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AP^® Chemistry

All nine College Board units, the six science practices, and the equation sheet: built for the MC and the 7 FRQs.

Start Unit 1 free. Unit 1: Atomic Structure & Properties is open to everyone, no account needed. Other topics are locked.

Unit 1: Atomic Structure & Properties

THE BIG PICTURE. Unit 1 establishes the quantitative and structural foundation for everything that follows. It is worth 7–9% of the exam but is referenced by every other unit. Two threads dominate: the mole concept (how chemists count atoms) and atomic structure (how electron arrangement determines reactivity). Master Avogadro's number, dimensional analysis, electron configuration, photoelectron spectroscopy (PES), and the periodic trends, and you have the toolkit to predict bonding (Unit 2), properties (Unit 3), and reaction behavior (Unit 4 onward).

THE MOLE AND DIMENSIONAL ANALYSIS

A mole is $6.022 \times 10^{23}$ entities (Avogadro's number, $N_A$ ). Molar mass is the mass of one mole, in grams per mole, numerically equal to the atomic/formula weight in amu. Every quantitative problem in AP Chemistry uses dimensional analysis to convert between mass, moles, particles, and (for gases) volume at STP.

Core conversions:

Moles ↔ Mass: $n = m / M$ (mole = grams / molar mass).
Moles ↔ Particles: $N = n \cdot N_A$ .
Moles of gas at STP (273.15 K, 1 atm): $V = n \cdot 22.4 \text{ L}$ (use ideal gas law for any other condition: see Unit 3).

MASS SPECTRA

A mass spectrum shows the abundance of each isotope of an element as peaks on a mass-to-charge axis. Average atomic mass is the weighted average: $M_{\text{avg}} = \sum (\text{isotope mass}) \cdot (\text{fractional abundance}).$ The MC routinely tests this: given a spectrum, compute the average or identify the element by comparing to a periodic table.

Mass spectrum of chlorine with peaks at mass-to-charge 35 (75.8 percent) and 37 (24.2 percent), beside a worked calculation of the average atomic mass, 35.48 amu. — **Mass spectrum of chlorine** Each peak is an isotope; peak height is its relative abundance. The weighted average lands nearer the more abundant isotope, which is why chlorine's atomic mass is about 35.5 rather than 36. On the exam you may be asked to work backward: given the average mass and the peak positions, find the abundances.

WORKED EXAMPLE: WORKING BACKWARD FROM AN AVERAGE MASS

Boron has two stable isotopes, boron-10 (10.013 amu) and boron-11 (11.009 amu). The periodic table lists 10.81 amu. What percentage of boron atoms are boron-10, and what would the mass spectrum look like?

Let $x$ be the fraction of boron-10, so $1 - x$ is boron-11: $10.013x + 11.009(1 - x) = 10.81$ .
Solve: $11.009 - 0.996x = 10.81$ , so $x = 0.199 / 0.996 = 0.200$ . About 20.0% boron-10 and 80.0% boron-11.
The spectrum has two peaks, at m/z 10 and 11, with the peak at 11 about four times taller.

Sanity check: 10.81 sits much closer to 11 than to 10, so the heavier isotope must dominate.

ELEMENTAL COMPOSITION AND EMPIRICAL FORMULAS

Percent composition by mass: $\%_X = (n_X \cdot M_X / M_{\text{compound}}) \cdot 100\%$ .

Empirical formula = smallest whole-number ratio of atoms. From percent composition: assume 100 g, convert each element's mass to moles, divide all moles by the smallest, multiply to whole numbers. Molecular formula = empirical formula multiplied by $M_{\text{molecular}} / M_{\text{empirical}}$ .

WORKED EXAMPLE: FROM PERCENT COMPOSITION TO MOLECULAR FORMULA

A white crystalline solid is 40.0% C, 6.7% H, and 53.3% O by mass, and its molar mass is 180 g/mol. Determine (a) the empirical formula and (b) the molecular formula.

Assume 100.0 g, so the percentages become grams: 40.0 g C, 6.7 g H, 53.3 g O.
Convert to moles. C: $40.0 / 12.01 = 3.33$ mol. H: $6.7 / 1.008 = 6.6$ mol. O: $53.3 / 16.00 = 3.33$ mol.
Divide by the smallest (3.33): C 1.00, H 2.0, O 1.00. Empirical formula CH $_2$ O (30.03 g/mol).
Scale up. $180 / 30.03 = 5.99 \approx 6$ , so the molecular formula is C $_6$ H $_{12}$ O $_6$ (glucose is one compound with this formula).

If a mole ratio comes out near 1.5 or 1.33, multiply every subscript by 2 or 3. Never round 1.5 to 2.

COMPOSITION OF MIXTURES

For mixtures, percent composition refers to the mass fraction of each component, and stoichiometry is applied per component: not per the mixture as a whole.

WORKED EXAMPLE: PURITY OF A MIXTURE

A 1.000 g sample is a mixture of NaCl and sugar (which contains no chlorine). Analysis finds 0.303 g of chlorine. What is the mass percent of NaCl in the sample?

Mass fraction of Cl in NaCl: $35.45 / 58.44 = 0.6066$ .
Mass of NaCl holding 0.303 g Cl: $0.303 / 0.6066 = 0.4995$ g.
Percent NaCl: $0.4995 / 1.000 \times 100\% = 50.0\%$ (three significant figures, set by 0.303 g).

The logic: only one component contains the element you measured, so that element "counts" that component. The same reasoning works for a mixture of two salts if you measure an ion found in only one of them.

ATOMIC STRUCTURE AND ELECTRON CONFIGURATION

Electrons occupy orbitals organized into shells, subshells, and individual orbitals. For AP, you must write configurations using the Aufbau order ( $1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, \ldots$ ), Hund's rule (singly occupy degenerate orbitals first with parallel spins), and the Pauli exclusion principle (no two electrons share the same set of quantum numbers). AP does NOT assess assigning a full set of four quantum numbers to a particular electron, but you should be fluent with shell, subshell, and orbital language.

Examples:

N (Z = 7): $1s^2 2s^2 2p^3$ .
Fe (Z = 26): $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6$ (or [Ar] $4s^2 3d^6$ ).
Fe $^{2+}$ : remove electrons from highest $n$ first → [Ar] $3d^6$ .
Aufbau exceptions (useful background, NOT assessed): Cr = [Ar] $4s^1 3d^5$ and Cu = [Ar] $4s^1 3d^{10}$ illustrate the extra stability of half-filled and filled $d$ subshells. The CED explicitly excludes writing electron configurations for Aufbau exceptions; the chemistry intuition is still worth knowing.

PHOTOELECTRON SPECTROSCOPY (PES)

PES bombards atoms with high-energy photons; ejected electrons' kinetic energies reveal the binding energy of each orbital. The spectrum has one peak per subshell. Peak interpretation:

Position (energy axis, MJ/mol): higher binding energy = closer to the nucleus (lower $n$ , or higher $Z_{\text{eff}}$ ).
Height/area: proportional to the number of electrons in that subshell (2 for $s$ , 6 for $p$ when full, 10 for $d$ ).
Order on a typical spectrum (College Board spectra usually plot binding energy decreasing left to right, but some sources reverse the axis: read axes carefully): $1s$ highest binding energy, valence electrons lowest.

Identifying an element from PES: count electrons (sum of areas), match to Z. The relative positions confirm shell structure.

Photoelectron spectrum of magnesium with four peaks: 1s at 126 MJ/mol, 2s at 9.07, 2p at 5.31 (tallest, six electrons), and 3s at 0.74, with binding energy decreasing left to right. — **Photoelectron spectrum of magnesium** Read a PES like a fingerprint: position gives binding energy (inner subshells far to the left), height gives electron count (the 2p peak is three times the 1s peak). The tiny 3s binding energy is the point of the diagram: two valence electrons shielded by ten core electrons, which is also why magnesium loses them so readily.

PERIODIC TRENDS

All trends derive from effective nuclear charge ( $Z_{\text{eff}}$ ) and shielding (and ultimately from Coulomb's law). The CED expects you to reason from these mechanisms, not just memorize directions:

Atomic radius: decreases left → right (greater $Z_{\text{eff}}$ pulls electrons in), increases top → bottom (more shells).
Ionic radius: cations smaller than parent atom; anions larger. Isoelectronic series: radius decreases as Z increases.
First ionization energy (IE $_1$ ): increases left → right, decreases top → bottom. Exceptions: dips at B/Al (new $p$ subshell) and O/S (paired $p$ electrons).
Successive IEs: large jump when removing a core electron after the valence shell is exhausted.
Electron affinity: the energy change when a gaseous atom gains an electron. Generally becomes more favorable (more negative) across a period, though the trend has notable exceptions (Group 2 elements are unfavorable due to their filled $s$ subshell, and Group 15 elements are less favorable than their neighbors due to their half-filled $p$ subshell). AP questions ask you to explain electron-affinity behavior using effective nuclear charge, electron-electron repulsion, and valence-shell structure, not to memorize numerical values.
Electronegativity: increases up and right; F is most electronegative.
Metallic character: opposite of electronegativity: increases down and left.

Simplified periodic table with arrows showing atomic radius increasing down and to the left and ionization energy and electronegativity increasing up and to the right, with cards explaining effective nuclear charge, shielding, and the Group 13 and Group 16 ionization-energy exceptions. — **Periodic trends and why they happen** The strongest trend explanations argue from Coulomb's law: charge and distance. Across a period the nuclear charge rises with no new shielding; down a group new shells add distance and shielding. Know the two ionization-energy exceptions (Groups 13 and 16) and why they happen.

Line graph of first ionization energy against atomic number for hydrogen through calcium, peaking at helium, neon, and argon, with dips at boron, oxygen, aluminum, and sulfur marked in amber and the drops at lithium, sodium, and potassium marked in blue. — **First ionization energy, Z = 1 to 20** IE rises across each period as effective nuclear charge grows, then plunges at each alkali metal, where a new shell begins. The small dips at B and Al (a new p subshell) and at O and S (the first paired p electron) are the exceptions the exam asks you to justify.

Reading the first-ionization-energy graph. Across a period IE $_1$ climbs because $Z_{\text{eff}}$ grows while the valence electrons stay in the same shell. The two dips are favorite "justify" prompts. B (801 kJ/mol) is lower than Be (900 kJ/mol) because boron's electron is removed from a $2p$ orbital, which is higher in energy and slightly more shielded than $2s$ . O (1314 kJ/mol) is lower than N (1402 kJ/mol) because the electron removed from oxygen shares an orbital with another electron, and that extra electron-electron repulsion makes it easier to remove. Down a group (Li, Na, K: 520, 496, 419 kJ/mol) the valence electron sits in a higher shell, farther from the nucleus, so the Coulombic attraction weakens.

Bar chart of the first five ionization energies of magnesium: 738, 1,451, 7,733, 10,542, and 13,630 kJ/mol, with a dashed line marking the large jump between the second and third. — **Successive ionization energies of magnesium** The first two electrons come from the 3s subshell; the third must come from the 2p core, which is closer to the nucleus and far less shielded. A jump after the second electron identifies a Group 2 element.

WORKED EXAMPLE: NAME THE ELEMENT FROM ITS IONIZATION ENERGIES

A period 3 element has successive ionization energies (kJ/mol) of 578, 1817, 2745, 11,577, and 14,842. Identify the element and justify your answer with Coulomb's law.

Find the jump. The ratios are IE $_2$ /IE $_1$ = 3.1, IE $_3$ /IE $_2$ = 1.5, and IE $_4$ /IE $_3$ = 4.2. The big jump (from 2,745 to 11,577 kJ/mol) comes after the third electron.
Claim: the element has three valence electrons, so it is in Group 13: aluminum.
Justify: the fourth electron comes from the $n = 2$ shell, which is much closer to the nucleus and far less shielded, so it feels a much larger effective nuclear charge. By Coulomb's law ( $F \propto q_1 q_2 / r^2$ ) the attraction is far stronger and far more energy is needed.

A rubric point usually requires BOTH ideas (smaller distance and less shielding, or larger $Z_{\text{eff}}$ ), not just "it is a core electron."

PRACTICE: TREND OR EXCEPTION?

Cover the last two columns and predict.

Comparison	Larger value	Why (Coulomb's law language)
IE $_1$ : Mg vs. Al	Mg	Al loses a $3p$ electron: higher in energy and more shielded than Mg's $3s$
IE $_1$ : P vs. S	P	S loses a paired $3p$ electron; electron-electron repulsion lowers the energy needed
Radius: Na vs. Na $^+$	Na	Na $^+$ has lost its entire $n = 3$ shell
Radius: F $^-$ vs. Na $^+$	F $^-$	Both have 10 electrons, but Na $^+$ has 11 protons vs. 9, pulling its electrons closer
IE $_1$ : Ne vs. Na	Ne	Na's electron is in a new shell ( $n = 3$ ), farther out and shielded by 10 core electrons
Electronegativity: O vs. S	O	O's valence shell is closer to the nucleus, so it attracts bonding electrons more strongly

VALENCE ELECTRONS AND IONIC COMPOUNDS

Main-group elements gain/lose electrons to achieve a noble-gas configuration. Group 1 → +1, Group 2 → +2, Group 13 → +3, Group 15 → −3, Group 16 → −2, Group 17 → −1. Transition metals can form multiple cations (Fe $^{2+}$ /Fe $^{3+}$ ; Cu $^+$ /Cu $^{2+}$ ). Ionic formula written so total charge is zero: e.g., Mg $^{2+}$ + Cl $^-$ → MgCl $_2$ .

EXAM CONNECTIONS. Unit 1 is mostly Practice 5 (mathematical routines) on the MC: mole conversions, percent composition, average atomic mass, simple electron configurations, periodic trend rankings. FRQs may pair PES with periodic trends ("Which spectrum is consistent with element X? Justify."). Empirical/molecular formula calculations can appear in both MC and FRQ parts. Avogadro's number and molar masses are provided on the exam (equation sheet and periodic table), so focus on using them fluently; memorize the periodic trends with their exceptions. AP does NOT require you to write Aufbau-exception configurations from memory or to assign quantum numbers to specific electrons.

Key Terms

Mole

A counting unit equal to Avogadro's number $N_A = 6.022 \times 10^{23}$ entities. Bridges the macroscopic (grams) and atomic (particles) scales.

Molar Mass ( $M$ )

Mass of one mole of a substance, in g/mol, numerically equal to the atomic or formula weight in amu. Use to convert between mass and moles: $n = m / M$ .

Mass Spectrum

A plot of relative isotope abundance versus mass-to-charge ratio. The weighted-average peak position gives the average atomic mass: $M_{avg} = \sum (\text{isotope mass}) \cdot (\text{fractional abundance})$ .

Empirical Formula

Smallest whole-number ratio of atoms in a compound. Derived from percent composition by converting to moles and dividing by the smallest mole value.

Electron Configuration

Arrangement of electrons in atomic orbitals, written using Aufbau order, Hund's rule, and Pauli exclusion. Cr ([Ar] $4s^1 3d^5$ ) and Cu ([Ar] $4s^1 3d^{10}$ ) are well-known Aufbau exceptions, but AP does not assess writing configurations for Aufbau exceptions.

Photoelectron Spectroscopy (PES)

Technique that ejects electrons via high-energy photons and measures their binding energies. Peak positions reveal shell/subshell structure; peak areas indicate the number of electrons in each subshell.

Effective Nuclear Charge ( $Z_{eff}$ )

Net positive charge experienced by a valence electron after accounting for shielding by inner electrons. Increases across a period (more protons, similar shielding); roughly constant down a group.

Ionization Energy (IE)

Energy required to remove one mole of electrons from one mole of gaseous atoms. IE $_1$ increases left → right and decreases top → bottom on the periodic table. Large jumps signal core-electron removal.

Electronegativity

Tendency of a bonded atom to attract shared electrons. Increases up and right; F most electronegative (4.0 on the Pauling scale). Drives bond polarity.

Periodic Trend Exceptions

IE dips at B/Al (new $p$ subshell starts) and O/S (first electron pair in $p$ creates repulsion). Atomic-radius and electron-affinity trends have similar small irregularities.

Electron Affinity

Energy change when a gaseous atom gains an electron. Generally becomes more favorable across a period (greater $Z_{eff}$ stabilizes the added electron), with exceptions for Group 2 (filled $s$ ) and Group 15 (half-filled $p$ ). AP asks you to explain electron-affinity behavior, not memorize values.

Isoelectronic Series

Atoms/ions with the same electron configuration (e.g., $N^{3-}$ , $O^{2-}$ , $F^-$ , Ne, $Na^+$ , $Mg^{2+}$ , $Al^{3+}$ all have Ne configuration). Within the series, radius decreases as $Z$ increases (greater pull on the same number of electrons).

Coulomb's Law (Atomic Context)

$F \propto q_1 q_2 / r^2$ . Binding energy of an electron increases with $Z_{eff}$ and decreases with $n$ : used to rationalise ionization energies, atomic radii, and PES peak positions.

Exam Tips

Avogadro's number is on the equation sheet and molar masses come from the provided periodic table, but knowing common values (H = 1.01, C = 12.01, N = 14.01, O = 16.00, Na = 23.0, Mg = 24.3, S = 32.1, Cl = 35.45, Fe = 55.85, Cu = 63.55) saves time. The Cr/Cu Aufbau exceptions are useful background but are NOT assessed on the AP exam.
For PES: count total electron area to identify Z, then check that peak positions and relative areas match the expected configuration. Peak position (binding energy) increases with $Z_{eff}$ and decreases with $n$ .
For periodic-trend MC, default to the standard trends but check for the Group 13 and Group 16 IE exceptions before locking in your answer.
For empirical formula problems, always start by assuming 100 g so percent composition becomes mass directly. Divide moles by the smallest, multiply if any number is non-integer ( $\times 2$ for 1.5, $\times 3$ for 1.33, etc.).
When writing cation configurations for transition metals, remove $ns$ electrons BEFORE $(n{-}1)d$ : a frequent FRQ trap.
Use Coulomb's law arguments ( $F \propto q_1 q_2 / r^2$ ) to JUSTIFY trend rankings in FRQ explanations: the rubrics reward the explicit mechanism, not just the direction.

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AP® Chemistry

Unit 1: Atomic Structure & Properties

THE MOLE AND DIMENSIONAL ANALYSIS

Core conversions:

MASS SPECTRA

WORKED EXAMPLE: WORKING BACKWARD FROM AN AVERAGE MASS

ELEMENTAL COMPOSITION AND EMPIRICAL FORMULAS

WORKED EXAMPLE: FROM PERCENT COMPOSITION TO MOLECULAR FORMULA

COMPOSITION OF MIXTURES

WORKED EXAMPLE: PURITY OF A MIXTURE

ATOMIC STRUCTURE AND ELECTRON CONFIGURATION

Examples:

PHOTOELECTRON SPECTROSCOPY (PES)

PERIODIC TRENDS

WORKED EXAMPLE: NAME THE ELEMENT FROM ITS IONIZATION ENERGIES

PRACTICE: TREND OR EXCEPTION?

VALENCE ELECTRONS AND IONIC COMPOUNDS

Key Terms

Exam Tips

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AP^® Chemistry