Chem 201 Fall 2025 - Worksheet 1 Key Terms (VOCABULARY Flashcards)

Page 1 — Particulate Nature, Atomic Structures, and Quick Concepts

• Particulate nature (describe with pictures):
  • Liquids
  • Definite volume, variable shape, flows and conforms to container.
  • Particles are close together with moderate attraction; some free movement, but not freely moving like gases.
  • Incompressible relative to gases; slight compressibility exists.
  • Gases
  • No definite shape or volume (fill space); highly compressible and expand to fill container.
  • Particles are far apart with mostly empty space; rapid, random motion; weak interparticle forces except during collisions.
  • Solids
  • Definite shape and volume; rigid structure; particles vibrate in place with strong interparticle forces.
• Question 2: Identify protons, neutrons, electrons in given ions/atoms
  • 79Br^-1 (Bromine-79 with -1 charge)
  • Protons: 35
  • Neutrons: 79 − 35 = 44
  • Electrons: 35 + 1 = 36 (since negative charge means extra electron)
  • 133Cs^+1 (Cesium-133 with +1 charge)
  • Protons: 55
  • Neutrons: 133 − 55 = 78
  • Electrons: 55 − 1 = 54 (lost one electron)
• Question 3: Element identification from data
  • Method: For each entity, use atomic number (protons) to identify the element; mass number (A) gives neutrons via N = A − Z; electrons = protons minus/plus the charge.
  • Note: The transcript lists a table prompt but specific elements are not provided in text here; apply the same method to any given Z and charge.
• Question 4: Matching exercise (Ernest Rutherford, John Dalton, J. J. Thomson, Robert Millikan)
  • Correct associations:
  • Ernest Rutherford → B. Nuclear atom
  • John Dalton → C. Atomic Theory
  • J. J. Thomson → D. Plum pudding model of the atom
  • Robert Millikan → F. Charge to mass ratio of the electron (e/m)
  • About A and E (noted in list):
  • A. Oil drop experiment – mass of the electron (also tied to Millikan via e/m)
  • E. Equal volumes of gas contain the same number of particles (Avogadro’s idea, concept not directly tied to the four scientists listed above)
• Question 5: True/False statements
  • In a neutral atom, protons = neutrons. False (Z ≈ N only for certain isotopes; many have N ≠ Z)
  • Electrons are less massive than neutrons. True (electrons are far lighter)
  • Charge(proton) + Charge(electron) = 0. True ( +1 + (−1) = 0 )
  • Charge(proton) + Charge(neutron) = +2. False (neutron charge is 0, so total is +1)
  • Protons are much more massive than neutrons. False (protons and neutrons have comparable masses; they differ by only a small amount)

Page 2 — Notation, Ions, Calculations, and Isotopes

• Question 6: Symbol for the element with six protons in the form ZX A

  • Answer: Yes. The element is Carbon (Z = 6). The common isotopic symbol is $^{A}{Z} ext{X}$. The most common isotope is Carbon-12, so $^{12}{6} ext{C}$.

• Question 7: Determine the charge of ions for alkali metals, alkaline earth metals, and halogens

  • Alkali metals (Group 1): +1 charge (lose one electron)
  • Alkaline earth metals (Group 2): +2 charge (lose two electrons)
  • Halogens (Group 17): −1 charge (gain one electron)
  • Rationale: Based on achieving a noble-gas electron configuration (valence electrons: 1, 2, 7 respectively).

• Question 8: Mass of 1,000 atoms of Co (cobalt)

  • Atomic mass of Co ≈ $M_{ ext{Co}} o 58.93$ g/mol
  • Number of moles in 1,000 atoms: n = rac{1000}{6.022 imes 10^{23}} ext{ mol}

  \,

  • Mass: m = n imes M_{ ext{Co}} \approx 58.93 imes rac{1000}{6.022 imes 10^{23}} \approx 9.8 imes 10^{-20} ext{ g}

• Question 9: How many moles and atoms are in 100 g of Ca (calcium)

  • Molar mass Ca ≈ 40.08 g/mol
  • Moles: n = rac{100}{40.08} \approx 2.50 ext{ mol}
  • Atoms: N = n imes N_A \approx 2.50 imes 6.022 imes 10^{23} \approx 1.51 imes 10^{24}

• Question 10: How many moles of N₂ in 550 g of N₂

  • Molar mass N₂ ≈ 28.02 g/mol
  • Moles: n = rac{550}{28.02} \approx 19.6 ext{ mol}

• Question 11: How many O atoms in 25 g of O₂

  • Molar mass O₂ ≈ 32.00 g/mol
  • Moles of O₂: n{O2} = rac{25}{32} \approx 0.78125 ext{ mol}
  • Each O₂ molecule contains 2 O atoms, so total O atoms: NO = 2 n{O2} NA \approx 2 imes 0.78125 imes 6.022 imes 10^{23} \approx 9.41 imes 10^{23}

• Question 12: What do all of these carbon atoms and ions have in common? 12C, 13C, 12C+, 13C-

  • All contain 6 protons (Z = 6). They differ in neutrons (N) and electrons (charge) but share the same nucleus species for the neutral isotopes.
  • 12C has Z = 6, N = 6, e⁻ = 6; 13C has Z = 6, N = 7, e⁻ = 6; 12C⁺ has Z = 6, N = 6, e⁻ = 5; 13C⁻ has Z = 6, N = 7, e⁻ = 7.

Page 3 — Magnesium, Molar Concept Checks, and Miscellanea

• Question 13: Periodic table magnesium information
  • Magnesium symbol: Mg; Atomic number Z = 12; The number just below the symbol (rounded to two decimals) is the average atomic mass: ~24.31 amu.
  • Comparison:
  • 24.31 amu ≈ the average mass of one Mg atom.
  • Mass of 6.022 × 10^23 Mg atoms (one mole): ~24.31 g.
• Question 14: Fun comparisons about elephants, chickens, dozens, and moles
  • a) Elephants in a dozen elephants: 12
  • b) Which has more animals: a dozen elephants vs a dozen chickens? They have the same number of animals (12 each).
  • c) Elephants in a mole of elephants: $6.022 imes 10^{23}$ elephants
  • d) Which has more animals: a mole of elephants vs a mole of chickens? They have the same number of animals ($6.022 imes 10^{23}$)
  • e) Which has more atoms: a dozen H atoms or a dozen Ar atoms? Both are 12 atoms; equal
  • f) Which has more atoms: a mole of H atoms or a mole of Ar atoms? Both have $6.022 imes 10^{23}$ atoms; equal

Page 4 — Mass, Moles, and Counting Atoms (Continuing Calculations)

• Question 15: Without a calculator
  • a) Which weighs more: 18 elephants or two dozen elephants? Two dozen elephants weigh more (24 vs 18).
  • b) Which weighs more: 5.136 × 10^23 sodium atoms or one mole of sodium atoms? One mole has $6.022 imes 10^{23}$ atoms; mass of Na (per mole) ≈ 22.99 g; 5.136 × 10^23 atoms ≈ 0.853 mol → mass ≈ 19.6 g. Therefore one mole of Na weighs more.
• Question 16: More atoms in 1.008 g H vs 39.95 g Ar
  • 1.008 g H corresponds to 1.0000 mol of H (approximately) → ~6.022 × 10^23 atoms
  • 39.95 g Ar corresponds to about 1.000 mol of Ar → ~6.022 × 10^23 atoms
  • Conclusion: They contain about the same number of atoms (one mole of each)
• Question 17: Mass in grams of 6.022 × 10^23 H and K atoms
  • 6.022 × 10^23 H atoms ≈ 1.008 g
  • 6.022 × 10^23 K atoms ≈ 39.10 g
• Question 18: Mass in grams of (a) 12.044 × 10^23 Na atoms and (b) 15.0 × 10^23 Na atoms
  • (a) moles = (12.044 × 10^23) / (6.022 × 10^23) ≈ 2.00 mol; mass ≈ 2.00 × 22.99 ≈ 45.98 g
  • (b) moles = (15.0 × 10^23) / (6.022 × 10^23) ≈ 2.492 mol; mass ≈ 2.492 × 22.99 ≈ 57.3 g
• Question 19: Difference between 35Cl and 37Cl
  • They are isotopes of chlorine with different neutron numbers: 35Cl has Z = 17, N = 18; 37Cl has Z = 17, N = 20
  • They have different masses and slightly different abundances; 35Cl is more abundant than 37Cl in nature
• Question 20: Boron isotope abundances on Krypton
  • Given: 10B 10.013 amu, 65.75%; 11B 11.009 amu, 25.55%; 12B 12.014 amu, 8.70%
  • Average atomic mass calculation:
  • ar{m} = 0.6575(10.013) + 0.2555(11.009) + 0.0870(12.014) \approx 10.4415 ext{ amu}
  • Krypton planet boron average ~ $10.44$ amu
• Question 21: Chlorine isotopic abundances (35Cl vs 37Cl)
  • Let f(35) = x, f(37) = 1 − x
  • Weighted mass: x(34.9689) + (1 − x)(36.9659) = 35.453
  • Solve: (-1.9970) x = -1.5129
    ightarrow x \,≈ 0.758
  • Therefore: approx 75.8% 35Cl and 24.2% 37Cl
• Question 22: Masses for one mole of helium and potassium atoms
  • a) One mole of He atoms: ≈ 4.00 g
  • b) One mole of K atoms: ≈ 39.10 g
• Question 23: The average mass, in grams, of one atom
  • a) He: mass per atom ≈ 4.00 amu → m{ ext{atom}} = rac{4.00}{NA} ext{ g} \, ext{(≈ 6.64 × 10^{-24} g)}
  • b) K: mass per atom ≈ 39.10 amu → m{ ext{atom}} = rac{39.10}{NA} ext{ g} \, ext{(≈ 6.49 × 10^{-23} g)}
• Question 24: Mass of 5.000 moles of carbon atoms
  • Molar mass C ≈ 12.01 g/mol
  • Mass: m = 5.000 imes 12.01 \approx 60.05 ext{ g}
• Question 25: How many sodium atoms in 6.000 moles?
  • N = 6.000 imes N_A \\approx 3.613 imes 10^{24}
• Question 26: How many sodium atoms in 100.0 g of sodium?
  • Molar mass Na ≈ 22.99 g/mol
  • Moles: n = rac{100.0}{22.99} \approx 4.348 ext{ mol}
  • Atoms: N = n imes N_A \approx 2.62 imes 10^{24}
• Question 27: Number of atoms in various masses
  • a) 50.7 g H: moles ≈ 50.7 / 1.008 ≈ 50.3 mol; atoms ≈ 50.3 × 6.022 × 10^23 ≈ 3.03 × 10^25
  • b) 1.00 mg Co: 0.001 g; moles ≈ 0.001 / 58.93 ≈ 1.70 × 10^−5 mol; atoms ≈ 1.70 × 10^−5 × 6.022 × 10^23 ≈ 1.02 × 10^19
  • c) 1.00 kg S: 1000 g; moles ≈ 1000 / 32.06 ≈ 31.2 mol; atoms ≈ 31.2 × 6.022 × 10^23 ≈ 1.88 × 10^25
  • d) 1.00 ton Fe (metric ton = 1,000 kg = 1.0 × 10^6 g): mol ≈ 1.0 × 10^6 / 55.85 ≈ 1.79 × 10^4 mol; atoms ≈ 1.79 × 10^4 × 6.022 × 10^23 ≈ 1.08 × 10^28
• Question 28: Element whose atoms have average mass 5.14 × 10^−23 g
  • Convert to amu: amu ≈ g per atom × (1 / 1.66054 × 10^−24) ≈ 5.14 × 10^−23 g corresponds to ≈ 31 amu
  • Element with ~31 amu average mass is Phosphorus (P, Z = 15)
• Question 29: Mass of iodine containing same number of atoms as 25.0 g Cl
  • Cl molar mass ≈ 35.453 g/mol; moles Cl in 25.0 g: n_{Cl} = rac{25.0}{35.453} \approx 0.706 ext{ mol}
  • Iodine molar mass ≈ 126.90 g/mol; mass for same moles: m{I} = n{Cl} imes 126.90 \approx 0.706 imes 126.90 \approx 89.6 ext{ g}

Page 5 — Isotopes, Averages, and Nuclear Questions

• Question 30: Neon isotopes and abundances

  • Given: 20Ne with mass 19.9924 amu and abundance 90.5%
  • Let x be the abundance of 22Ne (the remaining isotope)
  • If we use the known Neon average mass ≈ 20.1797 amu:
  • $0.905 imes 19.9924 + x imes 22.0 = 20.1797$ and $0.905 + x = 1$ → x ≈ 0.095
  • Therefore the other isotope is 22Ne (about 9.5%)

• Question 31: True/False with explanations

  • a) On average, one Li atom weighs 6.941 g. False. 6.941 is the atomic mass unit value (amu); 6.941 g per mole corresponds to 1 mole, not a single atom. Each Li atom weighs ~ (6.941 / N_A) g.
  • b) Every H atom weighs 1.008 amu. True as the standard atomic weight for H is 1.008 amu (noting natural isotopic variation exists but 1.008 is the conventional average).
  • c) A certain mass of solid Na contains fewer atoms than the same mass of gaseous Ne. True (Na has greater molar mass than Ne, so for the same mass there are fewer Na moles than Ne moles).
  • d) The average atomic mass of an unknown monatomic gas is 0.045 g/mol. False (even the lightest noble gas, He, is about 4.00 g/mol; 0.045 g/mol is far too light).

• Question 32: Periodic table entry for chlorine: symbol Cl and numbers 17 and 35.453 — four pieces of information

  • From 17 (atomic number Z):
  • Protons in Cl = 17
  • Electrons in neutral Cl = 17
  • From 35.453 (average atomic mass):
  • Mass per mole of Cl atoms = 35.453 g/mol
  • Mass per single Cl atom ≈ 35.453 / N_A g

• Question 33: Rhenium isotopes

  • Atomic mass of Re = 186.2
  • If 37.1% is Re-185, the remaining 62.9% is the other stable isotope
  • Solve for A in 0.371 × 185 + 0.629 × A = 186.2
  • A ≈ 187; therefore the other stable isotope is 187Re

• Key formulas and constants to remember

  • Avogadro’s number: N_A = 6.022 imes 10^{23} ext{ mol}^{-1}
  • Molar mass (g/mol) relates moles to mass: n = rac{m}{M}
  • Number of atoms: N = n imes N_A
  • Isotopic notation: ^{A}_{Z}X where A = mass number, Z = atomic number, X = symbol
  • Conversion between amu and grams per atom: 1 amu = 1.660 imes 10^{-24} ext{ g}; mass per atom = rac{ ext{amu}}{N_A} ext{ g}
  • For gases: molar mass in g/mol; for a gas, number of particles at a given mass follows mole concept (1 mole = 6.022 imes 10^{23} atoms/molecules)

• Notation and constants recap

  • Common element masses (examples):
  • C ≈ 12.01 g/mol
  • Ca ≈ 40.08 g/mol
  • Co ≈ 58.93 g/mol
  • Mg ≈ 24.31 g/mol
  • Cl ≈ 35.45 g/mol
  • H ≈ 1.008 g/mol (for a mole of H atoms)
  • He ≈ 4.00 g/mol
  • Ne ≈ 20.18 g/mol (average)

• Connections to foundational principles

  • Atomic number (Z) defines identity (element) and determines electron count in neutral atoms
  • Mass number (A) and neutron count determine isotopes and stability
  • Molar mass links atomic scale to macroscopic mass via Avogadro’s number
  • Isotopes differ in neutrons; chemical properties depend largely on Z, not on N, but physical properties (mass, stability, abundance) vary
  • Isotopic abundances drive average atomic masses; these values are used to compute weighted averages (as with chlorine, boron, neon, etc.)

• Practical implications and sample problems you can rehearse

  • From given Z and A, quickly compute N and electrons for ions
  • Use mass numbers and charges to deduce electron counts in ions
  • Convert between atoms, moles, and grams using M and N_A
  • Determine isotopic compositions from weighted averages; solve simple linear equations to find unknown abundances
  • Compare numbers of particles across different scales (dozens vs. moles) to reason about mass, count, and atomic scale