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Atomic Structure and Properties

AP Chemistry · Topic 1

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1.1

Moles and Molar Mass

Syllabus
Learning ObjectiveEssential Knowledge

1.1.A
Calculate quantities of a substance or its relative number of particles using dimensional analysis and the mole concept.

  • 1.1.A.1 One cannot count particles directly while performing laboratory work. Thus, there must be a connection between the masses of substances reacting and the actual number of particles undergoing chemical changes.
  • 1.1.A.2 Avogadro's number ($N_{\mathrm{A}} = 6.022 \times 10^{23}\ \mathrm{mol}^{-1}$) provides the connection between the number of moles in a pure sample of a substance and the number of constituent particles (or formula units) of that substance.
  • 1.1.A.3 Expressing the mass of an individual atom or molecule in atomic mass units (amu) is useful because the average mass in amu of one particle (atom or molecule) or formula unit of a substance will always be numerically equal to the molar mass of that substance in grams. Thus, there is a quantitative connection between the mass of a substance and the number of particles that the substance contains.
    • Equation: $n = m/M$

Source: College Board AP Course and Exam Description

Because atoms are far too small to count, chemists count in moles 摩尔. One mole is Avogadro's number 阿伏伽德罗常数 of particles, $N_A=6.022\times10^{23}$. The molar mass 摩尔质量 (grams per mole, read off the periodic table) converts between mass and moles:

$$n=\frac{m}{M}.$$
Moles are the bridge between the lab (grams you weigh) and the equation (particles that react).

The mole is the hub: convert between mass, particles, gas volume, and concentration The mole is the hub: convert between mass, particles, gas volume, and concentration

Worked example. How many moles, and how many molecules, are in $36.0\ \text{g}$ of water ($M=18.0\ \text{g/mol}$)?

$$n=\frac{m}{M}=\frac{36.0}{18.0}=2.0\ \text{mol},\qquad N=n\,N_A=2.0\times6.022\times10^{23}=1.2\times10^{24}\ \text{molecules}.$$

Explore

Link mass, moles and molar mass

Molar mass $M$ is the bridge between the mass you weigh and the number of moles: $m = M \times n$. Because $M$ is fixed for a substance, mass is proportional to moles — double the moles, double the mass.

Vocabulary Train
English Chinese Pinyin
moles 摩尔 mó ěr
Avogadro's number 阿伏伽德罗常数 ā fú gā dé luó cháng shù
molar mass 摩尔质量 mó ěr zhì liàng
1.2

Mass Spectra of Elements

Syllabus
Learning ObjectiveEssential Knowledge

1.2.A
Explain the quantitative relationship between the mass spectrum of an element and the masses of the element's isotopes.

  • 1.2.A.1 The mass spectrum of a sample containing a single element can be used to determine the identity of the isotopes of that element and the relative abundance of each isotope in nature.
  • 1.2.A.2 The average atomic mass of an element can be estimated from the weighted average of the isotopic masses using the mass of each isotope and its relative abundance.
    • Exclusion Statement: Interpreting mass spectra of samples containing multiple elements or peaks arising from species other than singly charged monatomic ions will not be assessed on the AP Exam.

Source: College Board AP Course and Exam Description

Mass spectrometry

A mass spectrometer 质谱仪 separates atoms by mass, giving a mass spectrum 质谱: peaks at each isotope 同位素 (same element, different neutron count) with heights showing their relative abundance 相对丰度. The element's average atomic mass is the abundance-weighted average of its isotope masses.

Relative atomic mass is the abundance-weighted average of the isotopes Relative atomic mass is the abundance-weighted average of the isotopes

The mass spectrum of chlorine: two isotopes with different abundances The mass spectrum of chlorine: two isotopes with different abundances

Worked example. Chlorine is $75.8\%$ chlorine-35 and $24.2\%$ chlorine-37. Its average atomic mass is

$$A_r=(35)(0.758)+(37)(0.242)=26.5+8.95=35.5.$$
The average lies closer to $35$ because that isotope is more abundant – which is why the periodic-table value is $35.5$, not a whole number.

Explore

Explore isotopes — same element, different mass

Keep the protons fixed but change the neutrons to build the two chlorine isotopes ($^{35}$Cl and $^{37}$Cl). Same element, different mass — exactly the peaks a mass spectrum shows.

Vocabulary Train
English Chinese Pinyin
mass spectrometer 质谱仪 zhì pǔ yí
mass spectrum 质谱 zhì pǔ
isotope 同位素 tóng wèi sù
relative abundance 相对丰度 xiāng duì fēng dù
Exercise sheet
1.3

Elemental Composition of Pure Substances

Syllabus
Learning ObjectiveEssential Knowledge

1.3.A
Explain the quantitative relationship between the elemental composition by mass and the empirical formula of a pure substance.

  • 1.3.A.1 Some pure substances are composed of individual molecules, while others consist of atoms or ions held together in fixed proportions as described by a formula unit.
  • 1.3.A.2 According to the law of definite proportions, the ratio of the masses of the constituent elements in any pure sample of that compound is always the same.
  • 1.3.A.3 The chemical formula that lists the lowest whole number ratio of atoms of the elements in a compound is the empirical formula.

Source: College Board AP Course and Exam Description

The percent composition 百分组成 of a compound is each element's mass fraction. From it you find the empirical formula 实验式 (simplest whole-number ratio of atoms) by converting each element's mass to moles and dividing by the smallest. The molecular formula 分子式 is a whole-number multiple of the empirical formula, found from the molar mass.

Finding the empirical formula: mass to moles, divide by the smallest, read the ratio Finding the empirical formula: mass to moles, divide by the smallest, read the ratio

Worked example. A compound is $40.0\%$ C, $6.7\%$ H, and $53.3\%$ O by mass. Assuming $100\ \text{g}$, convert each mass to moles and divide by the smallest:

$$\text{C}:\frac{40.0}{12}=3.33,\quad \text{H}:\frac{6.7}{1}=6.7,\quad \text{O}:\frac{53.3}{16}=3.33 \;\xrightarrow{\div 3.33}\; 1:2:1,$$
so the empirical formula is $\text{CH}_2\text{O}$. If the molar mass were $180\ \text{g/mol}$ ($=6\times30$), the molecular formula would be $\text{C}_6\text{H}_{12}\text{O}_6$ – glucose.

Vocabulary Train
English Chinese Pinyin
percent composition 百分组成 bǎi fēn zǔ chéng
empirical formula 实验式 shí yàn shì
molecular formula 分子式 fēn zǐ shì
1.4

Composition of Mixtures

Syllabus
Learning ObjectiveEssential Knowledge

1.4.A
Explain the quantitative relationship between the elemental composition by mass and the composition of substances in a mixture.

  • 1.4.A.1 Pure substances contain atoms, molecules, or formula units of a single type. Mixtures contain atoms, molecules, or formula units of two or more types, whose relative proportions can vary.
  • 1.4.A.2 Elemental analysis can be used to determine the relative numbers of atoms in a substance and to determine its purity.

Source: College Board AP Course and Exam Description

Unlike a pure compound, a mixture 混合物 has variable composition – its parts keep their own identities. Describe a mixture by the mass or mole fraction of each component; these do not follow a fixed formula. Spectroscopy (like PES or absorption) can measure how much of each component is present.

Vocabulary Train
English Chinese Pinyin
mixture 混合物 hùn hé wù
1.5

Atomic Structure and Electron Configuration

Syllabus
Learning ObjectiveEssential Knowledge

1.5.A
Represent the ground-state electron configuration of an atom of an element or its ions using the Aufbau principle.

  • 1.5.A.1 The atom is composed of negatively charged electrons and a positively charged nucleus that is made of protons and neutrons.
  • 1.5.A.2 Coulomb's law is used to calculate the force between two charged particles.
    • Equation: $F_{coulombic} \propto \dfrac{q_1 q_2}{r^2}$
  • 1.5.A.3 In atoms and ions, the electrons can be thought of as being in "shells (energy levels)" and "subshells (sublevels)," as described by the ground-state electron configuration. Inner electrons are called core electrons, and outer electrons are called valence electrons. The electron configuration is explained by quantum mechanics, as delineated in the Aufbau principle and exemplified in the periodic table of the elements.
    • Exclusion Statement: The assignment of quantum numbers to electrons in subshells of an atom will not be assessed on the AP Exam.
  • 1.5.A.4 The relative energy required to remove an electron from different subshells of an atom or ion or from the same subshell in different atoms or ions (ionization energy) can be estimated through a qualitative application of Coulomb's law. This energy is related to the distance from the nucleus and the effective (shield) charge of the nucleus.

Source: College Board AP Course and Exam Description

An atom is a tiny nucleus 原子核 (protons and neutrons) surrounded by electrons in shells and subshells (s, p, d, f). The electron configuration 电子排布 lists how electrons fill these, lowest energy first (e.g. $1s^2\,2s^2\,2p^6$). The outermost, highest-energy electrons – the valence electrons 价电子 – control chemistry. Coulomb's law explains their energies: electrons closer to, and less shielded from, the nucleus are held more tightly.

An atom: a tiny nucleus of protons and neutrons, with electrons in shells An atom: a tiny nucleus of protons and neutrons, with electrons in shells

A row of nine burning dishes, each flame a different vivid colour — crimson, orange, yellow, green, blue, violet Flame tests: each metal ion gives its own colour because heat excites its electrons and, as they drop back down, they emit light of specific wavelengths

Worked example. Write the electron configuration of sulfur ($Z=16$). Fill subshells in order until $16$ electrons are placed: $1s^2\,2s^2\,2p^6\,3s^2\,3p^4$. The $3s$ and $3p$ electrons (six in total) are the valence electrons, so sulfur tends to gain two electrons to complete its octet.

Explore

Explore how electrons fill the shells

Change the atomic number $Z$ and watch the electrons fill the shells lowest-energy-first (the Aufbau principle 构造原理). The outermost electrons are the valence electrons that drive bonding.

Vocabulary Train
English Chinese Pinyin
nucleus 原子核 yuán zǐ hé
electron configuration 电子排布 diàn zi pái bù
valence electrons 价电子 jià diàn zi
1.6

Photoelectron Spectroscopy

Syllabus
Learning ObjectiveEssential Knowledge

1.6.A
Explain the relationship between the photoelectron spectrum of an atom or ion and:
i. The ground-state electron configuration of the species.
ii. The interactions between the electrons and the nucleus.

  • 1.6.A.1 The energies of the electrons in a given shell can be measured experimentally with photoelectron spectroscopy (PES). The position of each peak in the PES spectrum is related to the energy required to remove an electron from the corresponding subshell, and the relative height of each peak is (ideally) proportional to the number of electrons in that subshell.

Source: College Board AP Course and Exam Description

Photoelectron spectroscopy 光电子能谱 (PES) measures the energy needed to remove electrons from each subshell. Each peak is a subshell: its position gives the binding energy (how tightly held) and its height gives the number of electrons in it. PES data let you read an element's electron configuration and confirm shell structure directly.

Successive ionisation energies reveal the shell structure through big jumps Successive ionisation energies reveal the shell structure through big jumps

The photoelectron spectrum of neon: one peak per subshell, height set by electron count The photoelectron spectrum of neon: one peak per subshell, height set by electron count

Vocabulary Train
English Chinese Pinyin
Photoelectron spectroscopy 光电子能谱 guāng diàn zi néng pǔ
1.7

Periodic Trends

Syllabus
Learning ObjectiveEssential Knowledge

1.7.A
Explain the relationship between trends in atomic properties of elements and electronic structure and periodicity.

  • 1.7.A.1 The organization of the periodic table is based on patterns of recurring properties of the elements, which are explained by patterns of ground-state electron configurations and the presence of completely or partially filled shells (and subshells) of electrons in atoms.
    • Exclusion Statement: Writing the electron configuration of elements that are exceptions to the aufbau principle will not be assessed on the AP Exam.
  • 1.7.A.2 Trends in atomic properties within the periodic table (periodicity) can be predicted by the position of the element on the periodic table and qualitatively understood using Coulomb's law, the shell model, and the concepts of shielding and effective nuclear charge. These properties include:
    • i. Ionization energy
    • ii. Atomic and ionic radii
    • iii. Electron affinity
    • iv. Electronegativity.
  • 1.7.A.3 The periodicity (in 1.7.A.2) is useful to predict/estimate values of properties in the absence of data.

Source: College Board AP Course and Exam Description

Trends across the periodic table follow from nuclear charge and shielding:

Electronegativity rises across a period and falls down a group Electronegativity rises across a period and falls down a group

  • Atomic radius 原子半径 decreases across a period (stronger pull) and increases down a group (more shells).
  • Ionization energy 电离能 (energy to remove an electron) increases across, decreases down – opposite to radius.
  • Electronegativity 电负性 (pull on shared electrons) increases across and up, toward fluorine.

Periodic trends: how radius, ionization energy, and electronegativity change across and down Periodic trends: how radius, ionization energy, and electronegativity change across and down

Explore

Explore atomic radius across a period

Step across Period 3 and watch the atomic radius shrink — each added proton raises the effective nuclear charge and pulls the same shell in tighter.

Vocabulary Train
English Chinese Pinyin
Atomic radius 原子半径 yuán zi bàn jìng
Ionization energy 电离能 diàn lí néng
Electronegativity 电负性 diàn fù xìng
1.8

Valence Electrons and Ionic Compounds

Syllabus
Learning ObjectiveEssential Knowledge

1.8.A
Explain the relationship between trends in the reactivity of elements and periodicity.

  • 1.8.A.1 The likelihood that two elements will form a chemical bond is determined by the interactions between the valence electrons and nuclei of elements.
  • 1.8.A.2 Elements in the same column of the periodic table tend to form analogous compounds.
  • 1.8.A.3 Typical charges of atoms in ionic compounds are governed by the number of valence electrons and predicted by their location on the periodic table.

Source: College Board AP Course and Exam Description

Atoms gain, lose, or share valence electrons to reach stable configurations. Metals (low ionization energy) lose electrons to form cations 阳离子; nonmetals gain electrons to form anions 阴离子. Oppositely charged ions attract into an ionic compound 离子化合物, whose formula balances the charges to make the whole neutral. For example, aluminium ($3+$) and oxide ($2-$) combine as $\text{Al}_2\text{O}_3$ so the $+6$ and $-6$ cancel.

Explore

Watch an ionic bond form by electron transfer

A metal gives up its valence electron(s) and a non-metal takes them, so both reach a full shell. The atoms become oppositely charged ions that attract — that electrostatic pull is the ionic bond.

Vocabulary Train
English Chinese Pinyin
cations 阳离子 yáng lí zi
anions 阴离子 yīn lí zi
ionic compound 离子化合物 lí zi huà hé wù
1.8

Exam tips

  • Use the mole as the hub: convert grams↔moles with $n=m/M$ and moles↔particles with Avogadro's number.
  • Relative atomic mass is the abundance-weighted average of the isotopes — it lies closer to the more abundant one, not halfway.
  • For an empirical formula, turn each element's mass into moles and divide by the smallest; scale up to whole numbers.
  • Read periodic trends from nuclear charge and shielding: atomic radius decreases across a period, ionisation energy and electronegativity increase across (and up).
  • Fill electron configurations in energy order; the outer valence electrons control the chemistry.

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