CHEM 111 — Exam 1 Study Guide

Chapters 1–3 · Fall 2026

Contents
  1. Chapter 1 — Matter & Measurement
  2. Chapter 2 — Atomic Theory & Nomenclature
  3. Chapter 3 — The Mole & Stoichiometry
  4. Exam Strategy Tips

Chapter 1 — Matter & Measurement

Classifying Matter

Matter splits first into pure substances and mixtures. Pure substances have a fixed composition; mixtures can vary.
Quick test: Ask "if I take a sample from two different spots, is it identical?" Yes → homogeneous. No → heterogeneous.

Uncertainty in Measurements

Estimating from instruments: Always record all certain digits plus one estimated digit (the smallest marked division, estimated to the nearest tenth of that division).

Significant Figures — Rules

RuleExample
Nonzero digits always count123 → 3 sig figs
Zeros between nonzero digits count1002 → 4 sig figs
Leading zeros never count0.0025 → 2 sig figs
Trailing zeros count only if there's a decimal point100. → 3 sig figs; 100 → 1 sig fig
Trailing zeros after a decimal always count1.500 → 4 sig figs

Math rules:

Metric Prefixes (memorize these four + base)

PrefixSymbolMeaning (× base unit)
kilok1,000 (10³)
— (base)1
centic0.01 (10⁻²)
millim0.001 (10⁻³)
microµ0.000001 (10⁻⁶)
Trick: Move the decimal point — converting to a smaller unit (e.g., m→cm) means a bigger number, and vice versa.

The 7 Base SI Units

QuantityUnitSymbol
Lengthmeterm
Masskilogramkg
Timeseconds
TemperaturekelvinK
Amount of substancemolemol
Electric currentampereA
Luminous intensitycandelacd

Unit Conversion — Dimensional Analysis

desired unit = given quantity × (conversion factor)

Chain factors so units cancel diagonally. Always double-check the final unit matches what's asked.

Density Conversions

Density (d) = mass (m) / volume (V)

Density acts as a conversion factor between mass and volume: e.g., g → mL using g/mL.

Example: A liquid has density 1.25 g/mL. Find the volume of 50.0 g.
50.0 g × (1 mL / 1.25 g) = 40.0 mL

Temperature Conversions

°F = (°C × 9/5) + 32   |   °C = (°F − 32) × 5/9
K = °C + 273.15

Know both directions (F→C and C→F) — you'll likely be given this formula on the exam, but practice using it quickly.


Chapter 2 — Atomic Theory & Nomenclature

Historical Atomic Theory

ScientistExperiment/IdeaConclusion
DaltonDalton's Atomic Theory (Laws of Definite/Multiple Proportions, Conservation of Mass)Matter is made of indivisible atoms; atoms of an element are identical; compounds form in fixed ratios
J.J. ThomsonCathode Ray Tube experimentDiscovered the electron — a negatively charged particle; proposed the plum pudding model (negative electrons scattered in a positive "pudding")
MillikanOil Drop experimentDetermined the charge of a single electron (and thus its mass) by balancing gravity vs. an electric field on charged oil droplets
RutherfordGold Foil experimentMost alpha particles passed through foil, but some deflected sharply → atom is mostly empty space with a small, dense, positively charged nucleus (disproved plum pudding model)

Subatomic Particles in Atoms & Ions

ParticleDetermines
# protonsAtomic number (Z) — defines the element; never changes for a given element
# neutronsMass number − atomic number (A − Z); varies → isotopes
# electrons= protons in a neutral atom. For ions: subtract charge from protons
Cation (+): fewer electrons than protons
Anion (−): more electrons than protons
Example: Fe³⁺ has 26 protons, so electrons = 26 − 3 = 23 electrons.

Isotopes & Nuclear Notation

AZX   (A = mass number, Z = atomic number, X = element symbol)

Isotopes = atoms of the same element (same Z) with different numbers of neutrons (different A).

Mass number (A) = protons + neutrons (always a whole number for a single isotope).

Natural abundance = the % of each isotope found in nature, used as weighting factors for average atomic mass.

Average Atomic Mass

Avg. mass = Σ (isotope mass × fractional abundance)
Example: Cl-35 (mass 34.97, abundance 75.76%) and Cl-37 (mass 36.97, abundance 24.24%):
(34.97 × 0.7576) + (36.97 × 0.2424) = 35.45 amu

Elements 1–86

You need to know symbol ↔ name for elements 1–86. Rather than reproduce all 86 here, drill yourself with these high-yield groups, then use flashcards/periodic table practice for the rest:

GroupElements
Common "tricky" symbolsNa (sodium), K (potassium), Fe (iron), Cu (copper), Ag (silver), Au (gold), Hg (mercury), Sn (tin), Pb (lead), Sb (antimony), W (tungsten)
Alkali metals (Grp 1)H, Li, Na, K, Rb, Cs
Alkaline earth (Grp 2)Be, Mg, Ca, Sr, Ba
Halogens (Grp 17)F, Cl, Br, I
Noble gases (Grp 18)He, Ne, Ar, Kr, Xe
Common transition metalsSc–Zn (row 4), plus Ag, Cd, Au, Hg, Pt

Study tip: print a blank periodic table (1–86) and quiz yourself filling in names/symbols until it's automatic.

Naming Ionic Compounds

Basic pattern: cation name + anion name (anion of a single element ends in -ide).

Example: NaCl → sodium chloride. CaO → calcium oxide.

Common Polyatomic Ions (memorize charges!)

IonNameIonName
NH₄⁺ammoniumOH⁻hydroxide
NO₃⁻nitrateNO₂⁻nitrite
SO₄²⁻sulfateSO₃²⁻sulfite
CO₃²⁻carbonateHCO₃⁻bicarbonate/hydrogen carbonate
PO₄³⁻phosphatePO₃³⁻phosphite
ClO⁻hypochloriteClO₂⁻chlorite
ClO₃⁻chlorateClO₄⁻perchlorate
C₂H₃O₂⁻ (or CH₃COO⁻)acetateCN⁻cyanide
MnO₄⁻permanganateCr₂O₇²⁻dichromate
CrO₄²⁻chromateC₂O₄²⁻oxalate
Pattern trick: "-ate" ions usually have one more O than the "-ite" version, same charge (e.g., NO₃⁻ vs NO₂⁻). "Per-...-ate" has one more O than "-ate"; "hypo-...-ite" has one fewer O than "-ite."

Naming with Transition Metals (variable charge)

Many transition metals form more than one charge, so you must indicate charge with a Roman numeral in parentheses (this is the "Stock system").

Example: Fe₂O₃ → each O is 2−, three O's = 6− total, so 2 Fe must total 6+ → each Fe is 3+ → iron(III) oxide.
FeO → iron(II) oxide.

Exceptions that don't need Roman numerals (fixed charge): Ag⁺, Zn²⁺, Cd²⁺.

Naming Molecular (Covalent) Compounds

Use Greek prefixes for both elements (mono- is dropped for the first element only).

#Prefix#Prefix
1mono-6hexa-
2di-7hepta-
3tri-8octa-
4tetra-9nona-
5penta-10deca-
Example: N₂O₄ → dinitrogen tetroxide. CO₂ → carbon dioxide (not "monocarbon").

Naming Acids

Acid typeRuleExample
Binary acid (H + one nonmetal)hydro- + root + -ic acidHCl → hydrochloric acid
Oxyacid from "-ate" ionroot + -ic acidH₂SO₄ (sulfate) → sulfuric acid
Oxyacid from "-ite" ionroot + -ous acidH₂SO₃ (sulfite) → sulfurous acid

Chapter 3 — The Mole & Stoichiometry

The Mole & Avogadro's Number

1 mole = 6.022 × 10²³ particles (Avogadro's number)

Molar Mass

Molar mass (g/mol) = sum of atomic masses (from the periodic table) of all atoms in a formula.

Example: H₂O = 2(1.008) + 16.00 = 18.02 g/mol

Mole ↔ Mass ↔ Particle Conversions

grams ⇄ moles: use molar mass (g/mol)
moles ⇄ particles: use Avogadro's number (6.022×10²³/mol)

Set up as a chain: grams → moles → particles (or reverse), canceling units at each step.

Percent Composition

% element = (mass of element in formula / molar mass of compound) × 100%
Example: % H in H₂O = (2.016/18.02) × 100 = 11.19%

Empirical Formula

The simplest whole-number ratio of atoms in a compound.

  1. Assume 100 g sample → convert each given % directly to grams.
  2. Convert grams of each element to moles (divide by atomic mass).
  3. Divide all mole values by the smallest mole value.
  4. If not whole numbers, multiply all by a small integer to get whole numbers.

Molecular Formula

Molecular formula = empirical formula × n,  where n = (molar mass given) / (empirical formula mass)
Example: Empirical formula CH₂O (mass 30 g/mol), actual molar mass = 180 g/mol → n = 6 → molecular formula = C₆H₁₂O₆

Combustion Analysis

Used to find empirical formulas of compounds containing C, H, (and O) by burning a sample completely in O₂ and measuring the CO₂ and H₂O produced.

  1. All C in sample ends up in CO₂ → mol C = mol CO₂ produced.
  2. All H in sample ends up in H₂O → mol H = 2 × mol H₂O produced.
  3. If the compound contains O: mass O = (mass sample) − (mass C) − (mass H); convert to moles.
  4. Find mole ratios → empirical formula (same method as above).

Balancing Chemical Equations

Law of Conservation of Mass: atoms of each element must be equal on both sides.
  1. Write correct formulas for all reactants/products first (don't touch subscripts).
  2. Balance one element at a time, usually starting with the most complex molecule.
  3. Balance elements that appear alone (like O₂ or H₂) last.
  4. Double-check every element's atom count matches on both sides.
  5. Reduce coefficients to smallest whole-number ratio if needed.
Example: C₃H₈ + O₂ → CO₂ + H₂O
Balance C: C₃H₈ + O₂ → 3CO₂ + H₂O
Balance H: C₃H₈ + O₂ → 3CO₂ + 4H₂O
Balance O (right side has 6+4=10 O): C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Exam Strategy Tips