Presentation is loading. Please wait.

Presentation is loading. Please wait.

Mark Rosengarten’s Amazing Chemistry Powerpoint Presentation!

Similar presentations

Presentation on theme: "Mark Rosengarten’s Amazing Chemistry Powerpoint Presentation!"— Presentation transcript:

1 Mark Rosengarten’s Amazing Chemistry Powerpoint Presentation!
Aligned to the New York State Standards and Core Curriculum for “The Physical Setting-Chemistry” Can be used in any high-school chemistry class! Please give the link to this file to your chemistry students! Enjoy it!!! A LOT of work has gone into bringing you this work, so please credit me when you use it! (c) 2006, Mark Rosengarten

2 Outline for Review 1) The Atom (Nuclear, Electron Config)
2) Matter (Phases, Types, Changes) 3) Bonding (Periodic Table, Ionic, Covalent) 4) Compounds (Formulas, Reactions, IMAF’s) 5) Math of Chemistry (Formula Mass, Gas Laws, Neutralization, etc.) 6) Kinetics and Thermodynamics (PE Diagrams, etc.) 7) Acids and Bases (pH, formulas, indicators, etc.) 8) Oxidation and Reduction (Half Reactions, Cells, etc.) 9) Organic Chemistry (Hydrocarbons, Families, Reactions) (c) 2006, Mark Rosengarten

3 The Atom 1) Nucleons 2) Isotopes 3) Natural Radioactivity 4) Half-Life
5) Nuclear Power 6) Electron Configuation 7) Development of the Atomic Model (c) 2006, Mark Rosengarten

4 Nucleons Protons: +1 each, determines identity of element, mass of 1 amu, determined using atomic number, nuclear charge Neutrons: no charge, determines identity of isotope of an element, 1 amu, determined using mass number - atomic number (amu = atomic mass unit) 3216S and 3316S are both isotopes of S S-32 has 16 protons and 16 neutrons S-33 has 16 protons and 17 neutrons All atoms of S have a nuclear charge of +16 due to the 16 protons. (c) 2006, Mark Rosengarten

5 Isotopes Atoms of the same element MUST contain the same number of protons. Atoms of the same element can vary in their numbers of neutrons, therefore many different atomic masses can exist for any one element. These are called isotopes. The atomic mass on the Periodic Table is the weight-average atomic mass, taking into account the different isotope masses and their relative abundance. Rounding off the atomic mass on the Periodic Table will tell you what the most common isotope of that element is. (c) 2006, Mark Rosengarten

6 Weight-Average Atomic Mass
WAM = ((% A of A/100) X Mass of A) + ((% A of B/100) X Mass of B) + … What is the WAM of an element if its isotope masses and abundances are: X-200: Mass = amu, % abundance = 20.0 % X-204: Mass = amu, % abundance = 80.0% amu = atomic mass unit (1.66 × kilograms/amu) (c) 2006, Mark Rosengarten

7 Most Common Isotope Co Ag S Pb
The weight-average atomic mass of Zinc is amu. What is the most common isotope of Zinc? Zn-65! What are the most common isotopes of: Co Ag S Pb FACT: one atomic mass unit (1.66 × kilograms) is defined as 1/12 of the mass of an atom of C-12. This method doesn’t always work, but it usually does. Use it for the Regents exam. (c) 2006, Mark Rosengarten

8 Natural Radioactivity
Alpha Decay Beta Decay Positron Decay Gamma Decay Charges of Decay Particles Natural decay starts with a parent nuclide that ejects a decay particle to form a daughter nuclide which is more stable than the parent nuclide was. (c) 2006, Mark Rosengarten

9 Alpha Decay The nucleus ejects two protons and two neutrons. The atomic mass decreases by 4, the atomic number decreases by 2. 23892U  (c) 2006, Mark Rosengarten

10 Beta Decay A neutron decays into a proton and an electron. The electron is ejected from the nucleus as a beta particle. The atomic mass remains the same, but the atomic number increases by 1. 146C  (c) 2006, Mark Rosengarten

11 Positron Decay A proton is converted into a neutron and a positron. The positron is ejected by the nucleus. The mass remains the same, but the atomic number decreases by 1. 5326Fe  (c) 2006, Mark Rosengarten

12 Gamma Decay The nucleus has energy levels just like electrons, but the involve a lot more energy. When the nucleus becomes more stable, a gamma ray may be released. This is a photon of high-energy light, and has no mass or charge. The atomic mass and number do not change with gamma. Gamma may occur by itself, or in conjunction with any other decay type. (c) 2006, Mark Rosengarten

13 Charges of Decay Particles
(c) 2006, Mark Rosengarten

14 Half-Life Half life is the time it takes for half of the nuclei in a radioactive sample to undergo decay. Problem Types: Going forwards in time Going backwards in time Radioactive Dating (c) 2006, Mark Rosengarten

15 Going Forwards in Time How many grams of a 10.0 gram sample of I-131 (half-life of 8 days) will remain in 24 days? #HL = t/T = 24/8 = 3 Cut 10.0g in half 3 times: 5.00, 2.50, 1.25g (c) 2006, Mark Rosengarten

16 Going Backwards in Time
How many grams of a 10.0 gram sample of I-131 (half-life of 8 days) would there have been 24 days ago? #HL = t/T = 24/8 = 3 Double 10.0g 3 times: 20.0, 40.0, 80.0 g (c) 2006, Mark Rosengarten

17 Radioactive Dating A sample of an ancient scroll contains 50% of the original steady-state concentration of C-14. How old is the scroll? 50% = 1 HL 1 HL X 5730 y/HL = 5730y (c) 2006, Mark Rosengarten

18 Nuclear Power Artificial Transmutation Particle Accelerators
Nuclear Fission Nuclear Fusion (c) 2006, Mark Rosengarten

19 Artificial Transmutation
4020Ca + _____ > 4019K + 11H 9642Mo + 21H > 10n + _____ Nuclide + Bullet --> New Element + Fragment(s) The masses and atomic numbers must add up to be the same on both sides of the arrow. (c) 2006, Mark Rosengarten

20 Particle Accelerators
Devices that use electromagnetic fields to accelerate particle “bullets” towards target nuclei to make artificial transmutation possible! Most of the elements from 93 on up (the “transuranium” elements) were created using particle accelerators. Particles with no charge cannot be accelerated by the charged fields. (c) 2006, Mark Rosengarten

21 Nuclear Fission 23592U n  9236Kr Ba n + energy The three neutrons given off can be reabsorbed by other U-235 nuclei to continue fission as a chain reaction A tiny bit of mass is lost (mass defect) and converted into a huge amount of energy. (c) 2006, Mark Rosengarten

22 Chain Reaction (c) 2006, Mark Rosengarten

23 Nuclear Fusion 21H + 21H  42He + energy
Two small, positively-charged nuclei smash together at high temperatures and pressures to form one larger nucleus. A small bit of mass is destroyed and converted into a huge amount of energy, more than even fission. (c) 2006, Mark Rosengarten

24 Electron Configuration
Basic Configuration Valence Electrons Electron-Dot (Lewis Dot) Diagrams Excited vs. Ground State What is Light? (c) 2006, Mark Rosengarten

25 Basic Configuration The number of electrons is determined from the atomic number. Look up the basic configuration below the atomic number on the periodic table. (PEL: principal energy level = shell) He: 2 (2 e- in the 1st PEL) Na: (2 e- in the 1st PEL, 8 in the 2nd and 1 in the 3rd) Br: (2 e- in the 1st PEL, 8 in the 2nd, 18 in the 3rd and 7 in the 4th) (c) 2006, Mark Rosengarten

26 Valence Electrons The valence electrons are responsible for all chemical bonding. The valence electrons are the electrons in the outermost PEL (shell). He: 2 (2 valence electrons) Na: (1 valence electron) Br: (7 valence electrons) The maximum number of valence electrons an atom can have is EIGHT, called a STABLE OCTET. (c) 2006, Mark Rosengarten

27 Electron-Dot Diagrams
The number of dots equals the number of valence electrons. The number of unpaired valence electrons in a nonmetal tells you how many covalent bonds that atom can form with other nonmetals or how many electrons it wants to gain from metals to form an ion. The number of valence electrons in a metal tells you how many electrons the metal will lose to nonmetals to form an ion. Caution: May not work with transition metals. EXAMPLE DOT DIAGRAMS (c) 2006, Mark Rosengarten

28 Example Dot Diagrams Carbon can also have this dot diagram, which it
has when it forms organic compounds. (c) 2006, Mark Rosengarten

29 Excited vs. Ground State
Configurations on the Periodic Table are ground state configurations. If electrons are given energy, they rise to higher energy levels (excited state). If the total number of electrons matches in the configuration, but the configuration doesn’t match, the atom is in the excited state. Na (ground, on table): 2-8-1 Example of excited states: 2-7-2, , 2-6-3 (c) 2006, Mark Rosengarten

30 What Is Light? Light is formed when electrons drop from the excited state to the ground state. The lines on a bright-line spectrum come from specific energy level drops and are unique to each element. EXAMPLE SPECTRUM (c) 2006, Mark Rosengarten

31 EXAMPLE SPECTRUM This is the bright-line spectrum of hydrogen. The top
numbers represent the PEL (shell) change that produces the light with that color and the bottom number is the wavelength of the light (in nanometers, or 10-9 m). No other element has the same bright-line spectrum as hydrogen, so these spectra can be used to identify elements or mixtures of elements. (c) 2006, Mark Rosengarten

32 Development of the Atomic Model
Thompson Model Rutherford Gold Foil Experiment and Model Bohr Model Quantum-Mechanical Model (c) 2006, Mark Rosengarten

33 Thompson Model The atom is a positively charged diffuse mass with negatively charged electrons stuck in it. (c) 2006, Mark Rosengarten

34 Rutherford Model The atom is made of a small, dense, positively charged nucleus with electrons at a distance, the vast majority of the volume of the atom is empty space. Alpha particles shot at a thin sheet of gold foil: most go through (empty space). Some deflect or bounce off (small + charged nucleus). (c) 2006, Mark Rosengarten

35 Bohr Model Electrons orbit around the nucleus in energy levels (shells). Atomic bright-line spectra was the clue. (c) 2006, Mark Rosengarten

36 Quantum-Mechanical Model
Electron energy levels are wave functions. Electrons are found in orbitals, regions of space where an electron is most likely to be found. You can’t know both where the electron is and where it is going at the same time. Electrons buzz around the nucleus like gnats buzzing around your head. (c) 2006, Mark Rosengarten

37 Matter 1) Properties of Phases 2) Types of Matter 3) Phase Changes
(c) 2006, Mark Rosengarten

38 Properties of Phases Solids: Crystal lattice (regular geometric pattern), vibration motion only Liquids: particles flow past each other but are still attracted to each other. Gases: particles are small and far apart, they travel in a straight line until they hit something, they bounce off without losing any energy, they are so far apart from each other that they have effectively no attractive forces and their speed is directly proportional to the Kelvin temperature (Kinetic-Molecular Theory, Ideal Gas Theory) (c) 2006, Mark Rosengarten

39 Solids The positive and negative ions alternate in the
ionic crystal lattice of NaCl. (c) 2006, Mark Rosengarten

40 Liquids When heated, the ions move faster and eventually
separate from each other to form a liquid. The ions are loosely held together by the oppositely charged ions, but the ions are moving too fast for the crystal lattice to stay together. (c) 2006, Mark Rosengarten

41 Gases Since all gas molecules spread out
the same way, equal volumes of gas under equal conditions of temperature and pressure will contain equal numbers of molecules of gas L of any gas at STP (1.00 atm and 273K) will contain one mole (6.02 X 1023) gas molecules. Since there is space between gas molecules, gases are affected by changes in pressure. (c) 2006, Mark Rosengarten

42 Types of Matter Substances (Homogeneous) Mixtures
Elements (cannot be decomposed by chemical change): Al, Ne, O, Br, H Compounds (can be decomposed by chemical change): NaCl, Cu(ClO3)2, KBr, H2O, C2H6 Mixtures Homogeneous: Solutions (solvent + solute) Heterogeneous: soil, Italian dressing, etc. (c) 2006, Mark Rosengarten

43 Elements A sample of lead atoms (Pb). All atoms in the sample consist of lead, so the substance is homogeneous. A sample of chlorine atoms (Cl). All atoms in the sample consist of chlorine, so the substance is homogeneous. (c) 2006, Mark Rosengarten

44 Compounds Lead has two charges listed, +2 and +4. This is a sample of lead (II) chloride (PbCl2). Two or more elements bonded in a whole-number ratio is a COMPOUND. This compound is formed from the +4 version of lead. This is lead (IV) chloride (PbCl4). Notice how both samples of lead compounds have consistent composition throughout? Compounds are homogeneous! (c) 2006, Mark Rosengarten

45 Mixtures A mixture of lead atoms and chlorine atoms. They exist in no particular ratio and are not chemically combined with each other. They can be separated by physical means. A mixture of PbCl2 and PbCl4 formula units. Again, they are in no particular ratio to each other and can be separated without chemical change. (c) 2006, Mark Rosengarten

46 Phase Changes Phase Change Types Phase Change Diagrams
Heat of Phase Change Evaporation (c) 2006, Mark Rosengarten

47 Phase Change Types (c) 2006, Mark Rosengarten

48 Phase Change Diagrams AB: Solid Phase BC: Melting (S + L)
CD: Liquid Phase DE: Boiling (L + G) EF: Gas Phase Notice how temperature remains constant during a phase change? That’s because the PE is changing, not the KE. (c) 2006, Mark Rosengarten

49 Heat of Phase Change How many joules would it take to melt 100. g of H2O (s) at 0oC? q=mHf = (100. g)(334 J/g) = J How many joules would it take to boil 100. g of H2O (l) at 100oC? q=mHv = (100.g)(2260 J/g) = J (c) 2006, Mark Rosengarten

50 Evaporation When the surface molecules of a gas travel upwards at a great enough speed to escape. The pressure a vapor exerts when sealed in a container at equilibrium is called vapor pressure, and can be found on Table H. When the liquid is heated, its vapor pressure increases. When the liquid’s vapor pressure equals the pressure exerted on it by the outside atmosphere, the liquid can boil. If the pressure exerted on a liquid increases, the boiling point of the liquid increases (pressure cooker). If the pressure decreases, the boiling point of the liquid decreases (special cooking directions for high elevations). (c) 2006, Mark Rosengarten

51 Reference Table H: Vapor Pressure of Four Liquids
(c) 2006, Mark Rosengarten

52 Bonding 1) The Periodic Table 2) Ions 3) Ionic Bonding
4) Covalent Bonding 5) Metallic Bonding (c) 2006, Mark Rosengarten

53 The Periodic Table Metals Nonmetals Metalloids Chemistry of Groups
Electronegativity Ionization Energy (c) 2006, Mark Rosengarten

54 Metals Have luster, are malleable and ductile, good conductors of heat and electricity Lose electrons to nonmetal atoms to form positively charged ions in ionic bonds Large atomic radii compared to nonmetal atoms Low electronegativity and ionization energy Left side of the periodic table (except H) (c) 2006, Mark Rosengarten

55 Nonmetals Are dull and brittle, poor conductors
Gain electrons from metal atoms to form negatively charged ions in ionic bonds Share unpaired valence electrons with other nonmetal atoms to form covalent bonds and molecules Small atomic radii compared to metal atoms High electronegativity and ionization energy Right side of the periodic table (except Group 18) (c) 2006, Mark Rosengarten

56 Metalloids Found lying on the jagged line between metals and nonmetals flatly touching the line (except Al and Po). Share properties of metals and nonmetals (Si is shiny like a metal, brittle like a nonmetal and is a semiconductor). (c) 2006, Mark Rosengarten

57 Chemistry of Groups Group 1: Alkali Metals
Group 2: Alkaline Earth Metals Groups 3-11: Transition Elements Group 17: Halogens Group 18: Noble Gases Diatomic Molecules (c) 2006, Mark Rosengarten

58 Group 1: Alkali Metals Most active metals, only found in compounds in nature React violently with water to form hydrogen gas and a strong base: 2 Na (s) + H2O (l)  2 NaOH (aq) + H2 (g) 1 valence electron Form +1 ion by losing that valence electron Form oxides like Na2O, Li2O, K2O (c) 2006, Mark Rosengarten

59 Group 2: Alkaline Earth Metals
Very active metals, only found in compounds in nature React strongly with water to form hydrogen gas and a base: Ca (s) + 2 H2O (l)  Ca(OH)2 (aq) + H2 (g) 2 valence electrons Form +2 ion by losing those valence electrons Form oxides like CaO, MgO, BaO (c) 2006, Mark Rosengarten

60 Groups 3-11: Transition Metals
Many can form different possible charges of ions If there is more than one ion listed, give the charge as a Roman numeral after the name Cu+1 = copper (I) Cu+2 = copper (II) Compounds containing these metals can be colored. (c) 2006, Mark Rosengarten

61 Group 17: Halogens Most reactive nonmetals
React violently with metal atoms to form halide compounds: 2 Na + Cl2  2 NaCl Only found in compounds in nature Have 7 valence electrons Gain 1 valence electron from a metal to form -1 ions Share 1 valence electron with another nonmetal atom to form one covalent bond. (c) 2006, Mark Rosengarten

62 Group 18: Noble Gases Are completely nonreactive since they have eight valence electrons, making a stable octet. Kr and Xe can be forced, in the laboratory, to give up some valence electrons to react with fluorine. Since noble gases do not naturally bond to any other elements, one atom of noble gas is considered to be a molecule of noble gas. This is called a monatomic molecule. Ne represents an atom of Ne and a molecule of Ne. (c) 2006, Mark Rosengarten

63 Diatomic Molecules Br, I, N, Cl, H, O and F are so reactive that they exist in a more chemically stable state when they covalently bond with another atom of their own element to make two-atom, or diatomic molecules. Br2, I2, N2, Cl2, H2, O2 and F2 The decomposition of water: 2 H2O  2 H2 + O2 (c) 2006, Mark Rosengarten

64 Electronegativity An atom’s attraction to electrons in a chemical bond. F has the highest, at 4.0 Fr has the lowest, at 0.7 If two atoms that are different in EN (END) from each other by 1.7 or more collide and bond (like a metal atom and a nonmetal atom), the one with the higher electronegativity will pull the valence electrons away from the atom with the lower electronegativity to form a (-) ion. The atom that was stripped of its valence electrons forms a (+) ion. If the two atoms have an END of less than 1.7, they will share their unpaired valence electrons…covalent bond! (c) 2006, Mark Rosengarten

65 Ionization Energy The energy required to remove the most loosely held valence electron from an atom in the gas phase. High electronegativity means high ionization energy because if an atom is more attracted to electrons, it will take more energy to remove those electrons. Metals have low ionization energy. They lose electrons easily to form (+) charged ions. Nonmetals have high ionization energy but high electronegativity. They gain electrons easily to form (-) charged ions when reacted with metals, or share unpaired valence electrons with other nonmetal atoms. (c) 2006, Mark Rosengarten

66 Ions Ions are charged particles formed by the gain or loss of electrons. Metals lose electrons (oxidation) to form (+) charged cations. Nonmetals gain electrons (reduction) to form (-) charged anions. Atoms will gain or lose electrons in such a way that they end up with 8 valence electrons (stable octet). The exceptions to this are H, Li, Be and B, which are not large enough to support 8 valence electrons. They must be satisfied with 2 (Li, Be, B) or 0 (H). (c) 2006, Mark Rosengarten

67 Metal Ions (Cations) Na: 2-8-1
Note that when the atom loses its valence electron, the next lower PEL becomes the valence PEL. Notice how the dot diagrams for metal ions lack dots! Place brackets around the element symbol and put the charge on the upper right outside! (c) 2006, Mark Rosengarten

68 Nonmetal Ions (Anions)
Note how the ions all have 8 valence electrons. Also note the gained electrons as red dots. Nonmetal ion dot diagrams show 8 dots, with brackets around the dot diagram and the charge of the ion written to the upper right side outside the brackets. F: 2-7 F-1: 2-8 O: 2-6 O-2: 2-8 N: 2-5 N-3: 2-8 (c) 2006, Mark Rosengarten

69 Ionic Bonding If two atoms that are different in EN (END) from each other by 1.7 or more collide and bond (like a metal atom and a nonmetal atom), the one with the higher electronegativity will pull the valence electrons away from the atom with the lower electronegativity to form a (-) ion. The atom that was stripped of its valence electrons forms a (+) ion. The oppositely charged ions attract to form the bond. It is a surface bond that can be broken by melting or dissolving in water. Ionic bonding forms ionic crystal lattices, not molecules. (c) 2006, Mark Rosengarten

70 Example of Ionic Bonding
(c) 2006, Mark Rosengarten

71 Covalent Bonding If two nonmetal atoms have an END of 1.7 or less, they will share their unpaired valence electrons to form a covalent bond. A particle made of covalently bonded nonmetal atoms is called a molecule. If the END is between 0 and 0.4, the sharing of electrons is equal, so there are no charged ends. This is NONPOLAR covalent bonding. If the END is between 0.5 and 1.7, the sharing of electrons is unequal. The atom with the higher EN will be d- and the one with the lower EN will be d+ charged. This is a POLAR covalent bonding. (d means “partial”) (c) 2006, Mark Rosengarten

72 Examples of Covalent Bonding
(c) 2006, Mark Rosengarten

73 Metallic Bonding Metal atoms of the same element bond with each other by sharing valence electrons that they lose to each other. This is a lot like an atomic game of “hot potato”, where metal kernals (the atom inside the valence electrons) sit in a crystal lattice, passing valence electrons back and forth between each other). Since electrons can be forced to travel in a certain direction within the metal, metals are very good at conducting electricity in all phases. (c) 2006, Mark Rosengarten

74 Compounds 1) Types of Compounds 2) Formula Writing 3) Formula Naming
4) Empirical Formulas 5) Molecular Formulas 6) Types of Chemical Reactions 7) Balancing Chemical Reactions 8) Attractive Forces (c) 2006, Mark Rosengarten

75 Types of Compounds Ionic: made of metal and nonmetal ions. Form an ionic crystal lattice when in the solid phase. Ions separate when melted or dissolved in water, allowing electrical conduction. Examples: NaCl, K2O, CaBr2 Molecular: made of nonmetal atoms bonded to form a distinct particle called a molecule. Bonds do not break upon melting or dissolving, so molecular substances do not conduct electricity. EXCEPTION: Acids [H+A- (aq)] ionize in water to form H3O+ and A-, so they do conduct. Network: made up of nonmetal atoms bonded in a seemingly endless matrix of covalent bonds with no distinguishable molecules. Very high m.p., don’t conduct. (c) 2006, Mark Rosengarten

76 Ionic Compounds (c) 2006, Mark Rosengarten

77 Molecular Compounds (c) 2006, Mark Rosengarten

78 Network Solids Network solids are made of nonmetal atoms covalently bonded together to form large crystal lattices. No individual molecules can be distinguished. Examples include C (diamond) and SiO2 (quartz). Corundum (Al2O3) also forms these, even though Al is considered a metal. Network solids are among the hardest materials known. They have extremely high melting points and do not conduct electricity. (c) 2006, Mark Rosengarten

79 Formula Writing The charge of the (+) ion and the charge of the (-) ion must cancel out to make the formula. Use subscripts to indicate how many atoms of each element there are in the compound, no subscript if there is only one atom of that element. Na+1 and Cl-1 = NaCl Ca+2 and Br-1 = CaBr2 Al+3 and O-2 = Al2O3 Zn+2 and PO4-3 = Zn3(PO4)2 Try these problems! (c) 2006, Mark Rosengarten

80 Formulas to Write Ba+2 and N-3 NH4+1 and SO4-2 Li+1 and S-2
Cu+2 and NO3-1 Al+3 and CO3-2 Fe+3 and Cl-1 Pb+4 and O-2 Pb+2 and O-2 (c) 2006, Mark Rosengarten

81 Formula Naming Compounds are named from the elements or polyatomic ions that form them. KCl = potassium chloride Na2SO4 = sodium sulfate (NH4)2S = ammonium sulfide AgNO3 = silver nitrate Notice all the metals listed here only have one charge listed? So what do you do if a metal has more than one charge listed? Take a peek! (c) 2006, Mark Rosengarten

82 The Stock System CrCl2 = chromium (II) chloride Try
CrCl3 = chromium (III) chloride Co(NO3)2 and CrCl6 = chromium (VI) chloride Co(NO3)3 FeO = iron (II) oxide MnS = manganese (II) sulfide Fe2O3 = iron (III) oxide MnS2 = manganese (IV) sulfide The Roman numeral is the charge of the metal ion! (c) 2006, Mark Rosengarten

83 Empirical Formulas Ionic formulas: represent the simplest whole number mole ratio of elements in a compound. Ca3N2 means a 3:2 ratio of Ca ions to N ions in the compound. Many molecular formulas can be simplified to empirical formulas Ethane (C2H6) can be simplified to CH3. This is the empirical formula…the ratio of C to H in the molecule. All ionic compounds have empirical formulas. (c) 2006, Mark Rosengarten

84 Molecular Formulas The count of the actual number of atoms of each element in a molecule. H2O: a molecule made of two H atoms and one O atom covalently bonded together. C2H6O: A molecule made of two C atoms, six H atoms and one O atom covalently bonded together. Molecular formulas are whole-number multiples of empirical formulas: H2O = 1 X (H2O) C8H16 = 8 X (CH2) Calculating Molecular Formulas (c) 2006, Mark Rosengarten

85 Types of Chemical Reactions
Redox Reactions: driven by the loss (oxidation) and gain (reduction) of electrons. Any species that does not change charge is called the spectator ion. Synthesis Decomposition Single Replacement Ion Exchange Reaction: driven by the formation of an insoluble precipitate. The ions that remain dissolved throughout are the spectator ions. Double Replacement (c) 2006, Mark Rosengarten

86 Synthesis Two elements combine to form a compound 2 Na + O2  Na2O
Same reaction, with charges added in: 2 Na0 + O20  Na2+1O-2 Na0 is oxidized (loses electrons), is the reducing agent O20 is reduced (gains electrons), is the oxidizing agent Electrons are transferred from the Na0 to the O20. No spectator ions, there are only two elements here. (c) 2006, Mark Rosengarten

87 Decomposition A compound breaks down into its original elements.
Na2O  2 Na + O2 Same reaction, with charges added in: Na2+1O-2  2 Na0 + O20 O-2 is oxidized (loses electrons), is the reducing agent Na+1 is reduced (gains electrons), is the oxidizing agent Electrons are transferred from the O-2 to the Na+1. No spectator ions, there are only two elements here. (c) 2006, Mark Rosengarten

88 Single Replacement An element replaces the same type of element in a compound. Ca + 2 KCl  CaCl2 + 2 K Same reaction, with charges added in: Ca0 + 2 K+1Cl-1  Ca+2Cl K0 Ca0 is oxidized (loses electrons), is the reducing agent K+1 is reduced (gains electrons), is the oxidizing agent Electrons are transferred from the Ca0 to the K+1. Cl-1 is the spectator ion, since it’s charge doesn’t change. (c) 2006, Mark Rosengarten

89 Double Replacement The (+) ion of one compound bonds to the (-) ion of another compound to make an insoluble precipitate. The compounds must both be dissolved in water to break the ionic bonds first. NaCl (aq) + AgNO3 (aq)  NaNO3 (aq) + AgCl (s) The Cl-1 and Ag+1 come together to make the insoluble precipitate, which looks like snow in the test tube. No species change charge, so this is not a redox reaction. Since the Na+1 and NO3-1 ions remain dissolved throughout the reaction, they are the spectator ions. How do identify the precipitate? (c) 2006, Mark Rosengarten

90 Identifying the Precipitate
The precipitate is the compound that is insoluble. AgCl is a precipitate because Cl- is a halide. Halides are soluble, except when combined with Ag+ and others. (c) 2006, Mark Rosengarten

91 Balancing Chemical Reactions
Balance one element or ion at a time Use a pencil Use coefficients only, never change formulas Revise if necessary The coefficient multiplies everything in the formula by that amount 2 Ca(NO3)2 means that you have 2 Ca, 4 N and 12 O. Examples for you to try! (c) 2006, Mark Rosengarten

92 Reactions to Balance ___NaCl  ___Na + ___Cl2 ___Al + ___O2  ___Al2O3
___SO3  ___SO2 + ___O2 ___Ca + ___HNO3  ___Ca(NO3)2 + ___H2 __FeCl3 + __Pb(NO3)2  __Fe(NO3)3 + __PbCl2 (c) 2006, Mark Rosengarten

93 Attractive Forces Molecules have partially charged ends. The d+ end of one molecule attracts to the d- end of another molecule. Ions are charged (+) or (-). Positively charged ions attract other to form ionic bonds, a type of attractive force. Since partially charged ends result in weaker attractions than fully charged ends, ionic compounds generally have much higher melting points than molecular compounds. Determining Polarity of Molecules Hydrogen Bond Attractions (c) 2006, Mark Rosengarten

94 Determining Polarity of Molecules
(c) 2006, Mark Rosengarten

95 Hydrogen Bond Attractions
A hydrogen bond attraction is a very strong attractive force between the H end of one polar molecule and the N, O or F end of another polar molecule. This attraction is so strong that water is a liquid at a temperature where most compounds that are much heavier than water (like propane, C3H8) are gases. This also gives water its surface tension and its ability to form a meniscus in a narrow glass tube. (c) 2006, Mark Rosengarten

96 Math of Chemistry 1) Formula Mass 2) Percent Composition
3) Mole Problems 4) Gas Laws 5) Neutralization 6) Concentration 7) Significant Figures and Rounding 8) Metric Conversions 9) Calorimetry (c) 2006, Mark Rosengarten

97 Formula Mass Gram Formula Mass = sum of atomic masses of all elements in the compound Round given atomic masses to the nearest tenth H2O: (2 X 1.0) + (1 X 16.0) = 18.0 grams/mole Na2SO4: (2 X 23.0)+(1 X 32.1)+(4 X 16.0) = g/mole Now you try: BaBr2 CaSO4 Al2(CO3)3 (c) 2006, Mark Rosengarten

98 Percent Composition The mass of part is the number of atoms of that element in the compound. The mass of whole is the formula mass of the compound. Don’t forget to take atomic mass to the nearest tenth! This is a problem for you to try. (c) 2006, Mark Rosengarten

99 Practice Percent Composition Problem
What is the percent by mass of each element in Li2SO4? (c) 2006, Mark Rosengarten

100 Mole Problems Grams <=> Moles Molecular Formula Stoichiometry
(c) 2006, Mark Rosengarten

101 Grams <=> Moles How many grams will 3.00 moles of NaOH (40.0 g/mol) weigh? 3.00 moles X 40.0 g/mol = 120. g How many moles of NaOH (40.0 g/mol) are represented by 10.0 grams? (10.0 g) / (40.0 g/mol) = mol (c) 2006, Mark Rosengarten

102 Molecular Formula Molecular Formula = (Molecular Mass/Empirical Mass) X Empirical Formula What is the molecular formula of a compound with an empirical formula of CH2 and a molecular mass of 70.0 grams/mole? 1) Find the Empirical Formula Mass: CH2 = 14.0 2) Divide the MM/EM: 70.0/14.0 = 5 3) Multiply the molecular formula by the result: 5 (CH2) = C5H10 (c) 2006, Mark Rosengarten

103 Stoichiometry Moles of Target = Moles of Given X (Coefficent of Target/Coefficient of given) Given the balanced equation N2 + 3 H2  2 NH3, How many moles of H2 need to be completely reacted with N2 to yield 20.0 moles of NH3? 20.0 moles NH3 X (3 H2 / 2 NH3) = 30.0 moles H2 (c) 2006, Mark Rosengarten

104 Gas Laws Make a data table to put the numbers so you can eliminate the words. Make sure that any Celsius temperatures are converted to Kelvin (add 273). Rearrange the equation before substituting in numbers. If you are trying to solve for T2, get it out of the denominator first by cross-multiplying. If one of the variables is constant, then eliminate it. Try these problems! (c) 2006, Mark Rosengarten

105 Gas Law Problem 1 A 2.00 L sample of N2 gas at STP is compressed to 4.00 atm at constant temp-erature. What is the new volume of the gas? V2 = P1V1 / P2 = (1.00 atm)(2.00 L) / (4.00 atm) = L (c) 2006, Mark Rosengarten

106 Gas Law Problem 2 To what temperature must a L sample of O2 gas at K be heated to raise the volume to L? T2 = V2T1/V1 = (10.00 L)(300.0 K) / (3.000 L) = K (c) 2006, Mark Rosengarten

107 Gas Law Problem 3 A 3.00 L sample of NH3 gas at kPa is cooled from K to K and its pressure is reduced to 80.0 kPa. What is the new volume of the gas? V2 = P1V1T2 / P2T1 = (100.0 kPa)(3.00 L)(300. K) / (80.0 kPa)(500. K) = 2.25 L (c) 2006, Mark Rosengarten

108 Neutralization 10.0 mL of 0.20 M HCl is neutralized by 40.0 mL of NaOH. What is the concentration of the NaOH? #H MaVa = #OH MbVb, so Mb = #H MaVa / #OH Vb = (1)(0.20 M)(10.0 mL) / (1) (40.0 mL) = M How many mL of 2.00 M H2SO4 are needed to completely neutralize 30.0 mL of M KOH? (c) 2006, Mark Rosengarten

109 Concentration Molarity Parts per Million Percent by Mass
Percent by Volume (c) 2006, Mark Rosengarten

110 Molarity What is the molarity of a mL solution of NaOH (FM = 40.0) with 60.0 g of NaOH (aq)? Convert g to moles and mL to L first! M = moles / L = 1.50 moles / L = 3.00 M How many grams of NaOH does it take to make 2.0 L of a M solution of NaOH (aq)? Moles = M X L = M X 2.0 L = moles Convert moles to grams: moles X 40.0 g/mol = 8.00 g (c) 2006, Mark Rosengarten

111 Parts Per Million 100.0 grams of water is evaporated and analyzed for lead grams of lead ions are found. What is the concentration of the lead, in parts per million? ppm = ( g) / (100.0 g) X = 1.0 ppm If the legal limit for lead in the water is 3.0 ppm, then the water sample is within the legal limits (it’s OK!) (c) 2006, Mark Rosengarten

112 Percent by Mass A 50.0 gram sample of a solution is evaporated and found to contain grams of sodium chloride. What is the percent by mass of sodium chloride in the solution? % Comp = (0.100 g) / (50.0 g) X 100 = 0.200% (c) 2006, Mark Rosengarten

113 Percent By Volume Substitute “volume” for “mass” in the above equation. What is the percent by volume of hexane if 20.0 mL of hexane are dissolved in benzene to a total volume of 80.0 mL? % Comp = (20.0 mL) / (80.0 mL) X100 = 25.0% (c) 2006, Mark Rosengarten

114 Sig Figs and Rounding How many Significant Figures does a number have?
What is the precision of my measurement? How do I round off answers to addition and subtraction problems? How do I round off answers to multiplication and division problems? (c) 2006, Mark Rosengarten

115 How many Sig Figs? Start counting sig figs at the first non-zero.
All digits except place-holding zeroes are sig figs. Measurement # of Sig Figs 0.115 cm 3 cm 2 cm cm cm 5 Measurement # of Sig Figs 234 cm 3 67000 cm 2 _ 45000 cm 4 560. cm cm 5 (c) 2006, Mark Rosengarten

116 What Precision? A number’s precision is determined by the furthest (smallest) place the number is recorded to. 6000 mL : thousands place 6000. mL : ones place mL : tenths place 5.30 mL : hundredths place 8.7 mL : tenths place mL : thousandths place (c) 2006, Mark Rosengarten

117 Rounding with addition and subtraction
Answers are rounded to the least precise place. (c) 2006, Mark Rosengarten

118 Rounding with multiplication and division
Answers are rounded to the fewest number of significant figures. (c) 2006, Mark Rosengarten

119 Metric Conversions Determine how many powers of ten difference there are between the two units (no prefix = 100) and create a conversion factor. Multiply or divide the given by the conversion factor. How many kg are in 38.2 cg? (38.2 cg) /( cg/kg) = km How many mL in dL? (0.988 dg) X (100 mL/dL) = 98.8 mL (c) 2006, Mark Rosengarten

120 Calorimetry This equation can be used to determine any of the variables here. You will not have to solve for C, since we will always assume that the energy transfer is being absorbed by or released by a measured quantity of water, whose specific heat is given above. Solving for q Solving for m Solving for DT (c) 2006, Mark Rosengarten

121 Solving for q How many joules are absorbed by grams of water in a calorimeter if the temperature of the water increases from 20.0oC to 50.0oC? q = mCDT = (100.0 g)(4.18 J/goC)(30.0oC) = J (c) 2006, Mark Rosengarten

122 Solving for m A sample of water in a calorimeter cup increases from 25oC to 50.oC by the addition of joules of energy. What is the mass of water in the calorimeter cup? q = mCDT, so m = q / CDT = (500.0 J) / (4.18 J/goC)(25oC) = 4.8 g (c) 2006, Mark Rosengarten

123 Solving for DT If a 50.0 gram sample of water in a calorimeter cup absorbs joules of energy, how much will the temperature rise by? q = mCDT, so DT = q / mC = ( J)/(50.0 g)(4.18 J/goC) = 4.8oC If the water started at 20.0oC, what will the final temperature be? Since the water ABSORBS the energy, its temperature will INCREASE by the DT: 20.0oC + 4.8oC = 24.8oC (c) 2006, Mark Rosengarten

124 Kinetics and Thermodynamics
1) Reaction Rate 2) Heat of Reaction 3) Potential Energy Diagrams 4) Equilibrium 5) Le Châtelier’s Principle 6) Solubility Curves (c) 2006, Mark Rosengarten

125 Reaction Rate Reactions happen when reacting particles collide with sufficient energy (activation energy) and at the proper angle. Anything that makes more collisions in a given time will make the reaction rate increase. Increasing temperature Increasing concentration (pressure for gases) Increasing surface area (solids) Adding a catalyst makes a reaction go faster by removing steps from the mechanism and lowering the activation energy without getting used up in the process. (c) 2006, Mark Rosengarten

126 Heat of Reaction Reactions either absorb PE (endothermic, +DH) or release PE (exothermic, -DH) Exothermic, PEKE, Temp Endothermic, KEPE, Temp Rewriting the equation with heat included: 4 Al(s) + 3 O2(g)  2 Al2O3(s) kJ N2(g) + O2(g) kJ  2 NO(g) (c) 2006, Mark Rosengarten

127 Potential Energy Diagrams
Steps of a reactions: Reactants have a certain amount of PE stored in their bonds (Heat of Reactants) The reactants are given enough energy to collide and react (Activation Energy) The resulting intermediate has the highest energy that the reaction can make (Heat of Activated Complex) The activated complex breaks down and forms the products, which have a certain amount of PE stored in their bonds (Heat of Products) Hproducts - Hreactants = DH EXAMPLES (c) 2006, Mark Rosengarten

128 Making a PE Diagram X axis: Reaction Coordinate (time, no units)
Y axis: PE (kJ) Three lines representing energy (Hreactants, Hactivated complex, Hproducts) Two arrows representing energy changes: From Hreactants to Hactivated complex: Activation Energy From Hreactants to Hproducts : DH ENDOTHERMIC PE DIAGRAM EXOTHERMIC PE DIAGRAM (c) 2006, Mark Rosengarten

129 Endothermic PE Diagram
If a catalyst is added? (c) 2006, Mark Rosengarten

130 Endothermic with Catalyst
The red line represents the catalyzed reaction. (c) 2006, Mark Rosengarten

131 Exothermic PE Diagram What does it look like with a catalyst?
(c) 2006, Mark Rosengarten What does it look like with a catalyst?

132 Exothermic with a Catalyst
The red line represents the catalyzed reaction. Lower A.E. and faster reaction time! (c) 2006, Mark Rosengarten

133 Equilibrium When the rate of the forward reaction equals the rate of the reverse reaction. (c) 2006, Mark Rosengarten

134 Examples of Equilibrium
Solution Equilibrium: when a solution is saturated, the rate of dissolving equals the rate of precipitating. NaCl (s)  Na+1 (aq) + Cl-1 (aq) Vapor-Liquid Equilibrium: when a liquid is trapped with air in a container, the liquid evaporates until the rate of evaporation equals the rate of condensation. H2O (l)  H2O (g) Phase equilibrium: At the melting point, the rate of solid turning to liquid equals the rate of liquid turning back to solid. H2O (s)  H2O (l) (c) 2006, Mark Rosengarten

135 Le Châtelier’s Principle
If a system at equilibrium is stressed, the equilibrium will shift in a direction that relieves that stress. A stress is a factor that affects reaction rate. Since catalysts affect both reaction rates equally, catalysts have no effect on a system already at equilibrium. Equilibrium will shift AWAY from what is added Equilibrium will shift TOWARDS what is removed. This is because the shift will even out the change in reaction rate and bring the system back to equilibrium NEXT (c) 2006, Mark Rosengarten

136 Steps to Relieving Stress
1) Equilibrium is subjected to a STRESS. 2) System SHIFTS towards what is removed from the system or away from what is added. The shift results in a CHANGE OF CONCENTRATION for both the products and the reactants. If the shift is towards the products, the concentration of the products will increase and the concentration of the reactants will decrease. If the shift is towards the reactants, the concentration of the reactants will increase and the concentration of the products will decrease. NEXT (c) 2006, Mark Rosengarten

137 Examples For the reaction N2(g) + 3H2(g)  2 NH3(g) + heat
Adding N2 will cause the equilibrium to shift RIGHT, resulting in an increase in the concentration of NH3 and a decrease in the concentration of N2 and H2. Removing H2 will cause a shift to the LEFT, resulting in a decrease in the concentration of NH3 and an increase in the concentration of N2 and H2. Increasing the temperature will cause a shift to the LEFT, same results as the one above. Decreasing the pressure will cause a shift to the LEFT, because there is more gas on the left side, and making more gas will bring the pressure back up to its equilibrium amount. Adding a catalyst will have no effect, so no shift will happen. (c) 2006, Mark Rosengarten

138 Solubility Curves Solubility: the maximum quantity of solute that can be dissolved in a given quantity of solvent at a given temperature to make a saturated solution. Saturated: a solution containing the maximum quantity of solute that the solvent can hold. The limit of solubility. Supersaturated: the solution is holding more than it can theoretically hold OR there is excess solute which precipitates out. True supersaturation is rare. Unsaturated: There are still solvent molecules available to dissolve more solute, so more can dissolve. How ionic solutes dissolve in water: polar water molecules attach to the ions and tear them off the crystal. (c) 2006, Mark Rosengarten

139 Solubility Solubility: go to the temperature and up to the desired line, then across to the Y-axis. This is how many g of solute are needed to make a saturated solution of that solute in 100g of H2O at that particular temperature. At 40oC, the solubility of KNO3 in 100g of water is 64 g. In 200g of water, double that amount. In 50g of water, cut it in half. (c) 2006, Mark Rosengarten

140 Supersaturated If 120 g of NaNO3 are added to 100g of water at 30oC:
1) The solution would be SUPERSATURATED, because there is more solute dissolved than the solubility allows 2) The extra 25g would precipitate out 3) If you heated the solution up by 24oC (to 54oC), the excess solute would dissolve. (c) 2006, Mark Rosengarten

141 Unsaturated If 80 g of KNO3 are added to 100g of water at 60oC:
1) The solution would be UNSATURATED, because there is less solute dissolved than the solubility allows 2) 26g more can be added to make a saturated solution 3) If you cooled the solution down by 12oC (to 48oC), the solution would become saturated (c) 2006, Mark Rosengarten

142 How Ionic Solutes Dissolve in Water
Water solvent molecules attach to the ions (H end to the Cl-, O end to the Na+) Water solvent holds the ions apart and keeps the ions from coming back together (c) 2006, Mark Rosengarten

143 Acids and Bases 1) Formulas, Naming and Properties of Acids
2) Formulas, Naming and Properties of Bases 3) Neutralization 4) pH 5) Indicators 6) Alternate Theories (c) 2006, Mark Rosengarten

144 Formulas, Naming and Properties of Acids
Arrhenius Definition of Acids: molecules that dissolve in water to produce H3O+ (hydronium) as the only positively charged ion in solution. HCl (g) + H2O (l)  H3O+ (aq) + Cl- Properties of Acids Naming of Acids Formula Writing of Acids (c) 2006, Mark Rosengarten

145 Properties of Acids Acids react with metals above H2 on Table J to form H2(g) and a salt. Acids have a pH of less than 7. Dilute solutions of acids taste sour. Acids turn phenolphthalein CLEAR, litmus RED and bromthymol blue YELLOW. Acids neutralize bases. Acids are formed when acid anhydrides (NO2, SO2, CO2) react with water for form acids. This is how acid rain forms from auto and industrial emissions. (c) 2006, Mark Rosengarten

146 Naming of Acids (polyatomic ion) -ate +ic acid
Binary Acids (H+ and a nonmetal) hydro (nonmetal) -ide + ic acid HCl (aq) = hydrochloric acid Ternary Acids (H+ and a polyatomic ion) (polyatomic ion) -ate +ic acid HNO3 (aq) = nitric acid (polyatomic ion) -ide +ic acid HCN (aq) = cyanic acid (polyatomic ion) -ite +ous acid HNO2 (aq) = nitrous acid (c) 2006, Mark Rosengarten

147 Formula Writing of Acids
Acids formulas get written like any other. Write the H+1 first, then figure out what the negative ion is based on the name. Cancel out the charges to write the formula. Don’t forget the (aq) after it…it’s only an acid if it’s in water! Hydrosulfuric acid: H+1 and S-2 = H2S (aq) Carbonic acid: H+1 and CO3-2 = H2CO3 (aq) Chlorous acid: H+1 and ClO2-1 = HClO2 (aq) Hydrobromic acid: H+1 and Br-1 = HBr (aq) Hydronitric acid: Hypochlorous acid: Perchloric acid: (c) 2006, Mark Rosengarten

148 Formulas, Naming and Properties of Bases
Arrhenius Definition of Bases: ionic compounds that dissolve in water to produce OH- (hydroxide) as the only negatively charged ion in solution. NaOH (s)  Na+1 (aq) + OH-1 (aq) Properties of Bases Naming of Bases Formula Writing of Bases (c) 2006, Mark Rosengarten

149 Properties of Bases Bases react with fats to form soap and glycerol. This process is called saponification. Bases have a pH of more than 7. Dilute solutions of bases taste bitter. Bases turn phenolphthalein PINK, litmus BLUE and bromthymol blue BLUE. Bases neutralize acids. Bases are formed when alkali metals or alkaline earth metals react with water. The words “alkali” and “alkaline” mean “basic”, as opposed to “acidic”. (c) 2006, Mark Rosengarten

150 Naming of Bases Bases are named like any ionic compound, the name of the metal ion first (with a Roman numeral if necessary) followed by “hydroxide”. Fe(OH)2 (aq) = iron (II) hydroxide Fe(OH)3 (aq) = iron (III) hydroxide Al(OH)3 (aq) = aluminum hydroxide NH3 (aq) is the same thing as NH4OH: NH3 + H2O  NH4OH Also called ammonium hydroxide. (c) 2006, Mark Rosengarten

151 Formula Writing of Bases
Formula writing of bases is the same as for any ionic formula writing. The charges of the ions have to cancel out. Calcium hydroxide = Ca+2 and OH-1 = Ca(OH)2 (aq) Potassium hydroxide = K+1 and OH-1 = KOH (aq) Lead (II) hydroxide = Pb+2 and OH-1 = Pb(OH)2 (aq) Lead (IV) hydroxide = Pb+4 and OH-1 = Pb(OH)4 (aq) Lithium hydroxide = Copper (II) hydroxide = Magnesium hydroxide = (c) 2006, Mark Rosengarten

152 Neutralization H+1 + OH-1  HOH
Acid + Base  Water + Salt (double replacement) HCl (aq) + NaOH (aq)  HOH (l) + NaCl (aq) H2SO4 (aq) + KOH (aq)  2 HOH (l) + K2SO4 (aq) HBr (aq) + LiOH (aq)  H2CrO4 (aq) + NaOH (aq)  HNO3 (aq) + Ca(OH)2 (aq)  H3PO4 (aq) + Mg(OH)2 (aq)  (c) 2006, Mark Rosengarten

153 pH A change of 1 in pH is a tenfold increase in acid or base strength.
A pH of 4 is 10 times more acidic than a pH of 5. A pH of 12 is 100 times more basic than a pH of 10. (c) 2006, Mark Rosengarten

154 Indicators At a pH of 2: Methyl Orange = red Bromthymol Blue = yellow
Phenolphthalein = colorless Litmus = red Bromcresol Green = yellow Thymol Blue = yellow Methyl orange is red at a pH of 3.2 and below and yellow at a pH of 4.4 and higher. In between the two numbers, it is an intermediate color that is not listed on this table. (c) 2006, Mark Rosengarten

155 Alternate Theories Arrhenius Theory: acids and bases must be in aqueous solution. Alternate Theory: Not necessarily so! Acid: proton (H+1) donor…gives up H+1 in a reaction. Base: proton (H+1) acceptor…gains H+1 in a reaction. HNO3 + H2O  H3O+1 + NO3-1 Since HNO3 lost an H+1 during the reaction, it is an acid. Since H2O gained the H+1 that HNO3 lost, it is a base. (c) 2006, Mark Rosengarten

156 Oxidation and Reduction
1) Oxidation Numbers 2) Identifying OX, RD and SI Species 3) Agents 4) Writing Half-Reactions 5) Balancing Half-Reactions 6) Activity Series 7) Voltaic Cells 8) Electrolytic Cells 9) Electroplating (c) 2006, Mark Rosengarten

157 Oxidation Numbers Elements have no charge until they bond to other elements. Na0, Li0, H20. S0, N20, C600 The formula of a compound is such that the charges of the elements making up the compound all add up to zero. The symbol and charge of an element or polyatomic ion is called a SPECIES. Determine the charge of each species in the following compounds: NaCl KNO3 CuSO4 Fe2(CO3)3 (c) 2006, Mark Rosengarten

158 Identifying OX, RD, SI Species
Ca0 + 2 H+1Cl-1  Ca+2Cl-12 + H20 Oxidation = loss of electrons. The species becomes more positive in charge. For example, Ca0  Ca+2, so Ca0 is the species that is oxidized. Reduction = gain of electrons. The species becomes more negative in charge. For example, H+1  H0, so the H+1 is the species that is reduced. Spectator Ion = no change in charge. The species does not gain or lose any electrons. For example, Cl-1  Cl-1, so the Cl-1 is the spectator ion. (c) 2006, Mark Rosengarten

159 Agents Ca0 + 2 H+1Cl-1  Ca+2Cl-12 + H20
Since Ca0 is being oxidized and H+1 is being reduced, the electrons must be going from the Ca0 to the H+1. Since Ca0 would not lose electrons (be oxidized) if H+1 weren’t there to gain them, H+1 is the cause, or agent, of Ca0’s oxidation. H+1 is the oxidizing agent. Since H+1 would not gain electrons (be reduced) if Ca0 weren’t there to lose them, Ca0 is the cause, or agent, of H+1’s reduction. Ca0 is the reducing agent. (c) 2006, Mark Rosengarten

160 Writing Half-Reactions
Ca0 + 2 H+1Cl-1  Ca+2Cl-12 + H20 Oxidation: Ca0  Ca+2 + 2e- Reduction: 2H+1 + 2e-  H20 The two electrons lost by Ca0 are gained by the two H+1 (each H+1 picks up an electron). PRACTICE SOME! (c) 2006, Mark Rosengarten

161 Practice Half-Reactions
Don’t forget to determine the charge of each species first! 4 Li + O2  2 Li2O Oxidation Half-Reaction: Reduction Half-Reaction: Zn + Na2SO4  ZnSO4 + 2 Na (c) 2006, Mark Rosengarten

162 Balancing Half-Reactions
Ca0 + Fe+3  Ca+2 + Fe0 Ca’s charge changes by 2, so double the Fe. Fe’s charge changes by 3, so triple the Ca. 3 Ca0 + 2 Fe+3  3 Ca Fe0 Try these: __Na0 + __H+1  __Na+1 + __H20 (hint: balance the H and H2 first!) __Al0 + __Cu+2  __Al+3 + __Cu0 (c) 2006, Mark Rosengarten

163 Activity Series For metals, the higher up the chart the element is, the more likely it is to be oxidized. This is because metals like to lose electrons, and the more active a metallic element is, the more easily it can lose them. For nonmetals, the higher up the chart the element is, the more likely it is to be reduced. This is because nonmetals like to gain electrons, and the more active a nonmetallic element is, the more easily it can gain them. (c) 2006, Mark Rosengarten

164 Metal Activity 3 K0 + Fe+3Cl-13 REACTION Metallic elements start out with a charge of ZERO, so they can only be oxidized to form (+) ions. The higher of two metals MUST undergo oxidation in the reaction, or no reaction will happen. The reaction 3 K + FeCl3  3 KCl + Fe WILL happen, because K is being oxidized, and that is what Table J says should happen. The reaction Fe + 3 KCl  FeCl3 + 3 K will NOT happen. Fe0 + 3 K+1Cl-1 NO REACTION (c) 2006, Mark Rosengarten

165 Voltaic Cells Produce electrical current using a spontaneous redox reaction Used to make batteries! Materials needed: two beakers, piece of the oxidized metal (anode, - electrode), solution of the oxidized metal, piece of the reduced metal (cathode, + electrode), solution of the reduced metal, porous material (salt bridge), solution of a salt that does not contain either metal in the reaction, wire and a load to make use of the generated current! Use Reference Table J to determine the metals to use Higher = (-) anode Lower = (+) cathode (c) 2006, Mark Rosengarten

166 Making Voltaic Cells More Info!!! Create Your Own Cell!!!!
(c) 2006, Mark Rosengarten

167 How It Works Since Zn is listed above Cu, Zn0 will be oxidized when it reacts with Cu+2. The reaction: Zn + CuSO4  ZnSO4 + Cu The Zn0 anode loses 2 e-, which go up the wire and through the load. The Zn0 electrode gets smaller as the Zn0 becomes Zn+2 and dissolves into solution. The e- go into the Cu0, where they sit on the outside surface of the Cu0 cathode and wait for Cu+2 from the solution to come over so that the e- can jump on to the Cu+2 and reduce it to Cu0. The size of the Cu0 electrode increases. The negative ions in solution go over the salt bridge to the anode side to complete the circuit. (c) 2006, Mark Rosengarten

168 You Start At The Anode (c) 2006, Mark Rosengarten

169 Make Your Own Cell!!! (c) 2006, Mark Rosengarten

170 Electrolytic Cells Use electricity to force a nonspontaneous redox reaction to take place. Uses for Electrolytic Cells: Decomposition of Alkali Metal Compounds Decomposition of Water into Hydrogen and Oxygen Electroplating Differences between Voltaic and Electrolytic Cells: ANODE: Voltaic (-) Electrolytic (+) CATHODE: Voltaic (+) Electrolytic (-) Voltaic: 2 half-cells, a salt bridge and a load Electrolytic: 1 cell, no salt bridge, IS the load (c) 2006, Mark Rosengarten

171 Decomposing Alkali Metal Compounds
2 NaCl  2 Na + Cl2 The Na+1 is reduced at the (-) cathode, picking up an e- from the battery The Cl-1 is oxidized at the (+) anode, the e- being pulled off by the battery (DC) (c) 2006, Mark Rosengarten

172 Decomposing Water 2 H2O  2 H2 + O2
The H+ is reduced at the (-) cathode, yielding H2 (g), which is trapped in the tube. The O-2 is oxidized at the (+) anode, yielding O2 (g), which is trapped in the tube. (c) 2006, Mark Rosengarten

173 Electroplating The Ag0 is oxidized to Ag+1 when the (+) end of the battery strips its electrons off. The Ag+1 migrates through the solution towards the (-) charged cathode (ring), where it picks up an electron from the battery and forms Ag0, which coats on to the ring. (c) 2006, Mark Rosengarten

174 Organic Chemistry 1) Hydrocarbons 2) Substituted Hydrocarbons
3) Organic Families 4) Organic Reactions (c) 2006, Mark Rosengarten

175 Hydrocarbons Molecules made of Hydrogen and Carbon
Carbon forms four bonds, hydrogen forms one bond Hydrocarbons come in three different homologous series: Alkanes (single bond between C’s, saturated) Alkenes (1 double bond between 2 C’s, unsaturated) Alkynes (1 triple bond between 2 C’s, unsaturated) These are called aliphatic, or open-chain, hydrocarbons. Count the number of carbons and add the appropriate suffix! (c) 2006, Mark Rosengarten

176 Alkanes CH4 = methane C2H6 = ethane C3H8 = propane C4H10 = butane
C5H12 = pentane To find the number of hydrogens, double the number of carbons and add 2. (c) 2006, Mark Rosengarten

177 Methane Meth-: one carbon -ane: alkane
The simplest organic molecule, also known as natural gas! (c) 2006, Mark Rosengarten

178 Ethane Eth-: two carbons -ane: alkane (c) 2006, Mark Rosengarten

179 Propane Prop-: three carbons -ane: alkane
Also known as “cylinder gas”, usually stored under pressure and used for gas grills and stoves. It’s also very handy as a fuel for Bunsen burners! (c) 2006, Mark Rosengarten

180 Butane But-: four carbons -ane: alkane
Liquefies with moderate pressure, useful for gas lighters. You have probably lit your gas grill with a grill lighter fueled with butane! (c) 2006, Mark Rosengarten

181 Pentane Pent-: five carbons -ane: alkane Draw Hexane: Draw Heptane:
Your Turn!!! Draw Hexane: Draw Heptane: (c) 2006, Mark Rosengarten

182 Alkenes C2H4 = Ethene C3H6 = Propene C4H8 = Butene C5H10 = Pentene
To find the number of hydrogens, double the number of carbons. (c) 2006, Mark Rosengarten

183 Ethene Two carbons, double bonded. Notice how each carbon has four bonds? Two to the other carbon and two to hydrogen atoms. Also called “ethylene”, is used for the production of polyethylene, which is an extensively used plastic. Look for the “PE”, “HDPE” (#2 recycling) or “LDPE” (#4 recycling) on your plastic bags and containers! (c) 2006, Mark Rosengarten

184 Propene Three carbons, two of them double bonded. Notice how each carbon has four bonds? If you flipped this molecule so that the double bond was on the right side of the molecule instead of the left, it would still be the same molecule. This is true of all alkenes. Used to make polypropylene (PP, recycling #5), used for dishwasher safe containers and indoor/outdoor carpeting! (c) 2006, Mark Rosengarten

185 Butene This is 1-butene, because the double bond is between the 1st and 2nd carbon from the end. The number 1 represents the lowest numbered carbon the double bond is touching. This is 2-butene. The double bond is between the 2nd and 3rd carbon from the end. Always count from the end the double bond is closest to. ISOMERS: Molecules that share the same molecular formula, but have different structural formulas. (c) 2006, Mark Rosengarten

186 Pentene This is 1-pentene. The double bond is on the first carbon from the end. This is 2-pentene. The double bond is on the second carbon from the end. This is not another isomer of pentene. This is also 2-pentene, just that the double bond is closer to the right end. (c) 2006, Mark Rosengarten

187 Alkynes C2H2 = Ethyne C3H4 = Propyne C4H6 = Butyne C5H8 = Pentyne
To find the number of hydrogens, double the number of carbons and subtract 2. (c) 2006, Mark Rosengarten

188 Ethyne Now, try to draw propyne! Any isomers? Let’s see!
Also known as “acetylene”, used by miners by dripping water on CaC2 to light up mining helmets. The “carbide lamps” were attached to miner’s helmets by a clip and had a large reflective mirror that magnified the acetylene flame. Used for welding and cutting applications, as ethyne burns at temperatures over 3000oC! (c) 2006, Mark Rosengarten

189 Propyne This is propyne! Nope! No isomers.
OK, now draw butyne. If there are any isomers, draw them too. (c) 2006, Mark Rosengarten

190 Butyne Well, here’s 1-butyne! And here’s 2-butyne!
Is there a 3-butyne? Nope! That would be 1-butyne. With four carbons, the double bond can only be between the 1st and 2nd carbon, or between the 2nd and 3rd carbons. Now, try pentyne! (c) 2006, Mark Rosengarten

191 Pentyne 1-pentyne 2-pentyne
Now, draw all of the possible isomers for hexyne! (c) 2006, Mark Rosengarten

192 Substituted Hydrocarbons
Hydrocarbon chains can have three kinds of “dingly-danglies” attached to the chain. If the dingly-dangly is made of anything other than hydrogen and carbon, the molecule ceases to be a hydrocarbon and becomes another type of organic molecule. Alkyl groups Halide groups Other functional groups To name a hydrocarbon with an attached group, determine which carbon (use lowest possible number value) the group is attached to. Use di- for 2 groups, tri- for three. (c) 2006, Mark Rosengarten

193 Alkyl Groups (c) 2006, Mark Rosengarten

194 Halide Groups (c) 2006, Mark Rosengarten

195 Organic Families Each family has a functional group to identify it.
Alcohol (R-OH, hydroxyl group) Organic Acid (R-COOH, primary carboxyl group) Aldehyde (R-CHO, primary carbonyl group) Ketone (R1-CO-R2, secondary carbonyl group) Ether (R1-O-R2) Ester (R1-COO-R2, carboxyl group in the middle) Amine (R-NH2, amine group) Amide (R-CONH2, amide group) These molecules are alkanes with functional groups attached. The name is based on the alkane name. (c) 2006, Mark Rosengarten

196 Alcohol On to DI and TRIHYDROXY ALCOHOLS (c) 2006, Mark Rosengarten

197 Di and Tri- hydroxy Alcohols
(c) 2006, Mark Rosengarten

198 Positioning of Functional Group
PRIMARY (1o): the functional group is bonded to a carbon that is on the end of the chain. SECONDARY (2o): The functional group is bonded to a carbon in the middle of the chain. TERTIARY (3o): The functional group is bonded to a carbon that is itself directly bonded to three other carbons. (c) 2006, Mark Rosengarten

199 Organic Acid These are weak acids. The H on the right side is the one that ionized in water to form H3O+. The -COOH (carboxyl) functional group is always on a PRIMARY carbon. Can be formed from the oxidation of primary alcohols using a KMnO4 catalyst. (c) 2006, Mark Rosengarten

200 Aldehyde Aldehydes have the CO (carbonyl) groups ALWAYS on a PRIMARY carbon. This is the only structural difference between aldehydes and ketones. Formed by the oxidation of primary alcohols with a catalyst. Propanal is formed from the oxidation of 1-propanol using pyridinium chlorochromate (PCC) catalyst.* (c) 2006, Mark Rosengarten

201 Ketone Ketones have the CO (carbonyl) groups ALWAYS on a SECONDARY carbon. This is the only structural difference between ketones and aldehydes. Can be formed from the dehydration of secondary alcohols with a catalyst. Propanone is formed from the oxidation of 2-propanol using KMnO4 or PCC catalyst.* (c) 2006, Mark Rosengarten

202 Ether Ethers are made of two alkyl groups surrounding one oxygen atom. The ether is named for the alkyl groups on “ether” side of the oxygen. If a three-carbon alkyl group and a four-carbon alkyl group are on either side, the name would be propyl butyl ether. Made with an etherfication reaction. (c) 2006, Mark Rosengarten

203 Ester Esters are named for the alcohol and organic acid that reacted by esterification to form the ester. If the alcohol was 1-propanol and the acid was hexanoic acid, the name of the ester would be propyl hexanoate. Esters contain a COO (carboxyl) group in the middle of the molecule, which differentiates them from organic acids. (c) 2006, Mark Rosengarten

204 Amine Component of amino acids, and therefore proteins, RNA and DNA…life itself! - Essentially ammonia (NH3) with the hydrogens replaced by one or more hydrocarbon chains, hence the name “amine”! (c) 2006, Mark Rosengarten

205 Amide Synthetic Polyamides: nylon, kevlar Natural Polyamide: silk!
For more information on polymers, go here. (c) 2006, Mark Rosengarten

206 Organic Reactions Combustion Fermentation Substitution Addition
Dehydration Synthesis Etherification Esterification Saponification Polymerization (c) 2006, Mark Rosengarten

207 Combustion Happens when an organic molecule reacts with oxygen gas to form carbon dioxide and water vapor. Also known as “burning”. (c) 2006, Mark Rosengarten

208 Fermentation Process of making ethanol by having yeast digest simple sugars anaerobically. CO2 is a byproduct of this reaction. The ethanol produced is toxic and it kills the yeast when the percent by volume of ethanol gets to 14%. (c) 2006, Mark Rosengarten

209 Substitution Alkane + Halogen  Alkyl Halide + Hydrogen Halide
The halogen atoms substitute for any of the hydrogen atoms in the alkane. This happens one atom at a time. The halide generally replaces an H on the end of the molecule. C2H6 + Cl2  C2H5Cl + HCl The second Cl can then substitute for another H: C2H5Cl + HCl  C2H4Cl2 + H2 (c) 2006, Mark Rosengarten

210 Addition Alkene + Halogen  Alkyl Halide
The double bond is broken, and the halogen adds at either side of where the double bond was. One isomer possible. (c) 2006, Mark Rosengarten

211 Etherification* Alcohol + Alcohol  Ether + Water
A dehydrating agent (H2SO4) removes H from one alcohol’s OH and removes the OH from the other. The two molecules join where there H and OH were removed. Note: dimethyl ether and diethyl ether are also produced from this reaction, but can be separated out. (c) 2006, Mark Rosengarten

212 Esterification Organic Acid + Alcohol  Ester + Water
A dehydrating agent (H2SO4) removes H from the organic acid and removes the OH from the alcohol. The two molecules join where there H and OH were removed. (c) 2006, Mark Rosengarten

213 Saponification The process of making soap from glycerol esters (fats).
Glycerol ester + 3 NaOH  soap + glycerol Glyceryl stearate + 3 NaOH  sodium stearate + glycerol The sodium stearate is the soap! It emulsifies grease…surrounds globules with its nonpolar ends, creating micelles with - charge that water can then wash away. Hard water replaces Na+ with Ca+2 and/or other low solubility ions, which forms a precipitate called “soap scum”. Water softeners remove these hardening ions from your tap water, allowing the soap to dissolve normally. (c) 2006, Mark Rosengarten

214 Polymerization A polymer is a very long-chain molecule made up of many monomers (unit molecules) joined together. The polymer is named for the monomer that made it. Polystyrene is made of styrene monomer Polybutadiene is made of butadiene monomer Addition Polymers Condensation Polymers Rubber (c) 2006, Mark Rosengarten

215 Addition Polymers Joining monomers together by breaking double bonds
Polyvinyl chloride (PVC): vinyl siding, PVC pipes, etc. Vinyl chloride polyvinyl chloride n C2H3Cl  (-C2H3Cl-)-n Polytetrafluoroethene (PTFE, teflon): TFE PTFE n C2F  (-C2F4-)-n (c) 2006, Mark Rosengarten

216 Condensation Polymers
Condensation polymerization is just dehydration synthesis, except instead of making one molecule of ether or ester, you make a monster molecule of polyether or polyester. (c) 2006, Mark Rosengarten

217 Rubber The process of toughing rubber by cross-linking the polymer strands with sulfur is called... (c) 2006, Mark Rosengarten

218 VULCANIZATION!!! (c) 2006, Mark Rosengarten

219 THE END (c) 2006, Mark Rosengarten

Download ppt "Mark Rosengarten’s Amazing Chemistry Powerpoint Presentation!"

Similar presentations

Ads by Google