⚛️ Unit 1: Atomic Structure and Properties

Explore the building blocks of matter — from counting atoms to predicting their behavior across the periodic table.

Exam Weight: 7–9%  |  Topics 1.1–1.8

📑 In This Unit

1.1 — Moles and Molar Mass

Chemistry deals with incredibly large numbers of atoms and molecules. A single drop of water contains roughly 1021 molecules. To make these numbers manageable, chemists use the mole (abbreviated mol), one of the seven SI base units. The mole bridges the gap between the atomic world (individual atoms and molecules) and the macroscopic world (grams, liters) that we can measure in the lab.

Avogadro’s Number: One mole of any substance contains exactly 6.022 × 1023 representative particles (atoms, molecules, ions, or formula units). This number is defined so that 12.000 g of carbon-12 contains exactly 1 mole of atoms.

The molar mass of an element is the mass of one mole of that element, expressed in grams per mole (g/mol). It is numerically equal to the element’s average atomic mass found on the periodic table (in amu). For compounds, the molar mass is the sum of the molar masses of all atoms in the chemical formula.

The Mole Roadmap — Dimensional Analysis

Dimensional analysis (factor-label method) is the core problem-solving strategy for mole conversions. Always start with what you know, then multiply by conversion factors until you reach the desired unit. Units must cancel properly.

ConversionFormulaUnits
Mass → Molesn = mass ÷ molar massg ÷ (g/mol) = mol
Moles → ParticlesParticles = n × 6.022 × 1023mol × (particles/mol)
Moles → Massmass = n × molar massmol × (g/mol) = g
Particles → Molesn = particles ÷ 6.022 × 1023particles ÷ (particles/mol) = mol
Example 1: How many moles are in 36.0 g of water (H₂O)?
Molar mass of H₂O = 2(1.008) + 16.00 = 18.016 g/mol
n = 36.0 g ÷ 18.016 g/mol = 2.00 mol
Example 2: How many molecules are in 2.50 mol of CO₂?
Molecules = 2.50 mol × 6.022 × 1023 molecules/mol = 1.51 × 1024 molecules

Follow-up: How many oxygen atoms are in that sample?
Each CO₂ has 2 oxygen atoms, so: 1.51 × 1024 × 2 = 3.01 × 1024 oxygen atoms
Example 3 (Multi-step): What is the mass of 3.01 × 1022 atoms of iron?
Step 1: Moles of Fe = 3.01 × 1022 ÷ 6.022 × 1023 = 0.0500 mol
Step 2: Mass = 0.0500 mol × 55.85 g/mol = 2.79 g
Common Mistakes on the AP Exam

1.2 — Mass Spectrometry of Elements

A mass spectrometer is an instrument that separates atoms or molecules based on their mass-to-charge ratio (m/z). In AP Chemistry, you will encounter mass spectra of elements to determine isotopic composition and calculate average atomic mass. Understanding mass spectrometry is essential because it provides direct experimental evidence for the existence of isotopes.

Block diagram of a mass spectrometer

Block diagram of a mass spectrometer — sample is ionized, accelerated, deflected by a magnetic field, and detected. (Wikimedia Commons)

How a Mass Spectrometer Works
  1. Vaporization: The sample is converted into a gas.
  2. Ionization: Atoms are bombarded with high-energy electrons to form positive ions (lose one electron). This gives a charge of +1 for most atoms, so m/z ≈ mass number.
  3. Acceleration: Ions are accelerated through an electric field. All ions with the same charge gain the same kinetic energy.
  4. Deflection: A magnetic field deflects the ions. Lighter ions are deflected more, heavier ions are deflected less. This separates isotopes.
  5. Detection: The detector records the relative abundance of each ion as a current. More ions → stronger signal → taller peak.
Reading a Mass Spectrum

The x-axis shows m/z (mass-to-charge ratio, which for singly-charged ions equals the mass number). The y-axis shows relative abundance (%). Each peak represents a different isotope of the element.

What the spectrum tells you:
Number of peaks = number of isotopes
Position of each peak (x-axis) = mass number of that isotope
Height of each peak (y-axis) = relative abundance of that isotope
• The tallest peak represents the most abundant isotope
• The average atomic mass on the periodic table will be closest to the most abundant isotope's mass
Calculating Average Atomic Mass:
Average atomic mass = Σ(isotope mass × fractional abundance)

Example — Chlorine:
Cl-35: mass = 34.97 amu, abundance = 75.76%
Cl-37: mass = 36.97 amu, abundance = 24.24%
Average = (34.97 × 0.7576) + (36.97 × 0.2424) = 26.50 + 8.96 = 35.45 amu

Notice: The average (35.45) is closer to 35 than 37, which makes sense because Cl-35 is about 3× more abundant.
Example — Estimating from a spectrum: If a mass spectrum shows two peaks at m/z = 63 and m/z = 65, with the peak at 63 being about twice as tall, you can estimate the average atomic mass is about (63 × 0.67) + (65 × 0.33) ≈ 63.7 amu. This matches copper (63.55 amu).
AP Exam Tips

1.3 — Elemental Composition of Pure Substances

Every pure substance has a fixed, definite composition by mass. This is described by the Law of Definite Proportions (also known as the Law of Constant Composition): a given compound always contains the same elements in the same proportions by mass, regardless of the sample size or source.

Percent Composition by Mass

Percent composition tells you the percentage of the total mass contributed by each element in a compound. This is useful for identifying unknown compounds and verifying purity.

Formula:
% by mass of element = (mass of element in 1 mol of compound ÷ molar mass of compound) × 100%

Example: Find the percent composition of H₂SO₄ (M = 98.09 g/mol).
%H = 2(1.008)/98.09 × 100 = 2.06%
%S = 32.07/98.09 × 100 = 32.69%
%O = 4(16.00)/98.09 × 100 = 65.25%
Check: 2.06 + 32.69 + 65.25 = 100.00%
Empirical vs. Molecular Formulas
Step-by-Step: Finding the Empirical Formula
  1. Start with percent composition (or mass data). Assume 100 g of the compound so percentages become grams directly.
  2. Convert grams to moles for each element (divide by atomic mass).
  3. Divide each by the smallest number of moles to get the mole ratio.
  4. If the ratios are not whole numbers, multiply all by the smallest integer that converts them to whole numbers (e.g., if you get 1.5, multiply all by 2).
Example: A compound is 40.0% C, 6.71% H, and 53.3% O. Find the empirical formula.
Assume 100 g: 40.0 g C, 6.71 g H, 53.3 g O
Moles: C = 40.0/12.01 = 3.33 mol; H = 6.71/1.008 = 6.66 mol; O = 53.3/16.00 = 3.33 mol
Ratio: 3.33/3.33 : 6.66/3.33 : 3.33/3.33 = 1 : 2 : 1
Empirical formula = CH₂O
Finding the Molecular Formula
n = molar mass of compound ÷ molar mass of empirical formula

Molecular formula = (empirical formula) × n

Example: If the empirical formula is CH₂O (M = 30.03 g/mol) and the actual molar mass is 180.2 g/mol:
n = 180.2 / 30.03 = 6
Molecular formula = C₆H₁₂O₆ (glucose)
Combustion Analysis

Combustion analysis is used to determine the empirical formula of organic compounds (containing C, H, and sometimes O). The compound is burned in excess O₂, and the masses of CO₂ and H₂O produced are measured.

Strategy:
• All carbon in the sample ends up as CO₂: mol C = mol CO₂
• All hydrogen ends up as H₂O: mol H = 2 × mol H₂O
• If oxygen is present: mass O = total sample mass − mass C − mass H
Then convert to moles and find the ratio as usual.

1.4 — Composition of Mixtures

Unlike pure substances, mixtures can vary in composition. They are classified as either homogeneous (uniform throughout, e.g., saltwater, air, brass) or heterogeneous (non-uniform, e.g., sand and water, granite). The key distinction from pure substances is that mixtures do not have a fixed composition.

Analyzing Mixtures

When a mixture contains substances with known formulas, you can use elemental analysis, mass data, or reaction data to determine the amount of each component present. These are often algebraic problems.

Strategy for mixture problems:
  1. Define variables: Let x = mass (or moles) of component A, and (total − x) = mass (or moles) of component B.
  2. Write an equation relating a measured quantity (e.g., total mass of a specific element, total moles of a product) to the variables.
  3. Solve the equation for x.
Example: A 10.0 g mixture of NaCl and KCl is dissolved in water and treated with excess AgNO₃, producing 18.0 g of AgCl. Find the mass percent of NaCl.

Let x = g NaCl, then (10.0 − x) = g KCl.
Mol Cl⁻ from NaCl = x / 58.44; mol Cl⁻ from KCl = (10.0 − x) / 74.55
Total mol Cl⁻ = mol AgCl = 18.0 / 143.32 = 0.1256 mol
x/58.44 + (10.0 − x)/74.55 = 0.1256
Solving: x = 4.3 g NaCl (43% NaCl by mass)
Separation Techniques

Mixtures can be separated based on differences in physical properties:

TechniqueSeparates Based OnExample Use
FiltrationParticle size (solid from liquid)Sand from saltwater
DistillationBoiling point differencesSeparating ethanol from water
ChromatographyDifferential affinity for stationary vs. mobile phaseSeparating pigments in ink
EvaporationVolatility (remove solvent)Recovering NaCl from saltwater
CentrifugationDensity differencesSeparating blood components
Pure Substance vs. Mixture — How to Tell

1.5 — Atomic Structure and Electron Configuration

Atoms consist of a dense, positively charged nucleus (containing protons and neutrons) surrounded by electrons occupying energy levels (shells) and subshells. Understanding electron configurations is foundational — it explains periodic trends, bonding, and reactivity.

Bohr model of the atom

Bohr model of the atom showing discrete energy levels (shells). (Wikimedia Commons)

Electron orbital shapes (s, p, d, f)

Shapes of s, p, d, and f orbitals. Each orbital holds a maximum of 2 electrons. (Wikimedia Commons)

Subatomic Particles
ParticleSymbolChargeMass (amu)Location
Protonp⁺+11.007Nucleus
Neutronn⁰01.009Nucleus
Electrone⁻−1~0 (5.49 × 10⁻⁴)Electron cloud

Atomic number (Z) = number of protons = defines the element. Mass number (A) = protons + neutrons. Isotopes have the same Z but different A (different numbers of neutrons).

Three Key Principles for Electron Configuration
Subshell Capacities
Subshell# of OrbitalsMax ElectronsShape
s12Spherical
p36Dumbbell (along x, y, z axes)
d510Cloverleaf (various orientations)
f714Complex multilobed
Writing Electron Configurations

List each occupied subshell with a superscript for the number of electrons. Noble-gas shorthand uses brackets around the previous noble gas core.

Examples:
Carbon (C, Z = 6): 1s² 2s² 2p²
Sodium (Na, Z = 11): 1s² 2s² 2p⁶ 3s¹  or  [Ne] 3s¹
Iron (Fe, Z = 26): [Ar] 3d⁶ 4s²  (note: 4s fills before 3d)
Krypton (Kr, Z = 36): [Ar] 3d¹⁰ 4s² 4p⁶

Important exceptions (memorize these):
Chromium (Cr, Z = 24): [Ar] 3d⁵ 4s¹ (not 3d⁴ 4s²) — half-filled d subshell provides extra stability
Copper (Cu, Z = 29): [Ar] 3d¹⁰ 4s¹ (not 3d⁹ 4s²) — fully-filled d subshell provides extra stability
Electron Configurations of Ions

This is a critical concept that appears frequently on the AP exam:

Example:
Fe²⁺ (Z = 26, lost 2e⁻): [Ar] 3d⁶  (remove the two 4s electrons, NOT the 3d electrons)
Fe³⁺ (Z = 26, lost 3e⁻): [Ar] 3d⁵  (remove 4s² first, then one 3d electron)
Cl⁻ (Z = 17, gained 1e⁻): [Ne] 3s² 3p⁶ = [Ar]  (isoelectronic with Ar)
Quantum Numbers (Supplemental)

Each electron is described by four quantum numbers:

1.6 — Photoelectron Spectroscopy (PES)

Photoelectron spectroscopy provides direct experimental evidence for the shell structure of atoms and validates electron configurations. It works by hitting atoms with high-energy photons (usually X-rays or UV light) and measuring the kinetic energy of the ejected electrons. The difference between the photon energy and the kinetic energy gives the binding energy of each electron.

Electromagnetic spectrum

The electromagnetic spectrum — PES uses high-energy UV or X-ray photons to eject core and valence electrons. (Wikimedia Commons)

Key Equation:
Ephoton = Ebinding + Ekinetic

The binding energy (ionization energy) of an electron = energy of the incoming photon − kinetic energy of the ejected electron.
Binding energy tells you how tightly an electron is held by the nucleus.
Reading a PES Diagram — Step by Step
Example — Nitrogen (Z = 7):
Configuration: 1s² 2s² 2p³
PES shows 3 peaks:
• Peak 1 (highest BE): relative height 2 → 1s²
• Peak 2 (medium BE): relative height 2 → 2s²
• Peak 3 (lowest BE): relative height 3 → 2p³
Example — Identifying an element from PES data:
A spectrum shows peaks with relative heights: 2, 2, 6, 2, 4.
Total electrons = 2 + 2 + 6 + 2 + 4 = 16 → Sulfur (S)
Configuration: 1s² 2s² 2p⁶ 3s² 3p⁴
What PES Reveals
Common AP Exam Mistakes

1.7 — Periodic Trends

The periodic table organizes elements so that properties change predictably. Understanding these trends is fundamental to predicting chemical behavior. The underlying explanation for all periodic trends is effective nuclear charge (Zeff) and electron shielding.

Periodic trends diagram

Major periodic trends: atomic radius, ionization energy, electron affinity, and electronegativity. (Wikimedia Commons)

Effective Nuclear Charge (Zeff):
Zeff = Z − S (where Z = atomic number, S = shielding by inner electrons)

Across a period: Z increases by 1 with each element, but electrons are added to the same shell (same shielding). So Zeff increases.
Down a group: New electron shells are added, greatly increasing shielding. Even though Z also increases, the effect of the new shell dominates, so Zeff felt by valence electrons does not increase as dramatically.
Atomic Radius
Ionic Radius Comparisons:
Cations are smaller than their parent atoms (lost electrons → fewer e⁻, same nuclear charge → stronger pull).
Anions are larger than their parent atoms (gained electrons → more e⁻ repulsion, same nuclear charge).
Isoelectronic series (same # of electrons, e.g., O²⁻, F⁻, Ne, Na⁺, Mg²⁺): The ion with more protons is smallest. So Mg²⁺ < Na⁺ < Ne < F⁻ < O²⁻.
Ionization Energy (IE)
Important exceptions to memorize:
1. Group 2 → Group 13: IE drops (e.g., Be > B). The electron removed from B is in a 2p orbital, which is higher in energy and easier to remove than the 2s electron in Be.
2. Group 15 → Group 16: IE drops (e.g., N > O). Oxygen has a paired electron in one 2p orbital, and the electron-electron repulsion makes it easier to remove.
Ionization energies of elements graph

First ionization energies of elements, showing the periodic pattern and notable exceptions. (Wikimedia Commons)

Successive Ionization Energies

Each successive ionization energy is larger than the previous one (removing an electron from a more positive ion is harder). A large jump in successive IEs indicates that a core electron is being removed (breaking into the next shell).

Example — Aluminum (Z = 13): 1s² 2s² 2p⁶ 3s² 3p¹
IE₁, IE₂, IE₃: remove the 3 valence electrons (gradual increase)
IE₄: HUGE jump — now removing a 2p core electron
This pattern reveals that Al has 3 valence electrons.
Electronegativity
Electron Affinity

1.8 — Valence Electrons and Ionic Compounds

Valence electrons are the electrons in an atom’s outermost energy level (highest principal quantum number). They determine an element’s chemical properties, bonding behavior, and reactivity. Elements in the same group have the same number of valence electrons, which is why they exhibit similar chemistry.

Finding Valence Electrons
Group (1–18 numbering)Valence ElectronsExamples
Group 1 (Alkali metals)1Na [Ne] 3s¹
Group 2 (Alkaline earth)2Mg [Ne] 3s²
Groups 13–18Group # − 10C (Group 14) = 4 VE
Transition metals (Groups 3–12)Varies; includes s + d electronsFe: 2 (4s²) shown commonly
Ionic Compounds

Ionic compounds form when metals transfer electrons to nonmetals. The result is oppositely charged ions held together by electrostatic attraction (the ionic bond). Both ions typically achieve a noble-gas electron configuration.

Predicting ionic charges:
• Group 1 metals form +1 cations (lose 1 e⁻)
• Group 2 metals form +2 cations (lose 2 e⁻)
• Aluminum forms +3 (loses 3 e⁻ to get [Ne] configuration)
• Group 16 nonmetals form −2 anions (gain 2 e⁻)
• Group 17 halogens form −1 anions (gain 1 e⁻)
• Transition metals can form multiple charges (e.g., Fe²⁺ and Fe³⁺)
Coulomb’s Law and Ionic Bond Strength
Coulomb’s Law: F = k · |q₁||q₂| / r²

The electrostatic force (and therefore the strength of the ionic bond) is:
Directly proportional to the product of the ion charges (higher charges = stronger bond)
Inversely proportional to the square of the distance between ion centers (smaller ions = stronger bond)
Lattice Energy

Lattice energy is the energy released when gaseous ions come together to form a solid ionic compound (or equivalently, the energy required to completely separate an ionic solid into gaseous ions). It directly correlates with the strength of ionic bonds in the crystal:

Example — Ranking lattice energies:
Rank: NaCl, MgO, KBr, CaO

CaO (2+/2−, small ions) ≈ MgO (2+/2−, even smaller ions) > NaCl (1+/1−, small) > KBr (1+/1−, larger ions)
MgO > CaO > NaCl > KBr

MgO has the highest lattice energy because Mg²⁺ is smaller than Ca²⁺, and both have 2+/2− charges.
Properties of Ionic Compounds

📝 Practice Multiple-Choice Questions

Test your knowledge of Unit 1. Click "Show Answer" to reveal the correct choice and explanation.

1. A sample of an element contains two isotopes: isotope A has a mass of 10.01 amu and a relative abundance of 19.9%, and isotope B has a mass of 11.01 amu and a relative abundance of 80.1%. What is the average atomic mass of the element?

Answer: (C)
Average = (10.01 × 0.199) + (11.01 × 0.801) = 1.99 + 8.82 = 10.81 amu. The average is weighted toward isotope B since it is far more abundant (~80%).

2. How many moles of oxygen atoms are present in 0.50 mol of Ca(NO₃)₂?

Answer: (C)
Ca(NO₃)₂ contains 6 oxygen atoms per formula unit (3 oxygens × 2 nitrate groups = 6). Moles of O = 0.50 mol × 6 = 3.0 mol.

3. A compound has the empirical formula CH₂O and a molar mass of approximately 180 g/mol. What is its molecular formula?

Answer: (C)
Empirical formula mass of CH₂O = 12.01 + 2(1.008) + 16.00 = 30.03 g/mol. Multiplier = 180 / 30.03 = 6. Molecular formula = C₆H₁₂O₆ (glucose).

4. Which of the following represents the correct ground-state electron configuration for chromium (Cr, Z = 24)?

Answer: (B)
Chromium is an exception: a half-filled 3d⁵ subshell is especially stable, so one electron from 4s moves to 3d, giving [Ar] 3d⁵ 4s¹.

5. A PES spectrum for an element shows three peaks. The peak with the highest binding energy has a relative height of 2, the middle peak has a relative height of 2, and the peak with the lowest binding energy has a relative height of 3. Which element is this most likely?

Answer: (C)
The peak heights indicate electron counts: 2 (1s²), 2 (2s²), 3 (2p³). Total = 7 electrons = Nitrogen. The 3 peaks correspond to 3 subshells.

6. Which of the following correctly ranks the atoms in order of increasing atomic radius?

Answer: (B)
All are in Period 3. Atomic radius decreases across a period (left to right) due to increasing Zeff. So Na is the largest and Si is the smallest: Si < Al < Mg < Na.

7. The first ionization energy of oxygen is lower than that of nitrogen. Which of the following best explains this observation?

Answer: (C)
Nitrogen has a half-filled 2p³ (one electron per orbital, no pairing). Oxygen's 2p⁴ forces one pair, and the electron-electron repulsion in that paired orbital makes it easier (less energy) to remove that electron. This is a classic AP exam exception.

8. Which of the following ionic compounds would be expected to have the highest lattice energy?

Answer: (C)
Lattice energy depends on charge and ionic radius (Coulomb's Law). MgO has 2+ and 2− charges with small ionic radii, producing a much stronger electrostatic attraction than any of the 1+/1− compounds. CsI would have the lowest (1+/1−, very large ions).

9. The successive ionization energies (in kJ/mol) of an unknown element X are: 578, 1817, 2745, 11,578, 14,831. In which group of the periodic table is element X most likely found?

Answer: (C)
The huge jump occurs between IE₃ (2745) and IE₄ (11,578) — a factor of >4. This means the first 3 electrons were valence electrons, and the 4th is a core electron. Three valence electrons → Group 3 (or Group 13 with the 1–18 system, but here the pattern of 3 relatively low IEs followed by a big jump matches Group 13/3 with 3 VE).

10. Which of the following elements has the highest electronegativity?

Answer: (D)
Electronegativity increases across a period (left → right). All four elements are in Period 2, and fluorine is the most electronegative element on the entire periodic table (3.98 on the Pauling scale).
💡 Study Tip for Unit 1

Unit 1 is the foundation for everything in AP Chemistry. Make sure you can write electron configurations quickly (including ions and exceptions like Cr and Cu), read PES spectra to identify elements, and explain every periodic trend using effective nuclear charge. These skills appear throughout Units 2, 3, 7, and 8. Practice dimensional analysis with mole conversions until it becomes second nature. For the AP exam, always show your units in calculations and watch for questions that ask about atoms vs. molecules.

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