Chapter 2: Atoms and Elements - Study Notes

Atoms and Elements: Fundamental Principles

Law of Conservation of Mass

• Definition: The sum of the masses of the reactants of a chemical equation is equal to the sum of the masses of the products.

• Principle: Mass is neither created nor destroyed in an "ordinary" chemical reaction.

  • This implies that atoms are rearranged, but their total quantity and mass remain constant.

Law of Definite Proportions

• Definition: All samples of a given compound have the same proportions of their constituent elements, regardless of the source.

• Example: Calcium Oxide (CaO)

  • 56.09 g of CaO decomposes to form 40.09 g of calcium (Ca) and 16.00 g of oxygen (O).

  • 100.00 g of CaO decomposes to form 71.47 g of calcium (Ca) and 28.53 g of oxygen (O).

  • Ratio of Ca to O: \frac{40.09}{16.00} = \frac{71.47}{28.53} = 2.506. This consistent ratio demonstrates the law.

Law of Multiple Proportions

• Definition: When two elements combine to form two or more different compounds, the mass ratios of one element that combines with a fixed mass of the other element can be expressed as small whole numbers.

• Example: Carbon Monoxide (CO) & Carbon Dioxide (CO$_2$)

  • For carbon dioxide (CO_2) reacting with 1 g of carbon (C), the mass of oxygen (O) is 2.67 g.

  • For carbon monoxide (CO) reacting with 1 g of carbon (C), the mass of oxygen (O) is 1.33 g.

  • Ratio of Oxygen Masses: \frac{\text{mass O in } CO_2}{\text{mass O in } CO} = \frac{2.67 \text{ g}}{1.33 \text{ g}} = 2:1

    • This 2:1 ratio is a small whole number ratio, consistent with the law.

Applying The Law of Multiple Proportions (Example Problem)

• Problem: Iron and oxygen react to form two different oxides. The first compound contains 0.2865 g of oxygen for every 1.0000 g of iron, whereas the second compound contains 0.4297 g of oxygen for every 1.0000 g of iron. Show that these results are consistent with the law of multiple proportions.

• Solution:

  1. Calculate the ratio of masses of oxygen that react with 1 g of iron. Always place the larger value on top to obtain a ratio greater than 1.

     • \frac{0.4297 \text{ g oxygen (for 2nd)}}{\text{0.2865 g oxygen (for 1st)}} = 1.50

  2. Convert the decimal to a ratio of small whole numbers:

     • 1.50 = \frac{3}{2}

  • Since the ratio is 3:2 (small whole numbers), the results are consistent with the Law of Multiple Proportions.

Practice Problem (Law of Multiple Proportions)

• Phosphorus and chlorine react to form two different compounds. The first compound contains 3.43 g of chlorine for every 1.00 g of phosphorus, whereas the second compound contains 5.72 g of chlorine for every 1.00 g of phosphorus. Using a calculation, show that these results are consistent with the law of multiple proportions.

Dalton's Atomic Theory (Early 19th Century)

1. Each element is composed of tiny, indestructible particles called atoms.

2. All atoms of a given element are identical in mass and have unique properties.

3. Atoms combine in simple, whole-number ratios to form compounds (e.g., AB & AB_2).

4. Atoms are neither created nor destroyed in chemical reactions; they only change how they are bound to other atoms.

Properties of Electric Charge

• Opposite charges (+, -) attract each other.

• Like charges (+,+ or -,-) repel each other.

• An object with no net charge is termed "neutral". This occurs when positive and negative charges of exactly the same magnitude sum to zero.

• When a charged particle crosses through a magnetic field, it is deflected by the field according to its charge:

  • Positively charged particles are deflected in one direction.

  • Negatively charged particles are deflected in the opposite direction.

Discovery of the Electron (J. J. Thomson)

• Using a cathode ray tube experiment, J. J. Thomson discovered the subatomic particle known as the electron.

• He also measured its charge-to-mass ratio (C/g) by balancing the electric and magnetic fields applied to the cathode ray.

  • The cathode ray consisted of negatively charged particles (electrons).

Millikan's Oil Drop Experiment

• Robert Millikan determined the fundamental charge of an electron (C) with his oil drop experiment.

  • Charged oil droplets were suspended in an electric field.

• The electron's mass was then calculated using Thomson's previously determined charge-to-mass ratio (e/m) and Millikan's absolute charge (e).

  • m = \frac{e}{e/m}

The Plum-Pudding Model of the Atom (J. J. Thomson)

• This model proposed that the atom consists of a sphere of uniformly distributed positive charge (the "pudding").

• Electrons (the "plums") were embedded within this positively charged sphere, balancing the charge to make the atom neutral.

Radioactivity and the Nuclear Atom (Henri Becquerel)

• Henri Becquerel discovered that some samples produce invisible radiation.

• Three main types of radiation were identified:

  • Beta ($\beta$) particles: High-energy electrons (e^-).

  • Alpha ($\alpha$) particles: Equivalent to helium nuclei (He^{2+}), consisting of 2 protons and 2 neutrons.

  • Gamma ($\gamma$) rays: High-energy electromagnetic radiation, effectively photons with no mass or charge.

Rutherford's Gold Foil Experiment

• Experiment Setup: Positively charged alpha particles were directed at an ultrathin sheet of gold foil.

• Thomson's Model Prediction: Predicted only slight deflection of the alpha particles, as the positive charge was thought to be diffuse (plum-pudding model).

• Actual Results:

  • Most alpha particles passed straight through with little or no deflection.

  • A few alpha particles were deflected through large angles, sometimes even bouncing directly back toward the source.

• Implications (leading to the Nuclear Theory of the Atom): The observed large deflections could only occur if a concentrated, positively charged mass was present within the atom, rather than a diffuse charge.

The Nuclear Theory of the Atom (Ernest Rutherford)

1. Most of the atom's mass and all of its positive charge (protons) are contained in a small, dense core called the nucleus.

2. Most of the volume of the atom is empty space, throughout which tiny, negatively charged electrons are dispersed.

3. Each atom has as many negatively charged electrons as positively charged protons, making it electrically neutral.

• Size Comparison: The radius of an atom is approximately 288 pm (10^{-10} m), while the nucleus is extremely small, about 0.01 pm (10^{-15} m) (comparable to a period in a book relative to a large stadium).

Practice Question: Early Atomic Theories

• Which statement below is INCORRECT?

  • A. J.J. Thomson's cathode ray experiments determined the charge-to-mass ratio of an electron.

  • B. The Plum-Pudding model of the atom proposed that positive charge is spread out evenly in an atom.

  • C. Millikan's famous Oil Drop experiment was able to directly measure the mass of one electron.

  • D. Alpha particles were used to bombard the gold foil in Rutherford's famous experiment.

  • E. The Nuclear Theory of the Atom proposes that all the positive charge is concentrated in a central nucleus.

  • Correct answer explanation for C: Millikan measured the charge of an electron. The mass was then calculated using Thomson's charge-to-mass ratio.

The Third Subatomic Particle: The Neutron

• Reason for Discovery: Mass measurements suggested that there was additional mass in atoms not accounted for by protons. For example, a helium atom (He) is approximately 4 times heavier than a hydrogen atom (H) but has only 2 times the number of protons.

• Definition: A neutron is an uncharged subatomic particle found in the nucleus of an atom.

• Atomic Mass Units (amu): A relative mass scale for atoms and subatomic particles.

  • Definition: 1 amu is defined as exactly 1/12 the mass of one atom of carbon with 6 protons and 6 neutrons (carbon-12).

Summary of Subatomic Particles

• Using the amu scale, both protons and neutrons have a mass value of approximately 1 amu.

• The mass of electrons is negligible (approximately 0 amu) in comparison.

Isotopes and the Mass Spectrometer

• Mass Spectrometer: A powerful analytical instrument used to measure precise masses and relative amounts of ions derived from atoms or molecules. It can effectively detect isotopes.

  • Mechanism: A sample is vaporized, ionized (electron beam), accelerated (electric field), separated by mass-to-charge ratio (magnetic field), and detected.

• Isotopes: Atoms of an element whose nuclei have the same number of protons but different numbers of neutrons.

  • Example: Three different kinds of neon atoms exist with mass-to-charge ratios corresponding to 20, 21, and 22 amu.

Terminology of Isotopes

• Atomic Number (Z):

  • Represents the number of protons in the nucleus.

  • Defines the element. Each element has a unique atomic number.

  • For neutral atoms, it is also equal to the number of electrons.

• Mass Number (A):

  • The sum of the protons and neutrons in one atom of an element.

Names & Symbols for Isotopes

• Notation:

  • \begin{smallmatrix} A \ Z \end{smallmatrix}X

  • X-A (where X is the chemical symbol or name and A is the mass number).

• Example: carbon-12 (carbon with a mass number of 12) is represented as \begin{smallmatrix} 12 \ 6 \end{smallmatrix}C

  • Here, Z=6 (protons) and A=12 (6 protons + 6 neutrons).

Atomic Mass

• Natural Abundance: The relative proportion of a specific isotope found in a natural sample of the element (applies to Earth's typical elemental composition).

• Atomic Mass (on Periodic Table): The weighted average of the naturally occurring isotopes' mass numbers for a given element.

  • This is the value listed on the Periodic Table.

Example of Weighted Averages (Analogy: Exam Grades)

• Formula for Atomic Mass: Atomic\ Mass = \sum (\text{isotopic abundance} \times \text{isotopic mass})

  • Am = (a1 \times m1) + (a2 \times m2) + (a3 \times m_3) + \dots

  • m1, m2, m_3 are the mass numbers (or exact isotopic masses) for specific isotopes.

  • a1, a2, a3 are the fractional abundances (expressed as decimals, not percentages), such that a1 + a2 + a3 + \dots = 1 (exactly).

Example Atomic Mass Calculation: Neon (Ne)

• Calculation:

  • Average atomic mass of neon = (19.9924 \times 0.904838) + (20.99395 \times 0.002696) + (21.9914 \times 0.092465)

  • = 18.08988 + 0.056600 + 2.03343

  • = 20.1799 \text{ amu}

Practice Problem: Atomic Mass Calculation

• Calculate the atomic mass for the element Surprisium (Su), using the data below:

• Hint: The sum of natural abundances must be 100\% or 1.00 as a decimal. So, the abundance of Su-313 is 100\% - (85.62\% + 10.41\%) = 3.97\%.

Mendeleev's Periodic Table & The Periodic Law

• Discovery: Dmitri Mendeleev noticed patterns of properties among elements.

• The Periodic Law (Mendeleev's version): "When the elements are arranged in order of increasing mass, certain sets of properties recur periodically."

• Arrangement: He arranged elements in the periodic table primarily by their chemical and physical properties, which led him to occasionally deviate from strict atomic mass order to align properties (e.g., Te before I).

  • Elements with similar properties fell into vertical columns (groups).

The Modern Periodic Table

• Arrangement: Based on physical and chemical properties, but arranged in order of increasing atomic number (Z).

• Horizontal Rows: Periods (1-7)

• Vertical Columns: Groups or Families (1-18 or 1A-8A and 1B-8B)

  • Elements within the same group generally have similar chemical properties.

• Major Divisions:

  • Metals: Located on the left-hand side of the table (most elements).

    • Properties: Shiny solids, good conductors of heat and electricity, malleable (can be hammered into sheets), and ductile (can be drawn into wires).

  • Nonmetals: Located in the upper right corner.

    • Properties: Many are gases at room temperature; have properties generally opposite to metals (e.g., poor conductors, brittle).

  • Metalloids (Semimetals): Located along the diagonal "stair-step divide" between metals and nonmetals.

    • Properties: Possess intermediate properties, often behaving as semiconductors (e.g., Silicon (Si) and Germanium (Ge)).

Major Groups of the Periodic Table

• Main Group Elements (Representative Elements): Groups 1A-8A or 1, 2, 13-18 (excluding the transition metals, lanthanides, and actinides).

  • Alkali Metals: Group 1A (1) (except Hydrogen).

  • Alkaline Earth Metals: Group 2A (2).

  • Halogens: Group 7A (17).

  • Noble Gases: Group 8A (18).

• Transition Metals: Groups 3B-2B or 3-12 (the block in the middle).

• Inner Transition Metals: Separate rows at the bottom.

  • Lanthanides: Elements 57-71.

  • Actinides: Elements 89-103.

Forming Ions

• Ions: Atoms can either lose or gain electrons during a chemical change, forming electrically charged particles.

• Cation: A positively charged ion formed when an atom loses electrons.

• Anion: A negatively charged ion formed when an atom gains electrons.

• Stability Principle (for Main Group Elements):

  • A Main Group metal tends to lose electrons, forming a cation with the same number of electrons (and thus the same electron configuration) as the nearest noble gas.

  • A Main Group nonmetal tends to gain electrons, forming an anion with the same number of electrons (and electron configuration) as the nearest noble gas.

  • This relates to the enhanced stability of the Noble Gas elements due to their full outer electron shells.

Metals Form Positive Ions (Cations)

• Metals lose electrons to form cations. The charge is predictable for many main group metals:

  • Group 1A (1) metals form ions with a +1 charge (e.g., Na^+).

  • Group 2A (2) metals form ions with a +2 charge (e.g., Ca^{2+}).

  • Group 3A (13) metals (like Aluminum) form ions with a +3 charge (e.g., Al^{3+}).

Nonmetals Form Negative Ions (Anions)

• Nonmetals gain electrons to form anions. The charge is predictable for many main group nonmetals:

  • Group 5A (15) nonmetals form ions with a -3 charge (e.g., N^{3-}, P^{3-}).

  • Group 6A (16) nonmetals form ions with a -2 charge (e.g., O^{2-}, S^{2-}).

  • Group 7A (17) nonmetals (halogens) form ions with a -1 charge (e.g., F^-$, ext{Cl}^-).

Predictable Ions Formed by Elements (Summary)

• Many Main Group elements form ions with predictable charges based on their group number and their tendency to achieve a stable noble gas electron configuration.

  • H^+ (Group 1A)

  • Li^+, Na^+, K^+, Rb^+, Cs^+ (Group 1A)

  • Mg^{2+}, Ca^{2+}, Sr^{2+}, Ba^{2+} (Group 2A)

  • Al^{3+} (Group 3A)

  • N^{3-}, P^{3-} (Group 5A)

  • O^{2-}, S^{2-}, Se^{2-}, Te^{2-} (Group 6A)

  • F^-, ext{Cl}^-, Br^-, I^- (Group 7A)

Practice Question: Predictable Ionic Charges

• Using the Periodic Table, which of the following "predictable ionic charges" is INCORRECT?

  • A. +2 for Sr

  • B. -1 for F

  • C. +3 for Al

  • D. -4 for C

  • E. +1 for Cs

  • Correct answer explanation for D: While carbon can sometimes form C^{4-} (carbide) in rare compounds, its common ionic behavior in simple predictions is not -4. It more typically forms covalent bonds.

The Mole: A Chemist's "Dozen"

• Rationale: We cannot count individual atoms in the laboratory, so we use a weight scale to measure macroscopic amounts related to a defined number of particles.

• Definition of 1 Mole: The number of atoms in exactly 12 g of pure carbon-12.

• Avogadro's Number (N_A): Experimentally determined to be 6.022 \times 10^{23}.

  • This means 1 mole of any substance contains 6.022 \times 10^{23} items (atoms, molecules, ions, etc.).

• Analogy: Just as 1 dozen equals 12 items, 1 mole equals 6.022 \times 10^{23} items.

• Abbreviation: mol

Conversions Using Avogadro's Number

• Avogadro's number serves as a crucial conversion factor to relate the number of particles (atoms, molecules, formula units, ions, etc.) to the number of moles.

• \text{Number of particles} \xleftrightarrow{\div 6.022 \times 10^{23}} \text{Number of moles} \xleftrightarrow{\times 6.022 \times 10^{23}} (This diagram shows \text{particles} = \text{moles} \times NA and \text{moles} = \frac{\text{particles}}{NA})

The Size of a Mole

• A mole represents a practical, measurable amount (in terms of volume or mass) in the laboratory, despite Avogadro's number being incredibly large.

  • Example 1: Twenty-two copper pennies contain approximately 1 mol of copper atoms.

  • Example 2: One tablespoon of water contains approximately 1 mole of water molecules.

Molar Mass

• Definition: The mass of one mole of the particles that comprise a substance, expressed in grams.

• Calculation: To calculate the molar mass (in grams per mole), sum up the atomic mass values (from the periodic table) and change the units from amu to grams.

  • Example: 1 atom of Helium (He) has a mass of 4.003 amu. Therefore, 1 mole of Helium has a mass of 4.003 g.

  • The molar mass (M) of helium is 4.003 \text{ g/mol}.

• Relationship to Avogadro's Number: Avogadro's number (6.022 \times 10^{23}) is the number of amu in 1 gram. This is why the numerical value of atomic mass in amu is the same as the molar mass in g/mol.

Examples (Molar Mass / Mole Conversions)

1. What is the mass of 1.06 \times 10^{24} atoms of zinc?

   • \text{Given Zinc's atomic mass is } 65.38 \text{ amu, so molar mass is } 65.38 \text{ g/mol}.

   • \text{Number of atoms} \rightarrow \text{moles} \rightarrow \text{mass}

2. How many atoms of mercury are present in 5.00 mL of mercury if its density is 13.55 \text{ g/mL}?

   • \text{Given Mercury's atomic mass is } 200.59 \text{ amu, so molar mass is } 200.59 \text{ g/mol}.

   • \text{Volume} \rightarrow \text{mass} \rightarrow \text{moles} \rightarrow \text{number of atoms}

Practice Problem: Moles from Mass

• How many moles of gold atoms are present in 10.0 g of gold (Au)?

  • Given: Atomic mass of Au is 196.967 amu, so molar mass is 196.967 \text{ g/mol}.

  • Calculation: \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{10.0 \text{ g}}{196.967 \text{ g/mol}} \approx 0.05077 \text{ mol}

  • Answer: E. 5.08 \times 10^{-2} \text{ mol}

Practice Problem: Volume to Atoms (Dimensional Analysis)

• If the density of copper is 8.96 \text{ g/mL}, what volume of copper represents 6.35 \times 10^{22} atoms?

  • Given: Atomic mass of Cu is 63.546 amu, so molar mass is 63.546 \text{ g/mol}.

  • Strategy: Use dimensional analysis:
    \text{Atoms} \xrightarrow{1/N_A} \text{Moles} \xrightarrow{\text{Molar Mass}} \text{Mass} \xrightarrow{1/\text{Density}} \text{Volume}

  • 6.35 \times 10^{22} \text{ atoms Cu} \times \frac{1 \text{ mol}}{6.022 \times 10^{23} \text{ atoms}} \times \frac{63.546 \text{ g}}{1 \text{ mol}} \times \frac{1 \text{ mL}}{8.96 \text{ g}} \approx 0.748 \text{ mL}

  • Answer: A. 0.748 \text{ mL}$$