Course Lessons

Module 0 — Chemistry Math, Representations, and Learning Toolkit

Estimated independent time: 3–5 hours
Purpose: Establish the mathematical and study habits used throughout the course.

Learning targets

You can:

rearrange a one-variable equation;
work with scientific notation and powers of ten;
interpret slope, intercept, direct proportion, and inverse proportion;
perform dimensional analysis;
distinguish a number, a unit, a measured quantity, and an exact count;
use retrieval practice and error analysis to study chemistry.

Core content

A quantity combines a number and a unit. Writing 25 is incomplete if the value represents 25 milliliters, 25 grams, or 25 kelvins. Units carry meaning and behave algebraically. Dimensional analysis uses conversion factors equal to one, arranged so unwanted units cancel.

For scientific notation, write a nonzero number as a × 10^n, where 1 ≤ |a| < 10. Multiplication adds exponents; division subtracts them. Addition and subtraction require the same power of ten before combining coefficients.

To rearrange an equation, perform inverse operations on both sides. In the ideal gas equation PV = nRT, solving for temperature gives T = PV/(nR). Parentheses protect the entire numerator or denominator.

A linear graph follows y = mx + b. Slope m = Δy/Δx carries units. A direct proportionality such as V ∝ T passes through the origin when the conditions and unit scale support the relationship. An inverse proportionality such as P ∝ 1/V becomes linear when P is plotted against 1/V.

Worked example: multi-step conversion

Convert 55.0 miles per hour to meters per second.

55.0 mi/h × (1609.344 m / 1 mi) × (1 h / 3600 s) = 24.6 m/s

The mile and hour units cancel. The original value has three significant figures, so the reported result has three.

Study method

Use a four-part cycle:

Retrieve: Close the notes and write what you remember.
Solve: Attempt a mixed problem without a model answer beside you.
Diagnose: Name the first incorrect decision, not merely the final arithmetic error.
Transfer: Solve a new problem that changes the surface details.

Common wrong turns

Treating units as labels added after the calculation. Units are algebraic constraints.
Typing a long expression into a calculator without estimating. First predict the sign, scale, and unit.
Rereading instead of retrieving. Familiarity with a page is not the same as recall.
Studying one problem type in a block. Exams require choosing a method, so practice should become mixed.

Practice

Write 0.0005080 in scientific notation and state its significant figures.
Solve q = mcΔT for c.
Convert 72.0 km/h to m/s.
If y doubles when x doubles, which relationship is suggested? What additional evidence is needed?

Answers

5.080 × 10^-4; four significant figures.
c = q/(mΔT).
20.0 m/s.
Direct proportionality is suggested, but more points and evidence of a near-zero intercept are needed.

Next moduleModule 1 — Matter, Measurement, and Chemical Reasoning

Module 1 — Matter, Measurement, and Chemical Reasoning

OpenStax reading: Chapter 1.1–1.6
Central question: How do chemists turn observations into defensible quantities and classifications?

Learning targets

Classify samples as elements, compounds, homogeneous mixtures, or heterogeneous mixtures.
Distinguish solids, liquids, gases, and phase changes at macroscopic and particle levels.
Distinguish intensive from extensive properties and physical from chemical properties.
Record measurements using SI units, significant figures, and uncertainty.
Calculate density and use it as a conversion factor.
Distinguish accuracy, precision, random variation, and systematic error.

Matter and models

Matter has mass and occupies space. A pure substance has a consistent composition. An element contains one kind of atom; a compound contains two or more elements joined in fixed proportions. A mixture combines substances without fixing their proportions. A homogeneous mixture is uniform on the scale observed; a heterogeneous mixture contains distinguishable regions.

Classification depends on scale and evidence. Milk appears uniform to the unaided eye but is a colloidal dispersion. Air is usually modeled as a homogeneous gas mixture, though local droplets or particles can make an atmospheric sample heterogeneous.

In a solid, particles occupy relatively fixed positions and vibrate. In a liquid, close particles rearrange and flow. In a gas, particles are far apart relative to their size and move throughout the container. A phase change rearranges particles and energy without changing chemical identity.

Properties and change

An extensive property depends on sample size, such as mass or total volume. An intensive property does not, such as temperature or density under specified conditions. A physical property can be observed without changing identity. A chemical property describes the capacity to undergo a transformation into different substances.

Evidence such as temperature change, gas formation, precipitate formation, or color change may suggest a chemical reaction, but evidence must be interpreted. Boiling produces a gas without producing a new substance. Dissolving can cause temperature change without necessarily creating new chemical species.

Measurement and uncertainty

Every measured value is an estimate. The final reported digit is uncertain. For an analog scale, record all certain digits plus one estimated digit. Digital devices display a resolution but may still have calibration error or limited accuracy.

For multiplication and division, round the result to the smallest number of significant figures among measured inputs. For addition and subtraction, round to the least precise decimal place. Keep guard digits during intermediate work and round once at the end.

Accuracy describes closeness to an accepted or reference value. Precision describes agreement among repeated results. Random effects tend to broaden a set of results; systematic effects tend to shift results in a consistent direction.

Density

density = mass/volume, or ρ = m/V.

Density is temperature dependent and should be reported with conditions when precision matters. It can act as a conversion factor between mass and volume.

Worked example: density and displacement

A metal sample has a mass of 43.862 g. Water in a graduated cylinder rises from 21.4 mL to 26.3 mL when the sample is submerged.

V_sample = 26.3 mL − 21.4 mL = 4.9 mL

ρ = 43.862 g / 4.9 mL = 8.9514... g/mL

The subtraction limits the volume to the tenths place and therefore two significant figures. Report 9.0 g/mL. The low precision of the displacement dominates the result.

Common wrong turns

Calling every uniform sample a compound. Solutions are homogeneous mixtures.
Assuming many decimal places guarantee accuracy. Instrument calibration and method matter.
Rounding every intermediate value. Carry guard digits until the final result.
Treating 1 mL = 1 g. That equality is not general; it requires a density near 1 g/mL.

Practice

Classify brass, distilled water, oxygen gas, and granite.
Which are intensive: mass, density, boiling temperature, energy, color?
Calculate the density of 12.43 g occupying 4.1 mL.
A balance reads 10.02, 10.03, and 10.02 g for a 10.50 g standard. Describe the result.

Answers

Brass: homogeneous mixture; distilled water: compound; oxygen gas: element; granite: heterogeneous mixture.
Density, boiling temperature under specified pressure, and color.
3.0 g/mL to two significant figures.
Precise but inaccurate; a systematic bias is plausible.

Next moduleModule 2 — Atoms, Isotopes, Ions, Formulas, and Names

Module 2 — Atoms, Isotopes, Ions, Formulas, and Names

OpenStax reading: Chapter 2.1–2.7
Central question: How does a chemical formula encode particle composition and charge?

Learning targets

Describe the evidence-based nuclear model of the atom.
Determine protons, neutrons, electrons, and charge from isotope notation.
Calculate average atomic mass from isotopic abundances.
Predict common monatomic ion charges using periodic position.
Distinguish molecular and ionic compounds.
Write and name common inorganic formulas.

Atomic structure

Atoms contain a small, positively charged nucleus with protons and neutrons, surrounded by electrons. Proton charge is +1, electron charge is −1, and a neutron is electrically neutral on this relative scale. Atomic number Z equals the number of protons and defines the element. Mass number A equals protons plus neutrons for a particular nuclide.

For ^A_ZX^charge:

protons = Z;
neutrons = A − Z;
electrons = Z − numerical charge, with signs handled algebraically.

Thus ^56_26Fe^3+ contains 26 protons, 30 neutrons, and 23 electrons.

Isotopes are atoms of the same element with different neutron counts. They have the same atomic number but different mass numbers. The decimal atomic mass on a periodic table is a weighted mean for naturally occurring isotopic composition, not the mass number of one atom.

Weighted average example

An element has two isotopes: 10.0129 u at 19.91% and 11.0093 u at 80.09%.

average = (10.0129)(0.1991) + (11.0093)(0.8009) = 10.811 u

Abundances must be expressed as fractions and sum to approximately one.

Ions and periodic patterns

Atoms become ions by gaining or losing electrons; ordinary chemical ion formation does not change the nucleus. Main-group metals often lose electrons to reach a stable valence configuration. Nonmetals often gain electrons. Common patterns include Group 1 +1, Group 2 +2, aluminum +3, Group 17 −1, Group 16 −2, and Group 15 −3, with important exceptions. Transition-metal charge must often be given by a Roman numeral in the name.

Ionic and molecular formulas

An ionic compound is electrically neutral overall. Formula subscripts give the lowest whole-number ratio of ions. Aluminum oxide combines Al^3+ and O^2−; the least common charge magnitude is six, so the formula is Al2O3.

Molecular compounds contain discrete covalently bonded molecules. Prefixes indicate atom counts: mono-, di-, tri-, tetra-, penta-, hexa-, hepta-, octa-, nona-, deca-. The first element usually omits mono-. N2O4 is dinitrogen tetroxide.

Naming essentials

Fixed-charge ionic: cation name + monatomic anion root + -ide; CaCl2, calcium chloride.
Variable-charge ionic: cation name + Roman numeral + anion; FeCl3, iron(III) chloride.
Polyatomic ionic: preserve polyatomic ion name; Na2SO4, sodium sulfate.
Binary molecular: use prefixes; PCl3, phosphorus trichloride.
Binary acid in water: hydro- + root + -ic acid; HCl(aq), hydrochloric acid.
Oxyacids: -ate → -ic acid, -ite → -ous acid; nitrate → nitric acid, nitrite → nitrous acid.

Essential polyatomic ions

Name	Formula	Name	Formula
ammonium	`NH4+`	hydroxide	`OH−`
nitrate	`NO3−`	nitrite	`NO2−`
sulfate	`SO4^2−`	sulfite	`SO3^2−`
hydrogen sulfate	`HSO4−`	carbonate	`CO3^2−`
hydrogen carbonate	`HCO3−`	phosphate	`PO4^3−`
hydrogen phosphate	`HPO4^2−`	dihydrogen phosphate	`H2PO4−`
acetate	`C2H3O2−`	cyanide	`CN−`
permanganate	`MnO4−`	chromate	`CrO4^2−`
dichromate	`Cr2O7^2−`	peroxide	`O2^2−`

Common wrong turns

Changing subscripts to balance an equation. Subscripts define identity; coefficients count entities.
Using a Roman numeral to count atoms. It reports cation charge.
Treating atomic mass as a whole-number proton-plus-neutron count. The listed value is generally a weighted average.
Dropping parentheses around repeated polyatomic ions: calcium nitrate is Ca(NO3)2, not CaNO32.

Practice

Find protons, neutrons, and electrons in ^79_35Br−.
Write the formula for iron(III) sulfate.
Name Cu2O, N2O5, and NH4Cl.
Explain why magnesium chloride is MgCl2 rather than Mg2Cl.

Answers

35 protons, 44 neutrons, 36 electrons.
Fe2(SO4)3.
Copper(I) oxide; dinitrogen pentoxide; ammonium chloride.
Neutrality requires one Mg^2+ for two Cl−; the formula is the lowest whole-number ratio.

Next moduleModule 3 — The Mole and Chemical Composition

Module 3 — The Mole and Chemical Composition

OpenStax reading: Chapter 3.1–3.5
Central question: How do chemists count microscopic particles by measuring macroscopic quantities?

Learning targets

Use the mole as an amount-of-substance unit.
Convert among particles, moles, and mass.
Calculate formula mass, molar mass, and percent composition.
Determine empirical and molecular formulas from composition data.
Connect subscripts in a formula to particle and mole ratios.

The mole

One mole contains exactly 6.02214076 × 10^23 specified elementary entities. Always name the entity: atoms, molecules, formula units, ions, electrons, or another defined group. The mole connects count to measurable mass through molar mass.

number of entities = amount in mol × N_A

mass = amount in mol × molar mass

The periodic-table atomic mass in atomic mass units per atom has the same numerical value as molar mass in grams per mole, within the intended chemical convention.

Conversion map

particles ↔ moles ↔ mass

Use Avogadro's constant between particles and moles. Use molar mass between moles and grams. Do not jump directly from particles to grams without showing the mole bridge.

Worked example: molecules to mass

What mass corresponds to 3.50 × 10^22 molecules of carbon dioxide?

3.50 × 10^22 molecules × (1 mol / 6.02214076 × 10^23 molecules) × (44.01 g / 1 mol) = 2.56 g CO2

Percent composition

For an element in a compound:

mass percent = (mass of that element in 1 mol compound / molar mass compound) × 100%

For H2O, hydrogen contributes approximately 2(1.008) = 2.016 g per mole and the molar mass is approximately 18.015 g/mol, giving about 11.19% H by mass.

Empirical and molecular formulas

An empirical formula gives the lowest whole-number atom ratio. A molecular formula gives the actual atom counts and is a whole-number multiple of the empirical formula.

From percent composition:

Assume a 100 g sample, converting percentages to grams.
Convert each element's grams to moles.
Divide every mole amount by the smallest.
Multiply all ratios by a small integer if necessary to obtain whole numbers.
Check by calculating the implied percentages.

If the empirical formula mass is M_emp and molecular molar mass is M_mol, then n = M_mol/M_emp, rounded only when the data justify a near-integer ratio.

Worked example: empirical formula

A compound is 40.00% C, 6.71% H, and 53.29% O.

For 100 g:

C: 40.00 g / 12.011 g mol−1 = 3.330 mol
H: 6.71 g / 1.008 g mol−1 = 6.66 mol
O: 53.29 g / 15.999 g mol−1 = 3.331 mol

Divide by about 3.330 to obtain 1 : 2 : 1, so the empirical formula is CH2O.

Common wrong turns

Using Avogadro's constant as a molar mass. It connects entity count and moles.
Counting atoms incorrectly in parentheses. One mole of Ca(NO3)2 contains two moles of N atoms and six moles of O atoms.
Rounding empirical ratios too early. A ratio near 1.50 suggests multiplying all ratios by two.
Reporting percent composition without checking that values total approximately 100%.

Practice

Find the number of formula units in 0.250 mol NaCl.
Find the moles in 9.00 g H2O.
Calculate the molar mass of Ca(NO3)2.
A compound has empirical formula NO2 and molar mass about 92.0 g/mol. Find its molecular formula.

Answers

1.51 × 10^23 formula units.
0.500 mol to three significant figures.
Approximately 164.10 g/mol.
Empirical mass ≈ 46.0 g/mol; multiplier 2; N2O4.

Next moduleModule 4 — Chemical Equations and Reaction Patterns

Module 4 — Chemical Equations and Reaction Patterns

OpenStax reading: Chapter 4.1–4.2 and 4.4
Central question: What information does a balanced chemical equation preserve and communicate?

Learning targets

Write and balance chemical equations.
Interpret coefficients as particle and mole ratios.
Include meaningful physical-state symbols.
Classify common reaction patterns.
Distinguish equation balancing from reaction prediction.
Assign oxidation numbers and identify oxidation and reduction.

Equations as conservation statements

A chemical equation represents rearrangement of atoms. It must conserve every element and, for ionic equations, net charge. Coefficients multiply whole formulas; subscripts remain unchanged.

For methane combustion:

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

The coefficients mean one molecule or mole of methane reacts with two of oxygen to produce one of carbon dioxide and two of water, under the represented conditions. They do not claim complete conversion, a particular rate, or a mechanism.

Balancing method

Write correct reactant and product formulas.
Inventory atoms on both sides.
Adjust coefficients, beginning with the most structurally complex species.
Leave free elements and often H/O until later.
Reduce to the smallest whole-number set.
Verify atoms, charge when relevant, and physical plausibility.

Common reaction patterns

Combination: multiple reactants form one product pattern.
Decomposition: one reactant produces multiple substances.
Combustion: a substance reacts with oxygen; complete hydrocarbon combustion yields carbon dioxide and water.
Precipitation: aqueous ions form a low-solubility solid.
Acid–base neutralization: acid and base species transfer protons; a common strong case forms water.
Gas evolution: reaction forms a gas such as CO2.
Oxidation–reduction: electrons are transferred formally, reflected by changes in oxidation number.

Pattern recognition helps predict possibilities but does not replace chemical data, conditions, or observation.

Oxidation numbers

Useful rules:

A free element has oxidation number 0.
A monatomic ion has its ionic charge.
Group 1 is usually +1; Group 2 usually +2.
F is −1 in compounds.
O is usually −2, with exceptions such as peroxides.
H is usually +1 with nonmetals and −1 in many metal hydrides.
The sum equals the species' total charge.

Oxidation is an increase in oxidation number; reduction is a decrease. In Zn(s) + Cu^2+(aq) → Zn^2+(aq) + Cu(s), zinc is oxidized and copper ion is reduced.

Worked example: balancing and classifying a combustion reaction

Balance the combustion of butane, C4H10(g) + O2(g) → CO2(g) + H2O(g), and classify the reaction.

Inventory the unbalanced equation: left side has 4 C, 10 H, 2 O; right side has 1 C, 2 H, 3 O (one from CO2, two from H2O) per formula unit as written.
Balance carbon first, since it appears in only one product: 4 carbons on the left means 4 CO2 on the right.

C4H10 + O2 → 4 CO2 + H2O

Balance hydrogen next: 10 hydrogens on the left means 5 H2O on the right (5 × 2 = 10).

C4H10 + O2 → 4 CO2 + 5 H2O

Balance oxygen last, since it's split across two products and easiest to adjust once everything else is fixed. The right side now has 4(2) + 5(1) = 13 oxygen atoms, so the left side needs 13 oxygen atoms — but O2 only comes in pairs, giving a fractional coefficient of 13/2:

C4H10 + 13/2 O2 → 4 CO2 + 5 H2O

Clear the fraction by multiplying every coefficient by 2, since a chemical equation's coefficients must be the smallest set of whole numbers:

2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g)

Verify: left side has 8 C, 20 H, 26 O; right side has 8 C, 20 H, 16 + 10 = 26 O. Every element balances.

This is a combustion reaction (a hydrocarbon reacting with oxygen to form carbon dioxide and water) and also a redox reaction: carbon's oxidation number rises from −5/2 (averaged across the molecule) to +4 in CO2 — oxidized — while oxygen's falls from 0 in O2 to −2 in both products — reduced. A fractional intermediate coefficient during balancing is normal and expected; only the final answer needs whole numbers.

Common wrong turns

Balancing oxygen by changing O2 into O3. That changes the reactant.
Believing a balanced equation proves a reaction occurs. It only conserves composition for the proposed process.
Assigning oxidation numbers as actual localized charges in every covalent bond. They are a formal accounting system.
Calling any oxygen-containing process combustion.

Practice

Balance Al + O2 → Al2O3.
Balance and classify C3H8 + O2 → CO2 + H2O.
Find the oxidation number of sulfur in SO4^2−.
Identify oxidized and reduced species in 2 Mg + O2 → 2 MgO.

Answers

4 Al + 3 O2 → 2 Al2O3.
C3H8 + 5 O2 → 3 CO2 + 4 H2O; combustion and redox.
+6.
Mg: 0 to +2, oxidized; O: 0 to −2, reduced.

Next moduleModule 5 — Stoichiometry, Limiting Reactants, and Yield

Module 5 — Stoichiometry, Limiting Reactants, and Yield

OpenStax reading: Chapter 4.4 and Chapter 3 stoichiometry review
Central question: How does a balanced equation constrain the quantities consumed and produced?

Learning targets

Use coefficients as mole conversion factors.
Solve mass, particle, and volume stoichiometry problems.
Identify limiting and excess reactants.
Calculate theoretical yield, actual yield, and percent yield.
Explain assumptions and sources of disagreement between theoretical and experimental yield.

Stoichiometric path

For most mass problems:

given unit → moles of given species → moles of wanted species → wanted unit

The balanced-equation coefficient ratio works only between mole amounts, not directly between grams unless molar masses happen to make a special coincidence.

Worked example: mass-to-mass

For 2 H2 + O2 → 2 H2O, what mass of water can form from 5.00 g H2 with excess oxygen?

5.00 g H2 × (1 mol H2 / 2.016 g H2) × (2 mol H2O / 2 mol H2) × (18.015 g H2O / 1 mol H2O) = 44.7 g H2O

Limiting reactant

When more than one reactant quantity is supplied, each reactant competes to determine the maximum product. The limiting reactant is consumed first under the assumed reaction and therefore limits theoretical yield.

Robust method:

Calculate the possible amount of the same product from each reactant.
The smaller product amount identifies the limiting reactant.
Use only the limiting reactant for theoretical yield.
To find excess remaining, calculate how much excess reactant is consumed and subtract from its initial amount.

Worked example: limiting reactant

N2 + 3 H2 → 2 NH3

Suppose 2.00 mol N2 and 5.00 mol H2 are available.

From N2: 2.00 mol N2 × 2/1 = 4.00 mol NH3
From H2: 5.00 mol H2 × 2/3 = 3.33 mol NH3

Hydrogen is limiting; theoretical yield is 3.33 mol ammonia. Nitrogen consumed is 5.00/3 = 1.67 mol, leaving 0.33 mol N2.

Yield

percent yield = (actual yield / theoretical yield) × 100%

Actual yield is measured. Theoretical yield is calculated under an idealized complete-reaction model. Loss during transfer, incomplete reaction, side reactions, impurities, wet product, and measurement bias can affect yield. A yield above 100% usually signals wet or contaminated product, an incorrect assumed formula, or measurement/calculation error.

Common wrong turns

Comparing initial grams to find the limiting reactant. Reaction ratios operate in moles.
Using the excess reactant to compute theoretical yield.
Subtracting unlike units or species.
Treating theoretical yield as a prediction of what an imperfect lab must obtain.

Practice

How many moles of O2 are required to react with 4.50 mol Al using 4 Al + 3 O2 → 2 Al2O3?
For 2 Na + Cl2 → 2 NaCl, identify the limiting reactant when 3.00 mol Na and 2.00 mol Cl2 are mixed.
If theoretical yield is 12.50 g and actual yield is 10.82 g, find percent yield.
Give two reasons an isolated solid could produce a calculated yield above 100%.

Answers

3.38 mol O2.
Na is limiting; it can form 3.00 mol NaCl, while Cl2 could form 4.00 mol.
86.56%, normally reported as 86.6% depending on measurement precision.
Retained solvent/water, contamination, wrong product formula, balance bias, or calculation error.

Next moduleModule 6 — Solutions and Aqueous Reactions

Module 6 — Solutions and Aqueous Reactions

OpenStax reading: Chapter 3.4 and 4.1–4.3
Central question: How do dissolved particles determine conductivity, reaction, and quantitative composition?

Learning targets

Describe dissolution at the particle level.
Distinguish strong, weak, and nonelectrolytes.
Calculate molarity and perform dilution calculations.
Use solubility rules to predict selected precipitates.
Write molecular, complete ionic, and net ionic equations.
Perform solution stoichiometry.

Solutions and concentration

A solution is a homogeneous mixture. The solvent is the component that establishes the medium; solutes are dispersed within it. Dissolution requires compatible interactions, but “like dissolves like” is only a starting heuristic, not a complete thermodynamic theory.

Molarity is moles of solute per liter of solution:

M = n/V

Volume must be in liters. Molarity changes with temperature if solution volume changes.

Dilution

When solute amount remains constant while solvent is added:

M1V1 = M2V2

This relationship is a conservation statement for solute moles, not a universal mixing formula. It does not apply when a reaction consumes solute or when combining arbitrary solutions without tracking total amount and volume.

Worked example: solution preparation

To prepare 250.0 mL of 0.1500 M NaCl:

n = MV = (0.1500 mol/L)(0.2500 L) = 0.03750 mol

m = nM_molar = (0.03750 mol)(58.44 g/mol) = 2.192 g

Dissolve the measured solute in less than the final volume, transfer quantitatively to a 250.0 mL volumetric flask, then dilute to the mark and mix. Do not add 250.0 mL of water.

Electrolytes and ionic equations

Strong electrolytes produce high concentrations of mobile ions in water. Soluble ionic compounds and strong acids/bases are common examples. Weak electrolytes ionize only partly. Nonelectrolytes may dissolve as neutral molecules.

For AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq):

Complete ionic:

Ag+(aq) + NO3−(aq) + Na+(aq) + Cl−(aq) → AgCl(s) + Na+(aq) + NO3−(aq)

Cancel unchanged spectator ions:

Ag+(aq) + Cl−(aq) → AgCl(s)

Do not split solids, liquids, gases, weak electrolytes, or molecular nonelectrolytes into aqueous ions.

Compact solubility guide

Usually soluble:

Group 1 and NH4+ salts;
nitrates and acetates;
most chlorides, bromides, and iodides, except important salts of Ag+, Pb^2+, and Hg2^2+;
most sulfates, with important exceptions including Ba^2+, Sr^2+, and Pb^2+, and limited solubility for some others.

Often low-solubility:

carbonates, phosphates, sulfides, and hydroxides, except Group 1 and ammonium; Group 2 hydroxides vary.

Use an instructor-approved table because condensed rules have exceptions.

Common wrong turns

Using milliliters directly in M = mol/L without conversion.
Splitting every compound into ions.
Canceling species that change phase or form weak products.
Confusing dilution with adding concentrations.

Practice

Find the molarity of 0.225 mol solute in 450.0 mL solution.
What volume of 2.00 M stock makes 100.0 mL of 0.250 M solution?
Predict whether mixing aqueous barium chloride and sodium sulfate forms a precipitate; write the net ionic equation.
How many moles of chloride ions are in 0.500 L of 0.200 M CaCl2, assuming complete dissociation?

Answers

0.500 M.
12.5 mL stock, then dilute to 100.0 mL total.
Yes; Ba^2+(aq) + SO4^2−(aq) → BaSO4(s).
0.200 mol Cl−.

Next moduleModule 7 — Acid–Base and Redox Stoichiometry

Module 7 — Acid–Base and Redox Stoichiometry

OpenStax reading: Chapter 4.2–4.4
Central question: How can proton and electron accounting make reaction quantities predictable?

Learning targets

Recognize acids and bases in introductory aqueous reactions.
Write net ionic equations for strong acid–strong base neutralization.
Use titration stoichiometry without relying blindly on M1V1 = M2V2.
Assign oxidation numbers and identify oxidizing and reducing agents.
Balance simple redox reactions by inspection when appropriate.

Acid–base reactions

At this course stage, an acid can be treated as a proton donor and a base as a proton acceptor. Strong acids and bases are represented as dissociated in aqueous ionic equations. For strong monoprotic acid and strong hydroxide base:

H+(aq) + OH−(aq) → H2O(l)

The broader Brønsted–Lowry model, weak equilibria, buffers, and pH calculations go beyond this module's scope — see the glossary for an orientation to each, and the readings and sources page for full quantitative treatment.

Titration logic

A titration uses a solution of known concentration to determine an unknown amount through reaction stoichiometry. The equivalence point is the stoichiometric point; an indicator endpoint is an observed signal intended to approximate it.

General method:

Write the balanced reaction.
Convert titrant volume and molarity to moles.
Apply the mole ratio.
Convert wanted moles to concentration, mass, or another requested quantity.

M_aV_a = M_bV_b works only for a 1:1 stoichiometric relationship. The balanced equation is the general rule.

Worked example: titration

25.00 mL of an HCl solution requires 18.64 mL of 0.1050 M NaOH for equivalence.

n_NaOH = (0.1050 mol/L)(0.01864 L) = 0.001957 mol

The reaction is 1:1, so n_HCl = 0.001957 mol.

M_HCl = 0.001957 mol / 0.02500 L = 0.07829 M

Redox agents

The species oxidized causes another species to be reduced and is the reducing agent. The species reduced is the oxidizing agent. These labels describe function in a specific reaction.

For Zn + Cu^2+ → Zn^2+ + Cu, Zn loses electron density formally and is the reducing agent; Cu^2+ gains electrons and is the oxidizing agent.

Common wrong turns

Assuming neutral pH and equivalence point are always identical. That holds for a strong acid–strong base case at ordinary conditions, not every titration.
Applying M1V1 = M2V2 without the reaction coefficients.
Saying the oxidizing agent is oxidized. The oxidizing agent is reduced.
Confusing indicator color persistence with greater measurement accuracy.

Practice

Write the net ionic equation for nitric acid and potassium hydroxide.
20.00 mL of 0.1500 M H2SO4 is fully neutralized by NaOH. How many moles NaOH are required?
In 2 Al + 3 Cu^2+ → 2 Al^3+ + 3 Cu, identify the oxidizing agent.
Explain the distinction between endpoint and equivalence point.

Answers

H+(aq) + OH−(aq) → H2O(l).
0.006000 mol NaOH, because one mole sulfuric acid supplies two stoichiometric acid equivalents in the represented full neutralization.
Cu^2+.
Equivalence is the calculated stoichiometric condition; endpoint is the experimental signal used to estimate it.

Next moduleModule 8 — Energy, Heat, Work, and Calorimetry

Module 8 — Energy, Heat, Work, and Calorimetry

OpenStax reading: Chapter 5.1–5.2
Central question: How can temperature measurements reveal energy transferred during a process?

Learning targets

Define system, surroundings, heat, work, temperature, and internal energy.
Apply the first law of thermodynamics with a stated sign convention.
Distinguish endothermic and exothermic processes.
Use q = mcΔT and calorimeter energy balances.
Interpret heating and cooling at macroscopic and particle levels.
Evaluate assumptions in calorimetry.

Energy language

Energy is the capacity to do work or transfer heat. A system is the part of the universe selected for study; everything else is the surroundings. Heat q is energy transferred because of a temperature difference. Work w is another mode of energy transfer, such as expansion against pressure. Temperature reflects the distribution of particle kinetic energy, not the total energy contained in a sample.

Using the common chemistry sign convention:

ΔE = q + w

q > 0: heat enters the system.
q < 0: heat leaves the system.
w > 0: work is done on the system.
w < 0: the system does work on the surroundings.

A process is endothermic when the system absorbs heat under the stated conditions and exothermic when the system releases heat. Always identify the system.

Temperature change and heat capacity

For a temperature interval without a phase change:

q = mcΔT

where m is mass, c is specific heat capacity, and ΔT = T_final − T_initial. A negative ΔT produces negative q for the object described: it lost heat.

Heat capacity C describes an entire object: q = CΔT. Specific heat capacity is per unit mass; molar heat capacity is per mole.

Worked example: warming a sample

How much heat is required to warm 125.0 g of water from 21.5 °C to 36.8 °C, using c = 4.184 J g−1 °C−1?

ΔT = 36.8 − 21.5 = 15.3 °C

q = (125.0 g)(4.184 J g−1 °C−1)(15.3 °C) = 8.00 × 10^3 J = 8.00 kJ

Coffee-cup calorimetry

In an ideal insulated calorimeter:

q_process + q_solution + q_calorimeter = 0

If calorimeter heat capacity is neglected, q_process ≈ −q_solution. This is an approximation. Heat exchange with the cup, probe, air, and delayed measurement can bias results.

Particle-level interpretation

Temperature changes when the distribution of kinetic energies changes. Potential energy also matters. Breaking an attractive interaction requires energy input; forming a stronger attractive arrangement releases energy relative to the separated state. The phrase “bonds store energy” can mislead if it suggests bond breaking releases energy. Breaking a bond requires energy; net reaction energy depends on both bonds broken and bonds formed.

Common wrong turns

Saying heat is “contained” in a body. Heat is energy in transfer.
Assigning endothermic or exothermic without defining the system.
Using T_initial − T_final in one problem and reversing it in another. Use ΔT = final − initial consistently.
Assuming the calorimeter absorbs no energy without evaluating the approximation.
Claiming cold flows. Energy transfers from higher to lower temperature spontaneously.

Practice

A system releases 75 J of heat and 20 J of work is done on it. Find ΔE.
Calculate the heat lost by 50.0 g copper cooling by 12.0 °C if c = 0.385 J g−1 °C−1.
In a coffee-cup reaction, solution temperature rises. What is the sign of q_solution and of q_reaction under the ideal approximation?
Why does equal heat added to equal masses of water and copper produce different temperature changes?

Answers

q = −75 J, w = +20 J; ΔE = −55 J.
q = (50.0)(0.385)(−12.0) = −231 J.
q_solution > 0; q_reaction < 0.
Their specific heat capacities differ; water requires more energy per gram per degree.

Next moduleModule 9 — Enthalpy and Thermochemical Equations

Module 9 — Enthalpy and Thermochemical Equations

OpenStax reading: Chapter 5.2–5.3
Central question: How can chemical equations be treated as algebraic statements about energy?

Learning targets

Interpret enthalpy change at constant pressure.
Scale and reverse thermochemical equations.
Apply Hess's law.
Calculate reaction enthalpy from standard enthalpies of formation.
Distinguish state functions from path-dependent transfers.
Connect energy diagrams to endothermic and exothermic processes.

Enthalpy

Enthalpy is defined as H = E + PV. At constant pressure, under conditions where only pressure-volume work is significant, the heat transferred equals enthalpy change: q_p = ΔH.

ΔH < 0 describes an exothermic process for the chosen system; products lie lower in enthalpy than reactants. ΔH > 0 describes an endothermic process.

A thermochemical equation includes balanced coefficients, physical states, and an enthalpy change for the reaction as written. If all coefficients are multiplied by a factor, ΔH is multiplied by the same factor. Reversing the equation reverses the sign of ΔH.

Hess's law

Enthalpy is a state function: its change depends on initial and final states, not the conceptual route. If equations add to a target equation, their enthalpy changes add as well.

Procedure:

Write the target equation.
Reverse and scale supplied equations so intermediates cancel.
Add equations and verify every species and state.
Add the correspondingly reversed and scaled ΔH values.

Standard enthalpy of reaction

Using standard enthalpies of formation:

ΔH°_rxn = ΣνΔH°_f(products) − ΣνΔH°_f(reactants)

The standard enthalpy of formation of an element in its standard state is zero by definition. Coefficients ν multiply formation values. Physical state matters: H2O(l) and H2O(g) have different formation enthalpies.

Worked example

For CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l), use approximate values:

ΔH°f[CH4(g)] = −74.8 kJ/mol
ΔH°f[CO2(g)] = −393.5 kJ/mol
ΔH°f[H2O(l)] = −285.8 kJ/mol
ΔH°f[O2(g)] = 0

ΔH°rxn = [−393.5 + 2(−285.8)] − [−74.8 + 2(0)] = −890.3 kJ

This value applies per stoichiometric reaction as written.

Energy diagrams

An energy diagram should distinguish the reactant-to-product enthalpy difference from the activation barrier. A catalyst lowers the activation pathway but does not change initial and final states and therefore does not change ΔH.

Common wrong turns

Forgetting coefficients in formation-enthalpy calculations.
Using reactants minus products instead of products minus reactants.
Ignoring physical states.
Believing a large negative ΔH guarantees a rapid reaction. Thermodynamics and rate are different questions.

Practice

What happens to ΔH when a reaction is reversed?
If 2 A → B has ΔH = +40 kJ, what is ΔH for B → 2 A?
Why is ΔH°f[O2(g)] = 0 but ΔH°f[O3(g)] is not zero?
Does a catalyst alter reaction enthalpy? Explain.

Answers

Its sign reverses.
−40 kJ.
O2(g) is oxygen's standard state under the reference conditions; ozone is not.
No. A catalyst changes the pathway and activation energy, not endpoint state values.

Next moduleModule 10 — Light, Quantization, and the Atomic Model

Module 10 — Light, Quantization, and the Atomic Model

OpenStax reading: Chapter 6.1–6.3
Central question: Why do atoms absorb and emit only particular energies?

Learning targets

Relate wavelength, frequency, and speed of electromagnetic radiation.
Calculate photon energy.
Interpret atomic line spectra as evidence of quantized energy differences.
Explain major limits of the Bohr model.
Describe orbitals as probability distributions rather than paths.
Identify quantum numbers and orbital capacities.

Electromagnetic radiation

In vacuum:

c = λν

where c = 2.99792458 × 10^8 m/s, λ is wavelength, and ν is frequency. Wavelength and frequency are inversely related.

Planck's relation connects radiation frequency to photon energy:

E = hν = hc/λ

where h = 6.62607015 × 10^-34 J s. Higher frequency and shorter wavelength mean greater photon energy.

Worked example: photon energy

Find the energy of a photon with wavelength 486 nm.

Convert: 486 nm = 4.86 × 10^-7 m.

E = hc/λ = [(6.62607015 × 10^-34 J s)(2.99792458 × 10^8 m/s)]/(4.86 × 10^-7 m)

E = 4.09 × 10^-19 J per photon

Per mole of photons, multiply by Avogadro's constant: approximately 246 kJ/mol.

Spectra and quantization

An excited atom can emit a photon when it moves to a lower-energy state. The photon energy equals the energy difference: ΔE = hν. Because allowed atomic energies are quantized, isolated atoms produce discrete line spectra rather than a continuous range.

The Bohr model successfully represents hydrogen energy levels but does not adequately describe multi-electron atoms or electron behavior as classical circular paths. In quantum mechanics, an orbital is associated with a wavefunction and describes a probability distribution for an electron in an atom. It is not a miniature planetary orbit.

Quantum numbers

Principal n = 1, 2, 3, ...: shell and energy/size trend.
Angular momentum l = 0 to n−1: subshell; 0=s, 1=p, 2=d, 3=f.
Magnetic m_l = −l ... +l: orbital orientation label.
Spin m_s = +1/2 or −1/2.

A subshell contains 2l + 1 orbitals. Each orbital can hold at most two electrons with opposite spins. Thus s, p, d, and f subshells hold 2, 6, 10, and 14 electrons.

Common wrong turns

Confusing wavelength with frequency.
Leaving nanometers unconverted when using c in meters per second.
Drawing quantum electrons as particles on known circular routes.
Saying an emitted photon contains the energy of an orbital rather than the difference between states.
Treating orbital drawings as hard surfaces.

Practice

Which has greater frequency: 400 nm or 700 nm light?
Calculate the frequency of 600. nm radiation.
How many orbitals are in a d subshell, and how many electrons can it hold?
Why does hydrogen show separate spectral lines?

Answers

400 nm.
5.00 × 10^14 s−1.
Five orbitals; ten electrons.
Only certain electronic energy states are allowed, so transitions release photons with specific energy differences.

Next moduleModule 11 — Electron Configurations and Periodic Trends

Module 11 — Electron Configurations and Periodic Trends

OpenStax reading: Chapter 6.4–6.5
Central question: How does electron arrangement produce recurring chemical patterns?

Learning targets

Write ground-state electron configurations and orbital diagrams.
Apply the Aufbau ordering, Pauli exclusion principle, and Hund's rule.
Identify valence and core electrons for main-group elements.
Write selected monatomic ion configurations.
Explain trends in atomic/ionic radius and ionization energy.
Use effective nuclear charge and shielding qualitatively.

Building configurations

Three organizing principles:

Pauli exclusion: no two electrons in an atom have the same four quantum numbers; an orbital holds at most two with opposite spins.
Hund's rule: degenerate orbitals are singly occupied with parallel spins before pairing in the ground-state model.
Aufbau pattern: electrons occupy available lower-energy orbitals before higher-energy orbitals, recognizing known exceptions for some atoms.

A common filling sequence is:

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p

Examples:

O: 1s2 2s2 2p4
Na: [Ne]3s1
Cl: [Ne]3s2 3p5
Ca: [Ar]4s2

When transition metals form cations, electrons are generally removed from the highest principal shell first. Iron is [Ar]4s2 3d6; Fe^2+ is represented as [Ar]3d6.

Periodic trends

Effective nuclear charge is the net attractive influence experienced by an electron after accounting qualitatively for shielding and electron-electron interactions.

Across a period, nuclear charge increases while shielding changes less dramatically for valence electrons, so atomic radius generally decreases and first ionization energy generally increases.
Down a group, electrons occupy shells with larger n and greater distance/shielding, so atomic radius generally increases and ionization energy decreases.
Cations are smaller than their neutral atoms after electron loss and reduced repulsion/possibly shell removal.
Anions are larger than their neutral atoms because added electrons increase repulsion within the valence shell.
In an isoelectronic series, more protons generally produce a smaller radius.

Trends include exceptions. Explanations should use configurations and interactions rather than treating arrows on a periodic table as causes.

Worked comparison

Order O^2−, F−, Ne, Na+, and Mg^2+ by decreasing radius.

All contain ten electrons. Nuclear charge increases from O (Z=8) to Mg (Z=12), so attraction becomes stronger and radius decreases:

O^2− > F− > Ne > Na+ > Mg^2+

Common wrong turns

Removing 3d electrons before 4s when forming many transition-metal ions.
Pairing p electrons before singly occupying all three p orbitals.
Claiming atoms “want” full shells as a causal explanation.
Treating periodic trends as exceptionless.
Confusing ionic charge with oxidation number in every setting.

Practice

Write the ground-state configuration of sulfur.
How many unpaired electrons appear in a ground-state nitrogen atom?
Which is larger, Na or Na+? Explain.
Which has the larger first ionization energy, Mg or Cl? Give a qualitative reason.

Answers

[Ne]3s2 3p4.
Three.
Na; forming Na+ removes the outer 3s electron and the occupied valence shell.
Cl generally has the larger value because its valence electron experiences greater effective nuclear attraction across the same period.

Next moduleModule 12 — Ionic and Covalent Bonding; Lewis Structures

Module 12 — Ionic and Covalent Bonding; Lewis Structures

OpenStax reading: Chapter 7.1–7.4
Central question: How do electrostatic interactions and electron distribution stabilize chemical structures?

Learning targets

Compare ionic, covalent, and metallic bonding models.
Relate lattice energy qualitatively to ionic charge and size.
Use electronegativity and bond dipoles.
Construct Lewis structures for common main-group species.
Calculate formal charges and use them to compare structures.
Draw resonance contributors and describe the resonance hybrid.

Bonding models

All chemical bonding ultimately involves electrostatic interactions among nuclei and electrons, but different models emphasize different organization.

Ionic bonding: extended attraction among cations and anions in a lattice; formulas give ratios, not isolated molecules.
Covalent bonding: electron density is shared between nuclei; localized bond models are useful for many molecules and polyatomic ions.
Metallic bonding: valence electrons are delocalized across many metal centers, helping explain conductivity and malleability.

Lattice energy magnitude generally increases with greater ionic charge and smaller ion separation. This helps compare related ionic solids but does not alone determine every physical property.

Electronegativity and polarity

Electronegativity describes an atom's attraction for shared electron density in a bond. A difference produces a bond dipole, conventionally pointing toward the more electronegative atom. Bond polarity is continuous; labels such as nonpolar covalent, polar covalent, and ionic are model categories rather than sharp natural boundaries.

Lewis structure procedure

Count total valence electrons; add for negative charge and subtract for positive charge.
Choose a plausible skeleton. Hydrogen is terminal; less electronegative atoms often occupy the center.
Connect atoms with single bonds.
Complete terminal atom octets, then place remaining electrons on the center.
If the center lacks an octet, form multiple bonds when chemically reasonable.
Calculate formal charges.
Evaluate resonance, octet exceptions, and the overall charge.

Formal charge:

FC = valence electrons − nonbonding electrons − 1/2(bonding electrons)

A preferred contributor often minimizes formal-charge magnitude and places negative formal charge on more electronegative atoms, but experimental evidence and known bonding patterns matter.

Resonance

When more than one valid Lewis contributor differs only in electron placement, the actual species is a resonance hybrid. Atoms do not rapidly switch between contributors. Equivalent resonance often makes bonds equivalent and intermediate in order.

For nitrate, three contributors place the N=O bond in different positions. The three N–O bonds are equivalent in the resonance hybrid.

Octet limitations

Hydrogen typically has a duet.
Electron-deficient species such as BF3 can have fewer than eight electrons at the central atom.
Odd-electron species exist.
Period 3 and heavier central atoms are often represented with expanded valence-shell structures in introductory Lewis accounting.

Worked example: carbonate

For CO3^2−, total valence electrons are 4 + 3(6) + 2 = 24. A structure with one C=O and two C–O single bonds gives carbon an octet. Each single-bonded O carries formal charge −1; the double-bonded O and C carry 0. Three equivalent contributors exist, so all C–O bonds are equivalent in the ion.

Common wrong turns

Adding electrons to satisfy octets without recounting the total.
Moving atoms when drawing resonance. Only electron placement changes.
Calling formal charge the measured partial charge.
Drawing ionic compounds as independent covalent molecules by default.
Assuming every atom must obey the octet rule without exception.

Practice

Count valence electrons in NH4+, CO2, and SO4^2−.
Draw the Lewis structure of HCN and identify formal charges.
How many equivalent major resonance contributors does NO3− have?
Which should have larger lattice-energy magnitude, NaCl or MgO? Explain qualitatively.

Answers

8, 16, and 32 electrons, respectively.
H–C≡N: with one lone pair on N; all formal charges 0.
Three.
MgO, because the ionic charge product is larger and the relevant ions are compact; a full comparison would use consistent structural data.

Next moduleModule 13 — Molecular Geometry, Polarity, and Valence Bond Models

Module 13 — Molecular Geometry, Polarity, and Valence Bond Models

OpenStax reading: Chapter 7.5–7.6 and 8.1–8.2
Central question: How does three-dimensional structure connect electron domains to molecular properties?

Learning targets

Count electron domains around a central atom.
Predict electron-domain and molecular geometry using VSEPR.
Estimate ideal bond angles and explain lone-pair distortions.
Determine molecular polarity from bond dipoles and geometry.
Assign common sp, sp2, and sp3 hybridization labels.
Distinguish sigma and pi bonding.

VSEPR model

Valence-shell electron-pair repulsion theory predicts geometry by arranging regions of electron density around a central atom to reduce repulsion. A single, double, or triple bond counts as one electron domain. Each lone pair counts as one domain.

Domains	Electron-domain geometry	Ideal angles	Common molecular geometries
2	linear	180°	linear
3	trigonal planar	120°	trigonal planar, bent
4	tetrahedral	109.5°	tetrahedral, trigonal pyramidal, bent
5	trigonal bipyramidal	90°, 120°, 180°	trigonal bipyramidal, seesaw, T-shaped, linear
6	octahedral	90°, 180°	octahedral, square pyramidal, square planar

Lone pairs usually repel neighboring electron density more strongly than bonding domains and compress adjacent bond angles. Multiple bonds can also alter angles. Treat ideal values as model baselines.

Geometry workflow

Draw a valid Lewis structure.
Identify the central atom.
Count bonding domains plus lone-pair domains.
Assign electron-domain geometry.
Ignore lone-pair positions when naming molecular geometry.
Estimate angles and distortions.

Molecular polarity

A molecule is polar when its bond dipole vectors do not cancel. A molecule can contain polar bonds and remain nonpolar because of symmetric geometry, as in CO2. Conversely, bent H2O is polar because its O–H bond dipoles reinforce rather than cancel.

Polarity workflow:

Identify polar bonds.
Determine three-dimensional molecular geometry.
Treat bond dipoles as vectors.
Decide whether their vector sum is zero.

Hybridization and bond types

In the introductory localized valence-bond model:

2 electron domains often correspond to sp hybridization.
3 correspond to sp2.
4 correspond to sp3.

A sigma (σ) bond has electron density along the internuclear axis. A pi (π) bond results from side-by-side overlap, with density above and below the axis in the common p-orbital description. Single bonds contain one sigma bond; double bonds contain one sigma and one pi; triple bonds contain one sigma and two pi.

Hybridization is a model for bonding and geometry, not a mechanical process an isolated atom must perform before a bond can form.

Worked example: `SO2`

A common Lewis representation has resonance and one lone pair on sulfur. Three electron domains around S give trigonal-planar electron-domain geometry. With two bonded atoms and one lone pair, the molecular geometry is bent. Because the S–O bond dipoles do not cancel, the molecule is polar. In the introductory model, sulfur is assigned sp2 hybridization.

Common wrong turns

Counting a double bond as two electron domains.
Naming electron-domain geometry when molecular geometry is requested.
Deciding polarity from electronegativity alone.
Drawing all molecules flat because Lewis structures are flat.
Applying hybridization labels without first establishing a structure.

Practice

For each species, give electron-domain geometry, molecular geometry, approximate angle, and polarity: CO2, NH3, H2O, BF3, and CH4.

Answers

CO2: linear; linear; 180°; nonpolar overall.
NH3: tetrahedral domains; trigonal pyramidal; about 107°; polar.
H2O: tetrahedral domains; bent; about 104.5°; polar.
BF3: trigonal planar; trigonal planar; 120°; nonpolar overall in the ideal symmetric molecule.
CH4: tetrahedral; tetrahedral; 109.5°; nonpolar.

Next moduleModule 14 — Gases, Intermolecular Forces, and Phases

Module 14 — Gases, Intermolecular Forces, and Phases

OpenStax reading: Chapter 9.1–9.5 and 10.1–10.4
Central question: How do particle motion and attractions determine pressure, gas behavior, and phase properties?

Learning targets

Define pressure and convert common pressure units.
Apply Boyle's, Charles's, Avogadro's, combined, and ideal gas relationships.
Use partial pressures and gas stoichiometry.
State the assumptions of kinetic molecular theory.
Explain non-ideal behavior qualitatively.
Identify major intermolecular forces and relate them to physical properties.
Interpret heating curves and phase diagrams.

Pressure

Gas pressure results from particle collisions with container walls. Pressure is force per area. Useful relationships include:

1 atm = 101.325 kPa = 760 torr = 760 mmHg

Use the unit attached to the selected gas constant R.

Empirical gas relationships

At constant temperature and amount, Boyle's law gives P1V1 = P2V2.

At constant pressure and amount, Charles's law gives V1/T1 = V2/T2.

At constant pressure and temperature, Avogadro's law gives V ∝ n.

The combined gas law for constant amount is:

P1V1/T1 = P2V2/T2

Temperature must be absolute. T(K) = T(°C) + 273.15.

Ideal gas law

PV = nRT

Common R values:

0.082057 L atm mol−1 K−1
8.314462618 J mol−1 K−1, equivalent to L kPa mol−1 K−1

Worked example: moles of gas

How many moles occupy 5.00 L at 98.0 kPa and 298.15 K?

Using R = 8.314 L kPa mol−1 K−1:

n = PV/RT = (98.0 kPa)(5.00 L)/[(8.314 L kPa mol−1 K−1)(298.15 K)] = 0.198 mol

Gas mixtures

For an ideal mixture, Dalton's law states:

P_total = ΣP_i

and P_i = X_iP_total, where X_i = n_i/n_total.

Gas collected over water contains water vapor. The dry gas pressure is P_gas = P_total − P_H2O at the collection temperature.

Kinetic molecular theory

The ideal model assumes particles have negligible volume, experience no intermolecular attraction except during elastic collisions, move randomly, and possess average kinetic energy determined by absolute temperature. Real gases deviate most at high pressure and low temperature, where particle volume and attractions become important.

At the same temperature, gases have the same average translational kinetic energy. Lighter particles have a greater root-mean-square speed on average, not greater average kinetic energy.

Intermolecular forces

London dispersion: present in all atoms and molecules; strength generally increases with polarizability, electron count, and contact area.
Dipole–dipole: attraction among permanent molecular dipoles.
Hydrogen bonding: particularly strong directional attraction when H is covalently bonded to N, O, or F and interacts with an available lone pair on N, O, or F in the standard introductory treatment.
Ion–dipole: attraction between an ion and a polar molecule; important in solvation.

Intermolecular forces are weaker than ordinary intramolecular covalent or ionic bonding, but collectively they strongly affect boiling point, vapor pressure, viscosity, surface tension, and solubility.

Stronger attractions generally correlate with higher boiling temperature, higher viscosity and surface tension, and lower vapor pressure when other factors are comparable.

Phase changes and curves

During a heating-curve slope, added energy raises average kinetic energy and temperature. During an idealized phase-change plateau at constant pressure, added energy changes potential-energy relationships and phase proportion rather than temperature.

A phase diagram maps stable phases against temperature and pressure. Phase boundaries represent coexistence. The triple point is where three phases coexist; the critical point terminates the liquid–gas boundary. Above the critical temperature, pressure alone cannot condense the substance into an ordinary liquid.

Common wrong turns

Using Celsius in a gas-law ratio.
Selecting R without matching pressure and volume units.
Saying gas particles expand when a gas expands. Average separation increases; particle size does not.
Treating hydrogen bonding as a covalent H–N/O/F bond itself.
Assuming the largest molar mass always has the highest boiling point without considering shape, polarity, and bonding.
Saying temperature rises during every addition of heat, including phase transitions.

Practice

A gas at 1.20 atm and 2.50 L expands isothermally to 4.00 L. Find final pressure.
Convert 25.0 °C to kelvins.
Find total pressure for partial pressures 0.250, 0.430, and 0.115 atm.
Rank the dominant intermolecular attraction categories for CH4, HCl, H2O, and Na+ in water.
Why do real gases deviate more at high pressure?

Answers

0.750 atm.
298.15 K.
0.795 atm.
CH4: dispersion; HCl: dipole–dipole plus dispersion; H2O: hydrogen bonding plus other forces; Na+ in water: ion–dipole.
Particles are closer, so their finite volume and intermolecular attractions matter relative to container volume and kinetic behavior.

End of the modulesWhere the glossary keeps going