Chapter 2: The Components of Matter Flashcards

Classification of Matter

  • Matter is classified based on its composition and uniformity:

    • Pure Substance: Composed of only a single type of atom or molecule.

    • Element: A pure substance that cannot be chemically broken down into simpler substances. Example: Helium (HeHe).

    • Compound: A pure substance composed of two or more elements in fixed, finite proportions. Example: Pure water (H2OH_2O).

    • Mixture: Composed of two or more different types of atoms or molecules.

    • Homogeneous Mixture: A mixture in which the composition is uniform and the same throughout. Example: Tea with sugar.

    • Heterogeneous Mixture: A mixture in which the composition varies from one region to another. Example: Wet sand.

Fundamental Laws of Matter

  • Law of Conservation of Mass:

    • Proposed by Antoine Lavoisier in 1789 based on experimental evidence.

    • States that matter is neither created nor destroyed in a chemical reaction.

    • Total mass of used reactants=Total mass of products produced\text{Total mass of used reactants} = \text{Total mass of products produced}

    • Total number of reactant atoms=Total number of product atoms\text{Total number of reactant atoms} = \text{Total number of product atoms}

  • Law of Definite Proportions:

    • Proposed by Joseph Proust in 1797.

    • States that all samples of a given compound, regardless of their source or how they are prepared, have the same proportions of their constituent elements.

    • Example: Composition analysis of a 20 g20\text{ g} sample of methane yields 15 g15\text{ g} of carbon and 5 g5\text{ g} of hydrogen.

    • Mass ratio: 5 g hydrogen:15 g carbon5\text{ g hydrogen} : 15\text{ g carbon}

    • Simplified mass ratio: 1 g hydrogen:3 g carbon1\text{ g hydrogen} : 3\text{ g carbon}

    • This ratio holds for any sample of methane regardless of its source.

  • Law of Multiple Proportions:

    • Published by John Dalton in 1804.

    • States that when two elements (designated A and B) form two different compounds, the masses of element B that combine with 1 g1\text{ g} of element A can be expressed as a ratio of small whole numbers.

    • Example: Carbon dioxide vs. Carbon monoxide:

    • Carbon dioxide: 2.67 g2.67\text{ g} of oxygen combines with 1 g1\text{ g} of carbon.

    • Carbon monoxide: 1.33 g1.33\text{ g} of oxygen combines with 1 g1\text{ g} of carbon.

    • Ratio of oxygen masses: 2.671.33=2.00\frac{2.67}{1.33} = 2.00

    • Underlying mechanical concept: Dalton suspected molecules were composed of small, indestructible particles. When elements A and B combine to form different compounds, an atom of A combines with 1, 2, 3, etc., atoms of B (ABAB, AB2AB_2, AB3AB_3, AB4AB_4). Consequently, the masses of B reacting with a fixed mass of A always relate as small whole-number ratios.

Dalton's Atomic Theory

  • Developed by John Dalton in 1808 to explain the laws of conservation of mass, definite proportions, and multiple proportions.

  • The theory incorporates the following key concepts:

    • Elements are composed of tiny, indestructible particles called atoms (derived from Greek atomos, meaning uncuttable).

    • The concept of discrete particles was popularized by the philosopher Democritus around 460 B.C., but lacked experimental evidence until centuries after his death.

    • Atoms of a given element are identical in size, mass, and other properties; atoms of different elements differ in size, mass, and other properties.

    • Atoms cannot be subdivided, created, or destroyed.

    • Atoms combine in simple, whole-number ratios to form compounds.

    • Atoms of one element cannot change into atoms of another element. In a chemical reaction, atoms re-arrange the way they are bound together with other atoms to form new substances.

Discovery of Subatomic Particles and the Electron

  • By the late 19th century, experimental observations demonstrated that atoms were composed of smaller, more fundamental subatomic particles.

  • Cathode Rays and J. J. Thomson:

    • Late 1800s: J. J. Thomson probed cathode rays using a cathode ray tube—a partially evacuated glass tube fitted with two electrodes connected to a high electric voltage.

    • Cathode rays are emitted by the negatively charged electrode (cathode) and travel to the positively charged electrode (anode).

    • Rays are detected when colliding with a fluorescent coating at the end of the tube.

    • Observed properties of cathode ray particles:

    • Travel in straight lines.

    • Independent of the composition of the source material.

    • Carry a negative electrical charge.

    • Thomson measured the charge-to-mass ratio of cathode ray particles by deflecting them using electric and magnetic fields:     Mass-to-charge ratio=1.76×108 C/g\text{Mass-to-charge ratio} = -1.76 \times 10^{8}\text{ C/g}

    • This ratio indicated that cathode ray particles were approximately 2000 times less massive than a hydrogen atom.

    • Thomson had discovered the electron—a low-mass, negatively charged subatomic particle. Practical applications include cathode ray tube televisions.

  • Millikan Oil Drop Experiment (1909):

    • Robert Millikan determined the charge of a single electron.

    • Reasoned that electric charge on any oil drop must be an integral multiple of an individual electron's charge.

    • Measured fundamental electric charge per electron: 1.60×1019 C-1.60 \times 10^{-19}\text{ C}.

    • Using Thomson's mass-to-charge ratio, the mass of an electron was calculated:     Mass of electron=1.60×1019 C×1 g1.76×108 C=9.10×1028 g\text{Mass of electron} = -1.60 \times 10^{-19}\text{ C} \times \frac{1\text{ g}}{-1.76 \times 10^{8}\text{ C}} = 9.10 \times 10^{-28}\text{ g}

Atomic Structure and the Nuclear Model

  • Plum-Pudding Model:

    • Most popular atomic model around 1900.

    • Pictured electrons distributed inside a uniform sphere of positive charge.

  • Rutherford's Gold Foil Experiment (1909):

    • Ernest Rutherford, alongside J. J. Thomson, directed positively charged α\text{α} (alpha) particles emitted by radioactive materials at an ultrathin gold foil surrounded by a detector.

    • Expected outcome under the plum-pudding model: All α\text{α} particles would pass through undeflected due to mass and charge being evenly spread.

    • Experimental outcome: Most α\text{α} particles passed straight through with little or no deflection, but a small fraction were deflected through large angles.

    • Rutherford described his shock: "about as credible as if you had fired a 15-inch shell at a piece of tissue paper and it came back and hit you."

  • Rutherford's Nuclear Theory of the Atom (Planetary Model):

    • Most of the atom's mass and all of its positive charge are concentrated in a tiny core called the nucleus.

    • Most of the volume of an atom is empty space, throughout which small, negatively charged electrons are dispersed.

    • An atom is electrically neutral because the number of negatively charged electrons outside the nucleus equals the number of positively charged particles (protons) inside the nucleus.

  • Discovery of Neutrons:

    • Discrepancy observed: A hydrogen atom has 1 proton and a helium atom has 2 protons. If protons were the only massive particles, the helium-to-hydrogen mass ratio should be 2:12:1. However, the actual mass ratio is 4:14:1.

    • Rutherford later demonstrated that the unaccounted mass was due to neutrons—neutral subatomic particles located in the nucleus with mass equal to protons.

Properties of Subatomic Particles

  • Proton: Positively charged; located in the nucleus; mass = 1.67262×1027 kg1.67262 \times 10^{-27}\text{ kg} (1 amu≈ 1\text{ amu}); charge = +1.60×1019 C+1.60 \times 10^{-19}\text{ C} (+1+1 relative atomic charge unit).

  • Neutron: Neutral charge; located in the nucleus; mass = 1.67493×1027 kg1.67493 \times 10^{-27}\text{ kg} (1 amu≈ 1\text{ amu}); charge = 0 C0\text{ C} (00 relative atomic charge unit).

  • Electron: Negatively charged; located in orbitals outside the nucleus; mass = 9.1×1031 kg9.1 \times 10^{-31}\text{ kg} (0.00055 amu0.00055\text{ amu}); charge = 1.60×1019 C-1.60 \times 10^{-19}\text{ C} (1-1 relative atomic charge unit).

  • Atomic Mass Unit (amu): Defined as exactly 112\frac{1}{12} the mass of a carbon atom containing 6 protons and 6 neutrons.

  • Distribution: The nucleus contains almost all of the mass in a tiny volume. Orbital electrons account for virtually all of the spatial volume, despite having negligible mass.

Elements, Isotopes, and Atomic Mass

  • Atomic Number (ZZ): The number of protons in an atom's nucleus. Determines elemental identity.

  • Chemical Symbols: One- or two-letter abbreviations.

    • Based on English names: Carbon (CC), Chlorine (ClCl).

    • Based on Latin names: Sodium (NaNa, Natrium), Gold (AuAu, Aurum).

    • Based on German names: Tungsten (WW, Wolfram).

    • Derived from properties: Argon (ArAr, from Greek argos, meaning inactive).

    • Named after mythology: Mercury (HgHg, Roman god of commerce and communication).

    • Named after places: Erbium (ErEr), Terbium (TbTb), Ytterbium (YbYb), Yttrium (YY) after Ytterby, Sweden; Polonium (PoPo) after Poland (birthplace of Marie Curie).

    • Named after scientists: Einsteinium (EsEs) after Albert Einstein; Bohrium (BhBh) after Niels Bohr.

  • Isotopes:

    • Atoms of the same element with identical numbers of protons (ZZ), but different numbers of neutrons.

    • Disproves Dalton's assertion that all atoms of an element have identical mass.

    • Example: Carbon always has 6 protons, but can have 6, 7, or 8 neutrons.

    • Natural Abundance: The relative percentage of a specific isotope in a naturally occurring sample of an element. Example: Natural bromine consists of 50.7 50.7\text{ }% 3579Br^{79}_{35}\text{Br} (44 neutrons) and 49.3 49.3\text{ }% 3581Br^{81}_{35}\text{Br} (46 neutrons).

    • Mass Number (AA): The total sum of protons and neutrons in an atom:     A=number of protons+number of neutronsA = \text{number of protons} + \text{number of neutrons}

  • Atomic Mass:

    • The weighted average mass of all naturally occurring isotopes of an element based on natural abundance.

    • Example 1 (Calculating Average Atomic Mass):

    • Iron (FeFe) isotopes:

      • 5.85 5.85\text{ }% of 54Fe^{54}\text{Fe} (53.94 amu53.94\text{ amu}) 0.0585×53.94 amu=3.15549 amu\rightarrow 0.0585 \times 53.94\text{ amu} = 3.15549\text{ amu}

      • 91.75 91.75\text{ }% of 56Fe^{56}\text{Fe} (55.93 amu55.93\text{ amu}) 0.9175×55.93 amu=51.315775 amu\rightarrow 0.9175 \times 55.93\text{ amu} = 51.315775\text{ amu}

      • 2.12 2.12\text{ }% of 57Fe^{57}\text{Fe} (56.94 amu56.94\text{ amu}) 0.0212×56.94 amu=1.207128 amu\rightarrow 0.0212 \times 56.94\text{ amu} = 1.207128\text{ amu}

      • 0.28 0.28\text{ }% of 58Fe^{58}\text{Fe} (57.93 amu57.93\text{ amu}) 0.00028×57.93 amu=0.162204 amu\rightarrow 0.00028 \times 57.93\text{ amu} = 0.162204\text{ amu}

    • Sum of contributions: 3.15549+51.315775+1.207128+0.162204=55.84 amu3.15549 + 51.315775 + 1.207128 + 0.162204 = 55.84\text{ amu}.

    • Example 2 (Calculating Isotopic Abundance):

    • Boron (BB) isotopes: 10B^{10}\text{B} (10.01 amu10.01\text{ amu}) and 11B^{11}\text{B} (11.01 amu11.01\text{ amu}); average atomic mass = 10.81 amu10.81\text{ amu}.

    • Let xx equal natural abundance of 10B^{10}\text{B}:       x(10.01 amu)+(1x)(11.01 amu)=10.81 amux(10.01\text{ amu}) + (1 - x)(11.01\text{ amu}) = 10.81\text{ amu}       10.01x+11.0111.01x=10.8110.01x + 11.01 - 11.01x = 10.81       -1x = -0.2 −−−> x = 0.2

    • Abundance percentages: 10B=20 ^{10}\text{B} = 20\text{ }%, 11B=80 ^{11}\text{B} = 80\text{ }%.

Ions and the Periodic Table

  • Ions:

    • Formed when neutral atoms gain or lose electrons during chemical changes.

    • Cation: Positively charged ion resulting from electron loss. Example: Sodium (NaNa) loses 1 electron to form Na+Na^+ (11 protons, 10 electrons).

    • Anion: Negatively charged ion resulting from electron gain. Example: Bromine (BrBr) gains 1 electron to form BrBr^- (35 protons, 36 electrons).

  • Development of the Periodic Table:

    • In 1869, Russian chemistry professor Dmitri Mendeleev organized 65 known elements in order of increasing mass.

    • Periodic Law: When elements are arranged in order of increasing mass, certain properties recur periodically.

    • Mendeleev left gaps for undiscovered elements and successfully predicted their existence and properties.

    • Modern periodic tables organize elements by increasing atomic number (ZZ) rather than atomic mass.

  • Classification Regions of the Periodic Table:

    • Metals: Located on the lower left side and middle. Conduct heat and electricity well, malleable, ductile, shiny, tend to lose electrons during chemical changes.

    • Nonmetals: Located on the upper right side. Poor conductors of heat and electricity, tend to gain electrons during chemical changes.

    • Metalloids: Lie along the zigzag diagonal line separating metals and nonmetals. Exhibit mixed metallic and nonmetallic properties.

  • Divisions and Groups:

    • Main-Group Elements: Properties are predictable based on position.

    • Group 1A: Alkali metals (extremely reactive).

    • Group 2A: Alkaline earth metals (fairly reactive).

    • Group 7A: Halogens (extremely reactive nonmetals).

    • Group 8A: Noble gases (rarely form compounds).

    • Transition Metals: Properties are less predictable based on position.

    • Lanthanides and Actinides: Inner transition elements.

  • Ion Charge Predictions:

    • Main-group metals lose electrons to form cations with the electron configuration of the nearest noble gas (e.g., Li+Li^+, Na+Na^+, K+K^+, Rb+Rb^+, Mg2+Mg^{2+}, Ca2+Ca^{2+}, Sr2+Sr^{2+}, Ba2+Ba^{2+}, Al3+Al^{3+}).

    • Main-group nonmetals gain electrons to form anions with the electron configuration of the nearest noble gas (e.g., N3N^{3-}, O2O^{2-}, S2S^{2-}, Se2Se^{2-}, Te2Te^{2-}, FF^-, ClCl^-, BrBr^-, II^-).

Chemical Bonds and Compounds

  • Chemical Bonds: Hold atoms together in compounds.

    • Ionic Bonds: Form between metals and nonmetals. Electrons are transferred from metal to nonmetal, forming cations and anions held together by electrostatic forces in a crystalline lattice.

    • Covalent Bonds: Form between two or more nonmetals. Electrons are shared between bonding atoms and interact with both nuclei.

  • Chemical Formulas and Representations:

    • Empirical Formula: Gives the relative whole-number ratio of atoms of each element in a compound (experimentally determined).

    • Molecular Formula: Gives the actual number of atoms of each element in a molecule. Example: Hydrogen peroxide empirical formula = HOHO; molecular formula = H2O2H_2O_2

    • Structural Formula: Represents how atoms in a molecule are physically connected or bonded.

    • Molecular Models: Ball-and-stick models represent atoms as spheres and bonds as rods; space-filling models represent atoms filling spatial volume.

  • Molecular vs. Atomic Forms:

    • Atomic Elements: Exist in nature as single individual atoms (e.g., HeHe, ArAr, NeNe).

    • Molecular Elements: Exist as diatomic molecules (e.g., O2O_2, N2N_2) or polyatomic molecules (e.g., S8S_8).

    • Molecular Compounds: Composed of two or more covalently bonded nonmetals.

    • Ionic Compounds: Composed of alternating cations and anions bonded in non-directional ionic lattices.

    • Formula Unit: The smallest electrically neutral collection of ions representing an ionic compound.

    • Charge balance rule: Sum of positive charges must equal sum of negative charges (Net charge=0\text{Net charge} = 0).

Chemical Nomenclature

  • Naming Binary Ionic Compounds:

    • For metals with invariant charges: [Name of Metal] [Base name of Nonmetal + -ide]

    • Example: CaF2CaF_2 = Calcium fluoride.

    • For metals with variable charges (Transition Metals): [Name of Metal](Roman Numeral of Charge) [Base name of Nonmetal + -ide]

    • Examples: FeCl2FeCl_2 = Iron(II) chloride; FeCl3FeCl_3 = Iron(III) chloride.

  • Common Metals forming Variable Cations:

    • Chromium: Cr2+Cr^{2+} (Chromium(II) / Chromous), Cr3+Cr^{3+} (Chromium(III) / Chromic)

    • Iron: Fe2+Fe^{2+} (Iron(II) / Ferrous), Fe3+Fe^{3+} (Iron(III) / Ferric)

    • Cobalt: Co2+Co^{2+} (Cobalt(II) / Cobaltous), Co3+Co^{3+} (Cobalt(III) / Cobaltic)

    • Copper: Cu+Cu^+ (Copper(I) / Cuprous), Cu2+Cu^{2+} (Copper(II) / Cupric)

    • Tin: Sn2+Sn^{2+} (Tin(II) / Stannous), Sn4+Sn^{4+} (Tin(IV) / Stannic)

    • Mercury: Hg22+Hg_2^{2+} (Mercury(I) / Mercurous), Hg2+Hg^{2+} (Mercury(II) / Mercuric)

    • Lead: Pb2+Pb^{2+} (Lead(II) / Plumbous), Pb4+Pb^{4+} (Lead(IV) / Plumbic)

  • Naming Polyatomic Ionic Compounds and Oxyanions:

    • Oxyanions contain oxygen and another element.

    • Two-oxyanion series: The anion with more oxygen gets suffix -ate; the anion with less oxygen gets suffix -ite.

    • Four-oxyanion series: per- prefix for maximum oxygen, hypo- prefix for minimal oxygen.

  • Hydrated Ionic Compounds:

    • Ionic compounds containing specific numbers of water molecules per formula unit (waters of hydration can be removed by heat).

    • Form: [Ionic compound name] [Greek prefix]hydrate

    • Example: MgSO47H2OMgSO_4 · 7H_2O = Magnesium sulfate heptahydrate.

  • Naming Acids:

    • Acids release H+H^+ ions in aqueous solution.

    • Binary Acids (HH + nonmetal):

    • hydro- + [Base name of nonmetal + -ic] + acid

    • Example: HBrHBr = Hydrobromic acid; HClHCl = Hydrochloric acid.

    • Oxyacids (HH + oxyanion):

    • Oxyanion ending in -ate \rightarrow [Base name of oxyanion + -ic] + acid (e.g., H2SO4H_2SO_4 = Sulfuric acid from sulfate; H3PO4H_3PO_4 = Phosphoric acid).

    • Oxyanion ending in -ite \rightarrow [Base name of oxyanion + -ous] + acid (e.g., H2SO3H_2SO_3 = Sulfurous acid from sulfite).

  • Naming Molecular Compounds:

    • Composed of nonmetals covalently bonded.

    • [Greek prefix][1st element] [Greek prefix][Base name of 2nd element + -ide]

    • Element with smallest group number (leftmost on periodic table) listed first.

    • Prefix mono- is omitted on the first element if singular, but kept on the second element.

    • Examples: P4S10P_4S_{10} = Tetraphosphorus decasulfide; CO2CO_2 = Carbon dioxide; COCO = Carbon monoxide.

  • Organic Chemistry:

    • The study of carbon-containing compounds (over 99% of 30+ million known compounds contain carbon).

    • Carbon always forms 4 chemical bonds, forming chains, branches, and ring backbones.

    • Hydrocarbons: Simple organic molecules containing only carbon and hydrogen. Simplest hydrocarbon = Methane (CH4CH_4).

Quantitative Worked Examples

  • Calculating Mass in Reactions (Conservation of Mass):

    • Antoine Lavoisier burned a 1.25 g1.25\text{ g} strip of magnesium (MgMg), resulting in 2.07 g2.07\text{ g} of magnesium oxide (MgOMgO).

    • Mass of oxygen consumed: 2.07 g MgO1.25 g Mg=0.82 g O2.07\text{ g } MgO - 1.25\text{ g } Mg = 0.82\text{ g } O.

  • Calculating Mass of an Element in a Compound (Definite Proportions):

    • A 78.80 g78.80\text{ g} sample of hematite contains 55.08 g55.08\text{ g} of iron (FeFe). Determine grams of FeFe and OO in a 260.82 g260.82\text{ g} rock.

    • Set up proportion:     55.08 g Fe78.80 g sample=x g Fe260.82 g rock\frac{55.08\text{ g Fe}}{78.80\text{ g sample}} = \frac{x\text{ g Fe}}{260.82\text{ g rock}}     78.80x = 14365.9656 −−−> x = 182.3\text{ g of Fe}

    • Mass of oxygen: 260.82 g182.3 g=78.5 g of O260.82\text{ g} - 182.3\text{ g} = 78.5\text{ g of O}.

  • Calculating Atomic Mass of an Unknown Metal:

    • A metal (MM) weighing 1.563 g1.563\text{ g} is burned in 127.4 mL127.4\text{ mL} of O2O_2 gas (density = 1.429 g/L1.429\text{ g/L}) to produce metal oxide MOMO.

    • Step 1: Calculate mass of O2O_2 gas consumed:     Mass of O2=127.4 mL×1 L1000 mL×1.429 g1 L=0.1820546 g O2\text{Mass of } O_2 = 127.4\text{ mL} \times \frac{1\text{ L}}{1000\text{ mL}} \times \frac{1.429\text{ g}}{1\text{ L}} = 0.1820546\text{ g } O_2

    • Step 2: Set up atomic ratio for formula MOMO (atomic mass of O=16.00 amu\text{atomic mass of } O = 16.00\text{ amu}):     Atomic mass of M16.00 amu=1.563 g M0.1820546 g O\frac{\text{Atomic mass of } M}{16.00\text{ amu}} = \frac{1.563\text{ g } M}{0.1820546\text{ g } O}     0.1820546x = 25.008 −−−> x = 137.4\text{ amu}

    • The calculated atomic mass (137.4 amu137.4\text{ amu}) identifies the metal MM as Barium (BaBa).

  • Calculating Molecular and Formula Mass:

    • Phosphorus Oxyfluoride (POF3POF_3):

    • 1×P=1×30.97 amu=30.97 amu1 \times P = 1 \times 30.97\text{ amu} = 30.97\text{ amu}

    • 1×O=1×16.00 amu=16.00 amu1 \times O = 1 \times 16.00\text{ amu} = 16.00\text{ amu}

    • 3×F=3×19.00 amu=57.00 amu3 \times F = 3 \times 19.00\text{ amu} = 57.00\text{ amu}

    • Total Molecular Mass=103.97 amu\text{Total Molecular Mass} = 103.97\text{ amu}

    • Copper(II) Phosphate (Cu3(PO4)2Cu_3(PO_4)_2):

    • 3×Cu=3×63.55 amu=190.65 amu3 \times Cu = 3 \times 63.55\text{ amu} = 190.65\text{ amu}

    • 2×P=2×30.97 amu=61.94 amu2 \times P = 2 \times 30.97\text{ amu} = 61.94\text{ amu}

    • 8×O=8×16.00 amu=128.00 amu8 \times O = 8 \times 16.00\text{ amu} = 128.00\text{ amu}

    • Total Formula Mass=380.59 amu\text{Total Formula Mass} = 380.59\text{ amu}