CHEM 1211 Exam #1 Study Guide - Vocabulary Flashcards

Chapter 1: Measurement, Matter, and Calculations

Memorize SI prefixes and how to use them. Example prefixes include nano (n = 10^{-9}), micro (μ = 10^{-6}), milli (m = 10^{-3}), kilo (k = 10^{3}), etc. Remember:
- 1 \text{ nm} = 10^{-9} \text{ m}
- 1 \text{ m} = 10^{9} \text{ nm}
- A common trap: 10^{-9} \text{ nm} = 1 \text{ m} is NOT correct.
Significant figures and rounding:
- Addition/Subtraction: the result has the same number of decimal places as the measurement with the fewest decimal places.
- Multiplication/Division: the result has the same number of significant figures as the measurement with the fewest sig figs.
- Mixed operations: keep track of sig figs and apply the rules to obtain the final correct figure count.
- Express final answers in scientific notation when appropriate.
- Examples:
- Addition/Subtraction: 12.3 + 0.45 = 12.75 \rightarrow 12.8 (least decimals = 1 from 12.3)
- Multiplication/Division: 3.40 \times 2.1 = 7.14 \rightarrow 7.1 (fewest sig figs = 2 from 2.1)
- Subtraction with decimals: 50.0 - 12.34 = 37.66 \rightarrow 37.7 (least decimals = 1 from 50.0)
Scientific notation: convert small/large numbers to scientific notation for clarity (e.g., 0.00052 = 5.2 \times 10^{-4}).
Temperature conversions (formulas provided with a periodic table sheet):
- K = C + 273.15
- F = C \times \frac{9}{5} + 32
- C = \frac{5}{9}(F - 32)
- C = K - 273.15
- K = \frac{5}{9}(F - 32) + 273.15
Dimensional analysis and conversions:
- Use conversion factors to cancel units and reach the desired unit.
- Example: convert 12.0\ \text{g} to kilograms:
  12.0\ \text{g} \times \frac{1\ \text{kg}}{1000\ \text{g}} = 0.0120\ \text{kg}
Accuracy vs precision:
- Accuracy: closeness of a measurement to the true value.
- Precision: reproducibility of measurements (consistency).
Density, mass, and volume (dimensional consistency):
- Density formula: \rho = \frac{m}{V}
- Common units: \text{g}/\text{mL}, \text{g}/\text{cm}^3, \text{kg}/\text{m}^3
- Example: 250 g in 125 mL -> \rho = \frac{250\ \text{g}}{125\ \text{mL}} = 2.00\ \text{g/mL}
- Always ensure the units match before plugging into the density formula.
Classification of matter:
- Atom: a single nucleus particle; Molecule: two or more atoms bonded together.
- Element: substance composed of one kind of atom.
- Compound: substance composed of two or more elements chemically bonded.
- Homogeneous mixture: uniform composition throughout (e.g., saltwater).
- Heterogeneous mixture: nonuniform composition (e.g., Italian salad dressing).
Physical vs chemical properties and changes:
- Physical properties: color, density, melting/boiling point, etc.
- Physical changes: changes in state (melting, freezing, sublimation) without new substances.
- Chemical properties: reactivity, flammability, acidity/basicity, etc.
- Chemical changes: formation of new substances (e.g., rusting, combustion).
Chapter 1 takeaways for exam readiness:
- Be able to perform unit conversions using dimensional analysis.
- Apply sig fig rules correctly in all calculation types.
- Convert between Celsius, Fahrenheit, and Kelvin accurately.
- Solve density problems with correct units and meaningful significant figures.
- Distinguish between atoms, molecules, elements, compounds, and mixtures; identify physical vs chemical properties/changes.

Chapter 2: Atomic Structure and the Mole Concept

Atomic structure basics:
- Nucleus contains protons (p) and neutrons (n); electrons (e) orbit around the nucleus.
- Atomic number Z = number of protons = number of electrons in a neutral atom; mass number A = Z + N (where N is neutrons).
Isotopes and notation:
- Isotopes are atoms of the same element with different neutron numbers (A varies, Z constant).
- Notation example: ^{A}_{Z}X where X is the element symbol, Z is atomic number, A is mass number.
- Calculations for a neutral atom: p = Z, \quad e = Z, \quad n = A - Z
- Differences among isotopes are in neutrons (N); protons (Z) remain the same for the element.
First 30 elements and symbols (memorize):
- 1 H, 2 He, 3 Li, 4 Be, 5 B, 6 C, 7 N, 8 O, 9 F, 10 Ne, 11 Na, 12 Mg, 13 Al, 14 Si, 15 P, 16 S, 17 Cl, 18 Ar, 19 K, 20 Ca, 21 Sc, 22 Ti, 23 V, 24 Cr, 25 Mn, 26 Fe, 27 Co, 28 Ni, 29 Cu, 30 Zn
Neutral atoms vs ions:
- Neutral atom: electrons = protons (Z).
- Cation: fewer electrons than protons (positive charge).
- Anion: more electrons than protons (negative charge).
Chemical structures: molecular vs empirical formulas
- Molecular formula: shows the exact number of each type of atom in a molecule.
- Empirical formula: the simplest whole-number ratio of atoms in a compound.
- Structural/constitutional isomers: same molecular formula but different connectivity (structure).
- Given a structure, you should be able to write either the molecular formula or the empirical formula; given formulas, you should be able to deduce the other.
From grams or percent to formulas:
- Determine empirical formula from mass data or percent composition:
  1) Convert percent to grams (assume 100 g sample if using percent).
  2) Convert grams to moles for each element.
  3) Divide all mole values by the smallest to get the simplest whole-number ratio.
  4) If needed, multiply to obtain whole numbers.
- To obtain molecular formula from empirical formula:
- Compute empirical formula mass: M{emp} = \sum (ni \times Mi) where $ni$ are the counts in the empirical formula and $M_i$ are atomic masses.
- Determine the multiplier: n = \dfrac{M{molar}}{M{emp}} where $M_{molar}$ is the molar mass of the molecular formula.
- Molecular formula = empirical formula multiplied by $n$.
The mole concept:
- One mole contains N_A = 6.022 \times 10^{23} entities (atoms, molecules, formula units, etc.).
- Key relationships:
- Mass to moles: n = \dfrac{m}{M} where M is molar mass in g/mol.
- Moles to molecules: N = n \times N_A
- Molecules to moles: n = \dfrac{N}{N_A}
Example problems and quick checks:
- Example 1: Water, \text{H}2\text{O}, molar mass M{H2O} \approx 18.015\ \text{g/mol}. If you have 18.0 g, n = \dfrac{18.0}{18.015} \approx 1.0008\ \text{mol}, so about N \approx 1.00 \times N_A = 6.022 \times 10^{23} molecules.
- Example 2: A compound with empirical formula CH$_2$O and molar mass 180.0 g/mol.
- Empirical mass: M_{emp} = (1\times 12.01) + (2\times 1.008) + (1\times 16.00) \approx 30.03\ \text{g/mol}.
- Multiplier: n = \dfrac{180.0}{30.03} \approx 6.
- Molecular formula: $(CH2O)6 = \mathrm{C}6\mathrm{H}{12}\mathrm{O}_6$.
Isomerism and connectivity:
- Structural (constitutional) isomers: same molecular formula, different connectivity.
- Examples to identify from a given structure or to draw from a given formula (practice from class/PowerPoints).
Practical exam connections:
- Expect questions on counting protons/electrons/neutrons from Z and A.
- Be able to classify substances into atoms, molecules, elements, compounds, and mixtures.
- Compute empirical and molecular formulas from mass/percent data.
- Perform mole-mass-molecule conversions with proper significant figures.
Exam format and day-of-exam logistics (summary):
- Exam format: 20 multiple choice questions (3 points each) = 60 points; Short answer questions = 40 points; Bonus questions = 5 points.
- Time: the exam will be held during the entire class period.
- Materials to bring: pencil and eraser; non-programmable scientific calculator; cell phones and sharing calculators are not allowed.
- Formulas will be provided with a periodic table on a sheet within this module.

Key Formulas and Quick Reference

Temperature conversions:
- K = C + 273.15
- F = C \times \frac{9}{5} + 32
- C = \frac{5}{9}(F - 32)
- C = K - 273.15
- K = \frac{5}{9}(F - 32) + 273.15
Density:
- \rho = \dfrac{m}{V}
Scientific notation and sig figs (guidelines):
- For multiplication/division: number of significant figures = smallest among factors.
- For addition/subtraction: decimal places = smallest among addends.
Mole concept:
- N_A = 6.022 \times 10^{23}
- n = \dfrac{m}{M}
- N = nN_A
- For molecules count: N = n \times N_A
Isotope notation:
- ^{A}_{Z}X where A = mass number, Z = atomic number, X = element symbol.
- In a neutral atom: e = p = Z,\; n = A - Z.
Empirical vs molecular formula:
- Empirical formula mass: M{emp} = \sumi ni Mi
- If molecular molar mass is $M{mol}$, then n = \dfrac{M{mol}}{M_{emp}} and Molecular Formula = Empirical Formula × n.

First 30 Elements (Symbols to Know)

H, He, Li, Be, B, C, N, O, F, Ne, Na, Mg, Al, Si, P, S, Cl, Ar, K, Ca, Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn

Study Strategy and Real-World Relevance

Why precision and accuracy matter: measurement quality affects experimental conclusions, safety, and reproducibility.
Dimensional analysis mirrors real lab practices: unit checks prevent misinterpretation and calculation errors.
Understanding isotopes helps in fields from radiometric dating to medical imaging.
Mastery of mole concept and stoichiometry underpins quantitative chemistry, reaction yield predictions, and material design.

Quick References for Exam Preparation

Memorize the first 30 elements and their symbols.
Be able to write and interpret isotope notation and determine numbers of p, e, n.
Be able to distinguish between molecular formulas and empirical formulas; identify structural isomers from given structures.
Be able to perform percent composition problems and derive empirical/molecular formulas.
Practice converting between grams, moles, and molecules using the mole concept, including conversions with Avogadro's number.
Practice density problems with correct unit handling and sig figs.
Practice temperature conversions and dimensional analysis problems.
Review the exam format and ensure you bring the allowed materials to class on exam day.

Hypothetical Worked Scenarios (Practice)

Dimensional analysis with multiple steps:
- Convert 7.50 cm to meters and then to millimeters:
- Step 1: 7.50\ \text{cm} \times \frac{1\ \text{m}}{100\ \text{cm}} = 0.0750\ \text{m}
- Step 2: 0.0750\ \text{m} \times \frac{1000\ \text{mm}}{1\ \text{m}} = 75.0\ \text{mm}
Sig figs in mixed operations:
- Compute (12.3 \text{ g} + 0.45 \text{ g}) \times 2.0
- Addition: 12.3 + 0.45 = 12.75 -> with 1 decimal place (due to 12.3), result becomes 12.8.
- Multiplication: 12.8 × 2.0 => 2 sig figs in the factor 2.0, so final result = 25.
Empirical/molecular formula example:
- A compound analysis shows 40.0% C, 6.7% H, and the remainder O by mass.
- Assume 100 g sample: 40.0 g C, 6.7 g H, 53.3 g O.
- 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.
- Divide by smallest (3.33): C ≈ 1.00, H ≈ 2.00, O ≈ 1.00 → empirical formula CH$_2$O.
- If molar mass of the compound is 180.0 g/mol, Memp = 30.03 g/mol, n ≈ 6, Molecular formula = C$6$H${12}$O$6$.