AP Chemistry: Composition of Mixtures, PES, Periodicity, and Mass Spectrometry - Vocabulary Flashcards

Composition of Mixtures (AP Chemistry 1.4)

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

Pure Substances vs. Mixtures 1.4.A.1

• Pure Substance: Made up of only one kind of particle (atom, molecule, or formula unit).

  • Examples: Pure water (H₂O), pure sodium chloride (NaCl), pure gold (Au).

  • Every sample has the same composition and properties.

• Mixture: Contains two or more different particles (atoms, molecules, or formula units).

  • Examples: Air (mixture of N₂, O₂, CO₂, etc.), saltwater (NaCl and H₂O).

  • The proportions of components can vary (you can have more or less salt in saltwater).

Quantitative Relationships: Elemental Composition by Mass

• Elemental Composition by Mass tells us what percentage of a substance’s mass comes from each element.

  • For example, in H₂O:

  • Mass % H = (mass of H in H₂O / mass of H₂O) × 100%

  • Mass % O = (mass of O in H₂O / mass of H₂O) × 100%

• In a mixture, the overall elemental composition depends on the amounts and types of substances present.

How to Calculate Elemental Composition in a Mixture

• Example: mixture of NaCl and KCl.

  • Find the mass of each component in the mixture.

  • Calculate the mass of each element in each compound:

  • NaCl: mass Na = (mass of Na in NaCl) × (mass NaCl in mixture) / (total mass of NaCl)

  • KCl: mass K = (mass of K in KCl) × (mass KCl in mixture) / (total mass of KCl)

  • Add up the total mass of each element from all sources.

  • Find the total mass of the mixture.

  • Calculate each element’s mass percent: (total mass of element / total mass of mixture) × 100%

• General formula (useful for any component i):

  • For an element E in compound i with mass fraction w{E,i} in that compound, and mass mi of compound i in the mixture:
    mE = igl(w{E,i}igr) \, m_i,

    ext{and}

    ext{Total } mE = \sumi m_E.

• Then percent composition:
  ext{Mass \% of E} = rac{m_E}{ ext{Total mass of mixture}} imes 100 on%

Elemental Analysis and Determining Purity 1.4.A.2

• Elemental Analysis: An experimental method for determining the amount of each element in a sample.

  • Used to find empirical formulas (the simplest whole-number ratio of atoms in a compound).

  • Can reveal if a substance is pure or a mixture/impure.

Purity Determination

• If the mass percentages match the expected values for a pure substance, it’s likely pure.

• If they don’t match, the sample may be a mixture or contaminated.

Worked Example: Carbon Content in a Mixture

• A sample contains 6.00 g of substance X and 4.00 g of substance Y.

  • Substance X is 50% C by mass, and Y is 20% C by mass.

  • What is the overall percent by mass of carbon in the mixture?

  • Step 1: Find mass of C in each component:

  • X: 6.00 g × 0.50 = 3.00 g C

  • Y: 4.00 g × 0.20 = 0.80 g C

  • Step 2: Total mass of mixture = 6.00 g + 4.00 g = 10.00 g

  • Step 3: Total mass of C = 3.00 g + 0.80 g = 3.80 g

  • Step 4: % C in mixture = (3.80 g / 10.00 g) × 100% = 38%

Key Takeaways

• Pure substances have a fixed composition; mixtures do not.

• The mass percent of an element in a mixture depends on both its abundance in each component and the relative amounts of those components.

• Elemental analysis helps determine both the formula and the purity of substances.

Summary of the Relationship Between Elemental Composition and Mixture Composition

• To understand the quantitative relationship, you need:

  • How to calculate mass percentages

  • Recognize the difference between pure substances and mixtures

  • Interpret elemental analysis data to assess composition and purity

Homogeneous vs. Heterogeneous (Topic 1.4 Overview)

• The AP Chemistry topic 1.4 covers composition of mixtures with emphasis on homogeneous vs. heterogeneous distinctions.

Photoelectron Spectroscopy (PES) and Electron Configuration (AP Chemistry 1.6)

Learning Objective 1.6.A

• Explain the relationship between the photoelectron spectrum (PES) of an atom or ion and:

  • The ground-state electron configuration of the species.

  • The interactions between the electrons and the nucleus.

What is Photoelectron Spectroscopy (PES)?

• PES is an experimental technique to determine the energies required to remove electrons from different subshells of an atom or ion.

• When high-energy photons (usually UV or X-ray) hit an atom, electrons are ejected.

• The kinetic energy of the ejected electrons is measured, and from this the binding energy (energy needed to remove an electron) is calculated.

Understanding the PES Spectrum

• X-axis: binding energy (energy to remove an electron).

  • Higher binding energy = electrons more tightly held by the nucleus.

• Y-axis: relative number of electrons detected at each energy (peak height/area).

Peaks and Subshells

• Each peak corresponds to electrons in a specific subshell (e.g., 1s, 2s, 2p, etc.).

• Position of the peak = energy required to remove an electron from that subshell.

• Height/area of the peak = number of electrons in that subshell.

Ground-State Electron Configuration via PES

• The PES spectrum directly reflects the ground-state electron configuration.

• Example: Carbon with configuration 1s² 2s² 2p² should show three peaks (1s, 2s, 2p).

  • 1s peak at the highest binding energy (closest to the nucleus), followed by 2s, then 2p.

  • Relative heights are proportional to the number of electrons in each subshell.

Reading Electron Configurations from PES

• Count the peaks: number of occupied subshells.

• Compare heights: number of electrons in each subshell.

• Order of binding energies: inner subshells (closer to nucleus) have higher binding energies.

Interpreting PES Data: Key Concepts

• Electron–Nucleus Interactions:

  • Greater nuclear charge (more protons) pulls electrons closer, increasing binding energy.

  • Inner shells shield outer electrons; inner electrons feel less shielding and have higher binding energies.

  • Subshell energy differences: 1s electrons have much higher binding energy than 2s/2p due to proximity and shielding.

• PES as experimental validation of electron configurations.

• Differences in peak positions/heights reveal electron interactions, shielding, and effective nuclear charge.

Essential Knowledge Points (1.6.A.1)

• PES measures energies of electrons in different shells.

• Peak position = energy to remove an electron from a subshell.

• Peak height = number of electrons in that subshell.

• PES provides a fingerprint for electron configuration and insight into forces holding electrons.

Reading PES: A Different Perspective

• The spectrum acts like a fingerprint for electron configuration and shows the effects of nuclear charge and shielding on binding energies.

Trends in Atomic Properties, Electronic Structure, and Periodicity (AP Chemistry 1.7)

Periodic Table Organization and Electron Configuration

• The periodic table is organized by recurring patterns (periodicity) in elemental properties.

• Elements in the same group have similar valence-shell configurations.

• The filling order of subshells (s, p, d, f) leads to repeating chemical behaviors.

• Noble gases (Group 18): full valence shells → high stability.

• Alkali metals (Group 1): one electron in the outermost s subshell → highly reactive.

Predicting Trends with the Periodic Table

• Trends are explained by Coulomb’s Law, the shell model, shielding, and effective nuclear charge (Z_eff).

Coulomb’s Law

• Formula: F = \frac{k q1 q2}{r^2}

  • Describes the force between charged particles.

  • In atoms, attraction is between nucleus (positive) and electrons (negative).

• Key intuition: closer electron to nucleus (smaller r) or greater nucleus charge (larger q₁) → stronger attraction.

Shell Model and Shielding

• Electrons are arranged in energy levels (shells) around the nucleus.

• Inner electrons shield outer electrons from the full nuclear charge.

Shielding and Effective Nuclear Charge (Z_eff)

• Shielding: Inner electrons reduce the pull on valence electrons.

• Z_eff = Z − S, where Z = number of protons, S = number of core electrons.

• Z_eff increases across a period (left to right) and is relatively constant down a group.

Trends in Ionization Energy

• Definition: Energy required to remove an electron from an atom in the gas phase.

• Trends:

  • Increases across a period (left to right) as Z_eff increases and electrons are held more tightly.

  • Decreases down a group as more shells are added and shielding increases.

Atomic and Ionic Radii

• Atomic radius: distance from nucleus to outermost electron.

• Ionic radius: size of an ion (cations are smaller; anions are larger than neutral atoms).

• Trends:

  • Atomic radius decreases across a period as Z_eff increases and pulls electrons closer.

  • Atomic radius increases down a group as more shells are added.

Electron Affinity and Electronegativity

• Electron Affinity: energy change when an atom gains an electron.

• Trend for Electron Affinity:

  • Generally becomes more negative across a period (left to right).

  • Becomes less negative down a group.

• Electronegativity: ability of an atom to attract electrons in a bond.

• Trend for Electronegativity:

  • Increases across a period.

  • Decreases down a group.

• Highest electronegativity: Fluorine; Lowest: Cesium/Francium.

Predicting Properties from Trends

• Periodic trends help predict reactivity, bond types, physical states, and chemical properties.

• Summary Table (Across a Period vs. Down a Group):

  • Ionization Energy: Increases across, Decreases down

  • Atomic Radius: Decreases across, Increases down

  • Electron Affinity: More negative across, Less negative down

  • Electronegativity: Increases across, Decreases down

Key Takeaways

• Periodic patterns arise from electronic structure and periodicity.

• Coulomb’s Law, the shell model, shielding, and Z_eff explain trends.

• Understanding trends allows prediction of chemical behavior and reactivity across elements.

Valence Electrons and Ionic Compounds (AP Chemistry 1.8)

What Drives Chemical Bond Formation?

• Chemical reactivity—the likelihood elements react and form bonds—depends on interactions between valence electrons and nuclei.

• Valence electrons: Outermost electrons involved in bonding.

• The nucleus’s positive charge attracts valence electrons, both its own and those of other atoms.

• The balance of attractions and repulsions (push and pull) determines:

  • Whether a bond forms

  • What kind of bond forms (ionic, covalent, metallic)

Periodic Table Columns and Analogous Compounds

• Elements in the same group have similar valence electron arrangements.

• Form analogous compounds across a group due to repeating electron configurations.

• Examples:

  • Group 1 alkali metals form compounds like NaCl, KCl (1:1 with halides).

  • Group 17 halogens form -1 ions and similar compounds with metals (e.g., NaF, NaCl, NaBr).

• Similarity arises from the repeating pattern of electron configuration down a group.

Predicting Charges in Ionic Compounds

• Typical ionic charges are driven by noble-gas configurations (full valence shell):

  • Metals (left) tend to lose electrons to form cations: Group 1 → +1, Group 2 → +2

  • Nonmetals (right) tend to gain electrons to form anions: Group 16 → −2, Group 17 → −1

• The pattern repeats across periods (periodicity).

Periodicity and Reactivity Trends

• Metals (left side): Reactivity increases down a group (Li < Na < K < Rb < Cs); valence electrons are farther from the nucleus and easier to lose.

• Metals: Reactivity decreases across a period (left to right).

• Nonmetals (right side): Reactivity decreases down a group (F > Cl > Br > I); atoms are less effective at pulling in electrons as they get bigger.

• Nonmetals: Reactivity increases across a period (left to right) as effective nuclear charge increases.

Why These Trends Matter

• Periodic trends help predict which elements are likely to bond and the formulas/charges of ionic compounds.

• The periodic table is not just a list—it maps how elements behave based on electron arrangements.

Takeaways

• Reactivity is tied to valence electron configuration and position on the periodic table.

• Elements in the same group behave similarly due to similar valence electron arrangements.

• Typical charges in ionic compounds reflect the drive to achieve a noble gas configuration, a direct result of periodicity.

Unit 1.2 Mass Spectrometry

What is Mass Spectrometry?

• An analytical tool for measuring the mass-to-charge ratio (m/z) of one or more molecules in a sample.

• Used to identify unknown compounds via molecular weight, quantify known compounds, and determine structure and properties.

Mass Spectrometer Components

• Ionization Source: Molecules are converted to gas-phase ions.

• Mass Analyzer: Ions are sorted by mass-to-charge ratios.

• Ion Detection System: Ions are detected; m/z ratios and relative abundances are recorded.

Mass-to-Charge Ratio (m/z)

• Definition: m/z = \frac{mass\ of\ ion}{charge\ of\ ion}

• For most AP Chemistry cases, the charge is +1, so m/z ≈ mass of the isotope.

• If an ion has a higher charge, its m/z is lower.

Mass Spectrum: What It Is

• A plot of m/z (x-axis) vs. relative abundance (y-axis).

• Each peak represents a different isotope or fragment; peak position identifies isotope mass; peak height/area indicates relative abundance.

Interpreting Isotopes in a Mass Spectrum

• Example: Magnesium with peaks at m/z 24, 25, 26 correspond to isotopes ²⁴Mg, ²⁵Mg, ²⁶Mg.

• The tallest peak indicates the most abundant isotope.

Relative Abundance and Average Atomic Mass

• To find percent abundance, compare peak heights/areas (or use provided percentages).

• Average atomic mass is the weighted average of isotopic masses: ext{Average atomic mass} = ext{isotopic mass} \ times \ ext{fractional abundance}

  • Fractional abundance = percent abundance ÷ 100

Step-by-Step Example: Chlorine

• Isotopes: ³⁵Cl mass = 34.97 u, abundance = 75.78%; ³⁷Cl mass = 36.97 u, abundance = 24.22%

• Convert to decimals: 0.7578 and 0.2422

• Compute weighted masses: 34.97 \times 0.7578 = 26.52, 36.97 \times 0.2422 = 8.96

• Average atomic mass: 26.52 + 8.96 = 35.48\,\text{u} (matches Cl on the periodic table)

Key Points and Common Mistakes

• The mass spectrum shows isotopes and their relative abundances.

• The average atomic mass is a weighted average and is not necessarily a whole number.

• Common mistakes include failing to convert percent abundance to decimal, confusing peak intensity with absolute abundance, and assuming the average equals a single isotope mass.

Practice Problems (Summary)

• Practice Problem 1: two peaks at 10 u (80%), 11 u (20%) → average mass = 10.2 u.

• Practice Problem 2: two peaks at m/z 62 (69.2%), 64 (30.8%) → identify isotopes (62 and 64) and compute average mass ≈ 62.62 u.

Connection to AP Chemistry

• You may be asked to interpret a mass spectrum, identify isotopes, calculate average atomic mass, or explain why the atomic mass on the periodic table is not a whole number.

Unit 1.3 Empirical Formulas and Elemental Composition by Mass

What Are Compounds Made Of?

• Pure substances can be:

  • Molecules (e.g., H₂O, CO₂) — discrete units bonded together.

  • Ionic compounds (e.g., NaCl, CaF₂) — composed of ions in fixed ratios.

  • Atomic solids (e.g., diamond, graphite) — large networks of atoms in a lattice.

• All pure substances have fixed composition: the ratios of atoms or ions are the same for a given compound.

Percent Composition

• Percent composition tells you what percentage of a compound’s mass comes from each element.

• Example: Percent C in CO₂:

  • Molar masses: C = 12.01 g/mol, O = 16.00 g/mol; CO₂ has mass 12.01 + 2×16.00 = 44.01 g/mol

  • Percent C: \%C = \left(\frac{12.01}{44.01}\right) \times 100\% \approx 27.3\%

Law of Definite Proportions

• In any pure sample of a compound, the ratio of the masses of constituent elements is fixed.

• Example: Water contains H and O in a fixed mass ratio; oxygen is about 16 times heavier than hydrogen, giving a fixed ratio by mass (approx. 1:8 by mass).

Empirical Formula

• Empirical formula shows the simplest whole-number ratio of atoms in a compound.

• For many molecular compounds, the empirical formula may be the same as or different from the molecular formula (e.g., glucose: empirical = CH₂O, molecular = C₆H₁₂O₆).

• For ionic compounds, the formula unit always reflects the empirical formula (e.g., MgCl₂).

Connecting Mass Percent to Empirical Formula

• Mass percentages describe element-by-element composition by mass.

• Convert mass percentages to moles (using atomic masses) to determine the simplest ratio of atoms (the empirical formula).

• The classic workflow is captured by the poem:

  • Percent to Mass → Mass to Mole → Divide by the Smallest → Multiply to get the Whole (for molecular formula)

Step-by-Step Example: C, H, O by Mass

• Given: 40.0% C, 6.7% H, 53.3% O in 100 g sample. 1) Assume 100 g sample: 40.0 g C, 6.7 g H, 53.3 g O. 2) Convert to moles:

  • C: 40.0 g / 12.01 g/mol = 3.33 mol

  • H: 6.7 g / 1.008 g/mol = 6.65 mol

  • O: 53.3 g / 16.00 g/mol = 3.33 mol
    3) Divide by smallest moles (3.33):

  • C: 3.33/3.33 = 1

  • H: 6.65/3.33 ≈ 2

  • O: 3.33/3.33 = 1
    4) Empirical formula: CH₂O

Key Takeaways: Empirical Formulas

• The empirical formula is the lowest whole-number ratio of atoms.

• Percent composition can be used to find the empirical formula for any pure compound.

• For ionic compounds, formula units reflect the empirical formula; for molecular compounds, the empirical formula may be a reduced version of the molecular formula.

Practice Problem: Empirical Formula from Percent Composition

• Sample contains 63.5% Ag, 8.2% N, 28.3% O by mass. 1) Assume 100 g sample: Ag 63.5 g, N 8.2 g, O 28.3 g. 2) Convert to moles:

  • Ag: 63.5 g / 107.87 g/mol ≈ 0.589 mol

  • N: 8.2 g / 14.01 g/mol ≈ 0.586 mol

  • O: 28.3 g / 16.00 g/mol ≈ 1.77 mol
    3) Divide by smallest (≈0.586):

  • Ag ≈ 1

  • N ≈ 1

  • O ≈ 3
    4) Empirical formula: AgNO₃

Going from Empirical to Molecular Formula

• Difference: Empirical formula shows the smallest ratio; molecular formula shows the actual numbers of atoms in a molecule.

• If the molar mass of the compound is M and the empirical formula mass is mₑ, compute the multiplier n: n = \frac{M}{m_\text{e}}

• The molecular formula is obtained by multiplying the empirical formula subscripts by n.

Example: From CH₂O (empirical, mass 30.03 g/mol) to a compound with molar mass 180 g/mol

• Step 1: Empirical formula mass mₑ = 12.01 + 2×1.008 + 16.00 = 30.03 g/mol

• Step 2: n = 180 / 30.03 ≈ 6

• Step 3: Molecular formula = (CH₂O) × 6 = C₆H₁₂O₆

Key Takeaways

• The empirical formula provides the simplest ratio; the molecular formula shows the actual composition.

• Use the compound’s molar mass to scale from empirical to molecular formula.

Unit 1.5 Atomic Structure and Electron Configuration

Atomic Structure

• Atoms consist of:

  • Nucleus: positively charged protons and neutral neutrons.

  • Electrons: negatively charged, orbiting outside the nucleus.

• The number of protons defines the element (atomic number Z).

• In a neutral atom, electrons = protons. Ions differ by electron gain/loss.

Coulomb's Law (conceptual, not just calculation)

• Formula: F = \frac{k q1 q2}{r^2}

  • F: force between charges

  • k: Coulomb's constant ≈ $8.99 \times 10^9\ \text{N·m}^2/\text{C}^2$

  • q₁, q₂: charges in Coulombs

  • r: distance between charges

• The law explains why electrons are attracted to the nucleus and how distance and charge influence that attraction.

• In atoms:

  • Nucleus is positive; electrons are negative.

  • Closer electrons (smaller r) or larger nuclear charge (larger q₁) increase the attractive force.

Electron Configuration, Shells, and Subshells

• Shells: n = 1, 2, 3, …

• Subshells within a shell: s, p, d, f

• Subshell capacities:

  • s: 1 orbital (max 2 electrons)

  • p: 3 orbitals (max 6 electrons)

  • d: 5 orbitals (max 10 electrons)

  • f: 7 orbitals (max 14 electrons)

Core vs. Valence Electrons

• Core electrons: inner electrons not involved in bonding.

• Valence electrons: outermost electrons that determine chemical properties.

Aufbau Principle and Electron Filling

• Aufbau Principle: Electrons fill the lowest energy orbitals first, in order of increasing energy.

• The order of filling is determined by energy (not strictly by shell number).

• The diagonal rule helps determine the filling order (e.g., 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, …).

Writing Electron Configurations

• Steps:
  1) Determine the number of electrons (use atomic number; adjust for ions).
  2) Fill orbitals following the Aufbau order.
  3) Use superscripts to indicate the number of electrons in each subshell.
  4) Bracket notation can be used for core electrons (noble gas shorthand).

Example Problems

• Sulfur (S, Z = 16): 16 electrons → 1s² 2s² 2p⁶ 3s² 3p⁴

  • Ground-state configuration: 1s² 2s² 2p⁶ 3s² 3p⁴

• Calcium ion, Ca²⁺ (Z = 20): Ca atom has 20 e⁻; Ca²⁺ has 18 e⁻

  • Configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ (or [Ar] core)

• Phosphorus (P, Z = 15): 1s² 2s² 2p⁶ 3s² 3p³ → Valence electrons are the ones in the highest principal energy level: 3s² 3p³ → 5 valence electrons

Ionization Energy Comparisons (Coulomb’s Law-Based Explanations)

• Na vs K (both alkali metals): Na has a higher first ionization energy because its outer electron is closer to the nucleus (smaller r) than K’s; stronger attraction requires more energy to remove the electron.

• F⁻ vs O²⁻ (isoelectronic, same electron count): F⁻ has more protons (9) than O²⁻ (8), so greater nuclear charge results in a stronger attraction on its electrons.

• Atomic radius across a period: increases across a period would pull electrons in closer due to increased Zeff even though distance r is similar; across a period, r is nearly constant but Zeff increases, so radius effectively decreases.

Summary: Why Coulomb's Law Matters

• It underpins atomic structure, periodic trends (radius, ionization energy), and electron filling in orbitals.

• Conceptual tool to explain trends and to reason about electron behavior in atoms.

Practice and Key Concepts Across Modules

• Always remember: higher Z_eff and shorter r yield higher binding energies and tighter electron controls.

• PES peaks tell you about subshell energies and electron counts; relate peak positions to subshell energies and peak heights to electron numbers.

• Mass percent calculations hinge on converting mass percentages to masses, then to moles, then to the simplest mole ratio for empirical formulas.

• From empirical to molecular formulas, use molar masses to determine the scaling factor, then multiply subscripts accordingly.

• Mass spectrometry provides real-world insight into isotopes, their abundances, and average atomic masses; use weighted averages to interpret the spectrum.

• In mixtures, elemental composition depends on both the composition of each component and the amounts of those components; use mass fractions to compute overall mass contributions.

Quick Reference Formulas (LaTeX)

• Elemental composition in a substance:
  \text{Mass \% of E} = \left(\frac{mE}{m\text{sample}}\right)\times 100\%

• Mass fraction in a compound:
  w{E,i} = \frac{nE ME}{Mi}

• Elemental contribution in a mixture:
  mE = \sumi w{E,i} \cdot mi

• Mass percent in a mixture:
  \%E = \left( \frac{mE}{m{\text{mixture}}} \right) \times 100\%

• Coulomb’s Law:
  F = \frac{k q1 q2}{r^2}

• Effective nuclear charge:
  Z_{\text{eff}} = Z - S

• Mass-to-charge ratio:
  m/z = \frac{\text{mass of ion}}{\text{charge of ion}}

• Average atomic mass (weighted):
  \text{Average mass} = \sum (mi \times fi)$$
  where $f_i$ is the fractional abundance of isotope i.

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