Comprehensive Study Notes: General Chemistry I

Course Overview, Policies, and Academic Structure

General Course Information
  • Course Title: General Chemistry I (CH210)

  • Course Sections & Meeting Times:

    • Section RU01: Monday and Wednesday 11:00 am – 12:15 pm; Friday 11:00 am – 11:50 am

    • Section RU04: Monday and Wednesday 12:30 pm – 1:45 pm; Friday 12:30 pm – 1:20 pm

  • Laboratory Co-requisite: Laboratory enrollment is a mandatory co-requisite. Students not registered must consult the instructor immediately.

  • Instructor Office Hours: Monday and Wednesday from 8:30 am to 11:00 am, or by appointment, located in the Chemistry Suite (Science Building, Room Sci 322).

  • Recitation Sessions: Fridays from 1:30 pm to 3:00 pm in Loyola 3.

Course Section Meeting Times
Grading Scheme and Assessment Policies
  • Grade Distribution:

    • Homework: 15%15\%

    • In-Class Worksheets: 15%15\%

    • Unit Examinations: 50%50\%

    • Comprehensive Final Exam: 20%20\%

  • Grading Scale:

    • A≥93%\text{A} \ge 93\%

    • A−=90%−93%\text{A}^- = 90\% - 93\%

    • B+=88%−89%\text{B}^+ = 88\% - 89\%

    • B=83%−87%\text{B} = 83\% - 87\%

    • B−=80%−82%\text{B}^- = 80\% - 82\%

    • C+=78%−79%\text{C}^+ = 78\% - 79\%

    • C=73%−77%\text{C} = 73\% - 77\%

    • C−=70%−72%\text{C}^- = 70\% - 72\%

    • D+=68%−69%\text{D}^+ = 68\% - 69\%

    • D=63%−67%\text{D} = 63\% - 67\%

    • D−=60%−62%\text{D}^- = 60\% - 62\%

    • F≤59%\text{F} \le 59\%

  • Assignment Submission Rules:

    • No late work is accepted under any circumstances.

    • Two lowest homework scores and two or more (2+2+) lowest worksheet scores are dropped automatically.

    • Homework hardcopies are distributed in class on Wednesdays and must be submitted as hardcopies the following Monday.

    • Worksheets are distributed as topics are introduced and are due on hardcopy at the conclusion of the class session in which they are finished.

    • If reprinting is required from online course files, original document files must be used.

    • Full mathematical work must be shown for all numerical problems to receive credit.

Matter: States, Properties, and Classification

States of Matter and Molecular Behavior

Matter is defined as anything that possesses mass and occupies physical space. It exists in three primary classical states:

  • Solid State:

    • Characterized by a dense, highly ordered arrangement of particles.

    • Maintains a definite shape and a fixed volume.

    • Particles vibrate about fixed positions; solids are only very slightly compressible.

  • Liquid State:

    • Characterized by a disordered particle arrangement that is usually slightly less dense than the solid state.

    • Has a fixed volume but conforms to the shape of its container with a free surface.

    • Particles can translate past one another; liquids are slightly compressible.

  • Gas State:

    • Characterized by high disorder and much lower density compared to liquid or solid forms.

    • Does not have a definite shape or volume; expands completely to fill the shape and volume of its container.

    • Particles move rapidly with large mean free paths; gases are highly compressible.

Properties of Matter: Physical versus Chemical

Properties of matter are categorized based on whether observing them involves changing the chemical identity of the substance:

  • Physical Properties: Characteristics measured or observed without altering the composition or identity of the substance.

    • Intensive Properties: Independent of the amount of substance present. Examples include boiling point, melting point, color, temperature, luster, hardness, and density.

    • Extensive Properties: Dependent on the quantity of matter present. Examples include mass, volume, length, weight, shape, and size.

  • Chemical Properties: Characteristics that describe the potential of a substance to undergo a chemical change that alters its molecular composition.

    • Examples include toxicity, oxidation states, heat of combustion, chemical stability, flammability, coordination number, reactivity toward air, acids, bases, or other reagents, and enthalpy of formation.

Physical versus Chemical Changes
  • Physical Change: A transformation that alters the state or physical appearance of matter without changing its chemical composition.

    • Phase Changes: Melting ice (H2O(s)→H2O(l)\text{H}_2\text{O}(s) \rightarrow \text{H}_2\text{O}(l)), boiling water (H2O(l)→H2O(g)\text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{O}(g)), sublimation of dry ice (CO2(s)→CO2(g)\text{CO}_2(s) \rightarrow \text{CO}_2(g)).

    • Mechanical Alterations: Crushing an aluminum can, breaking glass, shredding paper, chopping wood, grinding metal surfaces.

    • Dissolution: Dissolving sucrose or sodium chloride in water. In molecular dissolution, solute-solvent interactions overcome solute-solute interactions, but molecular compounds remain chemically intact as individual molecules in solution.

  • Chemical Change: A process in which chemical bonds are broken and/or formed, producing one or more new substances with different chemical formulas and properties.

    • Examples: Rusting of iron (4Fe+3O2→2Fe2O34\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3), combustion of wood or hydrocarbons, cellular metabolism, cooking an egg (denaturing/crosslinking proteins), baking a cake, electroplating, ripening of fruit, exploding fireworks, bleaching hair, chemical battery discharge, and reaction of sodium bicarbonate with acetic acid.

Classification of Matter

Matter is classified hierarchically based on its composition and uniformity:

  • Pure Substance: Composed of only a single type of atom or molecule; composition is invariant.

    • Elements: Substances that cannot be broken down chemically into simpler substances (e.g., pure helium gas, copper wire, gold).

    • Compounds: Substances composed of two or more elements chemically combined in fixed, definite proportions that can be chemically decomposed (e.g., pure water, sodium chloride, carbon dioxide).

  • Mixture: A combination of two or more pure substances in variable proportions, retaining their individual chemical identities.

    • Homogeneous Mixture (Solution): Uniform composition and properties throughout down to the molecular level (e.g., tea with dissolved sugar, brass, clean air, seawater).

    • Heterogeneous Mixture: Non-uniform composition; distinct regions or phases exist (e.g., wet sand, concrete, pizza, Italian dressing).

Separation Techniques for Matter

Mixtures are separated into pure components by exploiting differences in their physical properties:

  • Decanting: Separates an insoluble solid from a liquid, or two immiscible liquids of differing densities, by carefully pouring off the less dense upper liquid layer after the denser phase has settled.

  • Filtration: Separates an insoluble solid residue from a liquid filtrate using a porous barrier (such as filter paper supported by a filter funnel). The liquid passes through due to gravity or suction, while the solid is retained.

  • Distillation: Separates homogeneous mixtures containing liquids with different boiling points. The mixture is heated to vaporize the more volatile liquid; the vapor passes through a condenser cooled by running water, condenses back into liquid form, and is collected as the distillate.

Energy, Thermodynamics, and Chemical Stability

Fundamental Principles of Energy

Energy is defined as the capacity to perform work or produce heat.

  • Kinetic Energy (KEKE): Energy associated with motion.

  • Potential Energy (PEPE): Energy stored as a consequence of position, state, or chemical composition.

  • Thermal Energy: A form of kinetic energy associated with the random motion of atoms and molecules; higher molecular movement corresponds directly to higher temperature.

  • Law of Conservation of Energy: Energy can neither be created nor destroyed; it can only be transformed from one form to another. The total energy of an isolated system remains constant.

Molecular Forms of Kinetic and Potential Energy

At the molecular level, energy manifests in several distinct mechanical and electrostatic modes:

  • Translational Motion: Movement of the center of mass of a molecule through space.

  • Rotational Motion: Rotation of a molecule around an internal axis.

  • Vibrational Motion: Internal periodic motions of bonded atoms, including symmetric stretching, asymmetric stretching, and bending.

  • Potential Energy in Chemistry: Stored electrostatic potential energy within the chemical bonds holding atoms together (intramolecular potential) and between neighboring molecules (intermolecular potential).

Molecular Rotational Motion
Bond Energy and Enthalpy of Reaction

Chemical reactions involve breaking existing chemical bonds in reactants and forming new chemical bonds in products:

  • Bond Cleavage: Requires an input of energy (endothermic process; ΔH>0\Delta H > 0).

Breaking Molecular Bonds
  • Bond Formation: Releases energy to the surroundings (exothermic process; ΔH<0\Delta H < 0).

  • Enthalpy of Reaction (ΔH\Delta H): Determined by the difference between the total energy absorbed to break bonds and the total energy released upon bond formation:

ΔH=∑(Energy needed to break bonds)−∑(Energy released forming bonds)\Delta H = \sum (\text{Energy needed to break bonds}) - \sum (\text{Energy released forming bonds})

Worked Reaction Example: Combustion of hydrogen gas to form water vapor:

2H2(g)+O2(g)→2H2O(g)2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(g)

  • Energy needed to break reactant bonds:

(2×432 kJ/mol)+494 kJ/mol=1358 kJ/mol(2 \times 432\,\text{kJ/mol}) + 494\,\text{kJ/mol} = 1358\,\text{kJ/mol}

  • Energy released upon forming product bonds (44 moles of O−H\text{O}-\text{H} bonds):

4×459 kJ/mol=1836 kJ/mol4 \times 459\,\text{kJ/mol} = 1836\,\text{kJ/mol}

  • Net enthalpy change:

ΔH=1358 kJ/mol−1836 kJ/mol=−478 kJ/mol\Delta H = 1358\,\text{kJ/mol} - 1836\,\text{kJ/mol} = -478\,\text{kJ/mol}

The negative sign confirms that the overall reaction is exothermic.

Gibbs Free Energy and Reaction Spontaneity

Gibbs free energy (ΔG\Delta G) determines the thermodynamic spontaneity of a reaction at constant temperature and pressure:

  • Exergonic Reaction (ΔG<0\Delta G < 0):

    • Free energy is released to the surroundings.

    • The reaction proceeds spontaneously toward equilibrium without requiring continuous external energy input.

  • Endergonic Reaction (ΔG>0\Delta G > 0):

    • Free energy is absorbed from the surroundings.

    • The reaction is non-spontaneous under standard conditions and requires external energy input to proceed.

Potential Energy and Stability

Systems naturally evolve toward states of minimum potential energy and maximum stability:

  • Charges of the same sign brought near each other experience repulsive electrostatic forces, creating a state of high potential energy and low stability.

  • Opposite charges brought near each other experience attractive electrostatic forces, lowering the potential energy and establishing stable configurations.

Measurement, Units, and Mathematical Problem-Solving

The International System of Units (SI)

Scientific measurements rely on standard SI base units defined by fundamental physical constants:

Quantity

Unit Name

Symbol

Length

Meter

m\text{m}

Mass

Kilogram

kg\text{kg}

Time

Second

s\text{s}

Temperature

Kelvin

K\text{K}

Amount of substance

Mole

mol\text{mol}

Electric current

Ampere

A\text{A}

Luminous intensity

Candela

cd\text{cd}

  • 2019 SI Redefinition: Base units are locked to invariant physical constants:

    • The meter (m\text{m}) is defined by fixing the speed of light in a vacuum to exactly c=299792458 m/sc = 299792458\,\text{m/s}; light travels 1 m1\,\text{m} in 1299792458\frac{1}{299792458} seconds.

    • The kilogram (kg\text{kg}) is defined by fixing the Planck constant to exactly h=6.62607015×10−34 J sh = 6.62607015 \times 10^{-34}\,\text{J\,s}.

Common SI-Derived Units

Derived units are formed by algebraic combinations of the seven base units:

Derived Quantity

Quantity Symbol

Unit Name

Unit Symbol

Area

AA

Square meter

m2\text{m}^2

Volume

VV

Cubic meter

m3\text{m}^3

Speed, Velocity

vv

Meter per second

m/s\text{m/s}

Acceleration

aa

Meter per second squared

m/s2\text{m/s}^2

Wave number

ν~\tilde{\nu}

Reciprocal meter

m−1\text{m}^{-1}

Mass density

ρ\rho

Kilogram per cubic meter

kg/m3\text{kg/m}^3

Specific volume

vv

Cubic meter per kilogram

m3/kg\text{m}^3\text{/kg}

Current density

jj

Ampere per square meter

A/m2\text{A/m}^2

Magnetic field strength

HH

Ampere per meter

A/m\text{A/m}

Concentration

cc

Mole per cubic meter

mol/m3\text{mol/m}^3

Luminance

LvL_v

Candela per square meter

cd/m2\text{cd/m}^2

Mass fraction

ww

Kilogram per kilogram

kg/kg=1\text{kg/kg} = 1

Metric Prefixes

Multiples and submultiples of base units are indicated with standard prefixes:

Factor

Prefix

Symbol

Factor

Prefix

Symbol

10910^9

giga

G\text{G}

10−110^{-1}

deci

d\text{d}

10610^6

mega

M\text{M}

10−210^{-2}

centi

c\text{c}

10310^3

kilo

k\text{k}

10−310^{-3}

milli

m\text{m}

10210^2

hecto

h\text{h}

10−610^{-6}

micro

μ\mu

10110^1

deka

da\text{da}

10−910^{-9}

nano

n\text{n}

10010^0

(base)

—

10−1210^{-12}

pico

p\text{p}

Accuracy versus Precision in Measurements
  • Accuracy: Refers to how close a measured value is to the true or accepted reference value.

  • Precision: Refers to the agreement or repeatability among multiple measurements made under identical conditions.

Laboratory Measurement Protocols and Significant Figures
  1. Record all digits known with absolute certainty based on the graduation marks of the instrument.

  2. Estimate one final digit representing the uncertainty of the measurement.

  3. When reading volume from graduated glassware (graduated cylinders, burets), read the liquid level at the bottom of the concave meniscus at eye level.

Rules for Identifying Significant Figures
  • Non-zero Digits: All digits from 11 to 99 are significant.

  • Captive Zeros: Zeros located between non-zero digits are always significant (e.g., 50835083 has four significant figures; 94.072 g94.072\,\text{g} has five).

  • Leading Zeros: Zeros at the beginning of a number serve only to locate the decimal point and are never significant (e.g., 0.00508300.0050830 contains leading zeros that are not significant; 0.0076 mL0.0076\,\text{mL} has two significant figures; 0.083 μg0.083\,\mu\text{g} has two).

  • Trailing Zeros After a Decimal: Zeros at the end of a number that lie after a decimal point are significant because their inclusion signifies instrument precision (e.g., 3400.0 m23400.0\,\text{m}^2 has five significant figures; 32.0 mL32.0\,\text{mL} has three).

  • Trailing Zeros Before an Implied Decimal: Ambiguous when written without an explicit decimal point (e.g., 3400 kg3400\,\text{kg} may have two, three, or four significant figures; 125000125000 is ambiguous). An explicit decimal point removes ambiguity: 125000.125000. has six significant figures.

  • Exact Numbers: Numbers originating from exact counts (e.g., 33 apples) or defined conversion factors (e.g., 1 ft=12 in1\,\text{ft} = 12\,\text{in}, 1 kg=1000 g1\,\text{kg} = 1000\,\text{g}) possess infinite significant figures and do not limit calculation precision.

Significant Figures in Calculations
  • Multiplication and Division: The result cannot possess more significant figures than the measurement containing the fewest significant figures.

Multiplication and Division Significant Figures
  • Example 1: 22×2=44→4022 \times 2 = 44 \rightarrow 40 (11 significant figure due to 22).

  • Example 2: 22.56×11=248.16→25022.56 \times 11 = 248.16 \rightarrow 250 (22 significant figures due to 1111).

  • Example 3: 123.1×23=2831.3→2800123.1 \times 23 = 2831.3 \rightarrow 2800 (22 significant figures).

  • Example 4: 123.1÷23=5.352⋯→5.4123.1 \div 23 = 5.352\dots \rightarrow 5.4 (22 significant figures).

    • Addition and Subtraction: The result is limited by the measurement with the greatest uncertainty (fewest decimal places, or largest place value for integers).

Addition and Subtraction Significant Figures
  • Example 1: 33.25−1.1=32.15→32.233.25 - 1.1 = 32.15 \rightarrow 32.2 (limited to tenths place).

  • Example 2: 64.9767+11.56=76.5367→76.5464.9767 + 11.56 = 76.5367 \rightarrow 76.54 (limited to hundredths place).

  • Example 3: 3300+501=3801→38003300 + 501 = 3801 \rightarrow 3800 (hundreds place is least precise in 33003300).

  • Example 4: 3300+51=3351→34003300 + 51 = 3351 \rightarrow 3400 (hundreds place is least precise in 33003300).

    • Rounding Off Rules: If the digit following the last retained place is 55 or greater, round up; if it is 44 or less, leave the retained digit unchanged.

    • Order of Operations and Stepwise Calculations:

  • Order: Parentheses (P), Exponents (E), Multiplication and Division from left to right (M/D), Addition and Subtraction from left to right (A/S).

  • Example: 9/3+2(9+10)−82+4×6=3+2(19)−64+24=3+38−64+24=19/3 + 2(9+10) - 8^2 + 4 \times 6 = 3 + 2(19) - 64 + 24 = 3 + 38 - 64 + 24 = 1.

  • In multistep calculations, retain all intermediate digits in the calculator and round to the correct number of significant figures only at the final step.

Scientific Notation

Numbers are expressed in the form A×10nA \times 10^n, where 1≤∣A∣<101 \le |A| < 10 and nn is an integer.

  • Mean distance from Earth to Moon: 384000000 m=3.84×108 m384000000\,\text{m} = 3.84 \times 10^8\,\text{m}.

  • Resting mass of an electron: 9.11×10−31 kg9.11 \times 10^{-31}\,\text{kg}.

  • Addition and Subtraction: Exponents must be converted to match before adding or subtracting coefficients:

6.355×106−5.24×105=63.55×105−5.24×105=58.31×105=5.831×1066.355 \times 10^6 - 5.24 \times 10^5 = 63.55 \times 10^5 - 5.24 \times 10^5 = 58.31 \times 10^5 = 5.831 \times 10^6

  • Multiplication: Multiply coefficients and add exponents:

(6.355×106)×(5.24×105)=(6.355×5.24)×106+5=33.3002×1011=3.33×1012(6.355 \times 10^6) \times (5.24 \times 10^5) = (6.355 \times 5.24) \times 10^{6+5} = 33.3002 \times 10^{11} = 3.33 \times 10^{12}

  • Division: Divide coefficients and subtract exponents:

(6.355×106)÷(5.24×105)=(6.355÷5.24)×106−5=1.212786×101=1.21×101(6.355 \times 10^6) \div (5.24 \times 10^5) = (6.355 \div 5.24) \times 10^{6-5} = 1.212786 \times 10^1 = 1.21 \times 10^1

Problem-Solving and Dimensional Analysis

Dimensional analysis converts an initial quantity to a desired quantity using equivalent ratios:

Given Unit×Desired UnitGiven Unit=Desired Unit\text{Given Unit} \times \frac{\text{Desired Unit}}{\text{Given Unit}} = \text{Desired Unit}

  • Common Conversion Equivalencies:

    • Length: 1 ft=12 in1\,\text{ft} = 12\,\text{in}, 1 yd=3 ft1\,\text{yd} = 3\,\text{ft}, 1 mi=1760 yd=1.60934 km1\,\text{mi} = 1760\,\text{yd} = 1.60934\,\text{km}, 1 in=2.54 cm1\,\text{in} = 2.54\,\text{cm}, 1 m=100 cm1\,\text{m} = 100\,\text{cm}.

    • Mass/Weight: 1 lb=16 oz1\,\text{lb} = 16\,\text{oz}, 1 ton=2000 lb1\,\text{ton} = 2000\,\text{lb}, 1 lb=0.454 kg1\,\text{lb} = 0.454\,\text{kg}, 1 metric ton=1000 kg1\,\text{metric ton} = 1000\,\text{kg}.

    • Capacity: 1 cup=8 fl oz1\,\text{cup} = 8\,\text{fl oz}, 1 pt=2 cups1\,\text{pt} = 2\,\text{cups}, 1 qt=2 pt1\,\text{qt} = 2\,\text{pt}, 1 gal=4 qt1\,\text{gal} = 4\,\text{qt}, 1 L=1000 mL1\,\text{L} = 1000\,\text{mL}, 1 mL=1 cm31\,\text{mL} = 1\,\text{cm}^3.

  • Rate and Concentration Equivalencies: Ratios with "per" indicate direct conversion factors (e.g., 55 mph→55 miles=1 hour55\,\text{mph} \rightarrow 55\,\text{miles} = 1\,\text{hour}; 12.011 g/mol→12.011 g C=1 mol C12.011\,\text{g/mol} \rightarrow 12.011\,\text{g C} = 1\,\text{mol C}).

Temperature Scales and Calculations

Temperature measures the average kinetic energy of the particles within a sample.

  • Fahrenheit (^\nef): Water freezes at 32\,^\nef and boils at 212\,^\nef (180\,^\nef interval).

  • Celsius (^\nec): Water freezes at 0.00\,^\nec and boils at 100\,^\nec (100\,^\nec interval).

  • Kelvin (K\text{K}): Absolute temperature scale. Water freezes at 273.15 K273.15\,\text{K} and boils at 373.15 K373.15\,\text{K}. Absolute zero (0 K0\,\text{K}) corresponds to -273.15\,^\nec or -459.67\,^\nef.

  • Conversion Equations:

T(K)=T(∘C)+273.15T(\text{K}) = T(^\circ\text{C}) + 273.15

T(∘F)=95T(∘C)+32T(^\circ\text{F}) = \frac{9}{5} T(^\circ\text{C}) + 32

T(∘C)=59(T(∘F)−32)T(^\circ\text{C}) = \frac{5}{9} \left( T(^\circ\text{F}) - 32 \right)

ΔT=Tfinal−Tinitial\Delta T = T_{\text{final}} - T_{\text{initial}}

Density and Percentage Calculations
  • Density: Physical property relating mass to volume:

d=mVd = \frac{m}{V}

  • Density varies with temperature. Water reaches its maximum density of approximately 1.00000 g/cm31.00000\,\text{g/cm}^3 at 3.98\,^\nec (\sim 4\,^\nec).

    • Percentage Calculations:

%=PartWhole×100\% = \frac{\text{Part}}{\text{Whole}} \times 100

  • When converting a percentage into a conversion factor, assume a sample size of 100100 units (e.g., 1.68% S1.68\%\,\text{S} by mass means 1.68 kg S=100 kg coal1.68\,\text{kg S} = 100\,\text{kg coal}).

Scientific Methodology, Data Analysis, and Graphical Interpretation

The Scientific Method
  1. Observation / Question Formulation: Identifying an unexplained phenomenon.

  2. Background Research: Reviewing existing literature and physical principles.

  3. Hypothesis Formation: Constructing a testable, falsifiable explanatory model that defines independent and dependent variables.

  4. Experimental Testing: Performing controlled experiments holding extraneous variables constant.

  5. Data Analysis: Employing statistical and graphical evaluations.

  6. Conclusion and Reporting: Validating or revising the original hypothesis.

Types of Experimental Variables and Data
  • Quantitative (Numerical) Variables: Measured numerically.

    • Continuous Data: Measured on an unbroken scale (e.g., mass, height, temperature, discharge rate).

    • Discrete Data: Counted in integer steps (e.g., number of disease cases, bacteria colony counts).

  • Qualitative (Categorical) Variables: Expressed as classifications or labels.

    • Ordinal Data: Labels with a natural ranking or sequence (e.g., beverage size: small, medium, large; satisfaction ratings).

    • Nominal Data: Categories with no natural order (e.g., profession, pet species, candy color).

Graphing Conventions and Statistical Regression
  • Graph Construction: The dependent variable is plotted on the vertical yy-axis and the independent variable is plotted on the horizontal xx-axis (titled as "Dependent Variable vs. Independent Variable").

  • Linear Correlation Coefficient (rr): Quantifies linear relationship strength:

r=∑i=1n(xi−xˉ)(yi−yˉ)∑i=1n(xi−xˉ)2∑i=1n(yi−yˉ)2r = \frac{\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i=1}^n (x_i - \bar{x})^2} \sqrt{\sum_{i=1}^n (y_i - \bar{y})^2}}

  • r=+1r = +1 indicates perfect positive linear correlation.

  • r=−1r = -1 indicates perfect negative linear correlation.

  • r=0r = 0 indicates no linear association.

  • The coefficient of determination (R2R^2) represents the proportion of variance in the dependent variable explained by the regression line.

    • Interpolation versus Extrapolation:

  • Interpolation: Estimating values between measured data points within the tested domain (reliable).

  • Extrapolation: Projecting values beyond the extreme limits of the measured experimental data (unreliable due to unknown non-linear behaviors).

    • Analytical Figures of Merit:

  • Limit of Detection (LOD): The lowest quantity of analyte distinguishable from the blank baseline.

  • Limit of Quantitation (LOQ): The lowest concentration at which analyte can be quantitatively measured with acceptable precision and accuracy.

  • Linear Dynamic Range: The concentration interval between LOQ and the point where detector response deviates from linearity.

  • Sensitivity: The slope of the calibration curve (ΔSignalΔConcentration\frac{\Delta \text{Signal}}{\Delta \text{Concentration}}).

Atomic Structure and Modern Atomic Theory

Dalton's Atomic Theory (1803) and Modern Revisions
  1. Postulate 1: All matter is composed of tiny, indestructible particles called atoms.

    • Modern Amendment: Atoms are not indivisible; they are composed of subatomic particles (protons, neutrons, electrons) and can be split via nuclear fission.

  2. Postulate 2: All atoms of a given element have identical mass and other properties that distinguish them from other elements.

    • Modern Amendment: Atoms of the same element always possess the same number of protons (ZZ), but can possess different masses due to variable neutron numbers (isotopes).

  3. Postulate 3: Atoms combine in simple, whole-number ratios to form chemical compounds.

    • Modern Amendment: Valid for classical compounds (Law of Multiple Proportions and Law of Definite Proportions), though complex non-stoichiometric polymers and organometallics exist.

  4. Postulate 4: Atoms of one element cannot change into atoms of another element; chemical reactions involve only changes in how atoms are bonded.

    • Modern Amendment: Holds strictly true for chemical reactions, but atoms can transform into other elements through nuclear fusion and radioactive decay.

Subatomic Particles

Particle

Relative Mass (amu)

Exact Mass (kg)

Relative Charge

Absolute Charge (C)

Location in Atom

Proton (p+p^+)

11

1.67×10−27 kg1.67 \times 10^{-27}\,\text{kg}

+1+1

+1.6×10−19 C+1.6 \times 10^{-19}\,\text{C}

Dense atomic nucleus

Neutron (n0n^0)

11

1.67×10−27 kg1.67 \times 10^{-27}\,\text{kg}

00

0 C0\,\text{C}

Dense atomic nucleus

Electron (e−e^-)

0.00050.0005

9.11×10−31 kg9.11 \times 10^{-31}\,\text{kg}

−1-1

−1.6×10−19 C-1.6 \times 10^{-19}\,\text{C}

Diffuse orbitals / electron cloud

Fundamental Forces in the Atom
  • Electrostatic (Coulombic) Forces:

    • Opposite charges attract: Protons in the nucleus attract surrounding orbital electrons, binding them to the atom.

    • Like charges repel: Electrons repel other electrons, determining electron cloud geometry and molecular shapes according to Valence Shell Electron Pair Repulsion (VSEPR) theory.

  • Strong Nuclear Force:

    • Electrostatic repulsion between positively charged protons strains the nucleus.

    • The residual strong nuclear force acts over extremely short ranges (∼10−15 m\sim 10^{-15}\,\text{m}) between nucleons (protons and neutrons) to overcome electrostatic repulsion and hold the nucleus together.

Isotopic Abundance, Mass Spectrometry, and Atomic Weight Calculations

Atomic Number, Mass Number, and Isotopes
  • Atomic Number (ZZ): Number of protons in the nucleus; defines chemical identity.

  • Mass Number (AA or MM): Total sum of protons and neutrons in a specific nucleus:

A=Protons+NeutronsA = \text{Protons} + \text{Neutrons}

  • Standard Nuclear Symbol Notation:

ZAX^A_Z\text{X}

  • Examples include Hydrogen isotopes: Protium (11H^1_1\text{H}: 1 p+,0 n01\,p^+, 0\,n^0), Deuterium (12H^2_1\text{H}: 1 p+,1 n01\,p^+, 1\,n^0), Tritium (13H^3_1\text{H}: 1 p+,2 n01\,p^+, 2\,n^0).

Mass Spectrometry Operation

Mass spectrometry determines atomic or molecular masses and isotopic distributions through five stages:

  1. Vaporization: Sample is vaporized in a high-vacuum chamber.

  2. Ionization: Gaseous atoms are bombarded by a high-energy electron beam (∼70 eV\sim 70\,\text{eV}) generated by a heated filament, knocking out electrons to produce positive cations (A+e−→A++2e−\text{A} + e^- \rightarrow \text{A}^+ + 2e^-).

  3. Acceleration: Cations pass between electric accelerator plates across a potential difference (0.5−10 kV0.5 - 10\,\text{kV}) to form a high-velocity ion beam.

  4. Deflection: The accelerated ion beam enters a perpendicular magnetic field. Ions are deflected into curved trajectories based on their mass-to-charge ratio (m/zm/z):

    • Lighter ions and ions with higher positive charges deflect more sharply.

    • Heavier ions deflect less.

  5. Detection: An electrometer detects ions arriving at specific positions as electrical currents, generating a spectrum of signal intensity versus m/zm/z.

Calculating Atomic Weight

The atomic weight of an element is the weighted average mass of its naturally occurring isotopes:

Atomic Weight=∑(Isotopic Abundance×Isotopic Mass)\text{Atomic Weight} = \sum \left( \text{Isotopic Abundance} \times \text{Isotopic Mass} \right)

  • Carbon Example:

    • 12C^{12}\text{C}: Abundance =98.93%=0.9893= 98.93\% = 0.9893, Mass =12.0000 amu= 12.0000\,\text{amu}

    • 13C^{13}\text{C}: Abundance =1.07%=0.0107= 1.07\% = 0.0107, Mass =13.0034 amu= 13.0034\,\text{amu}

    • 14C^{14}\text{C}: Trace radioactive abundance

Atomic Weight of C=(0.9893×12.0000)+(0.0107×13.0034)=11.8716+0.1391=12.011 amu\text{Atomic Weight of C} = (0.9893 \times 12.0000) + (0.0107 \times 13.0034) = 11.8716 + 0.1391 = 12.011\,\text{amu}

  • Boron Example:

    • 10B^{10}\text{B} (19.8%19.8\%, 10.0129 amu10.0129\,\text{amu}) and 11B^{11}\text{B} (80.2%80.2\%, 11.0093 amu11.0093\,\text{amu}):

Atomic Weight of B=(0.198×10.0129)+(0.802×11.0093)=10.81 amu\text{Atomic Weight of B} = (0.198 \times 10.0129) + (0.802 \times 11.0093) = 10.81\,\text{amu}

  • Chlorine Example:

    • 35Cl^{35}\text{Cl} (75.77%75.77\%, 34.9689 amu34.9689\,\text{amu}) and 37Cl^{37}\text{Cl} (24.23%24.23\%, 36.9659 amu36.9659\,\text{amu}):

Atomic Weight of Cl=(0.7577×34.9689)+(0.2423×36.9659)=35.45 amu\text{Atomic Weight of Cl} = (0.7577 \times 34.9689) + (0.2423 \times 36.9659) = 35.45\,\text{amu}

The Mole Concept and Avogadro's Constant
  • The Mole (mol\text{mol}): The SI base unit for amount of substance. One mole contains exactly 6.02214076×10236.02214076 \times 10^{23} elementary entities (Avogadro's Constant, NAN_A).

  • Atomic Mass Unit versus Molar Mass: One atomic mass unit (1 amu1\,\text{amu}) is defined as 112\frac{1}{12} the mass of an unbound 12C^{12}\text{C} atom. Because of how the mole is defined, an element's relative atomic mass in amu\text{amu} is numerically identical to its molar mass in g/mol\text{g/mol}.

    • Example: A single 12C^{12}\text{C} atom has a mass of 12 amu12\,\text{amu}; 1 mol1\,\text{mol} of 12C^{12}\text{C} atoms has a mass of exactly 12.00 g12.00\,\text{g}.

Nuclear Transformations and Radiocarbon Dating
  • Nuclear Fusion: The combination of light nuclei to form a heavier nucleus, accompanied by the release of energy (e.g., hydrogen fusion in stellar cores):

12H+13H→24He+01n+Energy^2_1\text{H} + ^3_1\text{H} \rightarrow ^4_2\text{He} + ^1_0\text{n} + \text{Energy}

  • Nuclear Fission: The splitting of a heavy nucleus into lighter fragment nuclei triggered by neutron capture, releasing additional neutrons and binding energy:

01n+ 92235U→ 56141Ba+3692Kr+3 01n+Energy^1_0\text{n} + ^{235}_{\ 92}\text{U} \rightarrow ^{141}_{\ 56}\text{Ba} + ^{92}_{36}\text{Kr} + 3\,^1_0\text{n} + \text{Energy}

  • Radioactive Decay Modes:

    • Alpha Decay (α\alpha): Emission of a helium-4 nucleus (24α^4_2\alpha):

ZAX→Z−2A−4X′+24α^A_Z\text{X} \rightarrow ^{A-4}_{Z-2}\text{X}' + ^4_2\alpha

  • Beta Decay (β−\beta^-): Conversion of a neutron into a proton with emission of a high-speed electron (−10β^0_{-1}\beta):

ZAX→Z+1AX′+−10β^A_Z\text{X} \rightarrow ^A_{Z+1}\text{X}' + ^0_{-1}\beta

  • Gamma Emission (γ\gamma): De-excitation of an unstable nuclear state via release of high-energy photon (00γ^0_0\gamma).

    • Radiocarbon Dating: Cosmic ray neutrons strike atmospheric nitrogen atoms to produce carbon-14 (01n+714N→614C+11p^1_0\text{n} + ^{14}_7\text{N} \rightarrow ^{14}_6\text{C} + ^1_1\text{p}). Carbon-14 oxidizes to CO2\text{CO}_2 and enters biological systems through photosynthesis. Upon death, assimilation stops and 14C^{14}\text{C} decays back to 14N^{14}\text{N} via beta emission with a half-life of t1/2=5730 yearst_{1/2} = 5730\,\text{years}. Measuring remaining 14C/12C^{14}\text{C}/^{12}\text{C} ratios determines archeological age up to approximately 50000 years50000\,\text{years}.

The Periodic Table: Organization, Trends, and Families

Architecture of the Periodic Table
  • Periods: Seven horizontal rows (n=1n = 1 to 77), corresponding to the principal quantum shell being filled with electrons.

  • Groups (Families): Eighteen vertical columns representing elements with identical valence electron configurations and similar chemical reactivities.

  • Structural Blocks:

    • s-block: Groups 1 and 2 (plus Helium).

    • p-block: Groups 13 through 18.

    • d-block: Groups 3 through 12 (Transition Metals).

    • f-block: Lanthanides (elements 57−7157-71) and Actinides (elements 89−10389-103), positioned below the main grid.

Broad Elemental Classifications
  • Metals: Located on the left and center of the periodic table. Characteristic properties include metallic luster, malleability (hammerable into thin sheets), ductility (drawable into wires), high thermal and electrical conductivities, high densities, and a tendency to lose electrons to form cations.

  • Nonmetals: Located on the upper right side (with Hydrogen on the left). Poor conductors of heat and electricity, lack luster, exhibit diverse physical states at room temperature (gases, liquids, brittle solids), and tend to gain or share electrons in chemical reactions.

  • Metalloids (Semimetals): Lie along the border line separating metals and nonmetals: Boron (B\text{B}), Silicon (Si\text{Si}), Germanium (Ge\text{Ge}), Arsenic (As\text{As}), Antimony (Sb\text{Sb}), Tellurium (Te\text{Te}), Polonium (Po\text{Po}), and Astatine (At\text{At}). Display intermediate properties and function as semiconductors.

Primary Periodic Families
  • Alkali Metals (Group 1 / 1A): Valence configuration ns1ns^1. Soft, silvery, low-density metals with low melting points. Highly reactive; react vigorously with water to produce metal hydroxides and flammable hydrogen gas. Form +1+1 cations exclusively.

  • Alkaline Earth Metals (Group 2 / 2A): Valence configuration ns2ns^2. Harder, denser, and less reactive than alkali metals. Burn in oxygen to form basic oxides. Form +2+2 cations exclusively.

  • Transition Metals (Groups 3–12 / d-block): Hard, dense metals with high melting points. Characterized by variable oxidation states (e.g., Fe2+,Fe3+\text{Fe}^{2+}, \text{Fe}^{3+}), formation of colored coordination complexes, and catalytic properties.

  • Halogens (Group 17 / 7A): Valence configuration ns2np5ns^2 np^5. Highly reactive nonmetals that exist as diatomic volatile molecules (F2,Cl2,Br2,I2\text{F}_2, \text{Cl}_2, \text{Br}_2, \text{I}_2). Readily accept one electron to form halide anions (−1-1 charge).

  • Noble Gases (Group 18 / 8A): Valence configuration ns2np6ns^2 np^6 (Helium is 1s21s^2). Characterized by completely filled valence shells. Chemically inert, monatomic gases at room temperature with high ionization energies and low boiling points.

Electron Configurations and Valence Electrons

Electrons occupy quantized shells that fill from lowest to highest energy (Aufbau Principle):

  • Shell capacity is given by 2n22n^2 (n=1→2 e−n=1 \rightarrow 2\,e^-, n=2→8 e−n=2 \rightarrow 8\,e^-, n=3→18 e−n=3 \rightarrow 18\,e^-, n=4→32 e−n=4 \rightarrow 32\,e^-).

  • Valence Electrons: Electrons residing in the outermost principal energy shell (nn). Group numbers for main block elements directly yield valence electron counts:

    • Group 1 (1A): 11 valence electron (Li,Na,K→+1\text{Li}, \text{Na}, \text{K} \rightarrow +1 ion)

    • Group 2 (2A): 22 valence electrons (Be,Mg,Ca→+2\text{Be}, \text{Mg}, \text{Ca} \rightarrow +2 ion)

    • Group 13 (3A): 33 valence electrons (B,Al→+3\text{B}, \text{Al} \rightarrow +3 ion)

    • Group 14 (4A): 44 valence electrons (C,Si→±4\text{C}, \text{Si} \rightarrow \pm 4 or covalent)

    • Group 15 (5A): 55 valence electrons (N,P→−3\text{N}, \text{P} \rightarrow -3 ion)

    • Group 16 (6A): 66 valence electrons (O,S,Se→−2\text{O}, \text{S}, \text{Se} \rightarrow -2 ion)

    • Group 17 (7A): 77 valence electrons (F,Cl,Br,I→−1\text{F}, \text{Cl}, \text{Br}, \text{I} \rightarrow -1 ion)

    • Group 18 (8A): 88 valence electrons (Ne,Ar,Kr→0\text{Ne}, \text{Ar}, \text{Kr} \rightarrow 0, stable octet)

Chemical Bonding, Nomenclature, and Molecular Structure

The Octet Rule and Driving Forces of Bonding

Atoms bond to achieve stable electronic configurations, typically eight valence electrons identical to the nearest noble gas (or two electrons for Helium). This is achieved either through the transfer of electrons (ionic bonding) or the sharing of electron pairs (covalent bonding).

Comparison of Ionic and Covalent Compounds
Comparison of Ionic and Covalent Compounds

Property

Ionic Compounds

Covalent (Molecular) Compounds

Elements Involved

Metal + Nonmetal

Nonmetal + Nonmetal (or Metalloid + Nonmetal)

Bonding Mechanism

Electrons transferred to form ions

Electrons shared between nuclei

Formula Representation

Formula Unit (simplest whole-number ratio)

Molecular Formula (exact number of atoms)

Physical State at Room Temp

Crystalline solid (salt)

Gas, liquid, or low-melting solid

Melting / Boiling Points

Extremely high (very stable lattice)

Low to moderate (weak intermolecular forces)

Electrical Conductivity

Conducts in aqueous solution or molten state

Non-electrolyte (does not conduct electricity)

Ionic Structure and Formula Construction

Ionic compounds form rigid, three-dimensional crystal lattices composed of alternating cations and anions.

Sodium Chloride Crystal Lattice
  • Rules for Writing Formulas:

    1. The cation is written first, followed by the anion (e.g., NaCl\text{NaCl}, not ClNa\text{ClNa}).

    2. Overall electrical neutrality must be maintained; net charge must equal zero.

    3. Subscripts denote the lowest integer ratio of ions.

    4. Do not include individual ion charges in the final neutral chemical formula.

    5. Enclose polyatomic ions in parentheses when a subscript greater than one is required (e.g., Al2(SO4)3\text{Al}_2(\text{SO}_4)_3, not Al2SO43\text{Al}_2\text{SO}_{43}).

Multivalent Transition Metal Cations

Transition metals often form multiple stable cations, requiring Roman numerals in the Stock naming system:

Element

Ion Symbol

Stock System Name

Classic / Trivial 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

Polyatomic Ions Reference

Ion Name

Formula

Ion Name

Formula

Ammonium

NH4+\text{NH}_4^+

Hydroxide

OH−\text{OH}^-

Acetate

C2H3O2−\text{C}_2\text{H}_3\text{O}_2^-

Cyanide

CN−\text{CN}^-

Carbonate

CO32−\text{CO}_3^{2-}

Hypochlorite

ClO−\text{ClO}^-

Hydrogen carbonate (Bicarbonate)

HCO3−\text{HCO}_3^-

Chlorite

ClO2−\text{ClO}_2^-

Nitrite

NO2−\text{NO}_2^-

Chlorate

ClO3−\text{ClO}_3^-

Nitrate

NO3−\text{NO}_3^-

Perchlorate

ClO4−\text{ClO}_4^-

Sulfite

SO32−\text{SO}_3^{2-}

Permanganate

MnO4−\text{MnO}_4^-

Sulfate

SO42−\text{SO}_4^{2-}

Chromate

CrO42−\text{CrO}_4^{2-}

Hydrogen sulfate (Bisulfate)

HSO4−\text{HSO}_4^-

Dichromate

Cr2O72−\text{Cr}_2\text{O}_7^{2-}

Hydrogen sulfite (Bisulfite)

HSO3−\text{HSO}_3^-

Peroxide

O22−\text{O}_2^{2-}

Phosphate

PO43−\text{PO}_4^{3-}

Hydrogen phosphate

HPO42−\text{HPO}_4^{2-}

Dihydrogen phosphate

H2PO4−\text{H}_2\text{PO}_4^-



Hydrated Salts

Hydrated salts contain specific stoichiometric ratios of water molecules trapped within their crystal lattices.

  • Formula Format: Salt⋅nH2O\text{Salt} \cdot n\text{H}_2\text{O}, where nn is the integer number of water molecules per formula unit.

  • Naming: Named as the anhydrous ionic salt followed by a Greek prefix indicating nn and the term "hydrate".

    • CuSO4⋅5H2O\text{CuSO}_4 \cdot 5\text{H}_2\text{O}: Copper(II) sulfate pentahydrate

    • Na2CO3⋅10H2O\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}: Sodium carbonate decahydrate

    • BaCl2⋅2H2O\text{BaCl}_2 \cdot 2\text{H}_2\text{O}: Barium chloride dihydrate

  • Molar Mass Calculation for CuSO4⋅5H2O\text{CuSO}_4 \cdot 5\text{H}_2\text{O}:

    • Cu=63.55 g/mol\text{Cu} = 63.55\,\text{g/mol}

    • S=32.07 g/mol\text{S} = 32.07\,\text{g/mol}

    • 4×O=4×16.00 g/mol=64.00 g/mol4 \times \text{O} = 4 \times 16.00\,\text{g/mol} = 64.00\,\text{g/mol}

    • Anhydrous CuSO4=159.62 g/mol\text{Anhydrous } \text{CuSO}_4 = 159.62\,\text{g/mol}

    • 5×H2O=5×18.02 g/mol=90.10 g/mol5 \times \text{H}_2\text{O} = 5 \times 18.02\,\text{g/mol} = 90.10\,\text{g/mol}

    • Total Molar Mass=159.62+90.10=249.72 g/mol\text{Total Molar Mass} = 159.62 + 90.10 = 249.72\,\text{g/mol}

Binary Molecular Compound Nomenclature

Binary molecular compounds consist of two nonmetals:

Molecular Nomenclature Rules
  1. The more metallic element (further left or lower on the periodic table; lower electronegativity) is listed first.

  2. The second element is named using its root name with the suffix "-ide".

  3. Greek numerical prefixes specify the number of atoms present: mono- (11), di- (22), tri- (33), tetra- (44), penta- (55), hexa- (66), hepta- (77), octa- (88), nona- (99), deca- (1010).

  4. The prefix "mono-" is never used on the first element.

  5. If a prefix ends in "a-" or "o-" and the element name begins with a vowel, the trailing vowel of the prefix is dropped (e.g., carbon monoxide, not carbon monooxide; dinitrogen tetroxide, not dinitrogen tetraoxide).

    • Examples: N2O5\text{N}_2\text{O}_5 (dinitrogen pentoxide), BBr3\text{BBr}_3 (boron tribromide), SO3\text{SO}_3 (sulfur trioxide), SF6\text{SF}_6 (sulfur hexafluoride), N2O3\text{N}_2\text{O}_3 (dinitrogen trioxide), GeCl4\text{GeCl}_4 (germanium tetrachloride).

Acid Nomenclature

Acids are molecular compounds that ionize in water to release hydrogen ions (H+\text{H}^+):

  • Binary Acids (Hydrogen + Nonmetal): Named as "hydro-" + nonmetal root + "-ic acid".

    • HCl(aq)\text{HCl}(aq): Hydrochloric acid

    • HBr(aq)\text{HBr}(aq): Hydrobromic acid

    • HI(aq)\text{HI}(aq): Hydroiodic acid

    • H2S(aq)\text{H}_2\text{S}(aq): Hydrosulfuric acid

  • Oxyacids (Hydrogen + Oxyanion):

    • If the oxyanion ends in "-ate", the acid is named with root + "-ic acid".

    • HNO3\text{HNO}_3 (from nitrate): Nitric acid

    • H2SO4\text{H}_2\text{SO}_4 (from sulfate): Sulfuric acid

    • H3PO4\text{H}_3\text{PO}_4 (from phosphate): Phosphoric acid

    • H2CO3\text{H}_2\text{CO}_3 (from carbonate): Carbonic acid

    • HC2H3O2\text{HC}_2\text{H}_3\text{O}_2 (from acetate): Acetic acid

    • If the oxyanion ends in "-ite", the acid is named with root + "-ous acid".

    • HNO2\text{HNO}_2 (from nitrite): Nitrous acid

    • H2SO3\text{H}_2\text{SO}_3 (from sulfite): Sulfurous acid

Introduction to Organic Compounds

Organic chemistry encompasses carbon-containing compounds, characterized by covalent carbon backbones:

  • Hydrocarbon Prefixes: Based on the continuous carbon chain length: meth- (11), eth- (22), prop- (33), but- (44), pent- (55), hex- (66), hept- (77), oct- (88), non- (99), dec- (1010).

  • Primary Suffixes:

    • Saturated alkanes (single bonds only, C−C\text{C}-\text{C}): "-ane"

    • Alkenes (contain one double bond, C=C\text{C}=\text{C}): "-ene"

    • Alkynes (contain one triple bond, C≡C\text{C}\equiv\text{C}): "-yne"

    • Dienes (two double bonds): "-diene"

Quantitative Chemical Composition: Molar Relationships, Formulas, and Combustion Analysis

Mass Percent Composition

Mass percent defines the contribution of each element to a compound's total molar mass:

% Mass of Element=n×(Atomic Weight of Element)Molar Mass of Compound×100\% \text{ Mass of Element} = \frac{n \times (\text{Atomic Weight of Element})}{\text{Molar Mass of Compound}} \times 100

Example Calculation for Ethanol (C2H6O\text{C}_2\text{H}_6\text{O}):

  • Molar mass: (2×12.011)+(6×1.008)+(1×16.00)=24.022+6.048+16.00=46.07 g/mol(2 \times 12.011) + (6 \times 1.008) + (1 \times 16.00) = 24.022 + 6.048 + 16.00 = 46.07\,\text{g/mol}

  • C mass %=24.02246.07×100=52.14%\text{C mass } \% = \frac{24.022}{46.07} \times 100 = 52.14\%

  • H mass %=6.04846.07×100=13.13%\text{H mass } \% = \frac{6.048}{46.07} \times 100 = 13.13\%

  • O mass %=16.0046.07×100=34.73%\text{O mass } \% = \frac{16.00}{46.07} \times 100 = 34.73\%

Determining Empirical and Molecular Formulas
  • Empirical Formula: The simplest whole-number ratio of atoms in a compound.

  • Molecular Formula: The actual number of atoms of each element in a molecule:

Molecular Formula=(Empirical Formula)n\text{Molecular Formula} = (\text{Empirical Formula})_n

n=Molar Mass (Experimental MW)Empirical Formula Mass (EW)n = \frac{\text{Molar Mass (Experimental MW)}}{\text{Empirical Formula Mass (EW)}}

  • Stepwise Determination of Empirical Formula:

    1. Convert mass percentages to masses assuming a 100 g100\,\text{g} sample.

    2. Convert grams of each element into moles using atomic weights.

    3. Write a pseudoformula using the calculated moles as subscripts.

    4. Divide all subscripts by the smallest calculated mole value.

    5. If subscripts are not integers within rounding error, multiply all subscripts by the smallest integer that converts decimals into whole numbers:

Decimal Ending

Fractional Equivalent

Multiplier

Resulting Integer

0.10.1

1/101/10

1010

11

0.20.2

1/51/5

55

11

0.250.25

1/41/4

44

11

0.330.33

1/31/3

33

11

0.40.4

2/52/5

55

22

0.50.5

1/21/2

22

11

0.60.6

3/53/5

55

33

0.70.7

7/107/10

1010

77

0.750.75

3/43/4

44

33

0.80.8

4/54/5

55

44

0.90.9

9/109/10

1010

99

Aspirin Calculation:

  • Composition: 60.00% C60.00\%\,\text{C}, 4.48% H4.48\%\,\text{H}, 35.52% O35.52\%\,\text{O}

  • Moles in 100 g100\,\text{g}:

    • Moles C=60.00 g12.011 g/mol=4.995 mol\text{Moles C} = \frac{60.00\,\text{g}}{12.011\,\text{g/mol}} = 4.995\,\text{mol}

    • Moles H=4.48 g1.008 g/mol=4.444 mol\text{Moles H} = \frac{4.48\,\text{g}}{1.008\,\text{g/mol}} = 4.444\,\text{mol}

    • Moles O=35.52 g16.00 g/mol=2.220 mol\text{Moles O} = \frac{35.52\,\text{g}}{16.00\,\text{g/mol}} = 2.220\,\text{mol}

  • Divide by smallest (2.2202.220):

    • C:4.9952.220=2.25\text{C}: \frac{4.995}{2.220} = 2.25

    • H:4.4442.220=2.00\text{H}: \frac{4.444}{2.220} = 2.00

    • O:2.2202.220=1.00\text{O}: \frac{2.220}{2.220} = 1.00

  • Multiply by 44 to clear the 0.250.25 fraction: C2.25×4H2×4O1×4=C9H8O4\text{C}_{2.25 \times 4} \text{H}_{2 \times 4} \text{O}_{1 \times 4} = \text{C}_9\text{H}_8\text{O}_4.

Combustion Analysis

Combustion analysis determines the empirical and molecular formulas of organic compounds containing carbon, hydrogen, and oxygen by burning a known sample mass completely in oxygen gas:

  • All carbon is converted into CO2\text{CO}_2: absorbed and weighed.

  • All hydrogen is converted into H2O\text{H}_2\text{O}: absorbed and weighed.

  • Oxygen in the compound is determined by difference: Mass O=Mass sample−(Mass C+Mass H)\text{Mass O} = \text{Mass sample} - (\text{Mass C} + \text{Mass H}).

Comprehensive Worked Example:

  • Sample: 14.765 g14.765\,\text{g} of Compound X (contains only C,H,O\text{C}, \text{H}, \text{O}; experimental molar mass =300.44 g/mol= 300.44\,\text{g/mol}

  • Combustion products: 43.257 g CO243.257\,\text{g } \text{CO}_2 and 12.395 g H2O12.395\,\text{g } \text{H}_2\text{O}

  1. Determine Masses of Carbon and Hydrogen:

Mass C=43.257 g CO2×12.011 g C44.01 g CO2=11.805 g C\text{Mass C} = 43.257\,\text{g } \text{CO}_2 \times \frac{12.011\,\text{g C}}{44.01\,\text{g } \text{CO}_2} = 11.805\,\text{g C}

Mass H=12.395 g H2O×2.016 g H18.015 g H2O=1.387 g H\text{Mass H} = 12.395\,\text{g } \text{H}_2\text{O} \times \frac{2.016\,\text{g H}}{18.015\,\text{g } \text{H}_2\text{O}} = 1.387\,\text{g H}

  1. Determine Mass of Oxygen by Difference:

Mass O=14.765 g−(11.805 g+1.387 g)=14.765 g−13.192 g=1.573 g O\text{Mass O} = 14.765\,\text{g} - (11.805\,\text{g} + 1.387\,\text{g}) = 14.765\,\text{g} - 13.192\,\text{g} = 1.573\,\text{g O}

  1. Determine Moles of Each Element:

Moles C=11.805 g12.011 g/mol=0.98285 mol C\text{Moles C} = \frac{11.805\,\text{g}}{12.011\,\text{g/mol}} = 0.98285\,\text{mol C}

Moles H=1.387 g1.008 g/mol=1.3760 mol H\text{Moles H} = \frac{1.387\,\text{g}}{1.008\,\text{g/mol}} = 1.3760\,\text{mol H}

Moles O=1.573 g16.00 g/mol=0.09832 mol O\text{Moles O} = \frac{1.573\,\text{g}}{16.00\,\text{g/mol}} = 0.09832\,\text{mol O}

  1. Divide by Smallest Moles (0.098320.09832):

Ratio C=0.982850.09832=10.00\text{Ratio C} = \frac{0.98285}{0.09832} = 10.00

Ratio H=1.37600.09832=14.00\text{Ratio H} = \frac{1.3760}{0.09832} = 14.00

Ratio O=0.098320.09832=1.00\text{Ratio O} = \frac{0.09832}{0.09832} = 1.00

Empirical Formula: C10H14O\text{C}_{10}\text{H}_{14}\text{O}.

  1. Determine Molecular Formula:

  • Empirical Formula Mass (EWEW):

(10×12.011)+(14×1.008)+(1×16.00)=120.11+14.112+16.00=150.22 g/mol(10 \times 12.011) + (14 \times 1.008) + (1 \times 16.00) = 120.11 + 14.112 + 16.00 = 150.22\,\text{g/mol}

  • Molecular scaling factor:

n=Molar MassEmpirical Mass=300.44 g/mol150.22 g/mol=2.00n = \frac{\text{Molar Mass}}{\text{Empirical Mass}} = \frac{300.44\,\text{g/mol}}{150.22\,\text{g/mol}} = 2.00

  • Molecular Formula:

2×(C10H14O)=C20H28O22 \times (\text{C}_{10}\text{H}_{14}\text{O}) = \text{C}_{20}\text{H}_{28}\text{O}_2

Chemical Reactions, Balancing Equations, and Reaction Stoichiometry

Structure of Chemical Equations

Chemical equations represent chemical changes:

aA(s)+bB(aq)→cC(l)+dD(g)a\text{A}(s) + b\text{B}(aq) \rightarrow c\text{C}(l) + d\text{D}(g)

  • Reactants (Substrates): Initial substances written on the left.

  • Products: Final substances written on the right.

  • Coefficients (a,b,c,da, b, c, d): Stoichiometric integers balancing the number of atoms of each element.

  • Phase Descriptors: Solid (ss), liquid (ll), gas (gg), aqueous solution dissolved in water (aqaq).

Balancing Chemical Equations

Balancing satisfies the Law of Conservation of Mass; atoms cannot be created or destroyed.

Balancing Butane Combustion:

C4H10+O2→CO2+H2O\text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}

  1. Balance carbon: Add coefficient 44 to CO2\text{CO}_2:

C4H10+O2→4CO2+H2O\text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow 4\text{CO}_2 + \text{H}_2\text{O}

  1. Balance hydrogen: Add coefficient 55 to H2O\text{H}_2\text{O}:

C4H10+O2→4CO2+5H2O\text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow 4\text{CO}_2 + 5\text{H}_2\text{O}

  1. Balance oxygen: Total oxygen on right is (4×2)+(5×1)=13(4 \times 2) + (5 \times 1) = 13. This requires 132O2\frac{13}{2}\text{O}_2:

C4H10+132O2→4CO2+5H2O\text{C}_4\text{H}_{10} + \frac{13}{2}\text{O}_2 \rightarrow 4\text{CO}_2 + 5\text{H}_2\text{O}

  1. Clear fractional coefficients by multiplying all coefficients by 22:

2C4H10(g)+13O2(g)→8CO2(g)+10H2O(g)2\text{C}_4\text{H}_{10}(g) + 13\text{O}_2(g) \rightarrow 8\text{CO}_2(g) + 10\text{H}_2\text{O}(g)

Atom count verification:

  • Reactants: 8 C,20 H,26 O8\,\text{C}, 20\,\text{H}, 26\,\text{O}

  • Products: 8 C,20 H,26 O8\,\text{C}, 20\,\text{H}, 26\,\text{O}

Stoichiometric Mass-to-Mass Conversions

Mass A→÷MWAMoles A→×Mole RatioMoles B→×MWBMass B\text{Mass A} \xrightarrow{\div \text{MW}_A} \text{Moles A} \xrightarrow{\times \text{Mole Ratio}} \text{Moles B} \xrightarrow{\times \text{MW}_B} \text{Mass B}

Example Calculation: Calculate grams of Fe\text{Fe} produced from 1.00 g1.00\,\text{g} of Fe2O3\text{Fe}_2\text{O}_3:

2Fe2O3+3C→4Fe+3CO22\text{Fe}_2\text{O}_3 + 3\text{C} \rightarrow 4\text{Fe} + 3\text{CO}_2

Mass Fe=1.00 g Fe2O3×1 mol Fe2O3159.70 g Fe2O3×4 mol Fe2 mol Fe2O3×55.85 g Fe1 mol Fe=0.699 g Fe\text{Mass Fe} = 1.00\,\text{g } \text{Fe}_2\text{O}_3 \times \frac{1\,\text{mol } \text{Fe}_2\text{O}_3}{159.70\,\text{g } \text{Fe}_2\text{O}_3} \times \frac{4\,\text{mol Fe}}{2\,\text{mol } \text{Fe}_2\text{O}_3} \times \frac{55.85\,\text{g Fe}}{1\,\text{mol Fe}} = 0.699\,\text{g Fe}

Limiting Reactants, Theoretical Yield, and Percent Yield
  • Limiting Reactant: The reactant completely consumed first in a reaction, which sets the maximum possible yield of products.

  • Excess Reactant: The reactant that remains unreacted after the limiting reactant is completely consumed.

  • Theoretical Yield: The calculated maximum mass of product that can form from the limiting reactant.

  • Actual Yield: The mass of product actually isolated from an experiment.

  • Percent Yield:

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

Limiting Reactant Worked Example: Mixing 12.0 g C12.0\,\text{g C} with 64.0 g O264.0\,\text{g O}_2:

C(s)+O2(g)→CO2(g)\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g)

  • Moles of C\text{C}: 12.0 g12.011 g/mol=0.999 mol C\frac{12.0\,\text{g}}{12.011\,\text{g/mol}} = 0.999\,\text{mol C}

  • Moles of O2\text{O}_2: 64.0 g32.00 g/mol=2.00 mol O2\frac{64.0\,\text{g}}{32.00\,\text{g/mol}} = 2.00\,\text{mol O}_2

  • Stoichiometry is 1:11:1. Carbon yields 0.999 mol CO20.999\,\text{mol } \text{CO}_2, while oxygen could yield 2.00 mol CO22.00\,\text{mol } \text{CO}_2.

  • Carbon produces less product; therefore, C\text{C} is the limiting reactant, O2\text{O}_2 is in excess, and theoretical yield is 0.999 mol CO20.999\,\text{mol } \text{CO}_2 (44.0 g44.0\,\text{g}).

Solutions, Concentration Units, and Aqueous Behavior

Physical Nature of Solutions

A solution is a homogeneous mixture composed of a solute dispersed uniformly within a solvent:

  • Solute: The substance present in the smaller amount that is dissolved.

  • Solvent: The substance present in the larger amount that determines the physical state of the solution.

Classification of Solutions by Physical State

Solution Phase Type

Solute State

Solvent State

Real-World Example

Gas in Gas

Gas

Gas

Air (O2,N2,Ar\text{O}_2, \text{N}_2, \text{Ar})

Gas in Liquid

Gas

Liquid

Carbonated seltzer (CO2\text{CO}_2 in water)

Gas in Solid

Gas

Solid

Hydrogen gas (H2\text{H}_2) absorbed in palladium metal

Liquid in Liquid

Liquid

Liquid

Gasoline (homogeneous mixture of hydrocarbons)

Liquid in Solid

Liquid

Solid

Dental amalgam (liquid mercury dissolved in silver)

Solid in Liquid

Solid

Liquid

Seawater (solid NaCl\text{NaCl} dissolved in liquid water)

Solid in Solid

Solid

Solid

14-karat gold alloy (Au,Ag,Cu\text{Au}, \text{Ag}, \text{Cu})

Molecular Mechanics of Solvation: Like Dissolves Like

Solvation depends on the balance between solute-solute interactions, solvent-solvent interactions, and solute-solvent interactions:

  • Like Dissolves Like:

    • Polar solvents dissolve polar solutes and ionic salts via dipole-dipole, hydrogen bonding, and ion-dipole attractions.

    • Nonpolar solvents dissolve nonpolar solutes via London dispersion forces.

    • Polar and nonpolar substances (e.g., water and oil) are immiscible because nonpolar solute molecules cannot disrupt the strong hydrogen-bonding network of water.

Unique Aqueous Properties of Water
  • Molecular Structure: Water has a bent molecular geometry (∼104.5∘\sim 104.5^\circ bond angle) with polar covalent O−H\text{O}-\text{H} bonds. High electronegativity of oxygen creates large partial negative charges (δ−\delta^-) on oxygen and partial positive charges (δ+\delta^+) on hydrogen.

  • Hydrogen Bonding: Strong intermolecular forces between the oxygen lone pairs of one water molecule and the hydrogen atoms of adjacent molecules produce an extensive three-dimensional network.

Hydrogen Bonding in Liquid Water
  • Consequences of Hydrogen Bonding:

    • Unusually high melting and boiling points relative to molar mass (e.g., H2O\text{H}_2\text{O} is liquid at room temperature with a molar mass of 18.02 g/mol18.02\,\text{g/mol}, whereas heavier molecules like N2\text{N}_2, O2\text{O}_2, CO2\text{CO}_2, and Cl2\text{Cl}_2 are gases).

    • High surface tension.

    • Anomalous density behavior: Upon freezing into ice, water molecules expand into an open hexagonal crystal lattice (Ice Ih). Ice is less dense than liquid water and floats. Maximum liquid density occurs at 3.98\,^\nec.

Electrolytes and Ion Dissociation
  • Strong Electrolytes: Soluble ionic compounds and strong acids dissociate completely into mobile ions in aqueous solution, conducting electric current efficiently.

CaCl2(s)→Ca2+(aq)+2Cl−(aq)\text{CaCl}_2(s) \rightarrow \text{Ca}^{2+}(aq) + 2\text{Cl}^-(aq)

  • Weak Electrolytes: Solutes that ionize only partially in solution, existing in dynamic equilibrium between ions and intact molecules (e.g., acetic acid).

  • Non-electrolytes: Molecular solutes that dissolve without forming ions, existing entirely as neutral molecules (e.g., sucrose, ethanol).

  • Calculating Ion Concentrations: For a solution containing 0.315 mol0.315\,\text{mol} of CaCl2\text{CaCl}_2 in 2.75 L2.75\,\text{L}:

[Ca2+]=0.315 mol2.75 L=0.115 M[\text{Ca}^{2+}] = \frac{0.315\,\text{mol}}{2.75\,\text{L}} = 0.115\,\text{M}

[Cl−]=2×0.315 mol2.75 L=0.229 M[\text{Cl}^-] = \frac{2 \times 0.315\,\text{mol}}{2.75\,\text{L}} = 0.229\,\text{M}

Units of Concentration

Concentration Unit

Definition

Mathematical Expression

Weight/Weight Percent (m/m)%

Mass of solute per mass of solution

Mass of Solute (g)Total Mass of Solution (g)×100\frac{\text{Mass of Solute (g)}}{\text{Total Mass of Solution (g)}} \times 100

Weight/Volume Percent (m/v)%

Mass of solute per volume of solution

Mass of Solute (g)Volume of Solution (mL)×100\frac{\text{Mass of Solute (g)}}{\text{Volume of Solution (mL)}} \times 100

Volume/Volume Percent (v/v)%

Volume of solute per volume of solution

Volume of SoluteVolume of Solution×100\frac{\text{Volume of Solute}}{\text{Volume of Solution}} \times 100

Parts per Million (ppm)

Parts solute per million parts solution

Mass of SoluteMass of Solution×106\frac{\text{Mass of Solute}}{\text{Mass of Solution}} \times 10^6

Parts per Billion (ppb)

Parts solute per billion parts solution

Mass of SoluteMass of Solution×109\frac{\text{Mass of Solute}}{\text{Mass of Solution}} \times 10^9

Molarity (MM)

Moles of solute per liter of solution

Moles of Solute (mol)Volume of Solution (L)\frac{\text{Moles of Solute (mol)}}{\text{Volume of Solution (L)}}

Molality (mm)

Moles of solute per kilogram of solvent

Moles of Solute (mol)Mass of Solvent (kg)\frac{\text{Moles of Solute (mol)}}{\text{Mass of Solvent (kg)}}

Mole Fraction (XAX_A)

Ratio of moles of component to total moles

nAntotal\frac{n_A}{n_{\text{total}}}

Preparation of Solutions and Volumetric Calculations
  • Additivity Rules: Masses are additive (10.0 g solute+100.0 g solvent=110.0 g solution10.0\,\text{g solute} + 100.0\,\text{g solvent} = 110.0\,\text{g solution}). Volumes are not strictly additive due to differences in intermolecular packing (50.0 mL ethanol+50.0 mL water≠100.0 mL solution50.0\,\text{mL ethanol} + 50.0\,\text{mL water} \ne 100.0\,\text{mL solution}).

  • Preparing a Standard Solution from a Solid Solute:

    1. Weigh the calculated mass of solute on an analytical balance.

    2. Add solvent to a volumetric flask until approximately half full.

    3. Transfer the solute into the volumetric flask using a funnel.

    4. Rinse the weighing vessel and funnel with solvent to ensure quantitative transfer.

    5. Swirl the flask until the solute is completely dissolved.

    6. Add solvent dropwise until the bottom of the meniscus aligns with the calibration mark.

    7. Stopper the flask and invert repeatedly to ensure complete mixing.

Dilution and Serial Dilutions
  • Dilution Equation: Adding solvent preserves the total moles of solute:

M1V1=M2V2M_1 V_1 = M_2 V_2

  • M1,V1M_1, V_1: Molarity and volume of the concentrated stock solution.

  • M2,V2M_2, V_2: Molarity and volume of the diluted final solution.

    • Serial Dilution: Stepwise dilution of a substance across a series of tubes using a constant dilution factor.

  • For a 1:101:10 serial dilution, 1.0 mL1.0\,\text{mL} of stock is added to 9.0 mL9.0\,\text{mL} of solvent (total volume 10.0 mL10.0\,\text{mL}), giving a relative concentration of 0.10.1

  • Transferring 1.0 mL1.0\,\text{mL} of that mixture into 9.0 mL9.0\,\text{mL} produces a 0.010.01 dilution, followed by 0.0010.001, and so forth.

Spectrophotometry and the Beer-Lambert Law

Spectrophotometry measures the quantity of light absorbed by a dissolved solute as a function of concentration.

  • Beer-Lambert Law:

A=εclA = \varepsilon c l

  • AA: Absorbance (dimensionless quantity; formerly optical density).

  • ε\varepsilon: Molar absorptivity (molar absorption coefficient) in units of M−1cm−1\text{M}^{-1}\text{cm}^{-1}.

  • cc: Molar concentration of absorbing solute in units of M\text{M}.

  • ll: Optical path length through the sample cell (cuvette) in units of cm\text{cm} (standard cuvettes are 1.00 cm1.00\,\text{cm}).

    • Relationship Between Transmittance and Absorbance:

A=2−log⁡(%T)A = 2 - \log(\%T)

  • Transmittance (T)=II0\text{Transmittance } (T) = \frac{I}{I_0}

  • Percent Transmittance (%T)=II0×100\text{Percent Transmittance } (\%T) = \frac{I}{I_0} \times 100

    • Analytical Standard Curves: A plot of Absorbance (yy-axis) versus Concentration (xx-axis) generates a straight line through the origin (y=mx+by = mx + b) within the linear dynamic range. The slope of this line represents the sensitivity of the analytical method.