Chapter 1 Notes: Atoms and Matter

Matter From the Particulate Point of View

• Matter is composed of particles. Examples include subatomic particles (neutrons, protons, electrons) that make up atoms, and atoms make up molecules.

• How particles come together determines the physical properties of matter.

• Matter is defined as anything that has mass and occupies space (volume).

• Chemistry is the science that seeks to understand matter and its properties.

Elements, Molecules, and Mixtures

• Atoms: Basic submicroscopic particles that constitute the fundamental building blocks of ordinary matter.

• Molecules: Particles formed when two or more atoms bond together in specific geometric arrangements.

• Atoms and molecules determine how matter behaves.

The Classification of Matter

• Matter can be classified by state (solid, liquid, gas) based on physical properties, and by composition (types of particles).

• State changes: solid → liquid → gas with increasing temperature.

Solid Matter

• Atoms/molecules pack closely in fixed locations.

• They vibrate but do not move past one another.

• Fixed volume and rigid shape.

• Examples: ice, aluminum, diamond.

Liquid Matter

• Atoms/molecules pack about as closely as in solids but can move relative to each other.

• Fixed volume but not fixed shape; flows and takes the shape of its container.

• Examples: water, alcohol, gasoline.

Gaseous Matter

• Atoms/molecules have a lot of space between them and move freely.

• Gases are compressible.

Classification of Matter by Components (1 of 2)

• Matter can be classified as elements, compounds, and mixtures.

Classification of Matter by Components (2 of 2)

• Pure substances vs mixtures:

  • Pure substance: composed of a single component; composition is invariant.

  • Mixture: composed of two or more components in varying proportions between samples.

Classification of Pure Substances (1 of 2)

• Two types: Elements and Compounds.

• This classification depends on whether they can be decomposed into simpler substances.

Classification of Pure Substances (2 of 2)

• Element: cannot be chemically broken down into simpler substances; basic building blocks; composed of a single type of atom (e.g., helium).

• Compound: composed of two or more elements in fixed, definite proportions.

• Most elements are chemically reactive and combine with others to form compounds (e.g., water, sugar).

Classification of Mixtures

• Mixtures: heterogeneous vs homogeneous.

• This categorization depends on how uniformly substances mix.

Heterogeneous Mixture

• Composition varies from region to region; multiple substances visible (e.g., salt and sand).

• Portions of a heterogeneous mixture have different compositions and properties.

Homogeneous Mixture

• Made of multiple substances but appears as one substance.

• All portions have the same composition and properties; atoms or molecules mix uniformly.

Conceptual Connection 1.1 (1 of 2)

• In figures, blue circle represents one element, red square another.

• Question: Which image is a pure substance?

Conceptual Connection 1.1 (2 of 2)

• Reiterates: identify pure substance in the given images.

The Scientific Approach to Knowledge

• Empirical: based on observation and experimentation.

• Scientific method: observe nature and its behavior through experiments.

• Key characteristics:

  • Observations

  • Formulation of hypotheses

  • Experimentation

  • Formulation of laws and theories

Observations

• Observations (data) are descriptions about characteristics or behavior of nature.

• Observations, verification, and experimentation lead to hypotheses.

Hypothesis

• A tentative interpretation or explanation of observations.

• Example: Lavoisier proposed combustion involves substances combining with a component of air.

• A good hypothesis is falsifiable: experiments can support or refute it.

A Scientific Law

• Brief statement that summarizes past observations and predicts future ones.

• Example: Law of Conservation of Mass: In a chemical reaction, matter is neither created nor destroyed.

• Laws allow prediction and testing by experiments.

• Unlike laws in civil life, scientific laws cannot be violated by choice.

Theory

• A well-established hypothesis or set of hypotheses forming the basis for a theory.

• Theories explain not just what nature does but why.

• Examples: Dalton’s atomic theory, Big Bang theory.

• Theories are validated by experimental results and can predict future observations; they can be tested repeatedly.

• Theories can never be conclusively proven; new observations can reveal flaws.

Conceptual Connection 1.2 (1 of 2)

• Question about the difference between a law and a theory.

• Correct choice: (b) A law summarizes a series of related observations, while a theory gives the underlying reasons for them.

Conceptual Connection 1.2 (2 of 2)

• Reiterates the difference between law and theory.

Why Is Scientific Measurement Important?

• Data can be qualifiable (observational) or quantifiable (measurable).

  Qualifiable data: subjective (e.g., color, shape).

• Quantifiable data: objective; obtained with equipment and standardized units.

• Measurements can use English units or SI units.

Early Ideas About the Building Blocks of Matter (1 of 2)

• Leucippus and Democritus proposed matter is composed of small, indestructible particles.

• Democritus’ quote: “Nothing exists except atoms and empty space; everything else is opinion.”

• Atoms exist in different shapes/sizes and move through empty space.

• Plato and Aristotle rejected atomic ideas, proposing matter has no smallest parts and is made of fire, air, earth, and water in varying proportions.

Early Ideas About the Building Blocks of Matter (2 of 2)

• John Dalton offered convincing evidence for atomic ideas via Dalton’s atomic theory:

  • Law of conservation of mass

  • Law of definite proportions

  • Law of multiple proportions

The Law of Conservation of Mass

• In a chemical reaction, matter is neither created nor destroyed.

• The total mass of substances involved in a reaction remains constant.

• This supports the idea that matter is composed of small, indestructible particles.

Conceptual Connection 1.3 (1 of 2)

• Burning a log in a campfire: ash mass is less than log mass.

• Question: What happens to the matter that composed the log?

• Correct answer: (a) The matter that composed the log reacts to form gases released into the air.

Conceptual Connection 1.3 (2 of 2)

• Reiterates the burning log scenario and the mass balance question.

The Law of Definite Proportions

• Joseph Proust (1797) observed that all samples of a given compound have the same element proportions.

• Also called the law of constant composition.

An Example of the Law of Definite Proportions

• Decomposition of 18.0 g water yields 16.0 g O and 2.0 g H.

• Oxygen-to-hydrogen mass ratio = rac{16.0}{2.0} = 8:1.

• This same ratio should hold for all samples of water.

Conceptual Connection 1.4 (1 of 2)

• Mass ratio of nitrogen to hydrogen in ammonia is 4.7:1.

• If a sample of ammonia contains 10.0 g H, grams of N present = 4.7 g N.

Conceptual Connection 1.4 (2 of 2)

• Repeats the ammonia nitrogen-to-hydrogen ratio question.

Problem Solving: The Law of Definite Proportions (1 of 2)

• Example 1.1: Two samples of carbon dioxide decompose to O2 and C.

• First sample: 25.6 g O and 9.60 g C.

• Second sample: 21.6 g O and 8.10 g C.

• Show both have the same mass ratio of O to C after dividing by the larger mass.

• Result: Both give the same ratio (2.67:1 for O:C in the sample shown).

Problem Solving: The Law of Definite Proportions (2 of 2)

• Practice: Two samples of carbon monoxide decompose; compare the mass ratios to verify the law.

The Law of Multiple Proportions (1 of 2)

• Dalton’s Law (1804): When two elements A and B form two different compounds, the masses of B that combine with 1 g of A are in small whole-number ratios.

• If A binds with 1, 2, 3, … atoms of B, compounds AB, AB2, AB3, … exist.

The Law of Multiple Proportions (2 of 2)

• Carbon monoxide (CO) and carbon dioxide (CO2) both contain C and O.

• In CO2, mass ratio O:C = 2.67 g O per 1.00 g C; in CO, mass ratio O:C = 1.33 g O per 1.00 g C.

• The ratio of these two mass ratios is a small whole number (2.67 / 1.33 ≈ 2).

Problem Solving: The Law of Multiple Proportions (1 of 2)

• Example 1.2: Nitrogen forms NO2 and N2O; nitrogen to oxygen mass ratios are 2.28 g O per 1.00 g N (NO2) and 0.570 g O per 1.00 g N (N2O).

• Show these results are consistent with the law by taking the ratio of the larger to the smaller O-to-N masses. The ratio is a small whole number (4).

Problem Solving: The Law of Multiple Proportions (2 of 2)

• Practice: Hydrogen and oxygen form water (H2O) and hydrogen peroxide (H2O2); compare the mass ratios (O to H per 1 g H) for each compound.

Conceptual Connection 1.5 (1 of 2)

• Which statement best captures a difference between the law of definite proportions and the law of multiple proportions?

  • (a) The law of definite proportions applies to two or more samples of the same compound; the law of multiple proportions applies to two different compounds containing the same two elements.

  • (b) The law of definite proportions applies to two different compounds containing the same two elements; the law of multiple proportions applies to two or more samples of the same compound.

  • (c) None of the above.

Conceptual Connection 1.5 (2 of 2)

• Reiterates the same question and options.

The Discovery of the Electron (1 of 2)

• J. J. Thomson’s cathode ray experiments showed:

  • Cathode rays travel from the cathode (negative) to the anode (positive).

  • They travel in straight lines, are independent of the material of the cathode, and carry a negative charge.

• Thomson measured charge-to-mass ratio of the cathode ray particles by deflecting them with electric/magnetic fields; value reported as coulombs per gram (C/g).

• Result: Negative charge associated with very small mass; electrons exist inside all atoms.

John Dalton and the Atomic Theory

• Dalton’s atomic theory (summary of 4 postulates):
  1) Each element is composed of tiny, indestructible particles called atoms.
  2) All atoms of a given element have the same mass and other properties that distinguish them from atoms of other elements.
  3) Atoms combine in simple, whole-number ratios to form compounds.
  4) Atoms of one element cannot change into atoms of another element in a chemical reaction; atoms rearrange via bonding.

The Discovery of the Electron: J. J. Thomson’s Cathode Ray Experiment

• Visual/graphic content accompanying the discovery of the electron.

Millikan’s Oil Drop Experiment: Determining the Charge of an Electron

• Millikan measured the electric field needed to halt falling oil drops and determined drop masses from their radii/density.

• The charge on any oil drop was always a whole-number multiple of the fundamental charge of a single electron, e = 1.60218×10^-19 C.

Millikan’s Oil Drop Experiment: The Charge-to-Mass Ratio for an Electron

• From data of Millikan and Thomson’s electron mass-to-charge ratio, the electron mass is determined.

• The calculation ties the electron charge and mass together to yield m_e and e.

Conceptual Connection 1.6 (1 of 2)

• Suppose one oil drop has a charge of 19 × 4.8×10^-10 C. How many excess electrons does the drop contain? Choices include 1–4.

Conceptual Connection 1.6 (2 of 2)

• Repeats the electron-count question with the same context.

The Structure of the Atom: The Early Models

• J. J. Thomson’s Plum-Pudding Model: electrons embedded in a positively charged sphere.

Rutherford’s Model and the Gold Foil Experiment (1 of 2)

• In 1909, Rutherford tested Thomson’s model by directing positively charged particles at a thin gold foil.

Rutherford’s Model and the Gold Foil Experiment (2 of 2)

• Results: Most particles passed through; some were deflected; about 1 in 20,000 bounced back.

• Conclusion: Matter contains empty space with very dense regions (nucleus).

Building on the Rutherford Atomic Model: The Nuclear Atom Model

• Three basic parts:
  1) Most of the mass and all positive charge are in a tiny nucleus.
  2) Most of the atom’s volume is empty space with dispersed electrons.
  3) The atom is electrically neutral because as many electrons outside the nucleus as there are protons inside.

The Neutral Particles: Neutrons

• Rutherford’s model lacked the mass of neutrons in the nucleus.

• 1932: Rutherford and James Chadwick showed neutrons exist.

• Neutron: mass similar to proton; no electrical charge.

The Atom’s Subatomic Particles

• All atoms are composed of protons, neutrons, and electrons.

• Masses:

  • Proton: m_p \approx 1.67262 \times 10^{-27} \text{ kg}

  • Neutron: m_n \approx 1.67493 \times 10^{-27} \text{ kg}

  • Electron: m_e \approx 9.10938 \times 10^{-31} \text{ kg}

• Charge:

  • Proton: +e, Electron: −e (equal magnitude, opposite sign). Neutron: 0 charge.

Subatomic Particles Table 1.1 (Summary)

• Proton: Mass ≈ 1.00727\, \text{amu}; Charge = +1; Charge in Coulombs = +1.60218\times 10^{-19}\text{ C}

• Neutron: Mass ≈ 1.00866\, \text{amu}; Charge = 0; Charge in C = 0

• Electron: Mass ≈ 0.00055\, \text{amu}; Charge = -1; Charge in C = -1.60218\times 10^{-19}\text{ C}

Elements: Defined by Their Numbers of Protons

• The number of protons in a nucleus defines the element; this is the atomic number, symbolized by Z.

Elements & the Periodic Table

• Elements are arranged in the periodic table by their atomic number Z.

• Elements in the same column have very similar physical and chemical properties.

• Each element is represented by its symbol and its atomic number.

Conceptual Connection 1.7 (1 of 2)

• Question: What element contains 50 protons? Options: tin, vanadium, manganese, fermium.

• Answer: tin (Z = 50).

Conceptual Connection 1.7 (2 of 2)

• Repeats the same idea.

Isotopes: Elements With Varied Number of Neutrons

• All atoms of a given element have the same number of protons, but may have different numbers of neutrons.

• Example: Neon has 10 protons but 10, 11, or 12 neutrons; three isotopes exist with slightly different masses.

Isotopes: Representation (1 of 2)

• Mass number notation: A = Z + N where Z = number of protons, N = number of neutrons.

• Notation: ^{A}_{Z} ext{X} (X = chemical symbol).

Isotopes: Representation (2 of 2)

• Alternative notation: X with a dash and the mass number, e.g., Ar-40.

Isotopes: Varied Number of Neutrons

• Natural isotopes have a roughly constant relative abundance (natural abundance).

• Example table for neon: ^{20} ext{Ne} (abundance 90.48%), ^{21} ext{Ne} (0.27%), ^{22} ext{Ne} (9.25%).

Conceptual Connection 1.8 (1 of 2)

• Argon isotope with mass number 40 has how many neutrons? Given Z depends on the element, N = A − Z; for Ar (Z = 18), neutrons = 22.

Conceptual Connection 1.8 (2 of 2)

• Repeats the question.

Conceptual Connection 1.9 (1 of 2)

• Carbon has two naturally occurring isotopes: C-12 (abundance 98.93%, mass 12.0000 amu) and C-13 (abundance 1.07%, mass 13.0034 amu).

• Which depiction best represents C-13? (Protons = 6, Neutrons = 7 for C-13.)

Conceptual Connection 1.9 (2 of 2)

• Repeats the same prompt.

Example 1.3 Atomic Numbers, Mass Numbers, and Isotope Symbols (1 of 2)

• (a) Chlorine isotope with 18 neutrons: Z = 17 (Chlorine), A = 17 + 18 = 35. Symbol: ^{35}_{17} ext{Cl}.

• (b) For a neutral atom, electrons = protons = Z; neutrons = A − Z.

• Practice: Carbon isotope with 7 neutrons has Z = 6, A = 13, symbol ^{13}_{6} ext{C}.

The Ions: Charged Atoms Losing and Gaining Electrons

• Neutral atom has electrons equal to protons (Z).

• Ions form when electrons are lost or gained.

• Cations: positively charged (loss of electrons); common with metals.

• Anions: negatively charged (gain of electrons); common with nonmetals.

Conceptual Connection 1.10 (1 of 2)

• How many electrons are in the anion? Example: O^{2-} has 10 electrons (O neutral has 8; gained 2).

Conceptual Connection 1.10 (2 of 2)

• Repeats the anion electron-count question.

Conceptual Connection 1.11 (1 of 2)

• In light of the nuclear model, which statement is true?

• Options: (a) More neutrons -> larger isotope; (b) Size of atom is the same for all isotopes.

Conceptual Connection 1.11 (2 of 2)

• Answer: (b) The size of the atom is essentially the same for all isotopes (the nucleus mass changes, not the overall chemical size).

Atomic Mass: The Average Mass of an Element’s Atoms

• Atomic mass (also called atomic weight or standard atomic weight) is the average mass of the element’s atoms.

• It is a weighted value based on the natural abundances of isotopes.

Atomic Mass: Equation

• Atomic mass of element = \text{Atomic mass} = \sum{i} fi mi where fi = fractional abundance of isotope i, and m_i = mass of isotope i.

Atomic Mass: Problem

• Naturally occurring chlorine: 75.77% Cl-35 (mass 34.97 amu) and 24.23% Cl-37 (mass 36.97 amu).

• Calculate chlorine’s atomic mass.

Atomic Mass: Answer to the Problem

• Convert percents to decimals and sum: \text{Atomic mass} = (0.7577)(34.97) + (0.2423)(36.97) \approx 35.4\ \text{amu}.

Conceptual Connection 1.12 (1 of 2)

• Carbon has two naturally occurring isotopes: C-12 (abundance 98.93%, mass 12.0000 amu) and C-13 (abundance 1.07%, mass 13.0034 amu).

• Without calculations, which mass is closest to carbon’s atomic mass? (a) 12.00 amu, (b) 12.50 amu, (c) 13.00 amu

Conceptual Connection 1.12 (2 of 2)

• Repeats the carbon question.

Mass Spectrometry: Measuring the Mass of Atoms and Molecules

• Mass spectrometry separates particles according to mass.

• It measures masses and abundances of isotopes.

What Is a Mole?

• A mole (mol) is a counting unit (like a dozen) used for very large numbers.

• 1 mol contains the Avogadro number: 6.02214\times 10^{23} objects.

• Examples: 1 mol of marbles, 1 mol of sand grains, etc.

The Mole

• The mole is defined as the number of atoms in exactly 12 g of pure C-12.

• Therefore: 1\ \text{mol C atoms} = 6.022\times 10^{23}\ \text{atoms} and 12\ \text{g of C-12} = 1\ \text{mol C atoms}

Mole Conversions: Atoms to Moles or Moles to Atoms (1 of 3)

• Converting between number of moles and number of atoms uses the conversion factor 1\ \text{mol} = 6.022\times 10^{23} \text{ atoms}

• Form: \text{atoms} \times \frac{1\ \text{mol}}{6.022\times 10^{23}\ \text{atoms}} = \text{moles}

• Conversely: \text{moles} \times \frac{6.022\times 10^{23}\ \text{atoms}}{1\ \text{mol}} = \text{atoms}

Example 1.5: Converting between Number of Moles and Number of Atoms

• Problem: How many copper atoms are in 2.45 moles of copper?

• Strategy: Multiply by Avogadro’s number to convert moles to atoms:
  \text{atoms} = 2.45\ \text{mol} \times 6.022\times 10^{23}\ \text{atoms/mol}

Mole Conversions: Atoms to Moles or Moles to Atoms (2 of 3)

• Example continued: Result should be a large number of atoms, reflecting the large Avogadro’s number.

• Practice: 2.45 mol Cu → ~1.48×10^{24} Cu atoms (per slide).

Converting Between Mass and Amount (Moles)

• The molar mass is the mass of 1 mole of an element; numerically equal to the element’s atomic mass in amu.

• Molar mass (g/mol) is the conversion factor between mass (g) and amount (mol).

Converting Between Mass and Moles (1 of 2)

• Examples:

  • 12.01\ \text{g C} \,/\, 1\ \text{mol C}

  • 26.98\ \text{g Al} \,/\, 1\ \text{mol Al}

  • 4.003\ \text{g He} \,/\, 1\ \text{mol He}

Converting Between Mass and Moles (2 of 2)

• The molar mass is the conversion factor between mass and amount (moles).

• Example relationships:
  \frac{12.01\ \text{g C}}{1\ \text{mol C}} = \frac{12.01\ \text{g}}{1\ \text{mol}}

Mass to Moles to Number of Particles: The Conceptual Plan

• For an element, to go from mass to atoms, go mass -> moles -> atoms.

Converting Between Mass and Moles (1 of 5)

• Example 1.7: How many copper atoms are in a copper penny with mass 3.10 g?

• Plan: Convert mass to moles using molar mass, then convert moles to atoms using Avogadro’s number.

Converting Between Mass and Moles (2–5) and (3–5)

• Stepwise solutions shown for various examples (carbon, copper, aluminum, tungsten) illustrating the same plan.

• Typical calculation pattern:
  \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}
  \text{atoms} = \text{moles} \times 6.022\times 10^{23}\ \text{atoms/mol}

Converting Between Particles and Mass (1 of 4): Example 1.8

• The Mole Concept: An aluminum sphere contains aluminum atoms; density given; asked to find the sphere’s radius.

• Approach: Use number of atoms to find mass, then use density to find volume, then solve for radius via volume of a sphere.

• Provided data (illustrative): density of Al ≈ 2.70 g/cm³; number of atoms ≈ 8.55×10^23; molar mass of Al ≈ 26.98 g/mol; Avogadro’s number ≈ 6.022×10^23.

Converting Between Particles and Mass (2 of 4) and (3 of 4)

• Stepwise plan includes:
  1) atoms → moles via Avogadro’s number,
  2) moles → mass via molar mass,
  3) mass → volume via density, and
  4) volume → radius using the sphere volume formula.

• Example factor: 26.98 g Al per 1 mol Al and 6.022×10^23 atoms per mol.

Converting Between Particles and Mass (4 of 4)

• Final numerical results given with units in cm and g, culminating in a radius value (approximately 0.7 cm in the shown calculation).

Conceptual Connection 1.13 (1 of 2)

• Without calculations, which sample contains the most atoms given equal masses? (a) 1 g Cu, (b) 1 g C, (c) 1 g U)

Conceptual Connection 1.13 (2 of 2)

• Repeats the same question for emphasis.

Copyright

• Final slide notes the standard copyright statement.