FINAL CHEM 151 Review
FINAL CHEM 151 Review Chem 151
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This 71 page Study Guide was uploaded by Rodriguez Notetaker on Monday January 11, 2016. The Study Guide belongs to Chem 151 at Drexel University taught by Daniel A. Kleier in Summer 2015. Since its upload, it has received 36 views. For similar materials see General Chemistry in Chemistry at Drexel University.
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CHEM Final Review Unit 1: Fireworks: Reaction between magnesium & oxygen produces o Heat and light o Magnesium Oxide (MgO) o Chemical Equation: Mg + O (reactants) MgO (Products) Balanced: Mg + O → 2 MgO Balanced Eqn with physical state information 2 Mg(s) + O 2g)→ 2 MgO(s) Balanced Eqn including heat produced 2 Mg(s) + O 2g)→ 2 MgO(s) + Heat Quantifying the heat produced when Mg burns: ** The Basic of Matters: Atoms All matter is made of atoms. Atoms cannot be broken down by chemical means. There are different types of atoms. Each type corresponds to an element. An atom is the smallest unit of matter that when combined with other atoms of its type retains the properties of the element. The periodic table is a systematic arrangement of the elements that places elements with similar properties in families. Characteristics of Subatomic Particles: Name Symb Mass Charg ol relative e to the proton Electro e , or 0.0005 1− 0 n − ¿ 4 ¿ Proton p , or 1.0000 1+ 1 0 1p Neutro n or 1.0012 0 n 1 0 0 Electronic structure of the Hydrogen Atom: Rutherford-Bohr Model introduced in 1913 Nucleus consists of one proton (p ) with a mass of 1 amu Charge on nucleus is +1e o e is the magnitude of the charge on the electron One relatively light electron (e ) moves around the nucleus in a circular orbit Larger orbits are possible, but electron prefers the inner orbit Electronic Shell Structure of the Helium (He) Atom: Charge on nucleus is 2x that of an H nucleus Mass of He nucleus is ~4x that of an H nucleus 1 + Nucleus consists of 2 protons (p ) and 2 neutrons (n) 2 electrons (e ) circulate around the nucleus in the same orbit. Electronic Structure of the Lithium (Li) Atom: Nucleus consists of 3 protons (p ) and 3 or 4 neutrons (n) 2 electrons (e ) circulate around the nucleus in an inner orbit − 1 electron (e ) circulates in an outer orbit or valence shell Shorthand shell notation o Generally: [n1, n2, n3,..] o Specifically for Li: [2,1] Shell Structures for Elements in First Two Periods: First period= two e- Second period= Starting − with lithium (Li) one e is added to the second shell for each stepwise increase in the atomic number until neon (Ne) is reached Electronic Structure of the Neon (Ne) Atom: Nucleus consists of 10 protons (p ) and 10 neutrons (n) − 2 electrons (e ) circulate around the nucleus in an inner shell 8 electrons (e ) circulate in an outer orbit or valence shell Shorthand shell notation o Generally: [n ,1n ,2n .3,] o Specifically for Ne: [2,8] Shell Structures for Elements in Third Period: Pattern repeats Third Period − o Starting with sodium (Na) one e is added to the third shell for each stepwise increase in the atomic number until Argon (Ar) is reached Shell Structures for Elements in Fourth Period: Pattern repeats with a diversion Fourth Period 2 − o Starting with potassium (K), one e is added to the fourth shell for each stepwise increase in the atomic number until Krypton (Kr) is reached. o Ten e added after Ca are diverted into the third shell. Shell Structure for Ions: Metals commonly lose negatively charged electrons to form positively charge ions known as cations o Calcium is a metal that commonly loses 2 e to form a Ca 2+ion Shell structure for Ca: [2,8,8,2] 2+ Shell structure for Ca : [2,8,8] Nonmetals commonly gain negatively charged electrons to form negatively charged ions known as anions o Sulfur is a nonmetal that commonly gains 2 e to form a S 2−ion Shell structure of S: [2,8,6] 2− Shell structure of S : [2,8,8] The magic of the number 8: o Ar atoms, Ca 2+ and S 2−ions all have the same shell structure: [2,8,8] 3 ******* Writing a large quantity in scientific notation: Example: Write 56500 (or 56500.) in scientific notation Count the number of times that the decimal point bounces left while bringing the number into the range of 1 to 10 Each time that the point bounces left increase the power of 10 by one. Writing a small quantity in scientific notation: Example: Write 0.00078 in scientific notation Count the number of times that the decimal point bounces right while bringing the number into the range of 1 to 10 Each time that the point bounces right decrease the power of 10 by one. SI unit system: Systeme Internationale (SI) is standard unit system for science. Base unit of distance is the meter (m). The meter is a useful unit for measuring the length of a soccer field The meter is inconvenient for measuring distance between cities, thickness of a coins, the length of bonds or the diameter of atoms SI system is based on powers of ten for easy use (metric system). Prefixes adjust the scale to suit the object: 6 1 megameter = 1 Mm = 1,000,000 m = 10 m 1 kilometer = 1 km = 1,000 m = 10 3 m 1 centimeter = 1 cm = 0.01 m = 10 -2m 1 millimeter = 1 mm = 0.001 m = 10-3m -9 1 nanometer = 1 nm = 0.000 000 001 m = 10 -12 1 picometer = 1 pm = 0.000 000 000 001 m = 10 m These relationships can be used to convert a quantity from one unit to a more conveniently sized unit Using Conversion Factors to Express an atomic radius in More Convenient Units: 1 pm = 1.0 x 10 −12m = 10 −12m 10 12pm = 1 m Divide one number by the other and we have a conversion factor. Radius of Krypton atom is 8.8 x 10 -1m The meter is an inconvenient unit to use for the Krypton atom Therefore, convert radius to a more conveniently sized unit such as the pm Size of atoms: Atomic radii are measured in picometers: −12 1 pm = 1.0 x 10 m Periodic behavior of atomic radii Atoms shrink as we move from left to right within a period rLi rNe Atoms expand as we move from top to bottom of a group rLi rK Categories of Matter: Pure Substances: 4 o A pure substance is one that contains only one element or compound. o Example: Magnesium Mg Magnesium sulfate MgSO ∙4H O 2 Mixtures: o They are impure o Homogeneous mixture is uniform on a microscopic level (a solution). Copper sulfate solution Lemonade Salt water Maple syrup Tea air o Heterogeneous mixture is non- uniform on a microscopic level Milk of magnesia (MgO suspended in water). Mint chocolate chip ice cream Fruit salad Vetegable soup Smoke Tea with ice and a lemon slice Let’s Get Physical: Components of a mixture can be separated (purified) by physical means. Smoke is a heterogeneous mixture of small particles suspended in air o Particles can be removed by filtering. Sea water is a homogeneous solution of salt in water. o Water can be removed by evaporation. o Evaporation (physical) converts liquid water into gas. H 2(ℓ) → H 2(g) o Salts are left behind Physical and Chemical Changes: Physical changes alter a substance without any changes to the chemical composition. o Boiling water (physical change) o H2O(ℓ) → H 2(g) Chemical change involves breaking down a substance into other substances. o Chemical bonds are broken and reformed. o Electrolysis (chemical change) converts water into hydrogen and oxygen – new substances: o 2 H O(ℓ) → 2 H (g) + O (g) 2 2 2 Unit 2: Electronic Structure of Atoms and bonding in Molecules: Coinage Metals: Copper, Silver and Gold Coinage metals are a family of transition metals 5 Superstar Awards o Copper (Cu) for electrical conductivity o Silver (Ag) salts for photography o Gold (Au) for malleability and luster Expanding the Periodic Table: Transition metals: Shell occupancy limits 2 o 2, 8, 18, 32, ..., 2n Periodic table doesn’t expand to 18 columns until the 4 thperiod because the 3 rd shell, which can accommodate 18 electrons, has only accepted 8 when the 4 th shell begins to fill. Periodic Table Color Coded by Metallic Character: Metallic Bonding & Electrical Conductivity: Sodium has a single valence electron Electron dot structure for sodium: Na • Sodium has a low ionization potential so that it readily loses its valence electron Na → Na + e − In a lattice of Na atoms the lost electrons wander aimlessly in a sea of negative charge. + This sea of electrons glues the oppositely charged Na ions together. 6 The lost electrons will move in one direction under the influence of an applied electric field. Conductivity is a typical characteristic of metals and reaches its zenith in copper (Cu) Color coding by named groups: Noble Wannabes: The noble gases have stable electron configurations. o He, Ne[2,8], Ar[2,8,8], Kr[2,8,18,8], Xe[2,8,18,18,8] The other elements want to be noble too! Atoms gain, lose or share electrons to achieve nobility. o Fluorine[2,7] gains 1 electron to be just like Ne[2,8] o Oxygen [2,6] gains 2 electrons to be just like Ne[2,8] o Magnesium[2,8,2] loses 2 electrons to be just like Ne[2,8] Atoms gain or lose electrons as a result of stealing or sharing. In the process bonds are formed. Only valence electrons are involved in bonding. Electron Dot Structures: Counting valence electrons: Start at left of the period containing the element and count the number of “main group” elements in the period to reach the element of interest o C (group IVA): count 4 elements from left of table to reach C C has 4 valence electrons, C[2,4] o N (group VA) count 5 elements from left of table to reach N N has 5 valence electrons and a helium core, N[2,5] o O (group VIA) count 6 elements from left of table to reach O O has 6 valence electrons and a helium core, O[2,6] o F (group VIIA) count 7 elements from left of table to reach F F has 7 valence electrons and a helium core, F[2,7] Ionic Bonding: First Step is Ionization of the Metal to Form a Cation: Electron Dot Structure for Sodium o Na• Sodium has a low ionization potential (IP) so that it readily produces a cation by donating its valence electron t+ nei−hboring nonmetals. o Na• + 497 kJ/mol → Na + e IPs exhibit periodic behavior with increasing atomic number. 7 o IP increases from left to right in a period o IP decreases from top to bottom in a group Ionic Bonding: Second Step is Electron Capture by the Nonmetal to Form an Anion: Electron Dot Structure for Chlorine o Chlorine has a high electron affinity and readily produces an anion when it accepts a valence electron from a neighboring metal + e → + 349 kJ/mol Electron affinities exhibit periodic behavior with increasing atomic number. Ionic Bonding: Third Step is Attraction of Opposites: Oppositely charged sodium and chloride ions attract + − Na (g) + Cl (g) → NaCl(s) + 787 kJ/mol Net result is the exothermic formation of solid sodium chloride Na(g) + Cl(g) → NaCl(s) + 639 kJ/mol Crystal la+tice ha− cubic symmetry o Na and Cl ions are at alternate corners of a cube o Each Na cation is surrounded by 6 Cl anions at the corners of an octahedron. o Each Cl anion is surrounded by 6 Na cations at the corners of an octahedron. Extra notes: An exothermic process releases heat, and causes the temperature of the immediate surroundings to rise An endothermic process absorbs heat and cools the surroundings. A reaction releases KJ The energy level of the reactants is lower than that of the products The decomposition of phosphorus penta chloride Ionization energy is the energy needed to remove the first electron from a neutral atom. When we strip away the first electron, the result is a positive ion. The ionization energy is higher for atoms whose valence electrons are closer to the nucleus, and it decreases as the outer electrons get father from the nucleus. Number of protons in an atom (also called atomic number) does not change when an atom loses or gains electrons to form an ion. A neutral atom contains the same # of protons and electrons o Ion charge= # of protons- # of electrons Metallic bonding o Why conduct electricity The electrons in metals are fluid, and can move easily under the influence of an electric field Covalent Bonding: Covalent bond between two hydrogen atoms: Two H atoms each need an extra electron in their outer shell in order to satisfy the duet rule This can be achieved by sharing a pair The shared pair represents a covalent bond Shared pair is often represented by a line Covalent bond between two fluorine atoms: Two F atoms each need an extra electron in the valence shell in order to satisfy the octet rule. 8 This need can be satisfied by sharing a pair of electrons. The shared pair represents a covalent bond Abbreviated notations for fluorine molecule Covalent bond between Hydrogen and Fluorine: An H and an F atom each need an extra electron in the outer shell to satisfy the duet or octet rule. This can be achieved by sharing a pair. The shared pair represents a covalent bond. Abbreviated notations for Hydrogen Fluoride molecule Electronegativity: A measure of an element’s attractiveness to electrons: Electronegativity (EN) measures the tendency of an atom to attract electrons: o Metals (e.g., K) have low EN. o Hydrogen and Carbon have intermediate EN. o Several nonmetals (e.g., O and F) have exceptionally high EN. Electronegativity difference determines bond polarity: If ΔEN > 2.0, bond is ionic. o EN(F) – EN(K) = 4.19 – 0.73 = 3.46 > 2.0; Therefore, KF is ionic If 0< ΔEN < 2.0, bond is polar covalent . o EN(F) – EN(H) = 4.19 – 2.3 = 1.9 < 2.0; Therefore, HF is polar covalent If ΔEN ~ 0, bond is nonpolar covalent. o EN(F) – EN(F) = 4.19 − 4.19 = 0.0; Therefore F is 2onpolar covalent As bond order increases, bond length decreases. As bond order increases, bond strength increases. Covalent bonding in water (H O):2 The O atom needs 2 extra electrons in the outer shell to satisfy the octet rule Each H needs 1 extra electron to satisfy the duet rule Oxygen satisfies the octet rule, and each hydrogen satisfies the duet rule More abstract representations of water Accounting rules for electron dot structures: Rules were formulated in 1916 by G. N. Lewis o Structures are sometimes called Lewis structures Sample applications to o methane, CH 4 o ammonia, NH 3 o formaldehyde, CH O 2 o hydrogen cyanide, HCN Step 1: Count the Number of Valence Electrons in the Molecule: 9 Example Application: CH 4 Carbon atom contributes 4 valence electrons Each of 4 hydrogen atoms contributes 1 valence electron Methane (CH 4 has a total of 4 + 4 x 1 = 8 valence electrons Step 2: Determine the arrangement of atoms: The first non-hydrogen element in molecular formula is usually in the middle of the arrangement. Hydrogens are always terminal. For CH4, C is in the middle and the H’s are at the periphery. H Step 3: Draw the Molecular Scaffold H H Attach each terminal atom to the central atom with a line or C bond. Each bond represents a shared pair of electrons. Therefore, 8 electrons are being used for the bonds. H Total number of valence electrons was calculated to be 8 in the first step. There are no more electrons available. We are done! Determine Electron Dot Structure of ammonia, NH 3 Step 1: Calculate # of Valence Electrons o 5 + 3(1) = 8 Step 2: Arrange Atoms o N in the center surrounded by terminal Hs Step 3: Draw Scaffold o Each bond uses 2 electrons o 8 − 6 = 2 electrons remain Step 4: Satisfy Duet or Octet Rule for Terminal Atoms o All terminal Hs satisfy the duet rule Step 5: Satisfy Octet Rule for Central Atom o Add remaining pair of electrons to N We are done! Electron Dot Structures for Methane, Ammonia, and Water: In each of the molecules the central atom is surrounded by 4 pairs of electrons The central atoms satisfy the octet rule The hydrogens satisfy the duet rule Determine the Electron Dot Structure for formaldehyde, CH O: 2 Count the number of valence electrons o Each H contributes one electron o Oxygen contributes 6 electrons o Carbon contributes 4 electrons o Total number of valence electrons = 2(1) + 6 + 4 = 12 Determine arrangement of atoms o Carbon would be central element. It is the first non-hydrogen atom in the formula Draw scaffold by attaching terminal atoms to the central carbon atom 10 o Number of remaining electrons = 12 – 6 = 6 Place the remaining 6 electrons on oxygen atom to complete its H H C octet If there are electrons remaining, place them on the central carbon. o No action. All 12 valence electrons are accounted for. O Central carbon still does not satisfy octet rule. Therefore, move an unshared pair from the adjacent oxygen atom to form a bond with the central carbon atom. Determine the Electron Dot Structure for Hydrogen Cyanide, HCN: Count the number of valence electrons o H contributes one electron o Nitrogen contributes 5 electrons o Carbon contributes 4 electrons o Total number of valence electrons = 1 + 5 + 4 = 10 Determine arrangement of atoms o Carbon would be central element. It is the first non-hydrogen atom in the formula Attach each terminal atom to the central atom with a pair of electrons. o Number of valence electrons remaining = 10 – 4 = 6 Place the remaining electrons 6 electrons on N atom to satisfy its octet If there are electrons remaining, place them on the central carbon. o No action. All 10 valence electrons are accounted for. Central carbon still does not satisfy octet rule. Therefore, move two unshared pairs from the adjacent nitrogen atom to form two more bonds with the central carbon atom. Steps for writing Electron Dot Structures from Chemical Formulas: Count the number of valence electrons in the molecule Place the first non-hydrogen atom(s) in the chemical formula in the center of an arrangement. Other atoms are placed symmetrically around the central atom(s). Hydrogen atoms are always terminal. Draw scaffold by connecting terminal atoms to the central atom(s) and central atoms to one another with single bonds. Each connection uses two electrons. Count the remaining electrons. Distribute the remaining valence electrons on terminal atoms to satisfy the octet rule. If there are electrons remaining place them on the central atom(s) If a central atom lacks enough electrons to satisfy the octet rule, move available unshared pairs from an adjacent atom to form a bond with the electron deficient central atom. Unit 3: Chemical Accounting: The chemist’s dozen: The mole is a convenient unit for expressing large quantities of atoms or molecules Avogadro’s number of anything is a mole of that thing Avogadro’s Number, N A o NA= 602,200,000,000,000,000,000,000. o NA= 6.022 x 10 23 Subscripts and Moles: 11 The subscripts in a formula signify o The relative number of atoms in the formula. o The moles of each element in 1 mole of compound. Sample Problem: A. How many moles of O are in 0.150 mole aspirin C H 9 8 4 0.150 mole C H9O 8 4 4 moles O 1 mole C H9O8 4 = 0.600 mole O B. How many O atoms are in 0.150 mole aspirin C H O ? 9 8 4 0.150 mole C H9O 8 4 moles O x 6.02 x 10 23O atoms 1 mole C H O 1 mole O 23 9 8 4 = 3.61 x 10 O atoms Using Avogadro’s Number: Avogadro’s number is used to convert moles of an element to number of atoms. How many Cu atoms are in 0.50 mole Cu? 0.50 mole Cu x 6.02 x 10 23Cu atoms 1 mole Cu = 3.0 x 10 23Cu atoms Avogadro’s number is used to convert number of molecules of a substance to moles. 24 How many moles of CO are 2n 2.50 x 10 molecules CO ? 2 2.50 x 10 24molecules CO x 2 1 mole CO 2 23 6.02 x 10 molecules CO 2 = 4.15 moles CO 2 Sample calculation: The number of atoms in 2.0 moles Al is 2.0 moles Al x 6.02 x 10 23Al atoms 1 mole Al = 1.2 x 10 24Al atoms The number of moles of S in 1.8 x 10 24atoms S is 24 1.8 x 10 S atoms x 1 mole S 6.02 x 10 23 S atoms = 3.0 moles S atoms The Base Units in the SI System: Base units for important physical quantities in chemistry: Mass of a Mole of Atoms or Molecules: 12 Every element has an average atomic mass which appears below the element symbol in the periodic table. The mass of one mole of chlorine atoms is the atomic mass expressed in grams (g) o The molar mass of Cl is 35.453 g/mol o 6.022 x 10 23atoms of Cl weighs 35.453 g The mass of a mole of molecules is the sum of the molar masses of the constituent atoms. Molar mass of nitrogen trichloride, NCl 3 = (1 mol N)(14.01 g/mol) + (3 mol Cl)(35.45 g/mol) = 120.36 g o 1 mole of NCl m3lecules weighs 120.36 g o 6.022 x 10 23molecules of NCl w3ighs 120.36 g Sample Calculation: A. K 2 2 moles K (39.1 g/mole) + 1 mole O (16.0 g/mole) 78.2 g + 16.0 g = 94.2 g B. Al(OH) 3 1 mole Al (27.0 g/mole) + 3 moles O (16.0 g/mole) + 3 moles H (1.0 g/mole) 27.0 g + 48.0 g + 3.0 g = 78.0 g Calculate the number moles of oxygen in 100.0 g of CO 2 Molar mass of CO : 2 (1 mol of C )(12.01 g/mol) + (2 moles of O)(16.00 g/mol) = 12.01 g + 32.00 g = 44.01 g Moles of Oxygen in 100 g of CO : 2 (100.0 g CO )21 mol CO /42.01 g)(2 moles O/mole CO ) = 42544 moles O Calculate the mass of 5.00 moles of MgO Molar mass of MgO (1 mol of Mg/mol MgO)(24.31 g/mol of Mg) + (1 mol of O/mol MgO)(16.0 g/mol of O) = 40.31 g/mol MgO (5.00 mol MgO)(40.31 g/mol MgO) = 201.6 g/mol = 202 g/mol The Mole as a Convenient Unit for Compounds and Elements” One mole quantities of sugar, salt, carbon and copper sulfate are shown. One mole of carbon contains N atoAs of carbon, one mole of salt contains N of A sodium atoms and N atoAs of chlorine, ... 23. o Avogadro’s number = N = 6.A22 x 10 Isotopes of Carbon: Atomic number = number of protons (p ) + Atomic number determines the element type. Mass number (Z) is the sum of protons plus neutrons (n) Z = Number of neutrons + number of protons Isotopes of an element have the same number of protons, but different numbers of neutrons. Isotopes of an element have the same atomic number, but different mass number. The most common isotope of carbon has 6 protons & 6 neutrons Since Z = 6 + 6 = 12, the most common isotope is known as C-12 The isotope of carbon with 8 neutrons is radioactive! Since Z = 6 + 8 = 14, the radioactive isotope is known as C-14. 13 Atomic Mass: Why Neutrons Matter: Atoms of the same element contain the same number of protons. Isotopes of a given element contain the same number of protons but different number of neutrons. Isotopes have the same chemical properties. The most abundant form of carbon contains 6 protons and 6 neutrons and has an atomic mass of 12.000 units o Symbolized as Carbon-12 or 1C or 1C and 1C are less abundant isotopes of 1C o contains 6 protons and 7 neutrons o contains 6 protons and 8 neutrons Atomic mass appearing in periodic table is an average of the isotope mass numbers o Atomic mass of C is 12.011. Isotope Ratios: A Window Into History: Ratio of carbon-14 to carbon-12 in living organisms is constant. 1C/ C ~ 10 -12 14 Once dead, C decreases due to radioactive decay The smaller the relative amount of C-14 in an artifact, the older it is. Half of 1C decays in 5730 yrs. Radioactive decay (DR) rate can be determined using a Geiger counter The decay rate decreases exponentially with age. Age: Decay Rate DR (counts/min) 0 8,000 = DR 5,730 4,000 11,460 2,000 17,190 1,000 Sample Problem: Age of Ancient Ancestor’s War Club A sample of a wooden club found in a cave has a C-14 radioactive decay rate of 200 cpm. A sample of wood of the same mass from a living tree of the same type has a counting rate of 12,800 cpm. What is the age of the club? The war club’s decay rate is expected to be halved after every half-life (5730 yrs), so that after 6 half-lives the decay rate is expected to have fallen from 12,800 to 200 cpm: 12,800 cpm → 6,400 → 3,200 → 1,600 → 800 → 400 → 200 cpm (6 half-lives)(5730 yrs/half-life) ≈ 34,400 yrs. Nuclear Symbols: For an isotope, the n+clear symbol gives the o No. Protons (p ) = Atomic Number Mass Number o No. Neutrons (n) = Mass Number – Atomic Number o No. Electrons (e ) = No. Protons Element An atom of zinc has a mass number of 65. Symbol How many protons are in this zinc atom? 30 (atomic number 30) Atomic Number How many neutrons are in the zinc atom? 35 (65 – 30 = 35) What is the mass number of a zinc atom that has 37 neutrons? 67 (30 + 37 = 67) Introduction to Organic Chemistry: 14 Carbon and Hydrocarbons: The uniqueness carbon as family head Single, double, and triple bonds between carbon atoms Reactions of alkenes and alkynes Delocalized bonding in aromatic hydrocarbons Bond strength, bond length Heats of reaction of hydrocarbons Carbon is Smaller than Expected: Carbon is surprisingly smaller than would be expected based on the trend line for the heavier elements in group IVA. Generally, the 2drow representatives of each main group are smaller than expected Carbon is special: Small size of carbon enables close approach of other atoms. Bonds to carbon are stronger than expected. Four bonds to carbon in methane give central carbon an octet The 4 bonds point to the corners of a tetrahedron centered on the C atom. o Bold bond points above plane of drawing o Dashed bond points below plane of drawing Structure of Methane, CH 4 The Simplest Hydrocarbon: o Carbon forms four bonds in a tetrahedral arrangement Wedge bond indicates that bond points up Dashed bond indicates that bond points down H H o C-H bond distances are about 109 pm o H-C-H bond angle is 109.5 o H C C H Carbon forms single bonds to other carbons in alkanes: H H o The molecule ethane has two carbon atoms and six hydrogen atoms (C 2 )6 o Each carbon atom is bonded to three hydrogen atoms in each of two CH 3 groups. o A single bond is formed between the two carbons o Both carbons satisfy the octet rule after all bonds are formed o Ethane is a 3D-molecule as indicated by wedge structure Carbon forms double bonds to other carbons in alkenes: o The molecule ethene has two carbon atoms and four H hydrogen atoms (C H2)4 H o Each carbon atom is bonded to two hydrogens in each of two CH 2roups. C C o A double bond also forms between the two carbons H H o Both carbons satisfy the octet rule o Both carbons form four bonds o Ethene, the simplest alkene, is a flat and rigid molecule Carbon forms triple bonds to other carbons in alkynes: o The molecule ethyne has two carbon atoms and two hydrogen atoms (C H2) 2 o Each carbon atom is bonded to one hydrogen in each of two CH groups. o A triple bond forms between the two carbons 15 o Both carbons satisfy the octet rule o Both carbons form four bonds o Ethyne, the simplest alkyne, is a linear, and rigid molecule Carbon can form ring structures with double and single bonds: o The molecule benzene has six carbon atoms and six hydrogen atoms (C 6 6 o Each carbon is bonded to one hydrogen atom o Each carbon forms a double bond with one flanking carbon and a single bond with another flanking carbon o All carbons satisfy the octet rule o All carbons form four bonds o Benzene is a flat and rigid molecule Resonance and benzene: o Benzene contains alternating single and double bonds o Two equivalent ways to draw structure o Actual structure is average of two resonance hybrids Symbolized by double headed arrow, or as circle in ring o Carbon-carbon bonds have order 1.5 in benzene Summary of carbon bond types: Bonding to 4 atoms (all single bonds) Bonding to 3 atoms (2 single + 1 double) Bonding to 2 atoms (1 single + 1 triple or 2 double) Multiple bonds signal reactivity: Ethene is the simplest alkene Because of the double bond, ethene is classified as an unsaturated hydrocarbon The double bond can be converted to a single bond by hydrogenation. Addition of 2 to ethene results in the saturated hydrocarbon ethane which is a simple alkane. Reaction involves unpairing and repairing of electrons H H H H H H + H H H H C C C C C C H H H H Ethene H H Ethane Hydrogenation of Fatty Acids: Hydrogenation is an important industrial process. Oleic acid is a liquid used for oiling wood. Hydrogenation produces the saturated fatty acid known as stearic acid which is a solid. 16 Stearic acid is used for making candles, suppositories, pill coatings, etc. Oleic Acid Stearic Acid Bond lengths and strengths: Variation with bond order & atom size: As bond order increases, bond length decreases As bond order increases, bond strength increases Replacement of C with H shortens and strengthens bonds Replacement of C with Si lengthens and weakens bonds Estimation of Heat of a Reaction from Bond Energies: CH + H C=CH → CH CH CH 4 2 2 3 2 3 Energy Invested in breaking bonds o E invest(612 + 412) kJ/mol = 1024 kJ/mol Energy returned in making bonds o E return(2(348) + 412) kJ/mol = 1108 Bond Mean Bond Energy kJ/mol Type (kJ/mol) Heat of Rxn o ΔH rxn= Einvest E return− 84 kJ/mol o Negative value C ─ C 348 of ΔH rxnmeans that heat is generated by the C = C 612 reaction o The reaction is C ─ H 412 said to be exothermic Unit 4: Molecular Geometry: Treating Electron Pairs Like Balloons: 17 An atom A in a generic molecule, AX , ismsurrounded by pairs of electrons, which repel each other much like fat balloons tied tightly to a common point. The electron dot structure of a molecule displays these bonding and nonbonding pairs in a 2D representation. .. H ─ O ─ H ∙∙ The repulsion between electron pairs tied to the central atom A determines the 3D geometry of the AX m molecule. This theory is known as the Valence Shell Electron Pair Repulsion (VSEPR) theory For purposes of predicting molecular geometry, a multiple bond is treated like a single oversized electron pair. VSEPR Theory for: ∙∙ H− OH ∙∙ Water has 4 pairs of valence electrons o 2 bonding pairs o 2 nonbonding pairs Electron pairs are attracted to the oxygen, but repelled by one another o Best compromise is for electrons to reside at the corners of a tetrahedron with the O atom at the center. Electron pair geometry is tetrahedral Molecular geometry is bent VSEPR Theory: Molecular Geometries for Water (H O), Ammon2a (NH ) & Methane 3 (CH 4: Central atom in all 3 compounds is surrounded by 4 electron pairs. H2Ӧ: , H3N: and CH 4 Electron pair geometry is therefore tetrahedral in all three cases. Molecular geometries refer to the atom positions only. Carbon Dioxide (CO ) G2ometry: # valence e = 4 +2(6) = 16 - # valence e after scaffold = 16 – 4 = 12 Carbon has 2 double bonds and no lone pairs. Insofar as VSEPR is concerned, a multiple bond is counted as an oversized electron pair. Therefore, in VSEPR theory CO effectively has 2 electron pairs surrounding 2 carbon. CO 2s thus linear. Formaldehyde (H C=O) geometry: 2 18 In the VSEPR accounting system, the C atom in formaldehyde is effectively surrounded by 3 electron pairs. According to the theory, the best arrangement the electron pairs at the corners of a triangle Therefore, the molecular geometry of H C=O i2 trigonal planar with bond angles of ~120° The strange case of ozone, O Lewi3 Structure: Draw arrangement - Total number of valence electrons is 6(3) =18 e Attach terminal atoms to central atom Remaining electrons after drawing scaffold structure: - 18 – 4 = 14 e Add lone pairs to complete octets Convert a lone pair to bonding pair to satisfy octet for central atom The strange case of ozone, O 3D s3ructure from VSEPR: Draw Lewis electron dot structure Count electron pairs around central atom Insofar as VSEPR is concerned, a double bond is counted as a single oversized electron pair. Three electron pairs surround central oxygen According to VSEPR, the 3 electron pairs assume a flat triangular (trigonal planar) geometry Molecular Geometry is bent. Summary of VSEPR Rules: Number of electron “pairs” determines electron pair geometry o 2 linear, 3 flat triangular(trigonal planar), 4 tetrahedral o VSEPR accounting counts a multiple bond as a single electron pair. Molecular geometry is defined by atom geometry and ignores the position of lone pairs. Geometries of Six Common Electron Pair Atom Types: Geometry of Ethane, H C─CH 3 3: Both carbon atoms are tetrahedral Rotation about C─C is easy Many conformations are possible o Staggered Most stable 19 o Eclipsed Least stable Geometry of Ethene, H C=C2 2: Both carbon atoms are trigonal planar Rotation about C=C is difficult Only the planar conformation is observed at room temperature Geometry of Benzene, C H 6 6: All carbon atoms are trigonal planar Ring is rigid and planar Ring has shape of a regular hexagon Silicon: carbon’s big sister: Silicon has 4 valence electrons – forms 4 bonds Element silicon has same network structure as diamond Si – Si bonds are longer (and weaker) than C – C bonds Si does not form double and triple bonds like C Silicon does not form the wide range of chains and rings that carbon does. Bond lengths: Bond Type Bond length (pm) o C-C 154 o Si-Si 235 o Ge-Ge 245 Allotropes, Condensed Formulas and Skeletal Structures: Construction of the Diamond Lattice from Tetrahedral Carbon Fragments: Ethane has two tetrahedral carbons Replacing 3 Hs of lower C adds another layer of C Continue replacing Hs to build out within a layer, and down to the next layer. Diamond Structure: Diamond contains a tetrahedral framework Layers are rigidly held in place by a network of covalent bonds Graphite Structure: Graphite contains sheets of tiled hexagons Sheets are only weakly bound to one another Graphite is slippery Allotropes: different forms of the same element New allotrope of Carbon: Fullerene: Recently discovered allotrope of carbon contains 60 atoms o Structure is similar soccer ball Contains 12 pentagons, 10 hexagons, and 60 corners Bears resemblance to geodesic domes designed by Buckminster Fuller o Trivial name is buckminsterfullerene 20 Buckyballs or Fullerenes up close: Poyhedron of 12 pentagons and 10 hexagons with 60 corners forms a closed symmetrical structure Each carbon is bonded to 3 other carbons o Each carbon is involved in one double bond and two single bonds Alternating single and double bonds provide stability o Resonance stabilization How many bonds? Atoms complete octets by pairing up valence electrons with other atoms Soap Ingredients: Examples of Tetravalent Carbon and Divalent Oxygen: Lauric Acid: Methyl salicylate: A sodium salt of lauric acid is used for cleaning Methyl salicylate (Oil of wintergreen) is used for pleasant aroma Condensed formula conventions: Methyl and Methylene Groups: Methyl Groups H │ ─ C ─ H → ─ CH 3 │ H Expanded Condensed Structure Formula Methylene Groups H │ ─ C ─ → ─ CH ─ 2 21 │ H Expanded Condensed Structure Formula Condensed formula conventions: Carboxylic Acid Groups: Carboxylic Acid Groups O ││ ─ C ─ O ─ H → ─ C(=O)OH or ─ COOH or ─ CO H2 Expanded Condensed Structure Formulas Condensed Formulas for Large Molecules: Skeleton Structures: Hydrogen Suppression and Carbon Presumption: Hydrogen atoms are suppressed and carbon labels removed in skeletal structures Lauric Acid Skeleton Structure: Skeleton Structures Continued: Sample Problem: Skeletal Structure of DEET What is the molecular formula for DEET, the active ingredient of many mosquito repellants, which has the following skeletal structure? 22 Write the expanded structure o C’s @ each unlabeled vertex and terminus o H’s to fill out valence @ each C Count the atoms o Molecular Formula is C H 12 17 3D Structure of DEET: Methyl and methylene C’s are tetrahedral o o H-C-H & H-C-C angles are 109 Methyl groups are staggered with respect to methylene groups o H-C-C-H angles are 60 o Carbonyl C is trigonal planar Benzene C’s are trigonal planar o Benzene ring is planar Benzene ring is twisted with respect to the carbonyl group Unit 5: Salts Salt: The Staff of Life: Solid salt (NaCl) is a hard white powder. o Ions are arranged in a regular lattice Salt solution contains ions moving randomly with respect to one another Ions in solution can have powerful effects o Conduct currents, o Cause living cells to swell or collapse, o Involved in conduction of signals thru nerves Ionic solutions are essential for life. More than Morton’s: A salt is any ionic compound. It contains positive ions (cations) and negative ions (anions). Simple salts can be formed from elements Involves transfer of electrons from a metal to a non-metal. Crystal structure of KCl: Crystal lattice is cubic Each K ion has 6 Cl ions as nearest neighbors Each Cl ion has 6 K ions as nearest neighbors Densely packed ionic lattices make simple salts high melting point solids. Violet K Cl green Common Polyatomic Ions: 23 Naming Ionic Compounds from Formulas: The cation is named first followed by the anion name. NaNO 3 sodium nitrate K2SO 4 potassium sulfate (NH 4 3O 3 ammonium phosphite Fe(HCO 3 3 iron(III) bicarbonate or iron(III) hydrogen carbonate Writing Ionic Formulas from Names: The formula of an ionic compound o Containing a polyatomic ion must have a charge balance that equals zero(0). o Sodium nitrate Na + and NO 3− → NaNO 3 o With two or more polyatomic ions has the polyatomic ions in parentheses. Magnesium nitrate Mg 2+and 2NO 3 → Mg(NO )3 2 subscript 2 for charge balance Sample Problem: Select the correct formula for each: A. aluminum nitrate 1) AℓNO 3 2) Aℓ(NO) 3 3) Aℓ(NO 3 3 B. copper(II) nitrate 1) CuNO 3 2) Cu(NO 3 2 3) Cu 2NO )3 C. iron (III) hydroxide 1) FeOH 2) FeOH 3 3) Fe(OH) 3 D. tin(IV) hydroxide 1) Sn(OH) 4 2) Sn(OH) 2 3) Sn (4H) Write the correct formula for each: A. potassium bromate KBrO 3 B. calcium carbonate CaCO 3 C. sodium phosphate Na 3O 4 D. iron(III) oxide Fe 2 3 E. iron (II) nitrite Fe(NO ) 2 2 24 Polarity of Water: O is more electronegative than H and attracts electrons. H-O bonds are polarized with O slightly negative. H O Since water is a bent molecule, bond dipoles add constructively. Hydrogen Bonding: Electrostatic interaction between slightly positive (δ+) hydrogen of one water molecule and slightly negative (δ−) oxygen of another water molecule Hydrogen bonds are weak compared with covalent bonds Hydrogen bonds are responsible for the unusually high melting and boiling point of water when compared to H S, H2Se, 2r H Te 2 Generally, formed between one molecule that possesses an X ─ H bond, where X is a small electronegative atom from the second row of the periodic table (N, O, or F), and a second molecule that also possesses an electronegative Y atom from the same list of second row atoms. Why are ions attracted to water? Ion-dipole interactions Opposite charges attract Water and ions are drawn to one another by ion – dipole interactions o The negative O atoms on water attach to the positive ions. o The positive H atoms on water attach to the negative ions. Ion-dipole interactions aid the dissolution of a salt Aqueous Solubility Rules: 25 Sample Problem: Determine the water solubility of o Al(OH) 3 Water insoluble o NaOH water soluble (important exception to the rule) o AlCl 3 water soluble o NaCl water soluble Rule of Thumb Exceptions ─ Cl : All chlorides are AgCl and PbCl ar2 soluble insoluble ─ OH : All hydroxides are Group IA and ammonium insoluble hydroxides are soluble Hard Water and Soap: A Formula for Scum: Small highly charged ions make water hard: 2+ 2+ 3+ o Ca , Mg , Fe Soaps contain long chain carboxylate (CO ) anio2s: − o CH 3H C2 CH 2H C2 CH2CH 2H C2 CH2CH 2H C2 2 2 2 2 o CH 3CH )2 12 2− Reaction of soapy anions with “hard” cations: 2CH 3CH )2 n (a2) + Ca (aq) → Ca(CH (CH ) CO3) (s2 n 2 2 + Above reaction does not occur with soft ions like Na at concentrations usually seen in water Water softening works by ion exchange – hard for soft ions. Mechanics of Ion-Exchange: Ion-exchange resins consist of exchangeable sodium ions (Na ) that are+ attracted to negatively charged sulfite groups located at the ends of long chains which are in turn tethered to a solid support. + 2+ 2+ Exchange of Na ions with the Ca and Mg ions softens the water. Resins need to be recharged periodically with concentrated NaCl solution. Electrolytes and Solution Concentration: A Bright Idea: Ions Conduct Electricity through Salt Solutions: 26 Experiment o Materials placed in an electrical circuit in series with a light bulb to test conductivity. Observations: o Pure water does not conduct electricity. o Pure salt does not conduct electricity o Salt dissolved in water in conducts electricity. Hypothesis: o Salts ionize in water Would a sugar water conduct electricity? Model for Electrical Conductivity of Salt Solutions: A potential difference is applied across salt solution using a battery. Positively charged ions (cations) are attracted to the negatively charged electrode Negatively charged ions (anions) are attracted to the positively charged electrode Conductivity is expected to increase when more salt is added. Verde es CL y Amarillo Sodium Chemical Equations for the Dissolution of Electrolytes: Electrolytes such as sodium chloride (NaCl) o Dissociate in water producing positive and negative ions. o Conduct an electric current when dissolved in water. Solids are indicated by a “(s)” following the formula unit. Molecules or ions in aqueous solution are indicated by an “(aq)” following the formula unit. + − o NaCl(s) → Na (aq) + Cl (aq) (dissolution) o Ag (aq) + Cl (aq) → AgCl(s) (precipitation) Sample Problem: Complete the following equation for the dissolution of potassium phosphate in aqueous solution K3PO 4s) → ? 1) K PO (aq) 33+ 4 3− 2) K3 (aq) + P (aq) + 2O (a2) 3) 3K (aq) + PO 4−(aq) ------ ESTE Saturation and Limits on Solubility: Salts have solubility limits. Solubility of NaCl in water is 1.0 g/2.8 mL. Solutions are said to be Saturated when the solution concentration has reached the solubility limit. Salt added beyond the solubility limit will remain as a solid. 27 The Dynamic Equilibrium: At the macro scale little seems to be happening in a saturated solution. At the nanoscale there is constant exchange of ions on the solid surface with ions in the solution: o Rate of ions leaving solid = rate of ions entering solid. Dynamic equilibrium is indicated in a chemical equation by a double arrow o NaCl(s) ⇄ Na (aq) + Cl (aq) How Does Salt Preserve Mummies?: Embalming solutions contain a collection of salts including NaCl and Na CO 2 3 known as natron High salt concentration in a bathing solutions causes bacterial cells to lose water (dessication) by a process known as osmosis. Osmosis: Semi-permeable membrane allows passage of water, but prevents flow of ions. The direction of flow of water through a semi-permeable membrane is from a dilute to a more concentrated solution o Flow is known as osmosis What happens to water levels during osmosis? The movement of water from left to right by osmosis leads to a height difference in the solutions. This is tantamount to a higher pressure in the right arm. Water is sucked out of the left arm just like it is sucked out of biological cells in contact with a concentrated natron solution. Concentrations on either side of the membrane are equalized. Cells are thus desiccated Effect of salt on red blood cells: Isotonic solutions of sodium chloride are 0.9% (mass/Volume) NaCl. In an isotonic solution a red blood cell retains its normal volume. Excess salt outside cell causes passage of water out of the cell to equalize concentrations. Cell will shrivel and die. What will happen if there is excess salt inside the cell? Water will flow into the cell, and the cell could burst and die Molarity of a Saturated Salt Solution In chemistry, the most common unit for reporting concentration is molarity Base unit of molarity is M = mol/L. 28 mM means millimolar or 0.001 M. N mu ber m leso o f ute Example: What is concentration of a M lar oity saturated solution of sodium chloride that V luome o sf ution i L ters contains 1 g in 2.8 mL? First calculate molar mass of NaCl Molar mass of NaCl = (22.99 + 35.45) g/mol = 58.44 g/mol Then calculate number of moles per Liter Sample Problem: How many moles of NaOH are contained in 5.0 mL of 0.40 M NaOH? Molarity = (moles)/(L of solution) moles = (L of solution) x (Molarity) moles = (5.0 mL) x (1L/1000 mL) x (0.40 moles NaOH/L) = 0.0020 moles How would you calculate the number of g of NaOH in 5.0 mL of 0.40 M NaOH? Sample Problem: How many liters of 0.250 M NaOH are needed to deliver 2.00 moles of NaOH? Molarity = (moles)/(L of solution) L of solution = moles/Molarity Litersof solution=0moles=8.00L mol 0.250 L Preparation of a Solution of a Given Molarity by Weighing the Solute: How would you prepare 250 mL of a 0.500 M sodium nitrate solution from a bottle of the solid salt? o Determine formula for sodium nitrate: NaNO 3 o Determine molar mass of NaNO : 8530 g/mol o Determine number of moles, n, in 250 mL of 0.500 M NaNO solut3on n=V xM= 250mL 1L × 0.500mol =0.125mol ( (1000mL )( L ) o Determine mass of NaNO in 3.125 mol of NaNO 3 (0.125 mol)(85.0 g/mol) = 10.5 g Solution would be prepared by dissolving 10.5 g in enough water to give 250 mL of solution Preparation of a Solution of Given Molarity by Diluting a Stock Solution: How would you prepare 250 mL of a 0.500 M sodium nitrate solution by diluting a stock solution that is 2.00 M? o Determine number of moles, n, in the final solution of 250 mL of 0.500 M NaNO s3lution 1L mol n= ( f M =f)5(mL )(1000mL )(0.500 L )=0.125mol o Determine the volume, V , of the 2.00 M sodium nitrate stock solution that contains 0.125 mol of NaNOs 3 29 V (¿¿ f × f)= 0.125mol =0.0625L M mol s 2.00 L V s¿ o The final solution would be prepared by diluting 62.5 mL of the 2.00 M stock solution to 250 mL. Electron Dot Structures for Polyatomic Ions: Hydronium, Hydroxide and Nitrite Ions: Recipe is essentially the same as for neutral molecules except valence electron count needs to be adjusted to account for total charge on ion Ions in Pure Water: Even a sample of pure water contains a few ions. These ions are a product of auto-ionization: Forward reaction: o H 2 → H (aq) + OH (aq) Reverse reaction: + − o H (aq) + OH (aq)→ H O 2 Ion concentration in pure water compared with saturated salt solution: For+pure wat−r -7 [H ] = [OH ] = 0.0000001 M = 10 M For saturated NaCl solution + − [Na ] = [Cl ] = 6.1 M (See earlier slide) Concentration of ions in saturated NaCl solution is 6.1 x 10 = 61 million times larger than that in pure water Electrical conductivity of saturated salt solution is much higher than that of pure water Unit 6: Acids and Base: Proton Ping-Pong in Water: The pH Scale H 2 + H O 2 H O + 3H+ − (Auto-ionization) In pure water one molecule in 550,000,000 is auto-ionized Corresponds to a H O c3ncentration of 0.0000001 mol/L = 1.0 x 10 −7 M [A] is the symbol used for molar concentration of A 30
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