{-1}} = 0.1001\,mol; \quad \text{Moles } Cl_2 = \frac{4.05\,g}{70.90\,g\,mol^{-1}} = 0.05712\,mol$     $\text{From } Al: \quad 0.1001\,mol\,Al \times \frac{2\,mol\,AlCl_3}{2\,mol\,Al} \times 133.34\,g\,mol^{-1} = 13.35\,g\,AlCl_3$     $\text{From } Cl_2: \quad 0.05712\,mol\,Cl_2 \times \frac{2\,mol\,AlCl_3}{3\,mol\,Cl_2} \times 133.34\,g\,mol^{-1} = 5.08\,g\,AlCl_3$     $Cl_2$ is the limiting reagent, producing $5.08\,g\,AlCl_3$.     $\text{Moles } Al \text{ consumed} = 0.05712\,mol\,Cl_2 \times \frac{2\,mol\,Al}{3\,mol\,Cl_2} = 0.03808\,mol\,Al$     $\text{Excess } Al \text{ remaining} = (0.1001 - 0.03808)\,mol \times 26.98\,g\,mol^{-1} = 1.67\,g\,Al$

• Theoretical, Actual, and Percent Yield:

  • Theoretical Yield: The calculated maximum mass of product attainable assuming $100\%$ quantitative conversion of the limiting reactant.

  • Actual Yield: The experimentally recovered and weighed product mass.

  • Discrepancies between actual and theoretical yield arise due to incomplete conversions, competing side-reactions generating byproducts, mechanical losses during purification, or impure reactants.

  • Percent Yield Formula:     $\% \text{ Yield} = \frac{\text{Actual Yield (in grams or moles)}}{\text{Theoretical Yield (in grams or moles)}} \times 100\%$

  • Sample Calculations:

  • Production of $NH_3$ from $50.0\,g\,N_2$ and $50.0\,g\,H_2$ yielding $30.0\,g\,NH_3$:       $N_2 + 3H_2 \rightarrow 2NH_3$       $\text{Moles } N_2 = \frac{50.0\,g}{28.02\,g\,mol^{-1}} = 1.784\,mol \implies 3.568\,mol\,NH_3$       $\text{Moles } H_2 = \frac{50.0\,g}{2.016\,g\,mol^{-1}} = 24.80\,mol \implies 16.53\,mol\,NH_3$       $N_2$ limits the process; theoretical yield is $3.568\,mol \times 17.034\,g\,mol^{-1} = 60.78\,g\,NH_3$.       $\% \text{ Yield} = \frac{30.0\,g}{60.78\,g} \times 100\% = 49.4\%$

  • Mass of $F_2$ required to generate $120.0\,g\,PF_3$ ($M = 87.97\,g\,mol^{-1}$) at a $78.1\%$ yield:       $\text{Theoretical yield required} = \frac{120.0\,g}{0.781} = 153.65\,g\,PF_3 = 1.747\,mol\,PF_3$       $P_4 + 6F_2 \rightarrow 4PF_3$       $\text{Moles of } F_2 = 1.747\,mol\,PF_3 \times \frac{6\,mol\,F_2}{4\,mol\,PF_3} = 2.620\,mol\,F_2$       $\text{Mass of } F_2 = 2.620\,mol \times 38.00\,g\,mol^{-1} = 99.6\,g\,F_2$

• Sequential and Parallel Complex Reactions:

  • Consecutive Reactions: Chemical reactions carried out in a chronological stepwise pathway ($A \rightarrow B \rightarrow C$).

  • Intermediate: A transient chemical species produced during an early reaction step and subsequently consumed in a later reaction step; it does not appear in the overall balanced equation (e.g., $B$ in $A \rightarrow B \rightarrow C$).

  • Simultaneous Reactions: Distinct, independent chemical reactions that take place concurrently within a shared mixture.

  • Three-step Synthesis of Sodium Bromide from $Fe$ and Br_2$:*\n    Fe + Br_2 \rightarrow FeBr_2\n    3FeBr_2 + Br_2 \rightarrow Fe_3Br_8\n    Fe_3Br_8 + 4Na_2CO_3 \rightarrow 8NaBr + 4CO_2 + Fe_3O_4\n * Summing equations requires matching the intermediate FeBr_2$($\times 3):\n      3Fe + 4Br_2 + 4Na_2CO_3 \rightarrow 8NaBr + 4CO_2 + Fe_3O_4\n * Stoichiometric factor: 3\,mol\,Fe \equiv 8\,mol\,NaBr.\n * Mass of iron required for 250.0\,g\,NaBr$($M = 102.89\,g\,mol^{-1}):\n      \text{Mass } Fe = 250.0\,g\,NaBr \times \frac{1\,mol\,NaBr}{102.89\,g} \times \frac{3\,mol\,Fe}{8\,mol\,NaBr} \times 55.85\,g\,mol^{-1} = 50.89\,g\,Fe\n * *Simultaneous Reaction of a Binary Alloy (70.0\%\,Al$,$30.0\%\,Mg$) with excess$HCl$:     $2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2$     $Mg + 2HCl \rightarrow MgCl_2 + H_2$

  • In $12.9\,g$ of alloy: mass of $Al = 9.03\,g$; mass of $Mg = 3.87\,g$.     $\text{Moles } H_2 \text{ from } Al = 9.03\,g\,Al \times \frac{1\,mol\,Al}{26.98\,g} \times \frac{3\,mol\,H_2}{2\,mol\,Al} = 0.5020\,mol\,H_2$     $\text{Moles } H_2 \text{ from } Mg = 3.87\,g\,Mg \times \frac{1\,mol\,Mg}{24.305\,g} \times \frac{1\,mol\,H_2}{1\,mol\,Mg} = 0.1592\,mol\,H_2$     $\text{Total moles } H_2 = 0.5020 + 0.1592 = 0.6612\,mol\,H_2$     $\text{Total mass } H_2 = 0.6612\,mol \times 2.016\,g\,mol^{-1} = 1.33\,g\,H_2$

Aqueous Solutions, Electrolytes, and Concentration

• Solution Terminology:

  • Solution: A homogeneous macroscopic mixture containing two or more dispersed substances.

  • Solvent: The predominant medium in which solute particles disperse.

  • Solute: Any trace or secondary component dissolved within the solvent.

  • Aqueous Solution: A liquid solution where liquid water serves as the solvent.

• Molarity ($M$):

  • Defined as the moles of solute per liter of total solution:     $\text{Molarity } (M) = \frac{\text{moles of solute}}{\text{liters of solution}} \quad [\text{units: } mol\,L^{-1} \text{ or } M]$

  • Operates as an essential conversion factor between solution volume and moles of dissolved solute.

• Ion Dissociation and Concentrations in Solution:

  • Soluble ionic salts dissociate into independent hydrated ions:     $CuCl_2(aq) \rightarrow Cu^{2+}(aq) + 2Cl^-(aq)$

  • In a $0.30\,M\,CuCl_2$ solution: $[Cu^{2+}] = 0.30\,M$ and $[Cl^-] = 2 \times 0.30\,M = 0.60\,M$.

  • Iron(III) nitrate solution ($2.00\,mol\,Fe(NO_3)_3$ in $580\,mL$ water):     $[Fe(NO_3)_3] = \frac{2.00\,mol}{0.580\,L} = 3.45\,M$     $[Fe^{3+}] = 3.45\,M; \quad [NO_3^-] = 3 \times 3.45\,M = 10.3\,M$

  • Mass of $MgBr_2$ ($M = 184.11\,g\,mol^{-1}$) to prepare $5.0\,L$ of $0.100\,M\,Br^-$ solution:     $\text{Moles of } Br^- = 5.0\,L \times 0.100\,mol\,L^{-1} = 0.500\,mol\,Br^-$     $\text{Moles of } MgBr_2 = 0.500\,mol\,Br^- \times \frac{1\,mol\,MgBr_2}{2\,mol\,Br^-} = 0.250\,mol\,MgBr_2$     $\text{Mass} = 0.250\,mol \times 184.11\,g\,mol^{-1} = 46.0\,g\,MgBr_2$

• Dilutions:

  • Adding solvent increases total solution volume while solute moles remain constant, decreasing concentration:     $M_1 V_1 = M_2 V_2 = \text{moles of solute}$

  • Worked Examples:

  • Final concentration of $5.00\,L$ of $15.0\,M\,NaCl$ diluted with an additional $5.00\,L$ pure water:       $V_2 = 5.00\,L + 5.00\,L = 10.00\,L$       $M_2 = \frac{15.0\,M \times 5.00\,L}{10.00\,L} = 7.50\,M$

  • Volume of pure water to add to $25.0\,mL$ of $1.00\,M\,CaCl_2$ ($[Ca^{2+}] = 1.00\,M$) to achieve $[Ca^{2+}] = 0.0500\,M$:       $V_2 = \frac{1.00\,M \times 25.0\,mL}{0.0500\,M} = 500\,mL$       $\text{Water added} = V_2 - V_1 = 500\,mL - 25.0\,mL = 475\,mL$

• Hydration Process:

  • The dissolution of ionic compounds driven by electrostatic ion-dipole attractions between polar water molecules ($2\delta^-$ on oxygen, $\delta^+$ on hydrogens) and solid lattice ions.

  • Salts dissociate into individual solvated ions surrounded by hydration shells:     $NH_4NO_3(s) \xrightarrow{H_2O(l)} NH_4^+(aq) + NO_3^-(aq)$

• Electrolyte Classification and Conductivity:

  • Electrolytes: Substances that produce mobile ions upon dissolving in water and thus conduct electrical currents.

  • Strong Electrolytes: Solutes that completely dissociate ($100\%$) into individual ions (e.g., soluble ionic salts, strong acids, strong bases). Solutions conduct electricity efficiently, causing a conductivity testing bulb to glow brightly.

  • Weak Electrolytes: Solutes that dissociate only partially into ions in equilibrium with intact neutral molecules (e.g., weak acids, weak bases). Solutions conduct weakly, producing a dim light.

  • Nonelectrolytes: Molecular solutes that dissolve without generating free ions (remain exclusively as neutral molecules, e.g., sugars, alcohols). Solutions exhibit no electrical conductivity, leaving the bulb unlit.

Precipitation Reactions and Solubility Guidelines

• Precipitation Reactions (Exchange / Metathesis):

  • Occur when combining solutions of soluble salts brings together counter-ions that assemble into an insoluble solid known as a precipitate.

  • Follow the generic double-displacement format:     $AX + BY \rightarrow AY + BX$

  • The thermodynamic driving force is the formation of the insoluble precipitate lattice.

• Examples of Precipitate Formation Reactions:

  • Mixing lead(II) nitrate and potassium iodide:     $Pb(NO_3)_2(aq) + 2KI(aq) \rightarrow PbI_2(s) + 2KNO_3(aq)$

  • Mixing barium chloride and sodium sulfate:     $BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaCl(aq)$

• Distinct Colored Precipitates:

  • Lead(II) chromate ($PbCrO_4$, bright yellow) from $Pb(NO_3)_2$ and $K_2CrO_4$

  • Lead(II) sulfide ($PbS$, dense dark brown/black) from $Pb(NO_3)_2$ and $(NH_4)_2S$

  • Iron(III) hydroxide ($Fe(OH)_3$, orange-brown) from $FeCl_3$ and $NaOH$

  • Silver chromate ($Ag_2CrO_4$, brick-red) from $AgNO_3$ and $K_2CrO_4$

• Solubility Rules for Ionic Compounds in Water:

  • Soluble Compounds:

  • Group 1 alkali metals ($Li^+$, $Na^+$, $K^+$, $Rb^+$, $Cs^+$) and ammonium ($NH_4^+$) salts: universally soluble.

  • Nitrates ($NO_3^-$), chlorates ($ClO_3^-$), perchlorates ($ClO_4^-$), and acetates ($CH_3CO_2^-$ / $C_2H_3O_2^-$): soluble without exceptions.

  • Halides ($Cl^-$, $Br^-$, $I^-$): soluble, except those of $Ag^+$, $Hg_2^{2+}$, and $Pb^{2+}$.

  • Fluorides ($F^-$): soluble, except those of Group 2 ($Mg^{2+}$, $Ca^{2+}$, $Sr^{2+}$, $Ba^{2+}$) and $Pb^{2+}$.

  • Sulfates ($SO_4^{2-}$): soluble, except those of $Ca^{2+}$, $Sr^{2+}$, $Ba^{2+}$, $Pb^{2+}$, and $Ag^+$.

  • Insoluble Compounds:

  • Carbonates ($CO_3^{2-}$), phosphates ($PO_4^{3-}$), oxalates ($C_2O_4^{2-}$), chromates ($CrO_4^{2-}$), and sulfides ($S^{2-}$): insoluble, except when paired with $NH_4^+$ or Group 1 alkali metal cations.

  • Metal hydroxides ($OH^-$) and oxides: insoluble, except alkali metal hydroxides and heavy alkaline earths ($Ba(OH)_2$ and $Sr(OH)_2$; $Ca(OH)_2$ is slightly soluble).

• Comprehensive Reference Solubility Guidelines (Extended List):

  • Rule 1: All nitrates ($NO_3^-$), nitrites ($NO_2^-$), chlorates ($ClO_3^-$), and perchlorates ($ClO_4^-$) are soluble. (Exceptions: Silver nitrite and potassium perchlorate are slightly soluble.)

  • Rule 2: Essentially all alkali metal ($Li^+$, $Na^+$, $K^+$, $Rb^+$, $Cs^+$) and ammonium ($NH_4^+$) salts are soluble. (Exceptions: Some $Li^+$ salts are insoluble, notably lithium phosphate $Li_3PO_4$.)

  • Rule 3: Most halogen ($Cl^-$, $Br^-$, $I^-$) salts are soluble. (Exceptions: Salts of $Ag^+$, $Pb^{2+}$, $Hg_2^{2+}$, $Cu^+$, and $Tl^+$. Note that $Pb^{2+}$ halides dissolve in hot water; $HgBr_2$ is slightly soluble.)

  • Rule 4: Most acetate ($C_2H_3O_2^-$) salts are soluble. (Exceptions: Salts of $Ag^+$ and $Hg_2^{2+}$.)

  • Rule 5: Most sulfate ($SO_4^{2-}$) and fluoride ($F^-$) salts are soluble. (Exceptions: Sulfates/fluorides of $Ca^{2+}$, $Sr^{2+}$, $Ba^{2+}$, $Ra^{2+}$, $Pb^{2+}$, and $Hg^{2+}$. Calcium sulfate is frequently regarded as slightly soluble.)

  • Rule 6: Many sulfides ($S^{2-}$) are insoluble. (Exceptions: All alkali metal and alkaline earth ($Be^{2+}$, $Mg^{2+}$, $Ca^{2+}$, $Sr^{2+}$, $Ba^{2+}$, $Ra^{2+}$) sulfides and ammonium sulfide are soluble; some sources evaluate $MgS$, $CaS$, and $BaS$ as slightly soluble.)

  • Rule 7: Most borates ($BO_3^{3-}$), carbonates ($CO_3^{2-}$), chromates ($CrO_4^{2-}$), phosphates ($PO_4^{3-}$), and sulfites ($SO_3^{2-}$) are slightly soluble / insoluble. (Exceptions: $MgCrO_4$ is soluble; $MgSO_3$ is slightly soluble.)

  • Rule 8: Most hydroxide ($OH^-$) salts are insoluble. (Exceptions: Alkali metal hydroxides are soluble; $Ba^{2+}$, $Sr^{2+}$, $Ca^{2+}$, and $Tl^+$ are considered slightly soluble.)

• Writing Equations in Aqueous Solution:

  • Molecular Equation: Shows full, intact chemical formulas for all reactants and products without displaying dissociated ions.

  • Complete (Total) Ionic Equation: Explicitly writes all strong electrolytes in their dissociated ionic states as independent aqueous ions.

  • Spectator Ions: Ions present in the reaction vessel that do not directly engage in the bond-forming chemical transformation (they remain unchanged on both sides of the ionic equation).

  • Net Ionic Equation: Derived by canceling all spectator ions from the total ionic equation, revealing exclusively the participating reaction species.

  • Worked Practice: Aqueous copper(II) nitrate mixed with aqueous sodium hydroxide:

  • Molecular: $Cu(NO_3)_2(aq) + 2NaOH(aq) \rightarrow Cu(OH)_2(s) + 2NaNO_3(aq)$

  • Ionic: $Cu^{2+}(aq) + 2NO_3^-(aq) + 2Na^+(aq) + 2OH^-(aq) \rightarrow Cu(OH)_2(s) + 2Na^+(aq) + 2NO_3^-(aq)$

  • Spectator Ions: $Na^+(aq)$ and $NO_3^-(aq)$

  • Net Ionic Equation:       $Cu^{2+}(aq) + 2OH^-(aq) \rightarrow Cu(OH)_2(s)$

  • Worked Practice: Aqueous calcium chloride mixed with aqueous potassium perchlorate:

  • Potential products via exchange: $Ca(ClO_4)_2$ (soluble) and $KCl$ (soluble).

  • Because all combinations yield soluble species, no precipitate forms, spectator ions encompass all ions, and no net reaction occurs.