Chemistry: matter, atoms, and reactions
Start with observations and simple particle pictures before learning formulas and calculations.
GUIDE8 published lessons · Updated 2026-10-19
Before you begin
Start at lesson 1. No previous chemistry study is assumed. Read one idea at a time, work through the example on paper, then try the short exercise. Use a calculator when useful. Move on when you can explain the answer, not just repeat it. The opening lessons use everyday descriptions before formulas. Later calculations draw on the linked Mathematics foundations; pause for those prerequisites as needed.
See the pathway from novice to doctoral study · Review the mathematics foundations
Working toward: Interpret a chemical formula, count particles, balance a simple reaction, and calculate amount concentration.
Read each explanation, work through the example, and attempt the exercise before opening its answer. Published lessons are ready to study; unfinished roadmap topics remain planned.
Validation: Original calculation-based examples with explicit assumptions. The capstone's atom balance, dilution, aliquot and product-amount checks were verified arithmetically. These lessons do not report physical experiments.
1. Start by describing what you can observe
Before this lesson: No prior subject knowledge. Start here.
Chemistry studies matter and its transformations. Begin by separating an observation from an explanation. “The sample became liquid” describes an observed change. “Its particles are moving differently” proposes an explanation using a model. Useful observations say what was examined and under what conditions, rather than treating a guess about the cause as something directly seen.
Worked example
A written record says: “A piece of ice in a room became liquid water.” Becoming liquid is the observation. A particle account of melting is an explanation of that observation. No experiment is needed for this reading exercise.
Your turn
In “the metal surface became reddish-brown because it reacted with oxygen,” which part describes an observation and which part proposes an explanation?
Show answer and reasoning
Observation: the surface became reddish-brown. Explanation: it reacted with oxygen. Colour alone does not establish the cause; confirming the explanation requires further evidence.
Watch for: Everyday words such as “pure” and “chemical” can have different meanings from their scientific use. Name what you know rather than relying on the label.
Link to this lesson2. Elements, compounds, and mixtures
Before this lesson: Read 1. Start by describing what you can observe first; return to its example if the new terms are unfamiliar.
An element contains atoms with the same number of protons. A compound contains elements chemically combined in a definite proportion. A mixture combines substances without fixing a single chemical composition. Physical changes such as melting alter state or form; chemical changes produce different substances. Appearance alone does not reliably distinguish them.
Worked example
Pure copper is an element. Pure water is a compound with hydrogen and oxygen in a 2:1 atom ratio. Salt water is a mixture whose concentration can vary. Ice melting remains H₂O.
Your turn
Classify pure oxygen gas, pure carbon dioxide, and air as element, compound, or mixture.
Show answer and reasoning
Oxygen gas is an element: O₂ contains only oxygen atoms. Carbon dioxide is a compound: CO₂ combines carbon and oxygen in a fixed atom ratio. Air is a mixture of several gases.
Watch for: A molecule can be elemental, such as O₂. Molecule and compound are not interchangeable terms.
Link to this lesson3. A first picture of particles
Before this lesson: Read 2. Elements, compounds, and mixtures first; return to its example if the new terms are unfamiliar.
A particle model describes matter using entities too small to see individually with the unaided eye. An atom is a unit of an element; a molecule consists of atoms bonded together. In a solid, particles remain near relatively fixed positions while moving about them. In a liquid they can move past one another. Gas particles are much farther apart on average under ordinary conditions.
Worked example
A simple sketch of ice puts water molecules in an ordered arrangement; a sketch of liquid water has close but rearranging molecules. Melting does not turn each molecule into a tiny droplet. The state describes the collection.
Your turn
When ice melts into liquid water, do the water molecules themselves become miniature puddles? Explain.
Show answer and reasoning
No. A molecule is not a tiny sample with all the bulk properties of water. Melting changes the arrangement and motion of many molecules; they remain water molecules.
Watch for: Particle drawings are models, not photographs. Dot size and spacing in a sketch are usually not to scale.
Link to this lesson4. Read a chemical formula one symbol at a time
Before this lesson: Read 3. A first picture of particles first; return to its example if the new terms are unfamiliar.
A chemical symbol identifies an element. H means hydrogen; O means oxygen. Capital letters matter: a two-letter symbol starts with a capital and ends with a lowercase letter. A small number after a symbol counts that element’s atoms in a molecular formula. No written number means one. H₂O names a water molecule containing two hydrogen atoms and one oxygen atom.
Worked example
CO₂ has one carbon atom and two oxygen atoms per molecule. O₂ has two oxygen atoms per molecule. The large coefficient in 2H₂O counts two whole molecules: four H atoms and two O atoms altogether.
Your turn
How many hydrogen and oxygen atoms are represented by 3H₂O?
Show answer and reasoning
Each H₂O has 2 H and 1 O. Three molecules therefore contain 3 × 2 = 6 hydrogen atoms and 3 × 1 = 3 oxygen atoms.
Watch for: Co is the element cobalt; CO is a formula containing carbon and oxygen. Also, ionic formulas describe ratios rather than necessarily discrete molecules.
Link to this lesson5. Protons, isotopes, and ions
Before this lesson: Read 4. Read a chemical formula one symbol at a time first; return to its example if the new terms are unfamiliar.
Atomic number counts protons and identifies an element. Mass number counts protons plus neutrons in one isotope. A neutral atom has equal numbers of electrons and protons; gaining or losing electrons creates an ion. Isotopes differ in neutron count, while ions differ in electron count. Neither change alone changes the element’s identity.
Worked example
A sodium-23 ion with charge +1 has 11 protons, 12 neutrons, and 10 electrons. The positive charge means one fewer electron than proton.
Your turn
A chlorine-35 ion has atomic number 17 and charge −1. How many protons, neutrons, and electrons does it have?
Show answer and reasoning
17 protons; 35 − 17 = 18 neutrons; 18 electrons. One extra electron produces charge −1.
Watch for: The periodic table’s average atomic mass is not generally a whole-number mass number for an individual isotope.
Link to this lesson6. Balance atoms, then count moles
Before this lesson: Read 5. Protons, isotopes, and ions first; return to its example if the new terms are unfamiliar. Mathematics: fractions, multiplication, ratios, and equations. Review those linked foundations before calculating.
A balanced chemical equation conserves the number of atoms of each element in an ordinary chemical reaction. Adjust coefficients in front of formulas; changing a subscript changes the substance. Coefficients express particle ratios and mole ratios, not mass ratios. A mole counts a specified entity: one mole contains exactly 6.02214076 × 10²³ such entities.
Worked example
2H₂ + O₂ → 2H₂O has four hydrogen atoms and two oxygen atoms on each side. With enough oxygen and complete conversion, 3 mol H₂ can form 3 mol H₂O while consuming 1.5 mol O₂.
Your turn
Balance N₂ + H₂ → NH₃. How many moles of H₂ are required for 2 mol N₂ with complete conversion?
Show answer and reasoning
N₂ + 3H₂ → 2NH₃. The ratio is 3 mol H₂ per 1 mol N₂, so 2 mol N₂ requires 6 mol H₂. Both sides contain 2 N and 6 H atoms per balanced equation.
Watch for: An equation’s ratios do not guarantee completion or describe the reaction rate. Available reactant amounts can limit yield.
Link to this lesson7. Amount concentration and dilution
Before this lesson: Read 6. Balance atoms, then count moles first; return to its example if the new terms are unfamiliar. Mathematics: fractions, multiplication, ratios, and equations. Review those linked foundations before calculating.
Amount concentration is the amount of dissolved solute in moles divided by the final solution volume in litres: c = n/V. During dilution without solute loss or reaction, the amount of solute stays constant, so c₁V₁ = c₂V₂. Final solution volume is distinct from the volume of solvent added.
Worked example
A hypothetical solution containing 0.20 mol solute in 0.50 L has concentration 0.40 mol/L. Diluting a 0.10 L portion to a final 0.40 L gives 0.10 mol/L.
Your turn
A hypothetical 0.25 L solution has concentration 0.80 mol/L. What is its concentration after dilution to a final volume of 1.00 L without solute loss?
Show answer and reasoning
n = 0.80 × 0.25 = 0.20 mol. The final concentration is 0.20 / 1.00 = 0.20 mol/L. This is a calculation, not a preparation procedure.
Watch for: Use final solution volume, not the amount of water added. Keep all volume units consistent.
Link to this lesson8. Capstone: follow the amount through each change
Before this lesson: Complete chemistry lessons 4, 6 and 7. You need formula counts, balanced coefficients, c = n/V and millilitre-to-litre conversion. The readiness questions below point back to the relevant lessons.
This is an original paper calculation using hypothetical quantities, not a record of an experiment or a procedure for handling chemicals. Follow the named substance through three stages: a portion of stock solution, the diluted solution, then a reaction prediction. A portion taken from a well-mixed solution is called an aliquot; it has the same concentration as that solution, but contains only part of its total solute.
Write c for amount concentration in mol/L, V for solution volume in L, and n for amount in mol. First identify which concentration belongs to which volume. Then find n = cV. During a dilution with no loss or reaction, n for that solute stays fixed, while c changes. Use the final solution volume. Do not infer the volume of solvent added by assuming that separately measured liquid volumes always add exactly.
After the dilution stage, consider the specified reaction H₂O₂ → H₂O + O₂. Keeping those formulas, its smallest whole-number balance is 2H₂O₂ → 2H₂O + O₂. There are 4 H atoms and 4 O atoms on each side. The coefficients give 1 mol O₂ for every 2 mol H₂O₂ consumed. Multiply the reacting amount of H₂O₂ by 1/2 to predict O₂ under the stated complete-conversion assumption. This factor is a mole ratio, not a mass ratio or a ratio between the solution volume and a gas volume.
Keep the conservation statements separate. Dilution conserves the amount of H₂O₂ only while it neither reacts nor leaves the solution. Reaction consumes H₂O₂; atoms are conserved through the balanced equation instead. A balanced equation alone does not tell you how quickly the change happens or prove complete conversion. Calculating a gas volume would also require conditions and a suitable gas model.
Use two checks before accepting an answer: reconstruct the initial amount from the final concentration and volume, and compare the product amount with the equation's ratio. More solvent alone cannot increase the amount of O₂ available from the entire original aliquot in this idealized model. It may affect real reaction behaviour, which this calculation does not predict.
Worked example
Worked study — hypothetical values; assume a well-mixed solution, no H₂O₂ loss or reaction during dilution, and later complete conversion by the stated reaction. 1. A 50.0 mL portion of 0.600 mol/L H₂O₂ contains n = 0.600 × 0.0500 = 0.0300 mol H₂O₂. 2. Its final diluted solution volume is 150.0 mL = 0.1500 L. Therefore c = 0.0300 / 0.1500 = 0.200 mol/L. 3. Predict the O₂ from the entire diluted solution: 0.0300 mol H₂O₂ × (1 mol O₂ / 2 mol H₂O₂) = 0.0150 mol O₂. 4. Reverse check: 0.200 mol/L × 0.1500 L = 0.0300 mol H₂O₂. Tripling solution volume reduced concentration to one third; it did not change that amount. The balanced ratio makes the predicted O₂ amount half the consumed H₂O₂ amount.
Your turn
Independent study — use a new sheet and the same explicit assumptions, without copying the worked numbers. A 25.0 mL portion of 0.800 mol/L H₂O₂ is diluted to a final solution volume of 200.0 mL. Only a 50.0 mL aliquot of that well-mixed diluted solution is considered for complete decomposition. 1. Balance H₂O₂ → H₂O + O₂ and check both element counts. 2. Find the amount of H₂O₂ before dilution and the diluted concentration. 3. Find the amount of H₂O₂ in the 50.0 mL aliquot, then its predicted O₂ amount. 4. A notebook multiplies 0.800 mol/L by 0.0500 L and reports 0.0400 mol O₂. Identify both errors. 5. If the entire original diluted solution were considered instead, how would its O₂ prediction compare with the 50.0 mL aliquot? Explain without starting over. 6. Does this calculation establish reaction time, collected gas volume, or an experimentally observed yield? If some H₂O₂ reacted during dilution, which earlier conservation assumption would fail?
Show answer and reasoning
1. 2H₂O₂ → 2H₂O + O₂. H: 4 on each side. O: 4 on each side. 2. Initial amount: 0.800 mol/L × 0.0250 L = 0.0200 mol H₂O₂. Diluted concentration: 0.0200 mol / 0.2000 L = 0.100 mol/L. Reverse check: 0.100 × 0.2000 = 0.0200 mol. 3. The 50.0 mL aliquot contains 0.100 mol/L × 0.0500 L = 0.00500 mol H₂O₂. Its prediction is 0.00500 / 2 = 0.00250 mol O₂. 4. The notebook combines the stock concentration with a volume taken after dilution. It also uses an incorrect 1:1 H₂O₂:O₂ mole ratio instead of 2:1. Its result is sixteen times the correct prediction. 5. The entire 200.0 mL solution has four times the amount in its 50.0 mL aliquot, so the prediction is four times larger: 0.0100 mol O₂. Concentration is the same throughout the well-mixed solution; total amount depends on the portion considered. 6. None of those observations follows from the calculation. It predicts amount under assumptions, not reaction speed, collected gas volume or observed yield. If H₂O₂ reacted during dilution, its amount would no longer remain equal to the initial 0.0200 mol; the remaining amount would need new evidence. Ready to continue when you can show the units at every step, justify the 1/2 factor, distinguish total from sampled amount, and explain where the dilution conservation statement stops applying. If a step failed, return to its readiness review and solve the independent study again before moving on.
Watch for: Do not mix stock concentration with a diluted-solution aliquot, turn mole ratios into mass ratios, or confuse final solution volume with solvent added. Keep intermediate precision and round the reported results. An idealized prediction is not a measured chemical yield.
Lesson references
- OpenStax Chemistry 2e §3.3: concentration and dilution (checked 19 October 2026) ↗
- OpenStax Chemistry 2e §4.1: balancing with coefficients ↗
- OpenStax Chemistry 2e §4.3: reaction mole ratios ↗
- OpenStax Chemistry 2e §12.1: peroxide decomposition and reaction rate ↗
From first steps to doctoral study
Start slowly. Each stage builds on the previous one; later branches have their own prerequisites. The foundation lessons are available now. The remaining curriculum is a plan for future lessons, not a complete course.
1. First steps · FOUNDATION LESSONS AVAILABLE
Before you start: No prior chemistry; everyday reading and counting.
Observations, matter, particle models, formulas, atoms, ions, and introductory calculations. Mathematics is introduced or linked before use.
Ready to continue when you can: Distinguish observation from interpretation and explain a formula or calculation in words.
Start the foundation lessons →2. School foundations · PLANNED
Before you start: First-step chemistry and basic algebra.
Moles and stoichiometry in depth, bonding, periodic patterns, gases, solutions, energy changes, equilibrium, and acids and bases.
Ready to continue when you can: Track atoms, charge, quantities, and units through a multistep problem.
3. Undergraduate core · PLANNED
Before you start: School chemistry; algebra, calculus, and introductory physics as required.
General, organic, inorganic, analytical, and physical chemistry, with supervised measurement, uncertainty, and laboratory practice.
Ready to continue when you can: Interpret evidence from measurements, justify a proposed mechanism or model, and write a reproducible report.
4. Advanced undergraduate study · PLANNED
Before you start: Core chemistry and the mathematics required by the selected topic.
Spectroscopy, thermodynamics, kinetics, quantum chemistry, synthesis planning, instrumental analysis, and scientific computation.
Ready to continue when you can: Combine multiple forms of evidence, evaluate alternatives, and quantify uncertainty.
5. Graduate foundations · PLANNED
Before you start: Advanced core study plus relevant experimental or computational experience.
Advanced theory and methods chosen for a branch: organic, inorganic, physical, analytical, materials, chemical biology, or computational chemistry.
Ready to continue when you can: Critically read research papers and defend method selection and limitations.
6. Research specialization · PLANNED
Before you start: Relevant graduate methods and supervised practical competence where needed.
A focused literature review, specialist instruments or computational methods, validation, controls, and reproducibility.
Ready to continue when you can: Reproduce and critically evaluate a published result; design controls for a bounded research question.
7. Doctoral research practice · PLANNED
Before you start: Specialist knowledge and appropriate research supervision and facilities.
Original research design, evidence synthesis, method development, responsible data handling, publication, and thesis defence.
Ready to continue when you can: Produce and defend a new, well-supported contribution, including negative results and limits of inference.
Doctoral-level work involves specialization, original research, and expert critique. This pathway supports that preparation; completing site lessons does not confer a degree or establish research competence.
Experimental chemistry also requires supervised laboratory training and appropriate facilities. Reading and calculation exercises cannot replace that experience.
References
Original AUWEN lessons, with upstream documentation for further study and version checks.
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