9-5 CHEM 200 Notes - Atomic Theory & Electromagnetic Radiation

Atomic Mass, Isotopes, and Atomic Theory

• Lecture context

  • Course: CHEM 200

  • Topic: Atomic properties, structure, and formulas (Chapters 2 & 3).

  • Key idea: Atomic masses are averages of isotopes; chemical compounds use consistent naming rules.

Atomic Mass and Isotopes

• Atomic mass vs. mass number

  • Each proton/neutron is about $1\ \text{amu}$.

  • A single atom's atomic mass is roughly its mass number (a whole number).

  • Periodic table atomic masses are weighted averages of natural isotopes.

• Isotopes and natural abundance

  • Elements have isotopes (atoms with different masses).

  • Average atomic mass accounts for each isotope's mass and how common it is (abundance).

• Mathematical expression

  • Average atomic mass ($\bar{M}$) is the sum of (isotope mass $m<em>i$) times (its fractional abundance $f</em>i$).

  • $\bar{M} = \sum<em>i f</em>i m<em>i$ where $\sum</em>i f_i = 1$.

• Example: boron

  • Boron has 10B (mass $10.0129\ \text{amu}$, ~19.9% abundant) and 11B (mass $11.0093\ \text{amu}$, ~80.1% abundant).

  • Calculation: $\bar{M}_{B} = (0.199)(10.0129\ \text{amu}) + (0.801)(11.0093\ \text{amu}) \approx 10.81\ \text{amu}$ .

Percent Abundance by Isotopic Composition (Chlorine example)

• Chlorine (average mass $35.453\ \text{amu}$) has 35Cl (mass $34.96885\ \text{amu}$) and 37Cl (mass $36.96590\ \text{amu}$).

• Let 'x' be the fraction of 35Cl; then (1 - x) is the fraction of 37Cl.

• Equation: $\text{35.453\ \text{amu}} = x(34.96885\ \text{amu}) + (1 - x)(36.96590\ \text{amu})$.

  • Expanding: $\text{35.453} = 34.96885x + 36.96590 - 36.96590x$.

  • Solving for x: $x \approx 0.7576$ (fraction of 35Cl).

  • Fraction of 37Cl is $(1 - x) \approx 0.2424$.

• Resulting percent abundances:

  • $^{35}\text{Cl}: 75.76\% , ^{37}\text{Cl}: 24.24\%$.

Percent Abundance Calculation (practice problem)

• Unknown element X has isotopes 151X (mass = 150.99 amu) and 153X (mass = 153.03 amu).

• Average atomic mass of X is 151.69 amu.

• Let 'y' be the fraction of 153X; (1 - y) is the fraction of 151X.

• Equation: $\text{151.69} = (1 - y)(150.99) + y(153.03)$.

• Solve for y to get the percent abundance of 153X.

Chemical Language and Formulas

• Chemistry uses a special language:

  • Elements (Na, Cl) are letters.

  • Formulas (NaCl) are words.

  • Equations ($2\text{Na(s)} + \text{Cl}_2\text{(g)} \rightleftharpoons 2\text{NaCl(s)}$) are sentences.

Chemical Formulas: Representations of Molecules

• Methane ($\text{CH}_4$) can be shown as:

  • (a) Molecular formula ($\text{CH}_4$).

  • (b) Structural formula (shows how atoms are connected).

  • (c) Ball-and-stick model (shows 3D shape).

  • (d) Space-filling model (shows atom sizes and surface).

• Example: sulfur ($\text{S}_8$)

  • Sulfur exists as eight atoms linked in a ring ($\text{S}_8$ molecule); shown with structural, ball-and-stick, and space-filling models.

• Notation for entities:

  • H: one hydrogen atom.

  • H2: one hydrogen molecule.

  • 2H: two hydrogen atoms.

  • 2H2: two hydrogen molecules.

Molecular Formulas vs Empirical Formulas

• Molecular formula: Exact count of each atom in a molecule.

  • Examples: Water ($\text{H}<em>2\text{O}$); Hydrogen peroxide ($\text{H}</em>2\text{O}<em>2$); Glucose ($\text{C}</em>6\text{H}<em>{12}\text{O}</em>6$).

• Empirical formula: Simplest whole-number ratio of atoms in a molecule.

  • Examples: Water ($\text{H}<em>2\text{O}$); Hydrogen peroxide ($\text{HO}$); Glucose ($\text{CH}</em>2\text{O}$).

Empirical Formula and Elemental Analysis

• Elemental analysis finds mass percentages of elements.

• This gives the empirical formula, but not always the molecular formula.

• Example: Glucose ($\text{C}<em>6\text{H}</em>{12}\text{O}<em>6$) and formaldehyde ($\text{CH}</em>2\text{O}$) both have the empirical formula $\text{CH}_2\text{O}$).

Practice Question: Citric Acid Composition (conceptual)

• Given citric acid ($\text{C}<em>3\text{H}</em>5\text{O(CO}<em>2\text{H)}</em>3$).

• Determine the count of C, H, and O atoms in one molecule by expanding the formula.

Writing Molecular and Empirical Formulas from Structures (conceptual)

• From structural drawings, count atoms to find molecular and empirical (simplified ratio) formulas.

Empirical Formula Example: CH2O with molar mass 180 g/mol

• Step 1: Calculate empirical formula mass for $\text{CH}_2\text{O}$.

  • $M<em>{\text{CH}</em>2\text{O}} = (1\times 12.01) + (2\times 1.008) + (1\times 16.00) = 30.026\ \text{g/mol}$.

• Step 2: Find the ratio 'n' of molecular molar mass to empirical formula mass.

  • $n = \frac{180.00}{30.026} \approx 6$.

• Step 3: Multiply the empirical formula by 'n' to get the molecular formula.

  • Molecular formula = $\text{CH}<em>2\text{O} \times 6 = \text{C}</em>6\text{H}<em>{12}\text{O}</em>6$.

• Answer: $\text{C}<em>6\text{H}</em>{12}\text{O}_6$.

Structural Isomers vs Spatial Isomers

• Structural isomers

  • Same molecular formula, different atom connections.

  • Example: Acetic acid and methyl formate are both $\text{C}<em>2\text{H}</em>4\text{O}_2$ but have different structures.

• Spatial (stereoisomers)

  • Same connections, different 3D arrangements.

  • Example: Carvone enantiomers ($\text{C}<em>{10}\text{H}</em>{14}\text{O}$) have the same atoms linked but different spatial orientations.

The Formula Mass (Molar Mass) for Covalent and Ionic Compounds

• Covalent compounds: Mass of one mole of molecules.

  • Example: Aspirin ($\text{C}<em>9\text{H}</em>8\text{O}_4$) has molecular mass = $180.15\ \text{g/mol}$.

  • Calculation:

    • C: $9 \times 12.01\ \text{amu} = 108.09\ \text{amu}$

    • H: $8 \times 1.008\ \text{amu} = 8.064\ \text{amu}$

    • O: $4 \times 16.00\ \text{amu} = 64.00\ \text{amu}$

    • Total: $180.15\ \text{amu}\, (\text{g/mol})$.

• Ionic compounds: Formula mass is the mass of the repeating unit (not a discrete molecule).

  • Example: NaCl has formula mass = $58.44\ \text{g/mol}$ (Na $22.99$ + Cl $35.45$).

Example: Molar Mass of Ca(OH)2

• Calculation (two decimals):

  • Ca = $40.08\ \text{g/mol}$

  • O = $16.00\ \text{g/mol} \times 2 = 32.00\ \text{g/mol}$

  • H = $1.008\ \text{g/mol} \times 2 = 2.016\ \text{g/mol}$

  • Total = $74.10\ \text{g/mol}$

The Mole and Avogadro's Number

• The mole (mol): Amount of substance with the same number of entities as in 12 g of carbon-12.

  • One mole contains $N_A = 6.022 \times 10^{23}$ entities (Avogadro's number).

  • Symbolically: $1\ \text{mol} = N_A = 6.022 \times 10^{23}$.

• Illustration: 1 mol of any substance (e.g., Zn, C, Mg) has $6.022 \times 10^{23}$ atoms, connecting mass, moles, and particle count.

The Mole: Mass–Mole–Number Relationships

• Key relationships:

  • Mass to moles: $n = \frac{m}{M}$ (n=moles, m=mass, M=molar mass).

  • Moles to particles: $N = n \times N_A$

  • Particles to moles: $n = \frac{N}{N_A}$

  • Moles to mass: $m = n \times M$

• Applications:

  • Convert mass to moles to find substance amount.

  • Convert moles to molecules/atoms using $N_A$.

Chapter 3: Electronic Structure & Periodic Properties

• Major sections:

  • 3.1 Electromagnetic Energy

  • 3.2 The Bohr Model

  • 3.3 Development of Quantum Theory

  • 3.4 Electronic Structure of Atoms (Electron Configurations)

  • 3.5 Periodic Variations in Element Properties

  • 3.6 The Periodic Table

  • 3.7 Ionic and Molecular Compounds

Electromagnetic Radiation: Waves and Photons

• Matter vs. light

  • Matter: has mass, fixed position, particle-like.

  • Light: no mass, range of energies, acts like waves and particles.

• Waves: key quantities

  • Frequency ($\nu$) ($\text{s}^{-1}$ or Hz): cycles per second.

  • Wavelength ($\lambda$) (m): length of one full cycle.

  • Speed of light ($c$): $3.00 \times 10^8\ \text{m/s}$.

  • Relationship: $\lambda = \frac{c}{\nu}\quad\text{or}\quad \nu = \frac{c}{\lambda}$.

• Amplitude and intensity

  • Amplitude: wave height; higher amplitude means greater intensity (brightness).

The Electromagnetic Spectrum and Energy

• EM spectrum order: Higher energy means higher frequency and shorter wavelength.

  • Regions: radio, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays.

• Problem examples:

  • Use $\nu = \frac{c}{\lambda}$ and $E = h\nu = \frac{hc}{\lambda}$.

  • Planck’s constant: $h = 6.626 \times 10^{-34}\ \text{J}\cdot\text{s}$.

• Example 1: Sodium streetlight ($\lambda = 589\ \text{nm}$)

  • Convert to meters: $589\ \text{nm} = 5.89 \times 10^{-7}\ \text{m}$.

  • Frequency: $\nu = \frac{3.00 \times 10^8}{5.89 \times 10^{-7}} \approx 5.09 \times 10^{14}\ \text{s}^{-1}$.

• Problem: green light ($\lambda = 505\ \text{nm}$)

  • Frequency: $\nu = \frac{3.00 \times 10^8}{505 \times 10^{-9}} \approx 5.94 \times 10^{14}\ \text{s}^{-1}$.

The Electromagnetic Spectrum: Quick Takeaways

• Trends: More energy means more frequency and less wavelength.

• Visible light is between infrared (lower energy) and ultraviolet (higher energy).

• Problems: Identify EM radiation type from given wavelength or frequency.

Using E and \lambda to Find Wavelengths and Energies

• Fundamental relations:

  • $\nu = \frac{c}{\lambda}$

  • $E = h\nu = \frac{hc}{\lambda}$

• Example: Energy given ($E = 7.26 \times 10^{-19}\ \text{J}$), find wavelength.

  • Wavelength: $\lambda = \frac{hc}{E}$

  • Values: $h = 6.626 \times 10^{-34}\ \text{J}\cdot\text{s}, c = 3.00 \times 10^8\ \text{m/s}$.

  • Calculation: $\lambda = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{7.26 \times 10^{-19}} \approx 2.74 \times 10^{-7}\ \text{m} \approx 274\ \text{nm}$.

• Example: Red light vs ultraviolet (conceptual)

  • Red light has lower frequency and longer wavelength than ultraviolet light.

Quick Reference Formulas (Recap)

• Weighted average atomic mass: $\bar{M} = \sum<em>i f</em>i m_i$

• Isotope fraction (two isotopes): $\bar{M} = f<em>1 m</em>1 + (1 - f<em>1) m</em>2$

• Molecular vs empirical formulas:

  • Molecular: Actual counts (e.g., $\text{H}_2\text{O}$)

  • Empirical: Simplest ratio (e.g., $\text{HO}$ for hydrogen peroxide)

• Formula mass (molar mass) for covalent compounds: Sum of atomic masses.

  • Example: $\text{C}<em>9\text{H}</em>8\text{O}_4 \rightarrow M = 9(12.01) + 8(1.008) + 4(16.00) = 180.15\ \text{g/mol}$.

• Formula mass for ionic compounds: Mass of the repeating unit.

  • Example: NaCl $\rightarrow M = 22.99 + 35.45 = 58.44\ \text{g/mol}$.

• The mole and Avogadro’s number:

  • $1\ \text{mol} = N_A = 6.022 \times 10^{23}$

  • $N = n N_A, \ m = n M, \ n = \frac{m}{M}$

• Electromagnetic relations:

  • $c = \lambda\nu$ or $\nu = \frac{c}{\lambda}$

  • $E = h\nu = \frac{hc}{\lambda}$

  • Plank’s constant: $h = 6.626 \times 10^{-34}\ \mathrm{J\cdot s}$.