Chapters 1 and 2 Study Guide
1.1 Classifying matter (what it is and how we name it)
Physical state
Solid: fixed shape and volume (particles tightly packed; strong intermolecular forces).
Liquid: fixed volume but shape adapts to container (particles close but can move past each other).
Gas: no fixed shape or volume — fills container (particles far apart; weak interactions).
Pure substance vs mixture
Pure substance: one type of “stuff” with fixed composition.
Element: pure substance that cannot be chemically broken into simpler substances (e.g., O₂ gas consists of oxygen atoms).
Compound: pure substance made of two or more elements chemically bonded in fixed ratio (e.g., H₂O).
Mixture: two or more substances physically combined; components keep their chemical identities.
Homogeneous mixture: uniform throughout (solution like salt water).
Heterogeneous mixture: non-uniform (e.g., sand in water).
Particles
Atom: smallest unit of an element that retains element identity (e.g., one carbon atom).
Molecule: two or more atoms bonded together (could be same element, e.g., O₂, or different, e.g., CO₂).
Ion: atom or group of atoms with net electric charge (cation = positive, anion = negative).
How to classify — short checklist
Is composition fixed? yes → pure substance (element or compound). no → mixture.
Is it made of single type of atom? yes → element. If different elements bonded → compound.
Is it uniform? yes → homogeneous; no → heterogeneous.
1.2 SI units & common prefixes
Base SI units you need:
Length: meter (m)
Mass: kilogram (kg) — note: kg is base unit (not g), but we often use grams (g).
Time: second (s)
Temperature: kelvin (K)
Amount of substance: mole (mol)
Common prefixes (multipliers):
kilo- (k) = 10³ = 1,000
e.g., 1 km = 1,000 m
centi- (c) = 10⁻² = 0.01
e.g., 1 cm = 0.01 m
milli- (m) = 10⁻³ = 0.001
e.g., 1 mm = 0.001 m
Tip: memorize that kilo = 1000, centi = 1/100, milli = 1/1000.
1.3 Scientific notation
Why: makes very large or very small numbers easier and reduces error.
Format: a \times 10^n where 1 ≤ |a| < 10 and n is integer.
Examples
0.000345 = 3.45 \times 10^{-4}.
6,200,000 = 6.2 \times 10^{6}.
To convert: move decimal left for positive exponent, right for negative.
1.4 Significant figures (sig figs) & uncertainty
What sig figs mean: digits in a measurement that are known reliably plus one uncertain digit (last digit).
Rules for counting sig figs
Nonzero digits are always significant: 245 → 3 sig figs.
Zeros between nonzero digits are significant: 2005 → 4 sig figs.
Leading zeros (to left of first nonzero) are NOT significant: 0.0072 → 2 sig figs.
Trailing zeros in a number with a decimal are significant: 2.300 → 4 sig figs.
Trailing zeros in a whole number without a decimal are ambiguous — use scientific notation to show significance: 1200 (ambiguous) → write 1.200×10³ to show 4 sig figs or 1.2×10³ to show 2.
Relationship to uncertainty: the last sig fig is the estimated digit — it indicates the measurement’s uncertainty level.
1.5 Sig figs in calculations
Multiplication / Division: answer has same number of sig figs as factor with fewest sig figs.
Example: 2.5 \times 3.42 = 8.55 → 2 sig figs → round to 8.6.
Addition / Subtraction: align decimal places; answer has decimal places equal to the quantity with fewest decimal places.
Example: 12.11 + 0.3 = 12.41 → fewest decimals = 1 decimal → round to 12.4.
Always follow these rules and only round at the end of multi-step calculations (keeping extra guard digits during intermediate steps).
1.6 Accuracy vs precision
Accuracy: how close a measurement is to the true/accepted value.
Precision: how repeatable measurements are (how close they are to each other).
Illustration (dartboard):
All darts close to bullseye → accurate and precise.
All darts clustered but far from bullseye → precise but not accurate.
Darts spread out but centered on bullseye on average → accurate but not precise.
1.7 Derived units: volume & density
Volume: for regular shapes use geometry (e.g., m³ or L = dm³). Common lab units: mL (1 mL = 1 cm³).
Density: \rho = \dfrac{\text{mass}}{\text{volume}}. Common units: g/mL or g/cm³ for solids/liquids; kg/m³ in SI.
Rearrangements
\text{mass} = \text{density} \times \text{volume}
\text{volume} = \dfrac{\text{mass}}{\text{density}}
Example (lab / irregular object):
Object mass = 12.43 g. Volume by displacement = 3.10 mL.
Density = 12.43 g ÷ 3.10 mL = 4.0097 g/mL → with sig figs: three sig figs (3.10 has 3 sig figs; 12.43 has 4) → 4.01 g/mL.
1.8 Dimensional analysis (unit conversions)
Key idea: multiply by conversion factors that equal 1 (units cancel).
Worked example: convert 2.50 miles → meters.
1 mile = 1609.34 m.
2.50\ \text{mi}\times \frac{1609.34\ \text{m}}{1\ \text{mi}} = 4023.35\ \text{m}.
Sig figs: 2.50 has 3 sig figs → answer must have 3 sig figs → 4.02 × 10³ m.
1.9 Practice problems (with steps + sig figs)
Problem A — Scientific notation
Convert 0.000462 to scientific notation.
Move decimal 4 places right: 4.62 \times 10^{-4}.
Problem B — Multiplication with sig figs
Compute (3.60 \times 2.1).
Raw: 3.60\times2.1 = 7.56.
Sig figs: 3.60 (3 sig figs), 2.1 (2 sig figs) → result 2 sig figs → 7.6.
Problem C — Addition with sig figs
Compute 12.11 + 0.3 + 0.042.
Align decimals; fewest decimal places = 1 (from 0.3) → round final to 1 decimal place.
Sum = 12.11 + 0.3 + 0.042 = 12.452 → round to 12.5.
Problem D — Density (irregular object)
Mass = 24.68 g; initial water in graduated cylinder = 15.0 mL; final = 17.35 mL.
Volume displaced = 17.35 − 15.0 = 2.35 mL (note: 15.0 has 3 sig figs so difference has 3 sig figs).
Density = 24.68 ÷ 2.35 = 10.500 ≈ 4 sig figs? But check sig figs: mass 24.68 (4 sig figs), volume 2.35 (3 sig figs) → result to 3 sig figs → 10.5 g/mL.
Chapter 2 — Atomic structure, periodic table, bonding, naming
2.1 Historic experiments — what they showed and why they matter
J.J. Thomson (late 1800s) — cathode ray experiment
What he did: passed a beam (cathode ray) through electric & magnetic fields and measured deflection.
Observation: beam deflected toward positive plate → beam composed of negatively charged particles.
Discovery: existence of the electron — a very small, negatively charged particle present in atoms.
Model implication: atoms are not indivisible; they contain subatomic particles. Thomson proposed the “plum pudding” model: a positive “soup” with embedded electrons.
Significance: first discovery of subatomic particle; proved atoms have internal structure.
Ernest Rutherford (early 1900s) — gold foil experiment
What he did: fired alpha particles (positively charged) at very thin gold foil and detected scattering angles.
Observation: most alpha particles passed straight through, but a small fraction deflected at large angles; some bounced back.
Conclusion: atom is mostly empty space with a tiny, dense, positively charged nucleus that contains most mass — electrons orbit around that nucleus.
Model implication: replaced plum pudding with nuclear model (nucleus + orbiting electrons).
Why this matters: Rutherford explained the large-angle deflections that Thomson’s model couldn’t; introduced the nucleus concept — foundation for modern atomic structure.
2.2 Atomic number, mass number, isotopes
Atomic number (Z): number of protons in nucleus → defines the element.
Mass number (A): total number of protons + neutrons in nucleus (integer).
Isotopes: atoms of same element (same Z) with different numbers of neutrons (different A). Example: carbon-12 (^12C) and carbon-14 (^14C).
Isotopic notation: {}^{A}{Z}X^{\text{charge}}
Example: an ion of chlorine with 17 protons and 18 neutrons and −1 charge → {}^{35}{17}\text{Cl}^{-} (35 = 17+18).
Average atomic mass: weighted average of isotopic masses using natural abundances (found on periodic table). The periodic table lists average atomic mass (not integer mass numbers) because natural samples are mixtures of isotopes.
2.3 Example: average atomic mass (Neon)
Given isotopes (typical values):
^20Ne mass = 19.992440 amu, abundance = 90.48% (0.9048)
^21Ne mass = 20.993847 amu, abundance = 0.27% (0.0027)
^22Ne mass = 21.991386 amu, abundance = 9.25% (0.0925)
Average atomic mass:
(19.992440)(0.9048) + (20.993847)(0.0027) + (21.991386)(0.0925) = 20.1800\ \text{amu (approx.)}
This is the kind of number you’ll see on the periodic table: 20.180 amu.
Procedure (general): multiply each isotope mass × its fractional abundance, then sum.
2.4 Writing atomic/ionic symbols and counting particles
Given: protons, neutrons, electrons, and charge — decide isotope notation or ion symbol.
Example 1: 17 protons, 18 neutrons, 17 electrons → neutral chlorine atom ^35Cl (since A = 17 + 18 = 35). Symbol: {}^{35}_{17}\text{Cl}.
Example 2: 11 protons, 12 neutrons, 10 electrons → net charge +1 (lost 1 electron) → sodium ion {}^{23}_{11}\text{Na}^{+} (A = 23).
How to check:
protons = atomic number (Z) → identifies element.
electrons = protons − charge (if charge positive, fewer electrons).
mass number A = protons + neutrons.
2.5 Ionic vs covalent (molecular) compounds
Ionic compounds
Formed when electrons are transferred from a metal to a nonmetal (forming cations and anions).
Bonding characterized by electrostatic attraction between oppositely charged ions.
Usually full formula is a formula unit (empirical ratio). Often solids with high melting points and conduct electricity when molten or dissolved.
Example: NaCl (Na⁺ and Cl⁻).
Covalent (molecular) compounds
Formed when two nonmetals share electrons to achieve noble gas configuration.
Bonds are electron-sharing; molecules have discrete units.
Example: H₂O, CO₂.
Rule of thumb: metal + nonmetal → usually ionic. nonmetal + nonmetal → usually covalent.
2.6 Using the periodic table to get information
From an element’s location:
Atomic number → number of protons (top of box).
Atomic mass (average) → usually decimal number beneath symbol.
Group (column) number → similar chemical behavior and valence electrons.
Period (row) → number of electron shells occupied.
Metals/Nonmetals/Metalloids: left side metals, right side nonmetals; staircase demarcates metalloids.
Common ion charges: Groups 1A → +1, 2A → +2, 7A (halogens) → −1, 6A → −2, etc.
Transition metals: central block (d-block).
Halogens: Group 17 (F, Cl, Br, I…).
Noble gases: Group 18 (He, Ne, Ar…).
2.7 Writing chemical formulas
Ionic compounds (simple)
Balance total positive and negative charges to get neutral compound.
Example: Al³⁺ and O²⁻ → least common multiple of 3 and 2 = 6 → need 2 Al³⁺ (2×+3=+6) and 3 O²⁻ (3×−2=−6) → formula Al₂O₃.
With polyatomic ions
Treat polyatomic ion as a unit; balance charges.
Use parentheses when more than one polyatomic unit is needed: e.g., calcium nitrate = Ca²⁺ + NO₃⁻ → need two NO₃⁻ → Ca(NO₃)₂.
Naming
Ionic: cation name (metal) first, then anion name (nonmetal with −ide ending) or polyatomic ion name. For transition metals that can have multiple charges, use Roman numeral for charge (iron(III) chloride = FeCl₃).
Molecular (binary nonmetal compounds): use prefixes (mono-, di-, tri-, etc.) to show number of each atom (CO₂ = carbon dioxide).
Common polyatomic ions (memorize these)
NH₄⁺ ammonium
NO₃⁻ nitrate
SO₄²⁻ sulfate
CO₃²⁻ carbonate
OH⁻ hydroxide
PO₄³⁻ phosphate
ClO₄⁻ perchlorate
2.8 Hydrates
Definition: a compound that contains water molecules in its crystalline structure: written as \text{salt} \cdot x\text{H}_2\text{O}.
Example: copper(II) sulfate pentahydrate = CuSO₄·5H₂O.
Naming: name ionic compound then add prefix for water number + “hydrate” (e.g., decahydrate, pentahydrate).
Finding empirical hydrate formula (lab procedure)
Mass of hydrate (before heating) — measured.
Heat to remove water → mass of anhydrous salt measured.
Mass of water lost = mass hydrate − mass anhydrous.
Convert both masses to moles:
moles anhydrous = mass anhydrous ÷ molar mass of anhydrous salt.
moles water = mass water ÷ 18.015 g/mol.
Compute mole ratio: moles water ÷ moles anhydrous → round to nearest small whole number → that’s x in salt·x H₂O.
Worked lab example (complete): (This matches labs you described.)
Given:
Mass hydrate = 2.564 g
Mass anhydrous = 1.622 g
Assume the anhydrous formula is CuSO₄ (molar mass = 159.609 g/mol)
Steps:
Mass water = 2.564 − 1.622 = 0.942 g.
Moles anhydrous CuSO₄ = 1.622 g ÷ 159.609 g/mol = 0.010162 mol.
Moles water = 0.942 g ÷ 18.015 g/mol = 0.052290 mol.
Mole ratio water : salt = 0.052290 ÷ 0.010162 = 5.15 ≈ 5 → formula CuSO₄·5H₂O.
Why rounding to whole number: water molecules must be whole; experimental values near whole numbers are rounded to the nearest integer (if close enough — e.g., 2.99 → 3).
2.9 Stoichiometry & dimensional analysis reminders
Always write units and let them cancel.
Carry at least one extra guard digit through calculations; only round final answer to correct sig figs.
For atomic/molecular calculations you’ll often use Avogadro’s number: 6.022\times10^{23} particles/mol.
Calculations and practice problems from your list — solved
Identifying sig figs — quick answers
0.004500 → 4 significant figures (4500 with leading zeros not significant; trailing zeros after decimal are significant).
1200 → ambiguous; writing as 1.200×10³ shows 4 sig figs; 1.2×10³ shows 2.
Example: Putting numbers into/out of scientific notation
7,890,000 → 7.89 \times 10^{6} (3 sig figs if original had 3 sig figs).
3.40\times10^{-5} → 0.0000340.
Dimensional analysis multistep example (CH1 style)
Problem: Convert 45.0 km/h to m/s.
1 km = 1000 m, 1 h = 3600 s.
45.0\ \text{km/h} \times \dfrac{1000\ \text{m}}{1\ \text{km}} \times \dfrac{1\ \text{h}}{3600\ \text{s}} = \dfrac{45.0\times1000}{3600}\ \text{m/s}
= 12.5\ \text{m/s} (sig figs: 45.0 has 3 sig figs → answer 3 sig figs → 12.5 m/s).
Given isotopic data — calculate average atomic mass (worked)
(Neon example shown earlier; result 20.180 amu).
Using periodic table data to find p, n, e and determine if atom or ion
Problem: Given symbol: {}^{37}_{17}\text{Cl}^{-}.
Protons = 17 (by atomic number).
Mass number = 37 → neutrons = 37 − 17 = 20.
Charge −1 → electrons = protons + 1 = 18.
This is an anion (ion). Not a neutral atom.
Lab hydrate calculations — general example (walkthrough)
(We already did CuSO₄·5H₂O example; follow same steps for your lab values.)
Density equation & rearrangements (reminder)
\rho = \dfrac{m}{V},\qquad m=\rho V,\qquad V=\dfrac{m}{\rho}.
Essay-style questions (short, clear answers + reasoning)
1) Classify a substance as element/compound/mixture, pure or mixture, molecules or ions (how to explain)
Example substance: Table salt (NaCl) from a bag.
Is it pure substance or mixture? If chemically pure NaCl sample → pure substance (compound). If table salt has additives (iodide, anti-caking agents), it’s a mixture.
Element/compound? NaCl is a compound (sodium + chlorine chemically bonded).
Molecules or ions? Ionic compound made of Na⁺ and Cl⁻ ions (not discrete molecules), so it consists of ions in a crystal lattice.
Why: composition fixed for compound; ionic bonding indicates ions rather than molecules.
2) Explain why some elements form cations and others form anions (use noble gas concept)
Atoms tend to reach a lower energy, more stable electron configuration. Many atoms achieve stability by adopting the electron configuration of the nearest noble gas:
Metals (left side): have few valence electrons; losing them gives a noble gas configuration → form cations (positive).
Nonmetals (right side): have more valence electrons but are short of an octet; gaining electrons gives a noble gas configuration → form anions (negative).
Example: Na (1 valence electron) loses 1 → Na⁺ (like Ne). Cl (7 valence electrons) gains 1 → Cl⁻ (like Ar).
3) Describe J.J. Thomson and Rutherford experiments (short)
Thomson: cathode ray deflection → discovered electrons → atom contains negative particles → plum pudding model.
Rutherford: gold foil scattering → most of atom empty; tiny dense positive nucleus deflects alpha particles → nuclear model of atom.
Worked examples you specifically asked (Example 1 & 2)
Example 1:
Using cations: Li⁺ and Ba²⁺; anions: O²⁻ and ClO₄⁻ (perchlorate). Create four neutral ionic compounds and explain subscripts.
We want neutral compounds (total positive charge = total negative charge).
Possible pairings:
Li⁺ + O²⁻ → Li₂O.
Reason: O²⁻ has −2, Li⁺ is +1. Need two Li⁺ to balance one O²⁻ → Li₂O.
Ba²⁺ + O²⁻ → BaO.
Reason: Ba²⁺ (+2) and O²⁻ (−2) already balance 1:1 → BaO.
Li⁺ + ClO₄⁻ → LiClO₄.
Reason: perchlorate has −1, Li⁺ has +1 → 1:1 ratio.
Ba²⁺ + ClO₄⁻ → Ba(ClO₄)₂.
Reason: Ba²⁺ (+2) needs two ClO₄⁻ to neutralize → use parentheses for two polyatomic anions.
How subscripts determined: cross-balance charges to get net zero; the smallest whole-number ratio is used.
Example 2: Write formula or name for each
I’ll give the correct formula or name and a brief explanation.
K₂S — Name: potassium sulfide.
K⁺ and S²⁻ → need two K⁺ to balance one S²⁻.
(NH₄)₂SO₄ — Name: ammonium sulfate.
NH₄⁺ is +1; sulfate is SO₄²⁻ → need two ammonium to balance sulfate.
OBr₂ — Name: oxygen dibromide? Wait — check elements: O + Br is unusual: oxygen normally forms −2, bromine usually −1; a neutral binary molecular compound OBr₂ would be better thought as dibromine monoxide? This is a tricky one. But the formula given is OBr₂; if treated as a binary molecular compound (two elements, both nonmetals), use prefixes: oxygen dibromide (but molecule uncommon).
Important note: In practice, more common bromine-oxygen compounds have different formulas (e.g., BrO₂⁻ is chlorite analogs). For classwork, accept oxygen dibromide as name for OBr₂ (prefix naming: mono/di). If the intended was phosphorus triiodide etc., the list might have mixed types; follow molecular naming rules for two nonmetals.
Na₂CO₃ · 10H₂O — Name: sodium carbonate decahydrate.
Sodium carbonate with ten waters attached.
Strontium phosphate — Formula: Sr₃(PO₄)₂.
Strontium is Sr²⁺; phosphate is PO₄³⁻. LCM of 2 and 3 is 6 → need three Sr²⁺ (3×+2=+6) and two PO₄³⁻ (2×−3=−6).
Gold(III) bromide — Formula: AuBr₃.
Gold (III) = Au³⁺, bromide = Br⁻ → need three Br⁻ to neutralize.
Beryllium sulfate tetrahydrate — Formula: BeSO₄·4H₂O.
Beryllium is Be²⁺, sulfate SO₄²⁻ → BeSO₄, plus 4 waters.
Phosphorous triiodide — Formula: PI₃.
Binary molecular: phosphorus (P) + iodine (I) → prefixes: phosphorus triiodide.
Aluminum hydroxide — Formula: Al(OH)₃.
Al³⁺ and OH⁻ → need three OH⁻ to balance one Al³⁺
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