AP Chemistry: Atoms and Elements Study Guide

Unit 2 - Atoms and Elements: Key Concepts and Calculations

I. Fundamental Laws of Chemistry

Law of Conservation of Mass:
- States that in any closed system, mass is neither created nor destroyed during a chemical reaction.
- The total mass of the reactants must equal the total mass of the products.
- Example (Problem 2): If 1.50 ext{ g} of hydrogen reacts with 12.0 ext{ g} of oxygen to form water vapor (the only product), the mass of water vapor formed will be 1.50 ext{ g } + 12.0 ext{ g } = 13.5 ext{ g}.
Law of Definite Proportions (Law of Constant Composition):
- States that a given chemical compound always contains its component elements in fixed ratios (by mass), regardless of the source or method of preparation.
- Example (Problem 3):
  - Sample 1: 38.9 ext{ g} Carbon, 448 ext{ g} Chlorine. Ratio of Chlorine to Carbon = \frac{448 ext{ g}}{38.9 ext{ g}} \approx 11.516
  - Sample 2: 14.8 ext{ g} Carbon, 134 ext{ g} Chlorine. Ratio of Chlorine to Carbon = \frac{134 ext{ g}}{14.8 ext{ g}} \approx 9.054
  - These results are not consistent with the law of definite proportions because the mass ratios of chlorine to carbon are significantly different for the two samples. This suggests that the two samples might not be pure carbon tetrachloride or that there was an error in measurement, or they are different compounds.
Law of Multiple Proportions:
- States that if two elements can combine to form more than one compound, then the ratios of the masses of the second element that combine with a fixed mass of the first element are simple whole numbers.
- Example (Problem 4):
  - Compound 1: 0.168 ext{ g} oxygen per 1 ext{ g} osmium.
  - Compound 2: 0.3369 ext{ g} oxygen per 1 ext{ g} osmium.
  - Ratio of oxygen masses per gram of osmium = \frac{0.3369 ext{ g}}{0.168 ext{ g}} \approx 2.005
  - This ratio (approximately 2:1) is a simple whole number, demonstrating consistency with the law of multiple proportions.

II. Dalton's Atomic Theory Principles

Each element is composed of tiny, indestructible particles called atoms.
All atoms of a given element are identical in mass and other properties, but the atoms of one element are different from the atoms of all other elements.
Atoms combine in simple, whole-number ratios to form compounds.
Atoms of one element cannot be changed into atoms of a different element by chemical reactions; atoms are merely rearranged in chemical reactions (consistent with the Law of Conservation of Mass).

III. Structure of the Atom and Subatomic Particles

The Atom: Consists of a dense, positively charged nucleus surrounded by negatively charged electrons. Most of the atom's mass is concentrated in the nucleus.
Key Discoveries and Properties:

Particle	Location in Atom	Who discovered it?	Relative Charge	Relative Mass (amu)	Actual Mass (kg)
Electron	Outside nucleus (electron cloud)	J.J. Thomson (via Cathode Ray Tube experiments)	-1	0.00054858	9.109 \times 10^{-31}
Proton	Nucleus	Ernest Rutherford	+1	1.007276	1.673 \times 10^{-27}
Neutron	Nucleus	James Chadwick	0	1.008665	1.675 \times 10^{-27}

Cathode Ray Tube (CRT): Device used by J.J. Thomson to discover the electron. Cathode rays were observed to be streams of negatively charged particles (electrons) that were deflected by magnetic and electric fields.
Radioactivity: The spontaneous emission of radiation from atoms, leading to their decay into other elements.
Nuclear Theory (Rutherford's Model): Proposed that atoms have a small, dense, positively charged nucleus with electrons orbiting around it, based on the gold foil experiment. This replaced the 'plum pudding' model.

IV. Atomic Number, Mass Number, and Isotopes

Atomic Number (Z): The number of protons in an atom's nucleus. It defines the element. For a neutral atom, Z also equals the number of electrons.
Mass Number (A): The total number of protons and neutrons in an atom's nucleus (A = ext{protons} + ext{neutrons}).
Atomic Mass Unit (amu): Approximately one-twelfth the mass of a carbon-12 atom. Used to express the relative masses of atoms and subatomic particles.
Isotope: Atoms of the same element (same number of protons, Z) but with different numbers of neutrons, resulting in different mass numbers (A).
Nomenclature for Isotopes:
- Nuclear Symbol (_Z^A X): The chemical symbol (X) with the mass number (A) as a superscript and the atomic number (Z) as a subscript.
- Hyphen Notation (Name of element - A): The element's name followed by a hyphen and its mass number.
Examples (Problem 7):
- A. Copper isotope with 34 neutrons: Copper (Cu) has Z = 29 protons. Mass number A = 29 + 34 = 63. Nuclear symbol: _{29}^{63} ext{Cu}. Hyphen notation: Copper-63.
- B. Potassium isotope with 21 neutrons: Potassium (K) has Z = 19 protons. Mass number A = 19 + 21 = 40. Nuclear symbol: _{19}^{40} ext{K}. Hyphen notation: Potassium-40.
- C. Argon isotope with 22 neutrons: Argon (Ar) has Z = 18 protons. Mass number A = 18 + 22 = 40. Nuclear symbol: _{18}^{40} ext{Ar}. Hyphen notation: Argon-40.
Determining Protons, Electrons, and Neutrons (Problem 8):
- A. _{88}^{226} ext{Ra}
  - Protons: Z = 88
  - Electrons: 88 (neutral atom)
  - Neutrons: A - Z = 226 - 88 = 138
- B. Fluorine-19:
  - Fluorine (F) has Z = 9 protons.
  - Protons: 9
  - Electrons: 9 (neutral atom)
  - Neutrons: A - Z = 19 - 9 = 10
- C. _7^{15} ext{N}
  - Protons: Z = 7
  - Electrons: 7 (neutral atom)
  - Neutrons: A - Z = 15 - 7 = 8
- D. Tin-120:
  - Tin (Sn) has Z = 50 protons.
  - Protons: 50
  - Electrons: 50 (neutral atom)
  - Neutrons: A - Z = 120 - 50 = 70

V. Ions

Ions: Atoms or molecules that have gained or lost one or more electrons, resulting in a net electrical charge.
Cation: A positively charged ion (formed by losing electrons).
Anion: A negatively charged ion (formed by gaining electrons).
Electrical Charge: The net charge of an ion. For an atom with charge q, protons (p), and electrons (e), q = p - e
Determining Protons and Electrons in Ions (Problem 9):
- A. ext{Al}^{3+}
  - Aluminum (Al) has Z = 13 protons.
  - Protons: 13
  - Electrons: 13 - (+3 ext{ charge}) = 10
- B. ext{Se}^{2-}
  - Selenium (Se) has Z = 34 protons.
  - Protons: 34
  - Electrons: 34 - (-2 ext{ charge}) = 36
- C. ext{Ga}^{3+}
  - Gallium (Ga) has Z = 31 protons.
  - Protons: 31
  - Electrons: 31 - (+3 ext{ charge}) = 28
- D. ext{Sr}^{2+}
  - Strontium (Sr) has Z = 38 protons.
  - Protons: 38
  - Electrons: 38 - (+2 ext{ charge}) = 36
Predicting Ion Charge (Problem 10): Main-group elements tend to gain or lose electrons to achieve a stable electron configuration, typically resembling a noble gas.
- A. Ca (Calcium): Group 2, loses 2 electrons to form ext{Ca}^{2+}.
- B. Cl (Chlorine): Group 17, gains 1 electron to form ext{Cl}^{-}.
- C. S (Sulfur): Group 16, gains 2 electrons to form ext{S}^{2-}.
- D. P (Phosphorus): Group 15, gains 3 electrons to form ext{P}^{3-}.
- E. Li (Lithium): Group 1, loses 1 electron to form ext{Li}^{+}.

VI. The Periodic Table

Periodic Law: When elements are arranged in order of increasing atomic number, their chemical and physical properties show a periodic (repeating) pattern.
Classification of Elements:
- Metals: Typically solid, shiny, malleable, ductile, good conductors of heat and electricity. Located on the left and center of the periodic table.
- Nonmetals: Tend to be brittle, dull, poor conductors (insulators), can be gas, liquid, or solid. Located on the upper right side of the periodic table.
- Metalloids (Semiconductors): Have properties intermediate between metals and nonmetals. They lie along the zig-zag line separating metals and nonmetals.
Examples of Classification (Problem 12):
- A. Cesium (Cs): Metal (Group 1)
- B. Sulfur (S): Nonmetal (Group 16)
- C. Silicon (Si): Metalloid (Group 14)
- D. Cobalt (Co): Metal (Transition metal)
- E. Tellurium (Te): Metalloid (Group 16)
Element Groups/Families:
- Main-group elements: Elements in groups 1-2 and 13-18. Their chemistry is largely predictable.
- Transition elements (Transition metals): Elements in groups 3-12. Their chemistry is more complex.
- Group 1: Alkali Metals. Highly reactive metals (e.g., Li, Na, K).
- Group 2: Alkaline Earth Metals. Reactive metals, but less so than alkali metals (e.g., Mg, Ca, Sr).
- Group 16: Chalcogens. (e.g., O, S, Se, Te).
- Group 17: Halogens. Highly reactive nonmetals (e.g., F, Cl, Br, I).
- Group 18: Noble Gases. Very unreactive (inert) gases (e.g., He, Ne, Ar, Kr).
Classifying Element Type (Problem 14, 15):
- A. Chlorine (Cl): Main-group element (Group 17, Halogen)
- B. Mercury (Hg): Transition element (Group 12)
- C. Krypton (Kr): Main-group element (Group 18, Noble Gas)
- D. Silver (Ag): Transition element (Group 11)

VII. Atomic Mass

Atomic Mass: The weighted average mass of all naturally occurring isotopes of an element, based on their relative abundances.
Natural Abundance: The relative proportion of a particular isotope in a naturally occurring sample of an element.
Calculation of Atomic Weight (Problem 16):
- Atomic Weight = \Sigma [(\text{isotope mass}) \times (\text{natural abundance})]
- Magnesium isotopes:
  - Mg-24: Mass = 23.9850 ext{ amu}, Abundance = 0.7899
  - Mg-25: Mass = 24.9858 ext{ amu}, Abundance = 0.1000
  - Mg-26: Mass = 25.9826 ext{ amu}, Abundance = 0.1101
- Atomic Weight of Mg = (23.9850 imes 0.7899) + (24.9858 imes 0.1000) + (25.9826 imes 0.1101)
- Atomic Weight of Mg = 18.9458 + 2.49858 + 2.85908 (rounded)
- Atomic Weight of Mg = 24.303 ext{ amu}

VIII. Mole Concept and Stoichiometry

Mole (mol): The amount of a substance that contains Avogadro's number of particles (atoms, molecules, ions, etc.). \text{1 mol} = 6.022 \times 10^{23} \text{ particles}.
Avogadro's Number (N_A): 6.02214076 \times 10^{23} particles per mole.
Molar Mass: The mass in grams of one mole of a substance. Numerically equivalent to the atomic mass for elements (in g/mol).
Calculations using Mole Concept:
- Problem 17: Potassium atoms in 17.3 ext{ mol} of copper.
  - (Note: The question asks about potassium atoms, but gives moles of copper. Assuming it meant moles of potassium for consistency, or it's a trick question. We'll proceed as if it meant potassium atoms in 17.3 mol of potassium. If not, the answer would be 0.)
  - Potassium atoms = 17.3 ext{ mol K} \times (6.022 \times 10^{23} \text{ atoms K/1 mol K})
  - Potassium atoms = 1.042 \times 10^{25} \text{ atoms K}
- Problem 18: Moles of zinc in 2.82 \times 10^{22} zinc atoms.
  - Moles of Zn = (2.82 \times 10^{22} \text{ atoms Zn}) / (6.022 \times 10^{23} \text{ atoms Zn/1 mol Zn})
  - Moles of Zn = 0.0468 ext{ mol Zn}
- Problem 19: Amount (moles) of 11.8 ext{ g} argon (Ar).
  - Molar mass of Ar = 39.95 ext{ g/mol}
  - Moles of Ar = 11.8 ext{ g Ar} / (39.95 ext{ g Ar/1 mol Ar})
  - Moles of Ar = 0.295 ext{ mol Ar}
- Problem 20: Mass (grams) of 0.0355 ext{ mol} of barium (Ba).
  - Molar mass of Ba = 137.33 ext{ g/mol}
  - Mass of Ba = 0.0355 ext{ mol Ba} \times (137.33 ext{ g Ba/1 mol Ba})
  - Mass of Ba = 4.87 ext{ g Ba}
- Problem 21: Bismuth atoms in 1.87 ext{ g} bismuth (Bi).
  - Molar mass of Bi = 208.98 ext{ g/mol}
  - Moles of Bi = 1.87 ext{ g Bi} / (208.98 ext{ g Bi/1 mol Bi}) = 0.008948 ext{ mol Bi}
  - Atoms of Bi = 0.008948 ext{ mol Bi} \times (6.022 \times 10^{23} \text{ atoms Bi/1 mol Bi})
  - Atoms of Bi = 5.39 \times 10^{21} \text{ atoms Bi}
- Problem 22: Mass (grams) of 7.55 \times 10^{26} cadmium atoms.
  - Molar mass of Cd = 112.41 ext{ g/mol}
  - Moles of Cd = (7.55 \times 10^{26} \text{ atoms Cd}) / (6.022 \times 10^{23} \text{ atoms Cd/1 mol Cd}) = 1253.7 ext{ mol Cd}
  - Mass of Cd = 1253.7 ext{ mol Cd} \times (112.41 ext{ g Cd/1 mol Cd})
  - Mass of Cd = 140974 ext{ g Cd} or 1.410 \times 10^{5} ext{ g Cd}