Chemistry Lecture Notes: Atomic Structure, Isotopes, and Practice Problems
Schedule and Assessments
- Quiz: scheduled for 09:11 on the next Thursday (online); must be completed before it closes.
- Exam: scheduled for 09:18, one week after the quiz; will cover material up to the section 1a/1b; 2a and 2b are each worth 10 points.
- 1a and 1b: already posted; you have enough information to complete both.
- The instructor will pull some problems from the summertime test to illustrate typical question styles.
- Homework and practice on iCollege:
- There are four practice sets labeled as practice (not the homework). These are due at 05:00 on test day.
- The homework (actual graded) is due at 05:00 on test day and is split into part a and part b, with separate point values (part 1 and part 2, around 10 points each).
- It’s common for questions to use two-part problems where you can deduce one part from the other if you know one of the quantities; always ensure the two pieces sum to 100% when dealing with isotopic abundances.
Review of Key Concepts from Today
- Dalton and Rutherford are two central figures in early atomic theory.
- Dalton
- Law of definite proportions: Elements in a given compound are always in the same proportion by mass.
- Law of multiple proportions: When two elements form more than one compound, the ratios of the masses of the second element, which combine with a fixed mass of the first element, are ratios of small whole numbers.
- Thomson
- Cathode ray experiment led to the discovery of the electron.
- Rutherford
- Gold foil experiment led to the nuclear model: a dense positively charged nucleus in the center with electrons surrounding it.
- Nuclear model components
- Nucleus contains protons (positive charge) and neutrons (neutral).
- Electrons (negative charge) are located outside the nucleus.
- Protons, neutrons have masses ≈ 1 amu each; electrons have negligible mass in comparison.
- Subatomic particles
- Proton: positive charge, ~1 amu.
- Neutron: neutral charge, ~1 amu.
- Electron: negative charge, ~5.485 × 10^-4 amu (negligible for most mass calculations).
- Atomic structure terminology
- Atom consists of a nucleus (protons + neutrons) and surrounding electrons.
- Atomic number Z: number of protons in the nucleus.
- Mass number A: total number of protons and neutrons in the nucleus (A = Z + N).
- Isotopes: atoms of the same element (same Z) with different numbers of neutrons (different A).
- Isotopes and Abundances
- Isotopes have the same Z but different N, hence different masses.
- Natural isotopes: those found in nature; some elements have multiple natural isotopes; some isotopes are radioactive (e.g., carbon-14) while others are stable (e.g., carbon-12, carbon-13).
- Atomic mass unit (amu or u): defined as 1/12 the mass of a carbon-12 atom, i.e., 1~ ext{amu} = rac{1}{12} m( ext{^{12}C}). Carbon-12 is defined to be exactly 12 amu, so 1 amu is exactly one twelfth of the carbon-12 mass. Other elements have atomic masses that are weighted averages of their naturally occurring isotopes.
- The proton and neutron masses are approximately 1 amu; the electron mass is about 1/1836 of a proton mass (much smaller).
- The periodic table arrangement places elements by increasing Z; symbols have 1 or 2 letters, with the first letter capitalized and the second (if present) lowercase.
- The symbol W stands for tungsten (from the German word Wolfram).
- Many very old element names/symbols are from historical origins; the standard two-letter symbols reflect those origins and capitalization rules.
Atomic Mass Unit and Isotopes: Foundational Details
- Isotopes of an element
- Have the same number of protons (Z) but different numbers of neutrons (N).
- This leads to different atomic masses for each isotope.
- Carbon isotopes as a canonical example
- Common natural isotopes: ${}^{12} ext{C}, {}^{13} ext{C}, {}^{14} ext{C}$.
- Carbon-12 is exactly 1 amu by definition (one twelfth of the mass of a ${}^{12} ext{C}$ atom).
- Carbon-13 and carbon-14 masses are close to 13 amu and 14 amu respectively but are defined relative to the carbon-12 standard.
- Atomic mass of an element on the periodic table
- It is a weighted average of the masses of the naturally occurring isotopes, using their fractional abundances.
- Example concept: if an element has two naturally occurring isotopes with masses $ma$ and $mb$ and abundances $fa$ and $fb$ (with $fa + fb = 1$), then the atomic mass $M$ is M = fa ma + fb mb. When a third isotope is present, the same weighted-average approach applies with more terms.
- Practical point about notation
- Notation like ${}^{A}_{Z} ext{X}$ indicates an isotope of element X with mass number A and atomic number Z (i.e., A = Z + N).
- If the charge is shown (e.g., ${}^{A}_{Z} ext{X}^{n+}$), the charge indicates the net number of electrons relative to protons.
- Isotope abundance problem-solving approach
- Given two isotopes with masses $ma$ and $mb$ and a measured average mass $M$, solve for the unknown fraction (abundance) of one isotope using the relation M = fa ma + fb mb, ext{ with } fa + fb = 1. A convenient rearrangement is $fb = 1 - fa$ and $fa = rac{M - mb}{ma - mb}$ (assuming $ma
eq mb$).
- If you’re given $M$, $ma$, and $mb$, you can compute the unknown fraction directly. Then the other fraction is the complement.
Worked Examples from Today’s Transcript
- Boron isotopes: ${}^{10} ext{B}$ and ${}^{11} ext{B}$
- Masses: $m{10} = 10$ amu, $m{11} = 11$ amu.
- Atomic mass of boron in nature (weighted average): $M = 10.81$ amu.
- Let $f{10}$ be the fraction of ${}^{10} ext{B}$ and $f{11} = 1 - f_{10}$ for ${}^{11} ext{B}$.
- Solve: M = f{10} imes 10 + (1 - f{10}) imes 11 = 11 - f{10} \ herefore f{10} = 11 - M = 11 - 10.81 = 0.19.
- Thus: f{10} = 0.19 ext{ (19%), } f{11} = 0.81 ext{ (81%)}.
- Bromine isotopes: ${}^{79} ext{Br}$ and ${}^{81} ext{Br}$
- Masses: $m{79} = 79$, $m{81} = 81$ amu.
- Weighted average mass given: $M ext{ (approximately }79.9 ext{ amu)}$.
- Solve: M = f{79} imes 79 + (1 - f{79}) imes 81 = 81 - 2f{79} \ herefore f{79} = rac{81 - M}{2}.
- If $M = 79.9$, then $f{79} = rac{81 - 79.9}{2} = rac{1.1}{2} = 0.55$ (55%), and $f{81} = 0.45$ (45%).
- The heavier isotope ${}^{81} ext{Br}$ mass is around 81 amu, as inferred from the calculation.
- Lithium isotopes: ${}^{6} ext{Li}$ and ${}^{7} ext{Li}$
- Given that the atomic mass of lithium is about $6.94$ amu and there are two naturally occurring isotopes with masses 6 and 7, determine which is more abundant.
- Since the average is closer to 7, ${}^{7} ext{Li}$ is the more abundant isotope.
- If you know the average and the masses, you can deduce the fractional abundances similarly to the boron example.
- Carbon isotopes and dating context
- Carbon has isotopes ${}^{12} ext{C}$, ${}^{13} ext{C}$, and ${}^{14} ext{C}$.
- ${}^{12} ext{C}$ is exactly 1 amu by definition. ${}^{13} ext{C}$ and ${}^{14} ext{C}$ have masses near 13 and 14 amu, respectively, but the atomic mass of carbon on the periodic table is a weighted average of these naturally occurring isotopes.
- ${}^{14} ext{C}$ is radioactive and used for dating (carbon dating).
- The statement that there are “three isotopes” of carbon refers to the three naturally occurring isotopes, among which ${}^{12} ext{C}$ and ${}^{13} ext{C}$ are stable while ${}^{14} ext{C}$ is radioactive.
- Isotope notation conventions and real-world relevance
- In practice, you’ll encounter isotope notation in the form of ${}^{A}_{Z} ext{X}$ or sometimes ${}^{A} ext{X}$ for shorthand, where A is the mass number and Z is the atomic number.
- In neutral atoms, the number of electrons equals the number of protons (Z). If the atom is ionized, the electron count differs by the net charge.
- Understanding isotopes is crucial for fields like archaeology (carbon dating), geology, medicine, and environmental science.
Periodic Table Essentials (Context from Today)
- The periodic table arrangement is described as elements being placed along rows and columns with charges shown on the table for some students.
- Tungsten’s symbol is W, derived from the German word Wolfram.
- Element symbols are often one or two letters; the first letter is capitalized and the second (if present) is lowercase.
- When naming and classifying elements, historical origins influence symbols and naming conventions.
Isotopes: Core Relationships and Notation
- A (mass number) and Z (atomic number)
- ${}^{A}_{Z} ext{X}$ denotes an isotope of element X with mass number A and atomic number Z.
- Determining neutrons from A and Z
- Charge and electron count
- For a neutral isotope, the number of electrons equals Z; for an ion, electrons = Z ± |charge| depending on whether it’s an anion or cation.
Quick Reference: Key Concepts to Remember
- Atomic mass unit (amu)
- Defined via carbon-12: 1~ ext{amu} = rac{1}{12} m( ext{^{12}C}).
- Protons and neutrons ~ 1 amu; electrons ~ 5.485 × 10^-4 amu (negligible for most mass calculations).
- Isotopes have the same Z but different N; masses differ accordingly.
- The atomic mass shown on the periodic table is a weighted average of natural isotopes.
- The nucleus contains protons and neutrons; electrons orbit the nucleus.
- Isotope abundance problems require solving simple linear equations with the constraint that abundances sum to 1 (or 100%).
Practical Tips and Exam Strategies
- When solving isotope-abundance problems, always check that abundances sum to 1 (or 100%). If you’re given one abundance, you can find the other using the weighted-average mass.
- For problems with two isotopes, use the relation M = fa ma + (1 - fa) mb to solve for the fractional abundance $f_a$.
- Express abundances as decimals (fractions) or percentages consistently; remember to convert to the opposite form if needed.
- Recognize that many exam questions will present the problem in the form of A/Z notation and ask you to extract Z, or to determine N from A and Z.
- Be comfortable with unit conversions (mm → cm → m → km) and with dimensional analysis to avoid errors in the number of significant figures.
Worked Practice Concept Recap
- Example: Density calculation from a cube
- Given edge length in millimeters, convert to centimeters, compute volume in cm^3, and then density from mass/volume.
- For 11.4 mm edge length:
- Edge in cm: 11.4~ ext{mm} o 1.14~ ext{cm}.
- Cube volume: V = (1.14~ ext{cm})^3 \ = 1.48~ ext{cm}^3 ext{ (approximately)}.
- Mass: 25 g.
- Density:
ho = rac{m}{V} = rac{25~ ext{g}}{1.48~ ext{cm}^3} \
ho
oughly 16.9~ ext{g/cm}^3.
- Example: Millimeters to kilometers
- Given 1.23 mm: convert to kilometers via meters and then to kilometers:
- 1.23~ ext{mm} = 1.23 imes 10^{-3}~ ext{m} = 1.23 imes 10^{-6}~ ext{km}.$$
Final Note on Relevance and Ethics (Practical Implications)
- Understanding isotopes and atomic structure informs real-world applications like medical imaging, radiometric dating, environmental tracing, and materials science.
- Accurate unit handling, notation, and problem-solving skills are foundational for exams and for responsible laboratory work.
- Ethical and philosophical reflections may arise in contexts such as nuclear energy, radiometric dating’s role in archaeology, and the societal implications of scientific knowledge; these were not covered explicitly in today’s lecture but are important to consider in broader coursework.