Chemistry Textbook - The Nature of Chemistry, PDF

John W. Moore Conrad L. Stanitski http://academic.cengage.com/chemistry/moore http://academic.cengage.com/chemistry/moore Chapter 1 The Nature of Chemistry Stephen C. Foster Mississippi State University Why Care about Chemistry? Chemistry The sc...

John W. Moore Conrad L. Stanitski http://academic.cengage.com/chemistry/moore http://academic.cengage.com/chemistry/moore Chapter 1 The Nature of Chemistry Stephen C. Foster Mississippi State University Why Care about Chemistry? Chemistry The science of matter and the transformations it can undergo. Why should you study it? It helps us understand our surroundings and the way we function. It plays a central role in medicine, engineering and many sciences. Identifying Matter: Physical Properties Physical properties can be measured without changing the composition of a substance. Examples Temperature Melting point Pressure Boiling point Mass Density Volume Color State (solid, liquid or gas) Shape of crystals Physical Change The same substance is present before and after a physical change. A physical state may change. ice can melt solid water → liquid water A gross shape may change. a lump of lead can be hammered into a sheet. Size may change a piece of wood can be cut in two. Melting & Boiling Point Temperature (T) Measures relative energy (E) content of an object. E transfers from high-T to low-T objects. U.S.: often in degrees Fahrenheit (°F). Elsewhere: degrees Celsius (°C). T (°F) T (°C) Water freezes 32 0 Water boils 212 100 Normal body T 98.6 37.0 Melting and Boiling Point water boils T (°C) = [T (°F) – 32] x 100 180 100°C 212°F or T (°C) = [T (°F) – 32] x 5 100 180 9 steps steps and T (°F) = 9 [T (°C)] + 32 0°C 32°F 5 water freezes Density A physical property. Substance Density* Ethanol 0.789 mass density = Water 0.998 Volume Magnesium 1.74 d= m Aluminum 2.70 V Titanium 4.50 Copper 8.93 Lead 11.34 Mercury 13.55 Gold 19.52 Water, copper and gold *g/mL at 20 °C Measurements, Units & Calculations A measurement is the comparison of a quantity (e.g. volume) with a unit (e.g. mL). Scientists use SI units (Système International d'unités) Physical Quantity Name of Unit Symbol Length Meter m Mass Kilogram kg Time Second s Temperature Kelvin K Amount of substance Mole mol Electrical current Ampere A Luminous intensity Candela cd SI Units Prefixes multiply or divide a unit by multiples of ten. Prefix Meaning Example mega M 106 1 megaton = 1 x 106 tons kilo k 103 1 kilometer (km) = 1 x 103 meter (m) deci d 10-1 1 deciliter (dL) = 1 x 10-1 liter (L) centi c 10-2 1 centimeter (cm) = 1 x 10-2 m milli m 10-3 1 milligram (mg) = 1 x 10-3 gram (g) micro μ 10-6 1 micrometer (μm) = 1 x 10-6 m nano n 10-9 1 nanogram (ng) = 1 x 10-9 g pico p 10-12 1 picometer (pm) = 1 x 10-12 m Significant Digits & Rounding Numbers All measurements involve some uncertainty. What’s the volume of water in this measuring cylinder? It’s between 8.3 mL and 8.4 mL. 8.30 mL ? 8.31 mL ? The last digit is uncertain, but is a good estimate (it is a significant digit). Significant Digits & Rounding Numbers Number Sig. figs. Comment on Zeros 2.12 3 4.500 4 Not placeholders. Significant. 0.002541 4 Placeholders (not significant). 0.00100 3 Only the last two are significant. 500 1, 2, 3 ? Ambiguous. May be placeholders or may be significant. 500. 3 Add a decimal point to show they are significant. 5.0 x 102 2 No ambiguity. Significant Figures in Calculations Addition and subtraction Find the decimal places (dp) in each number answer dp = smallest input dp Add: 17.245 dp = 3 + 0.1001 dp = 4 17.3451 Rounds to: 17.345 (dp = 3) Significant Figures in Calculations Subtract 6.72 x 10-1 from 5.00 x 101 Use equal powers of 10: 5.00 x 101 dp = 2 – 0.0672 x 101 dp = 4 4.9328 x 101 Rounds to: 4.93 x 101 dp = 2 Significant Figures in Calculations Multiplication and Division Answer sig. fig = smallest input sig. fig. 17.245 sig. fig. = 5 x 0.1001 sig. fig. = 4 1.7262245 Rounds to: 1.726 sig. fig. = 4 Multiply 2.346, 12.1 and 500.99 = 14,221.402734 Rounds to: 1.42 x 104 (3 sig. fig.) Rules for Rounding Examine the 1st non-significant digit. If it is: > 5, round up. < 5, round down. = 5, check the 2nd non-significant digit. ▪ round up if absent or odd; round down if even Round 37.663147 to 3 significant figures. 2nd non-sig. last Rounds up to 37.7 digit retained 1st non-sig. digit digit Rules for Rounding Round the following numbers to 3 sig. figs. 1st non-sig. 2nd non-sig. Rounded Number digit digit Number 2.123 2.123 2.12 51.379 51.379 51.379 51.4 131.5 131.5 132. 24.752 24.752 24.752 24.7 24.751 51.379 24.751 24.8 0.06744 0.06744 0.0674 51.379 51.379 51.379 51.4 Rules for Rounding Answer dp = 3. 92.803 is the sig.result. dp = 5 dp = 3 (5 sig. figs). 99.12444 – 6.321 92.80344 = = 3.37153195571 27.5256 27.5256 6 sig. figs. Significant figures? = 3.3715 (5 sig. figs.) Rules for Rounding To avoid rounding errors Carry additional digits through a calculation. Use the correct number of places in the final answer. Note Exact conversion factors: (100 cm / 1 m) or (2H / 1 H2O) Have an infinite number of sig. figs. Calculations Involving Density Example Determine the mass of 1 gallon (3274 mL) of mercury. m = V x d = 3274 mL x 13.55 g = 4.436 x 104 g 1 mL A proportionality (or conversion) factor was used. known units x desired units = desired units known units Dimensional Analysis Since 1 lb = 453.59 g we can write: 453.59 g = 1 and 1 lb =1 1 lb 453.59 g Example What is the mass in grams of a 2011 lb car? 2011. lb x 453.59 g = 9.122 x 105 g 1 lb Multiplication by 1! The quantity doesn’t change – just the units! Chemical Changes & Chemical Properties Chemical property A chemical reaction that a substance can undergo. Chemical Reaction? Reactants change into different substances. Sucrose will caramelize, then form carbon on heating. sucrose heat carbon + water reactant products Chemical Properties Describe these changes as a chemical or physical: (a) A cup of household bleach changes the color of your favorite T-shirt from purple to pink. Chemical change (b) Fuels in the space shuttle (hydrogen and oxygen) combine to give water and provide energy to lift the shuttle into space. Chemical change (c) An ice cube in your glass of lemonade melts. Physical change Separation & Purification Mixtures can be separated by physical methods. e.g. magnetic separation of iron filings from sulfur powder. Classifying Matter: Elements & Compounds Elements Cannot be decomposed into new substances Compounds Can be decomposed Sucrose is composed of carbon, hydrogen and oxygen. Have specific composition Sucrose is always 42.1% C, 6.5% H and 51.4% O by mass. Have specific properties Water always melts at 0.0°C and boils at 100.0°C (1 atm.) Types of Matter Matter (may be solid, liquid, or gas): anything that occupies space and has mass Physically Heterogeneous matter: separable into Homogeneous matter: nonuniform composition uniform composition throughout Physically Substances: fixed separable into Solutions: homogeneous composition; cannot mixtures; uniform compositions be further purified that may vary widely Chemically separable into Compounds: elements Elements: cannot be subdivided united in fixed ratios by chemical or physical changes Combine chemically to form Nanoscale Theories & Models macroscale objects are large enough to be seen, measured and handled without any aids. microscale objects require a microscope to view them. nanoscale objects have dimensions ≈ an atom. (nano: SI prefix for 10-9, so 1 nm = 1x10-9 m) States of Matter: Solids, Liquids & Gases Kinetic-Molecular Theory “Matter consists of tiny particles in constant motion”. Solid Closely-packed particles often in regular arrays. Fixed locations. Vibrate back & forth. Rigid materials. Small fixed volume. External shape often reflects inner structure. States of Matter: Solids, Liquids & Gases Liquid Very similar to a solid, BUT: Slightly more open. Slightly larger volume. More randomly arranged. Less confined; molecules can move past each other. Gas Continuous rapid motion. Particles are widely spaced. Travel large distances before colliding. No fixed volume or shape. The Atomic Theory All matter is made up of extremely small atoms. All atoms of a given element are chemically identical. Compounds form when atoms of two or more elements combine. usually combine in the ratio of small whole numbers. Chemical reactions join, separate, or rearrange atoms. Atoms are not created, destroyed or converted into other kinds of atoms during a chemical reaction. The Chemical Elements Elements have unique names and symbols. From people, places, mythology… Symbols start with a capital letter. Extra letters are lower-case. Most symbols are obvious abbreviations Helium = He Hydrogen = H Titanium = Ti Zinc = Zn “Old”-element symbols come from ancient names. Gold = Au (aurum) Tin = Sn (stannum) Silver = Ag (argentum) Lead = Pb (plumbum) The Chemical Elements Element/Symbol Discovery Name’s Origin Copernicium Cn 2009: S. Hofmann et al. Honoring Nicolaus Copernicus Curium Cm 1944: G. Seaborg et al. Honoring Marie & Pierre Curie Hydrogen H 1766: H. Cavendish Gr. hydro (water) + genes (maker) Mercury Hg Ancient Mythology: messenger of the gods Gr. hydrargyrum (liquid silver) Titanium Ti 1791: W. Gregor L. Titans (1st sons of the earth) Neon Ne 1898: W. Ramsey & M. Gr. neos (new) Travers Polonium Po 1898: M. Curie & P. Curie Honoring Poland (Marie’s home) Types of Elements More than 110 elements are currently known 90 occur naturally on earth. the rest are man-made (synthetic). most are metals (only 24 are not). Metals solids (except mercury – a liquid). conduct electricity. ductile (can be drawn into wires). malleable (can be rolled into sheets). Iron, aluminum, copper & gold Types of Elements Nonmetals Occur in all physical states. solids: sulfur, phosphorus, carbon. liquid: bromine gases: oxygen, helium, nitrogen. Non metals do not conduct electricity. graphite (a form of carbon) is an exception. Types of Elements Six are metalloids: boron silicon germanium arsenic antimony tellurium They exhibit metallic and nonmetallic properties: Look like metals (shiny). Conduct electricity (not as well as metals). semiconductors. Periodic Table International 1A (1) USA: “A” = main block… 1… 18 8A (18) 1 2 1 H 2A 3A 4A 5A 6A 7A He 1.008 (2) (13) (14) (15) (16) (17) 4.003 3 4 …“B” = transition 5 6 7 8 9 10 2 Li Be element. B C N O F Ne 6.94 9.01 10.81 12.01 14.01 16.00 19.00 20.18 11 12 13 14 15 16 17 18 3 Na Mg 3B 4B 5B 6B 7B 8B 1B 2B Al Si P S Cl Ar 22.99 24.30 (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) 26.98 28.09 30.97 32.07 35.45 39.95 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 39.10 40.08 44.96 47.88 50.94 52.00 54.94 55.85 58.93 58.69 63.55 65.39 69.72 72.59 74.92 78.96 79.90 83.80 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 85.47 87.62 88.91 91.22 92.91 95.94 (98) 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 6 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 132.9 137.3 138.9 178.5 180.9 183.9 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 (209) (210) (222) 87 88 89 104 105 106 107 108 109 110 111 112 114 116 7 Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Fl Lv (223) 226.0 227.0 261.1 262.1 263.1 262.1 (265) (266) (271) (272) (285) (289) (293) 58 59 60 61 62 63 64 65 66 67 68 69 70 71 Lanthanides Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 140.1 140.9 144.2 (145) 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.0 175.0 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Actinides Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 232.0 231.0 238.0 237.0 (244) (243) (247) (247) (251) (252) (257) (258) (259) (260) Periodic Table Noble gases 1A Alkali metals 8A (1) Alkaline Halogens (18) (not H) earth metals 1 2 1 H 2A 3A 4A 5A 6A 7A He 1.008 3 (2) 4 Group = vertical column (13) 5 (14) 6 (15) 7 (16) 8 (17) 9 4.003 10 2 Li Be B C N O F Ne 6.94 9.01 Period = horizontal row 10.81 12.01 14.01 16.00 19.00 20.18 11 12 13 14 15 16 17 18 3 Na Mg 3B 4B 5B 6B 7B 8B 1B 2B Al Si P S Cl Ar 22.99 24.30 (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) 26.98 28.09 30.97 32.07 35.45 39.95 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 39.10 40.08 44.96 47.88 50.94 52.00 54.94 55.85 58.93 58.69 63.55 65.39 69.72 72.59 74.92 78.96 79.90 83.80 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 85.47 87.62 88.91 91.22 92.91 95.94 (98) 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 6 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 132.9 137.3 138.9 178.5 180.9 183.9 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 (209) (210) (222) 87 88 89 104 105 106 107 108 109 110 111 112 114 116 7 Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Fl Lv (223) 226.0 227.0 261.1 262.1 263.1 262.1 (265) (266) (271) (272) (285) (289) (293) 58 59 60 61 62 63 64 65 66 67 68 69 70 71 Lanthanides Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 140.1 140.9 144.2 (145) 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.0 175.0 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Actinides Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 232.0 231.0 238.0 237.0 (244) (243) (247) (247) (251) (252) (257) (258) (259) (260) Periodic Table 1A 8A (1) Main group metal (18) 1 2 1 H 2A Transition metal 3A 4A 5A 6A 7A He 1.008 (2) (13) (14) (15) (16) (17) 4.003 3 4 Metalloid 5 6 7 8 9 10 2 Li Be B C N O F Ne 6.94 9.01 Nonmetal 10.81 12.01 14.01 16.00 19.00 20.18 11 12 13 14 15 16 17 18 3 Na Mg 3B 4B 5B 6B 7B 8B 1B 2B Al Si P S Cl Ar 22.99 24.30 (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) 26.98 28.09 30.97 32.07 35.45 39.95 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 39.10 40.08 44.96 47.88 50.94 52.00 54.94 55.85 58.93 58.69 63.55 65.39 69.72 72.59 74.92 78.96 79.90 83.80 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 85.47 87.62 88.91 91.22 92.91 95.94 (98) 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 6 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 132.9 137.3 138.9 178.5 180.9 183.9 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 (209) (210) (222) 87 88 89 104 105 106 107 108 109 110 111 112 114 116 7 Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Fl Lv (223) 226.0 227.0 261.1 262.1 263.1 262.1 (265) (266) (271) (272) (285) (289) (293) 58 59 60 61 62 63 64 65 66 67 68 69 70 71 Lanthanides Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 140.1 140.9 144.2 (145) 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.0 175.0 Inner transition elements 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Actinides Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 232.0 231.0 238.0 237.0 (244) (243) (247) (247) (251) (252) (257) (258) (259) (260) Elements that Consist of Molecules Most non-metal elements form molecules. A chemical formula shows the composition: Diatomic examples: H2 O2 N2 F2 Cl2 Br2 l2 Polyatomic examples: O3 P4 S8 Communicating Chemistry: Symbolism Chemical formulas show: Number and type of atoms in the molecule. Relative ratio of the atoms in a compound. C12H22O11 CH3OH NaCl sucrose methanol table salt Chemical equations show: How reactants convert into products. heat C12H22O11 12 C + 11 H2O sucrose carbon + water Biological Periodic Table 1A 8A (1) (18) 1 1 H 2A 3A 4A 5A 6A 7A 1.008 (2) (13) (14) (15) (16) (17) 5 6 7 8 9 2 B C N O F 10.81 12.01 14.01 16.00 19.00 11 12 14 15 16 17 3 Na Mg 3B 4B 5B 6B 7B 8B 1B 2B Si P S Cl 22.99 24.30 (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) 28.09 30.97 32.07 35.45 19 20 23 24 25 26 27 28 29 30 32 33 34 4 K Ca V Cr Mn Fe Co Ni Cu Zn Ge As Se 39.10 40.08 50.94 52.00 54.94 55.85 58.93 58.69 63.55 65.39 72.59 74.92 78.96 42 53 5 Mo I 95.94 126.9 6 Building block element: 99.36% of atoms in the body Major minerals: most of the remaining atoms Trace element Biological Periodic Table Element in Symbol Abundance the body atoms/106 atoms Hydrogen H 628,000 Oxygen O 257,000 98.0% 99.4% Carbon C 95,000 Nitrogen N 13,600 Calcium Ca 2,400 Phosphorus P 2,100 Sulfur S 500 0.6% Sodium Na 420 Potassium K 330 Chlorine Cl 270 Magnesium Mg 130  John W. Moore Conrad L. Stanitski http://academic.cengage.com/chemistry/moore http://academic.cengage.com/chemistry/moore Chapter 2 Chemical Compounds Stephen C. Foster Mississippi State University Atomic Structure & Subatomic Particles Atoms are composed of three subatomic particles: electrons (e-) protons (p+) neutrons (n0). These particles were discovered in a series of important experiments which began in the 1890’s. Radioactivity Becquerel (1896) U ore emits rays that “fog” a photographic plate. Marie and Pierre Curie (1898) Isolated new elements (Po & Ra) that did the same. Marie Curie called the phenomenon radioactivity. Are these “rays” electrically charged? Scientists used a well known property: Like charges repel; opposite charges attract. Uncharged particles are unaffected. Radioactivity Three distinct types of radiation were discovered: β-particles “–” charge Gamma ray (γ) α-particle No charge “+” charge No deflection Heavier, deflected less than β Atoms must contain smaller sub-units. Electrons Thomson (1897) discovered the e-: fluorescent screen – high voltage + cathode ray “Cathode rays” Travel from cathode (-) to anode (+). Negative charge (e−). Emitted by cathode metal atoms. Electric and magnetic fields deflect the beam. Gives mass/charge of e- = −5.60 x 10-9 g/C. Coulomb (C) = SI unit of charge. Electrons Millikan (1911) studied electrically-charged oil drops. Mist of oil Oil droplet droplets injector (+) electrically charged plate Oil droplets fall (small central hole) through the hole A telescope is X-rays ionize the used to observe air; droplets add 1 the drops. or more e-. (-) charged plate (+) and (-) plate charges are adjusted until a drop is stationary. Electrons Droplet charges obeyed: multiple e- n (−1.60 x 10-19 C) with n = 1, 2, 3,… per drop n (e- charge) Modern value = −1.602176487 x 10-19 C. = −1 “atomic units”. mass me = charge x charge = (-1.60 x 10-19 C)(-5.60 x 10-9 g/C) = 8.96 x 10-28 g Modern value = 9.10938215 x 10-28 g Protons Atoms gain a positive charge when e- are lost. Implies a positive fundamental particle. Hydrogen ions had the lowest mass. Hydrogen nuclei assumed to have “unit mass” Called protons. Modern science: mp = 1.672621637 x 10-24 g mp ≈ 1800 x me Charge = -1 x (e- charge) = +1.602176487 x 10-19 C = +1 atomic units The Nuclear Atom How are the p+ and e- arranged? Thompson: Ball of uniform positive charge, with small negative dots (e-) stuck in it. The “plum-pudding” model. Ernest Rutherford 1910 Rutherford fired α-particles at thin metal foils Expected them to pass through with minor deflections. The Nuclear Atom But … some had large deflections. α particles Rutherford “It was about as credible as if you had fired a 15-inch shell at a piece of paper and it came back and hit you.” The Nuclear Atom Most of the mass and all “+” charge is concentrated in a small core, the nucleus. α particles A gold nucleus is ≈10,000 x smaller in diameter than a gold atom. e- occupy the remaining space. Neutrons Measured atomic masses were too large: Larger than the sum of the p+ and e- masses. Rutherford proposed a neutral particle. Chadwick (1932) fired -particles at Be atoms. Neutral particles, neutrons (n0), were ejected. The n0 beam ejected p+ from other atoms, allowing detection. mn ≈ mp (0.1% larger). mn = 1.674927211 x 10-24 g. n0 are present in all atoms (except normal H). The Nuclear Atom Nucleus Contains p+ and n0. Most of the atomic mass. Compact. Positive (each p+ has +1 charge). Electrons Tiny low mass particles surrounding the nucleus. Occupy most of the volume. Charge = -1. Atoms are neutral: Number of e− = Number of p+ Atomic sizes How many copper atoms lie across the diameter of a penny? A penny has a diameter of 1.90 cm, and a copper atom has a diameter of 256 pm. 1 pm = 1 x 10-12 m ; 1 cm = 1 x 10-2 m 1.90 cm x 1 x 10 -2 m 1 pm = 1.90 x 1010 pm x 1 cm 1 x 10-12 m Number of atoms across the diameter: 1.90 x 1010 pm x 1 Cu atom = 7.42 x 107 Cu atoms 256 pm Atomic Numbers & Mass Numbers All atoms of the same element have the same number of protons: Atomic number (Z) = number of p+ The mass number (A) = number of p+ + number of n0 A 12 For element X, write: ZX e.g. 6C A 12 or X e.g. C or X-A e.g. carbon-12 (Z is constant for a given element) Atomic Numbers & Mass Numbers How many p+, n0 and e- are in the following elements: 63 29 p+ = 29 e- (neutral atom: e- = p+) 29 Cu 63−29 = 34 n0 25 12 p+ = 12 e- (periodic table; neutral) Mg 25−12 = 13 n0 27 13 p+ = 13 e- Al 27−13 = 14 n0 Isotopes & Average Atomic Mass Absolute atomic masses refer to the mass of a single 12C atom: or amu or dalton (Da) 1 Unified atomic mass unit (u) = ¹∕12(mass of 12C) 1 u = 1.66054 x 10-24 g Particle Mass Charge g u ~u a.u. e- 9.1094 x 10-28 0.000548579 0 -1 p+ 1.6726 x 10-24 1.00728 1 +1 n0 1.6749 x 10-24 1.00866 1 0 Mass Spectrometer A mass spectrometer measures the mass to charge ratio of charged atoms and molecules. Pure neon result: Three peaks: 3 different types of Ne. Peak areas give relative amounts of each. Isotopes & Average Atomic Mass Atoms of the same element with different mass (different A) are called isotopes. They have: equal numbers of p+ different numbers of n0 20 Neon isotopes: 10 Ne 10 p+ and 10 n0 21 10 Ne 10 p+ and 11 n0 22 10 Ne 10 p+ and 12 n0 Isotopes & Atomic Weight Most elements occur as a mixture of isotopes. Magnesium is a mixture of: 24Mg 25Mg 26Mg Number of p+ 12 12 12 Number of n0 12 13 14 Mass (amu) 23.985 24.986 25.983 Abundance 79 % 10 % 11 % Atomic weight 24.305 Atomic wt = ∑(fractional abundance)(isotope mass) Sum of… fraction of each isotope Average Atomic Mass Boron contains 10B and 11B (10.0129; 11.0093 u). 10B is 19.91% abundant. Calculate the atomic mass of B. 10B 19.91(10.0129 u) = 1.994 u fraction 100 mass Abundance of 11B = 100% - 19.91% = 80.09% 11B 80.09 (11.0093 u) = 8.817 u 100 5 fraction mass Atomic weight of B = 1.994 + 8.817 u B Boron = 10.811 u 10.811 Ions & Ionic Compounds Ions - charged units - formed by transfer of e- between elements. Cation = positive ion. Metals form cations Na Na+ + e- Anion = negative ion. Nonmetals form anions S + 2 e- S2- Monatomic Ions Main group elements Add/lose enough e- to “get to” the nearest noble gas. Charge on ion = group A# or (grpA# - 8) Each e- lost produces one positive charge Each e- gained adds one negative charge S → S2- 16 e- → 18 e- (like Ar) Na → Na+ 11 e- → 10 e- (like Ne) P → P3- 15 e- → 18 e- (like Ar) Sr → Sr2+ 38 e- → 36 e- (like Kr) Monatomic Ions Transition metals: lose varying number of e-. old (and new) group number not very helpful. Ti Ti2+ (grp 4B) Cr Cr2+ or Cr3+ (grp 6B) Fe Fe2+ or Fe3+ (grp 8B) Cu Cu+ or Cu2+ (grp 1B) Mn Mn2+ Mn5+ or Mn7+ (grp 7B) Monatomic Ions Number of e- lost varies. Polyatomic Ions Multiple atom “units” with a net electrical charge. Ion Name NH4+ ammonium ion OH- hydroxide ion SO42- sulfate ion HSO4- hydrogen sulfate ion CN- cyanide ion NO3- nitrate ion Memorize all the ions in Table 2.2 Oxoanions A series of related names: more O ClO4- perchlorate PO43- phosphate SO42- sulfate NO3- nitrate ClO3- chlorate PO33- phosphite SO32- sulfite NO2- nitrite ClO2- chlorite less O ClO- hypochlorite If they begin with H, add a prefix “hydrogen” HSO4- hydrogen sulfate ion HCO3- hydrogen carbonate ion (common name: bisulfate ion) (common name: bicarbonate ion) Ionic Compounds Ionic compounds are: Held together by electrostatic forces. Always electrically neutral. Cation Anion Compound Charge Balance Mg2+ F- MgF2 (+2) + 2(-1) = 0 Mg2+ SO42- MgSO4 (+2) + (-2) = 0 Mg2+ PO43- Mg3(PO4)2 3(+2) + 2(-3) = 0 NH4+ CO32- (NH4)2CO3 2(+1) + (-2) = 0 Recognizing Ionic Compounds Compounds are ionic when they contain: A metal cation + a non-metal anion One (or more) polyatomic ion(s) Ionic non ionic Compound CaCl2 CH3OH Na2SO3 BrF5 Sr3(PO4)2 H2O NH4NO3 SO3 Naming Ions & Ionic Compounds Positive ions Most are metal ions (exception: ammonium NH4+ ). metal ion with only one charge state? Use metal name + ion. metal ion with multiple charge states? Use metal name + (Roman numeral) to show charge. Na+ sodium ion Ca2+ calcium ion Fe2+ iron(II) ion Fe3+ iron(III) ion no space Naming Ions & Ionic Compounds Negative ions Monatomic ion? Add “-ide” to the name stem. Polyatomic ion? Memorize these. P phosphorus P3- phosphide ion S sulfur S2- sulfide ion SO32- sulfite ion Naming Ionic Compounds Name the ions and add together… … cation then anion (drop “ion” from both). NaCl Single charge metal-ion examples NaCl sodium chloride MgCO3 magnesium carbonate SrO strontium oxide Mg(OH)2 magnesium hydroxide K2Cr2O7 K2Cr2O7 potassium dichromate Naming Ionic Compounds Multiple charge examples FeCl2 iron(II) chloride NaCl CaF2 FeCl3 iron(III) chloride CuBr2 Fe2O3 Cu2O copper(I) oxide CuO copper(II) oxide Fe2O3 iron(III) oxide Cu2O CuO Naming Ionic Compounds When are Roman numerals used? Main block metals form one type of ion: equal to A group number omit Roman numerals. exceptions: lead (Pb2+, Pb4+) and tin (Sn2+, Sn4+). Transition metals form multiple ions use Roman numerals. exceptions: silver (Ag+) and zinc (Zn2+). Naming Compounds Li2SO3 lithium sulfite CuSO4 copper(II) sulfate AlCl3 aluminum chloride AgF silver fluoride Na2S sodium sulfide PbO2 lead(IV) oxide Naming Compounds sodium hypochlorite NaClO chromium(III) sulfate Cr2(SO4)3 potassium dichromate K2Cr2O7 ammonium perchlorate (NH4)ClO4 potassium chloride KCl Ionic Compound Properties Ionic compounds are generally: Solids at room temperature with high melting and boiling temperatures. Hard Brittle (and easily cleaved). Poor heat conductors. Poor electrical conductors (unless molten). The Crystal Lattice Ionic compounds exist as Crystal lattices Each ion is surrounded by many others. Sodium chloride: Formula unit = smallest ratio of anions to cations Properties of Ionic Compounds Ionic crystals have distinctive shapes and can be cleaved: External force displaces layers Repulsion occurs Na+ Cl- Properties of Ionic Compounds Ionic compounds are electrical insulators when SOLID will conduct if molten Many are soluble in water Molecular Compounds Ionic compounds contain metal ions. Compounds formed exclusively with non-metal atoms are usually molecular. At the nanoscale: 2 or more elements combine Individual, independent units (molecules) form The molecular formula shows the number and kind of elements combined: H2O NH3 C8H18 Molecular & Ionic Compounds Property Molecular Ionic Compounds Compounds Composition Mostly non-metal Metal/non-metal combinations combinations Nanoscale Individual molecules Ions in a crystal lattice Physical state Gas, liq. or solid Crystalline solid Brittle & weak or Hard & brittle soft & waxy Melting point Low High Boiling point Low High Heat transport Low Low Electrical Low Low conductivity (but high if molten) Molecular Compounds Inorganic compounds Do not contain C or (C and H) e.g. water = H2O ammonia = NH3 carbon dioxide = CO2 Organic compounds Always contain C, usually H May contain many other elements e.g. benzene = C6H6 ethanol = C2H6O Most (but not all) are molecular. Molecular Formulas Ethanol has the formula C2H6O … Doesn’t show atom connections. A structural formula does. C2H6O may not be ethanol. Two C2H6O structural formulas: H H H H | | | | H–C–C–O–H H–C–O–C–H | | | | H H H H ethanol dimethyl ether Molecular Formulas Condensed formula Similar information, but in a more compact form. C, what’s attached to it, C … CH3CH2OH CH3OCH3 ethanol dimethyl ether Groups of atoms attached to C (like OH) are called functional groups –OH is the alcohol group. Molecular Formulas More elaborate models of ethanol: Naming Binary Molecular Compounds Binary compounds contain two different elements. If one element is hydrogen: It is written first in the formula and named first The other element is renamed with an “-ide” ending HCl hydrogen chloride H2S hydrogen sulfide HF hydrogen fluoride Naming Binary Molecular Compounds Other binary compounds (without H): # Prefix Name elements in formula order. 1 Mono The 2nd element’s name ends in “-ide” 2 Di Prefixes show the number of each 3 Tri atom present. 4 Tetra 5 Penta 6 Hexa 7 Hepta 8 Octa N2F4 = dinitrogen tetrafluoride 9 Nona 10 Deca Naming Binary Molecular Compounds Mono is optional for the 1st element (usually omitted): CO carbon monoxide* NO2 nitrogen dioxide N2O dinitrogen monoxide P2O5 diphosphorus pentaoxide PBr5 phosphorus pentabromide SF6 sulfur hexafluoride P4O10 tetraphosphorus decaoxide *Monooxide would also be correct. Naming Binary Molecular Compounds There are several common names in wide-spread use that you should know: H2O water NO nitric oxide NH3 ammonia N2O nitrous oxide PH3 phosphine N2H4 hydrazine Hydrocarbons Binary molecules containing (C and H) are known as hydrocarbons. Alkanes Hydrocarbons with C-C single bonds only: Use an –ane ending. Linear and branched molecules formula CnH2n+2 Cyclic molecules formula: CnH2n Butane (C4H10) Hydrocarbons Boiling points (°C) -162 -88 -42 -1 Larger mass = higher b.p. Hydrocarbons C8H18 = octane # of C Prefix C5H12 = pentane 1 Meth 2 Eth 3 Prop Rings use a “cyclo-” prefix. 4 But Examples: 5 Pent 6 Hex 7 Hept 8 Oct 9 Non 10 Dec Isomers Isomers: Two or molecules with the same formula, but with different atom arrangements. Branched alkanes are isomers of the linear forms. methylpropane C4H10 butane C4H10 H H H H H H H | | | | | | | H ―C―C―C―C― H isomers H―C―C―C― H | | | | Ι Ι H H H H H H H C H | H Alkanes & Their Isomers Alkyl functional groups An alkane with a H atom removed. Named by replacing “-ane” with “-yl” methylpropane C4H10 -CH3 methyl H H H | | | -CH2CH3 ethyl H―C―C―C― H Ι Ι H H -CH2CH2CH3 propyl H C H | H CH3CHCH3 isopropyl a methyl group Alkanes & Their Isomers Formula Isomers Formula Isomers CH4 1 C9H20 35 C2H6 1 C10H22 75 C3H8 1 C11H24 159 C4H10 2 C12H26 355 C5H12 3 C15H32 4,347 C6H14 5 C20H42 366,319 C7H16 9 C30H62 4.1 x 109 C8H18 18 C40H82 6.9 x 1013 Amount of Substance: The Mole A counting unit – a familiar counting unit is a “dozen”: 1 dozen eggs = 12 eggs 1 dozen peas = 12 peas 1 mole (mol) = Number of atoms in 12 g of 12C Latin for “heap” or “pile” 1 mol = 6.02214179 x 1023 “units” Avogadro’s number 1 mole eggs = 6.02 x 1023 eggs 1 mole peas = 6.02 x 1023 peas Amounts of Substances: The Mole A green pea has a ¼-inch diameter. 48 peas/foot. (48)3 / ft3 ≈ 1 x 105 peas/ft3. V of 1 mol ≈ (6.0 x 1023 peas)/(1x 105 peas/ft3) ≈ 6.0 x 1018 ft3 U.S. surface area = 3.0 x 106 mi2 = 8.4 x 1013 ft2 height = V / area, 1 mol would cover the U.S. to: 6.0 x 1018 ft3 =7.1 x 104 ft = 14 miles ! 8.4 x 1013 ft2 Amounts of Substances: The Mole 1 mole of an atom = atomic weight in grams. 1 Xe atom has mass = 131.29 u 1 mol of Xe atoms has mass = 131.29 g 1 He atom has mass = 4.0026 u 1 mol of He has mass = 4.0026 g There are 6.022 x 1023 atoms in 1 mol of He and 1 mol of Xe – but they have different masses. … 1 dozen eggs is much heavier than 1 dozen peas! Amounts of Substances: The Mole Gram-Mole Calculations Example How many moles of copper are in a 320.0 g sample? Cu-atom mass = 63.546 g/mol (periodic table) Conversion factor: 1 mol Cu = 1 63.546 g 1 mol Cu nCu = 320.0 g x = 5.036 mol Cu 63.546 g n = number of moles Gram-Mole Calculations Calculate the number of atoms in a 1.000 g sample of boron. nB = (1.000 g) 1 mol B = 0.092507 mol B 10.81 g 6.022  1023 atoms B atoms = 0.092507 mol B 1 mol = 5.571  1022 B atoms Moles of Compounds A mole of XmYn contains: m moles of atom X and n moles of atom Y 1 mol of H2O contains: 2 mol of H atoms and 1 mol of O atoms Molar mass = sum of the atomic masses Mass of 1 water molecule: = 2(1.008 u) + 1(15.999 u) = 18.015 u Molar mass of water: = 2(1.008 g/mol) + 1(15.999 g/mol) = 18.015 g/mol Molar Mass of Ionic Compounds Ionic compounds do not contain molecules. Formula weight (or molar mass) should be used to describe the mass of an ionic compound. Compound Atomic Masses, Formula mass Molar mass NaCl 22.99 u + 35.45 u = 58.44 u 58.44 g/mol Ca(NO3)2 40.08+2(14.01)+6(16.00) =164.10 u 164.10 g/mol Ionic Hydrates Ionic hydrate: ionic compound with water trapped in the crystal the water of hydration. use “hydrate” with a Greek prefix for the number. heat can remove some, or all, of this water. CuSO4 (white) CuSO4 5H2O CuSO4 5H2O (blue) copper(II) sulfate pentahydrate Heat Gram-Mole Calculations How many moles of Ca3(PO4)2 are in 10.0 g of the compound? Molar mass = 3(40.08) + 2(30.97) + 8(16.00) g = 310.18 mol Moles of Ca3(PO4)2 = 10.0 g 1 mol = 0.0322 mol 310.2 g nCa3(PO4)2 = 0.0322 mol Gram-Mole Calculations Find the mass of cobalt in 3.49 g of cobalt(II) sulfate. Molar mass CoSO4 = 58.93 + 32.07 + 4(16.00) = 155.00 g mol-1 1 mol CoSO4 1 Co nCo = 3.49 g CoSO4 155.0 g CoSO4 1 CoSO4 nCo = 0.02252 mol Co 58.93 g Co mass of Co = 0.02252 mol Co = 1.33 g Co 1 mol Co Composition & Chemical Formula Two names used: percent composition by mass, or mass percent of the compound. Example What is the mass percent of each element in sodium chlorite, NaClO2? Molar mass = 22.990 + 35.453 + 2(15.999) g mol-1 = 90.441 g mol-1 Percent Composition mass of Na in 1 mol NaClO2 %Na = x 100 % mass of NaClO2 in 1 mol NaClO2 22.990 g = x 100 % = 25.42% 90.441 g %O = mass of O … x 100 % mass of NaClO2 … = 2(15.999) g x 100 % = 35.38% 90.441 g Percent Composition %Cl = mass of Cl … x 100 % mass of NaClO2 … = 35.453 g x 100 % = 39.20% 90.441 g Check your work: %Na + %O + %Cl = 25.42 + 35.38 + 39.20 = 100% Empirical & Molecular Formulas Last example: molecular formula percent composition The process can be reversed: percent composition empirical formula Not molecular formula Empirical formula = the simplest ratio of atoms in a compound. Empirical & Molecular Formulas Compound Mol. formula Emp. formula Hydrogen peroxide H2O2 HO Borane (boron trihydride) BH3 BH3 Diborane (diboron hexahydride) B2H6 BH3 Octene C8H16 CH2 Butene C4H8 CH2 Empirical & Molecular Formulas Example An orange compound is 26.6% K, 35.4% Cr and 38.0% O. Determine its empirical formula. Assume a 100.0 g sample. % becomes mass in grams Divide each mass by its atomic mass. Gives the number of moles of each (in 100 g). Divide each by the smallest answer found. The smallest integer ratio = empirical formula. Empirical & Molecular Formulas Unknown: 26.6% K 35.4% Cr 38.0% O In 100.0 g 1 mol K = 0.6803 mol K 26.6 g K 39.10 g K 35.4 g Cr 1 mol Cr = 0.6808 mol Cr 52.00 g Cr 38.0 g O 1 mol O = 2.375 mol O 16.00 g O Empirical & Molecular Formulas Empirical formula = smallest integer ratio. Divide by the smallest (ratios stay the same!) 0.6803 mol = 1.000 x2 K 0.6803 mol 2 0.6808 mol = 1.001 x2 2 Cr 0.6803 mol 2.375 mol = 3.491 x2 7 O 0.6803 mol Choose a multiplier to make integer Empirical formula: K2Cr2O7 Empirical & Molecular Formulas The molecular formula can be determined if the molecular mass is known. Example Vitamin C has the empirical formula C3H4O3 and molecular mass = 175 g/mol. Empirical mass: 3(12.01) + 4(1.008) + 3(15.99) = 88.03 g/mol Empirical mass ≈ ½(molecular mass) Mol. formula = 2(emp. formula) = C6H8O6  John W. Moore Conrad L. Stanitski http://academic.cengage.com/chemistry/moore http://academic.cengage.com/chemistry/moore Chapter 3 Chemical Reactions Stephen C. Foster Mississippi State University Chemical Equations Reactants Products sodium + hydrogen carbon + water + sodium hydrogen carbonate chloride dioxide chloride NaHCO3(s) + HCl(aq) → CO2(g) + H2O(ℓ) + NaCl(aq) Physical states are often listed: (g) gas (s) solid (ℓ) liquid (aq) aqueous (dissolved in water) Chemical Equations Any equation can be interpreted as referring to the nanoscale or macroscale: NaHCO3(s) + HCl(aq) → CO2(g) + H2O(ℓ) + NaCl(aq) 1 formula unit + 1 molecule → 1 molecule + 1 molecule + 1 form. unit 4 form. units + 4 molecules → 4 molecules + 4 molecules + 4 form. units NA form. units + NA molecules → NA molecules + NA molecules + NA form. units 1 mol + 1 mol → 1 mol + 1 mol + 1 mol Chemical Equations “Mass is neither created nor destroyed in an ordinary chemical reaction” Atoms must balance have equal total numbers on each side of the equation: NaHCO3(s) + HCl(aq) → CO2(g) + H2O(ℓ) + NaCl(aq) Na 1 0 → 0 0 1 H 1 1 → 0 2 0 C 1 0 → 1 0 0 O 3 0 → 2 1 0 Cl 0 1 → 0 0 1 Balancing Chemical Equations A simple list of each reactant converting to single product molecules is usually unbalanced. The stoichiometry must be adjusted. Stoichiometry The relationship between the number of reactant and product molecules in a chemical equation. H2(g) + Cl2(g) → 2 HCl(g) Stoichiometric coefficient Balancing Chemical Equations 1. Write an unbalanced equation with correct formulas for all substances. 2. Balance the atoms of one of the elements. i. Start with the most complex molecule. ii. Change the stoichiometric coefficients. iii. Do NOT alter the chemical formulas. 3. Balance the remaining elements. Balancing Chemical Equations Balance : Al + Fe2O3 → Al2O3 + Fe step 1 Al + Fe2O3 → Al2O3 + Fe 1 Al (2Fe + 3O) (2Al + 3O) 1Fe not balanced not balanced balance Fe from Fe2O3 step 2 Al + Fe2O3 → Al2O3 + 2 Fe 1 Al (2Fe + 3O) (2Al + 3O) 2Fe balanced not balanced step 3 2 Al + Fe2O3 → Al2O3 + 2 Fe 2Al (2Fe + 3O) (2Al + 3O) 2Fe Balancing Chemical Equations Combustion of rocket fuel: Not C2H8N2 + N2O4 → N2 + H2O + CO2 balanced 2C + 8H + 4N + 4O 2N + 2H + 3O + 1C Balance C and N in C2H8N2 first: C2H8N2 + N2O4 → N2 + 4 H2O + 2 CO2 2C + 8H + 4N + 4O 2N + 8H + 8O + 2C Still not balanced. Adjust N and O Balanced C2H8N2 + 2 N2O4 → 3 N2 + 4 H2O + 2 CO2 Balancing Chemical Equations Polyatomic ion on both sides of an equation? Balance as “units”. NaNO3(s) + H2SO4(aq) → Na2SO4(aq) + HNO3(aq) Na + NO3 + 2H + SO4 2Na + SO4 + H + NO3 Balance Na 2 NaNO3(s) + H2SO4(aq) → Na2SO4(aq) + HNO3(aq) 2Na + 2NO3 + 2H + SO4 2Na + SO4 + H + NO3 Balance H & NO3 2 NaNO3(s) + H2SO4(aq) → Na2SO4(aq) + 2 HNO3(aq) 2Na + 2NO3 + 2H + SO4 2Na + SO4 + 2H + 2NO3 The Mole & Chemical Reactions N2O4(g) + 2 N2H4(g) 3 N2(g) + 4 H2O(g) 1 mole of N2O4 reacts with 2 moles of N2H4 2 moles of N2H4 produce 3 moles of N2 1 mol N2O4 ≡ 2 mol N2H4 2 mol N2H4 ≡ 4 mol H2O etc. Mole ratios: 2 mol N2H4 3 mol N2 =1 =1 4 mol H2O 1 mol N2O4 The Mole & Chemical Reactions How many moles of oxygen gas and solid sulfur (S8) are produced when 10.0 moles of sulfur trioxide gas decompose? Write an unbalanced equation: SO3(g) → S8(s) + O2(g) Balance it: 8 SO3(g) → S8(s) + 12 O2(g) Stoichiometric ratios: 8 SO2 ≡ 1 S8 8 SO3 ≡ 12 O2 The Mole & Chemical Reactions Moles of S8 and O2 from 10 mol SO3? 8 SO3(g) → S8(s) + 12 O2(g) 1 mol S8 nO2 = 10.0 mol SO3 = 1.25 mol S8 8 mol SO3 Abbreviation for 8 SO3 ≡ 1 S8 “number of moles” 12 mol O2 nO2 = 10.0 mol SO3 = 15.0 mol O2 8 mol SO3 8 SO3 ≡ 12 O2 The Mole & Chemical Reactions Mass of product(s) can be calculated from mass of reactant(s): For: xA→yB Mass of Mass of A Use molar B Use molar mass of A mass of B Moles of Moles of A B Use mole ratio y/x The Mole & Chemical Reactions What mass of O2 and Br2 is produced by the reaction of 25.0 g of TiO2 with excess BrF3? 3 TiO2(s) + 4 BrF3(ℓ) → 3 TiF4(s) + 2 Br2(ℓ) + 3 O2(g) Notes: Check the equation is balanced! Stoichiometric ratios: 3 TiO2 ≡ 3 O2 ; 3 TiO2 ≡ 2 Br2 etc. Excess BrF3 (enough BrF3 to react all the TiO2). The Mole & Chemical Reactions Mass of O2 and Br2 from 25.0g of TiO2 and excess BrF3? 3 TiO2(s) + 4 BrF3(ℓ) → 3 TiF4(s) + 2 Br2(ℓ) + 3 O2(g) nTiO2 = mass TiO2 / molar mass TiO2 = 25.0 g x 1 mol = 0.3130 mol TiO 2 79.88 g 3 TiO2 ≡ 3 O2 3 mol O2 nO2 = 0.3130 mol TiO2 = 0.3130 mol O2 3 mol TiO2 massO2 = nO2MO2 = 0.3130 mol 32.00 g 1 mol massO2 = 10.0 g The Mole & Chemical Reactions Mass of O2 and Br2 from 25.0g of TiO2 and excess BrF3? 3 TiO2(s) + 4 BrF3(ℓ) → 3 TiF4(s) + 2 Br2(ℓ) + 3 O2(g) 2 Br2 ≡ 3 TiO2 2 Br2 nBr2 = 0.3130 mol TiO2 = 0.2087 mol Br2 3 TiO2 massBr2 = 0.2087 mol 159.81 g = 33.4 g Br2 mol Br2 The Mole & Chemical Reactions The purity of Mg can be found using the reaction… Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) Calculate the % Mg in a 1.72-g sample that produced 6.46 g of MgCl2 when reacted with excess HCl. More difficult – What should you calculate? How much pure Mg will make 6.46 g of MgCl2? Express as a % of the original mass. Practice Problem 4.8 % Mg in 1.72 g that produced 6.46 g of MgCl2 (excess HCl). FW of MgCl2 = 24.31 + 2(35.45) = 95.21 g/mol 1 mol nMgCl2 = 6.46 g MgCl2 = 0.06785 mol MgCl2 95.21 g Mg + 2 HCl → MgCl2 + H2 1 Mg nMg required = 0.06785 mol MgCl2 1 MgCl2 = 0.06785 mol of pure Mg Practice Problem 4.8 % Mg in 1.72 g that produced 6.46 g of MgCl2 (excess HCl). Calculate mass of pure Mg needed (massMg) massMg = 0.06785 mol Mg 24.31 g = 1.649 g 1 mol Given 1.72 g of impure Mg. 1.649 g Purity (as mass %) = x 100% = 95.9 % 1.72 g Limiting Reactant Given 10 slices of cheese and 14 slices of bread. How many sandwiches can you make? Balanced equation 1 cheese + 2 bread 1 sandwich 1 cheese ≡ 2 bread 1 cheese ≡ 1 sandwich 2 bread ≡ 1 sandwich Limiting Reactant Two methods can be used: Mass Method Calculate the same product from each reactant. Reactant producing the smallest amount is limiting. mass or moles 10 cheese x 1 sandwich = 10 sandwiches 1 cheese 14 bread x 1 sandwich = 7 sandwiches 2 bread Limiting Correct (Used up first) answer Limiting Reactant Bread is limiting… …base all other calculations on the limiting reactant. Sandwiches made 14 bread x 1 sandwich = 7 sandwiches 2 bread Cheese remaining 14 bread x 1 cheese = 7 cheese used 2 bread Used Initially Excess cheese = 10 – 7 = 3 slices Limiting Reactant Mole Ratio Method Calculate the amount (mol) of two reactants and then their mole ratio. If: actual < theoretical: numerator reactant is limiting. actual > theoretical: numerator reactant is in excess. Example If 374 g of NH3 and 768 g of O2 are mixed, what mass of NO will form? 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) Limiting Reactant 374 g NH3 + 768 g O2, massNO formed? 4 NH3 + 5 O2 → 4 NO + 6 H2O Balanced? Yes 1 mol nNH3 = 374 g = 21.96 mol 17.03 g 1 mol nO2 = 768g = 24.00 mol 32.00 g Mole ratio of reactants nO2 24.00 5 actual = 1.093 theoretical = 1.25 nNH3 21.96 4 actual < theo. O2 is limiting Limiting Reactant Mass of NO formed? 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) excess 24.00 mol ? O2 is limiting. Base all calculations on O2. nNO = 24.00 mol O2 4 NO = 19.20 mol 5 O2 massNO = 19.20 mol 30.01g = 576 g 1 mol Limiting Reactant How much water will be produced by the combustion of 25.0 g of H2 in the presence of 100. g of O2? Write a balanced equation: 2 H2(g) + O2(g) → 2 H2O(ℓ) 1 mol H2 nH2 = 25.0 g 2.016 g = 12.40 mol H2 1 mol O2 nO2 = 100. g 32.00 g = 3.125 mol O2 Limiting Reactant 2 H2 + O2 → 2 H 2O n 12.40 3.125 Larger amount. Excess H2 Mass Method 2H2O From H2 nH2O = 12.40 mol H2 = 12.40 mol 2H2 2H2O From O2 nH2O = 3.125 mol O2 = 6.250 mol 1O2 Smaller amount O2 is limiting Use O2 in all calculations. 18.02 g massH2O = 6.250 mol H2O = 113. g 1 mol Limiting Reactant 2 H2 + O2 → 2 H 2O Moles available: 12.40 3.125 Mole Ratio Method nO2 3.125 1 actual = 0.252 theoretical = 0.5 nH2 12.40 2 actual < theo. O2 is limiting H2O made: 2 H2 O 3.125 mol O2 = 6.250 mol H2O = 113. g 1 O2 Limiting Reactant What mass of MgI2 is made by the reaction of 75.0 g of Mg with 75.0 g of I2? Mg + I2 → MgI2 Balanced? YES nMg = 75.0g/(24.31 g mol-1) = 3.085 mol nI2 = 75.0g/(253.9 g mol-1) = 0.2955 mol limiting 1Mg ≡ 1MgI2 (1Mg ≡ 1I2) 0.2955 mol I2 → 0.2955 mol MgI2 massMgI2 = 0.2955 mol 278.2 g = 82.2 g 1 mol Percent Yield Theoretical yield The amount of product predicted by stoichiometry. Actual yield The quantity of desired product actually obtained. Percent yield Actual yield % yield = Theoretical yield x 100% Percent Yield Few chemical reactions have 100% yield. Possible reasons Side reactions may produce undesired product(s). Incomplete reaction due to poor mixing or reaching equilibrium… Product loss during isolation and purification. Percent Yield 2.50 g of copper heated with an excess of sulfur made 2.53 g of copper(I) sulfide 16 Cu(s) + S8(s) → 8 Cu2S(s) What was the percent yield for this reaction? 1 mol nCu used: 2.50 g = 0.03934 mol Cu 63.55g Theoretical yield 16 Cu ≡ 8 Cu2S 0.03934 mol Cu 8 Cu2S = 0.01967 mol Cu2S 16 Cu Percent Yield 2.50 g Cu + S8 (excesss) made 2.53 g Cu2S… What was the %-yield? Theoretical yield (0.01967 mol Cu2S) = 0.01967 mol Cu2S 159.2 g = 3.131 g Cu2S 1 mol Actual yield = 2.53 g Cu2S (in problem) Percent yield = 2.53 g x 100% = 80.8% 3.131 g Atom Economy Examines the fate of all starting-material atoms. Sum the mass of all reactant atoms (∑Mi,reactants) and the mass of useful product atom(s) (∑Mi,useful prods): ∑Mi,useful prods Atom economy = ∑Mi,reactants High atom economy = low waste production Ideal reaction: high % yield & high % atom economy. Composition & Empirical Formulas Combustion analysis can be used to determine the empirical formula for organic compounds: H2O CO2 absorber absorber O2 Sample in a furnace Mg(ClO4)2 NaOH C and H are converted to CO2 & H2O. Both are trapped and the weight gain measured. Other elements (e.g. O) can be found by mass difference. massO = masssample – massC – massH Composition & Empirical Formulas Butyric acid contains C, H & O only. If 1.20 g is burned in O2, 2.41 g CO2 and 0.982 g H2O form. Determine its empirical and molecular formulas. M = 88.10 g mol-1 1 mol CO2 1 mol C 12.01 g C 2.41 g CO2 44.01 g CO2 1 mol CO2 1 mol C massC = 0.658 g C or 12.01 g C massC = 2.41 g CO2 = 0.658 g C 44.01 g CO2 2 H in H2O 2.016 g H massH = 0.982 g H2O = 0.110 g H 18.02 g H2O Composition & Empirical Formulas Combustion of 1.20 g butyric acid gave 2.41 g CO2 and 0.982 g H2O massO = masssample – massC – massH = 1.20 g – 0.658 g – 0.110 g = 0.432 g Convert to moles: 0.658 g C = 0.0548 mol C 12.01 g/mol 0.110 g H = 0.109 mol H 1.008 g/mol 0. 432 g O = 0.0270 mol O 16.00 g/mol Composition & Empirical Formulas Find the mole ratio (divide by smallest…): C 0.0548 / 0.0270 = 2.03 H 0.109 / 0.0270 = 4.03 O 0.0270 / 0.0270 = 1.00 Close to 2 : 4 : 1 (C : H : O) Empirical formula is C2H4O Composition & Empirical Formulas …butyric acid contains C, H & O only… … molar mass of butyric acid is 88.10 g/mol, find its empirical and molecular formula. Empirical formula = C2H4O Empirical mass = 2(12) + 4(1) + 1(16) g = 44 g Molar mass = 88.10 g Molar mass ≈ 2 x (empirical mass) Butyric acid has the molecular formula C4H8O2 Solution Concentration Relative amounts of solute and solvent. ▪ solute – substance dissolved. ▪ solvent – substance doing the dissolving. There are several concentration units. Most important to chemists: Molarity Molarity Molarity = nsolute mol/L units Vsolution(in L) V of solution not solvent. M ≡ mol/L Shorthand: IUPAC c(solute) = 5.12 M commonly seen [solute] = 5.12 M Brackets [ ] represent “molarity of ” Molarity 36.0 g of sodium sulfate are dissolved in enough water to make 750.0 mL of solution. Calculate the molarity of the Na2SO4. 36.0 g = 0.2534 mol nNa2SO4 = 142.0 g/mol 0.2534 mol c(Na2SO4) = Unit change! 0.7500 L mL → L c(Na2SO4) = 0.338 mol/L = 0.338 M Molarity 6.37 g of Al(NO3)3 are dissolved to make a 250. mL aqueous solution. Calculate (a) c(Al(NO3)3) (b) c(Al3+) and c(NO3-). (a) Al(NO3)3 molar mass = 26.98 + 3(14.00) + 9(16.00) = 213.0 g mol 6.37 g nAl(NO3)3= = 2.991 x 10-2 mol 213.0 g/mol 2.991 x 10 -2 mol c(Al(NO3)3) = = 0.120 M 0.250 L Molarity 6.37 g of Al(NO3)3 in a 250. mL solution. (a) c(Al(NO3)3)? (b) c(Al3+) and c(NO3-)? (b) Molarity of Al3+ , NO3-? Al(NO3)3(aq) → Al3+(aq) + 3 NO3-(aq) 1 Al(NO3)3 ≡ 1 Al3+ 1 Al 3+ c(Al3+) = 0.120 M Al(NO3)3 = 0.120 M Al3+ 1 Al(NO3)3 1 Al(NO3)3 ≡ 3 NO3- -) 3 NO - c(NO3 = 0.120 M Al(NO3)3 3 = 0.360 M NO3- 1 Al(NO3)3 Solution Preparation Solutions are prepared either by: 1. Dissolving a measured amount of solute and diluting to a fixed volume. or… 2. Diluting a more concentrated solution. Solution Preparation from Pure Solute Prepare a 0.5000 M solution of KMNO4 (potassium permanganate) in a 250.0 mL volumetric flask. Mass of KMnO4 required nKMnO4 = c(KMnO4) x V = 0.5000 M x 0.2500 L (M ≡ mol/L) = 0.1250 mol massKMnO4 = 0.1250 mol x 158.03 g/mol = 19.75 g Solution Preparation from Pure Solute 1. Weigh exactly 19.75 g of pure KMnO4. Transfer to a volumetric flask. 2. Rinse all solid from the weighing dish into the flask. 3. Fill the flask ≈ ⅓ full 4. Swirl to dissolve the solid 5. Fill the flask to the mark on the neck. 6. Shake to thoroughly mix Solution Preparation by Dilution cconcVconc = amount of solute = cdilVdil Example Commercial concentrated sulfuric acid is 17.8 M. If 75.0 mL of this acid is diluted to 1.00 L, what is the final concentration of the acid? cconc = 17.8 M Vconc = 75.0 mL cdil = ? Vdil = 1000. mL cconcVconc 17.8 M x 75.0 mL = 1.34 M cdil = = Vdil 1000. mL Stoichiometry in Aqueous Solution Mass Mass A Use molar B Use molar mass of B mass of A Amount Amount A B Use mole ratio Use solution Use solution molarity of A molarity of B Volume Volume B solution A solution nA = c(A) x V c(product) = nproduct / (total volume). Stoichiometry in Aqueous Solution 25.0 mL of 0.234 M FeCl3 and 50.0 mL of 0.453 M NaOH are mixed. Which reactant is limiting? How many moles of Fe(OH)3 will form? FeCl3(aq) + 3 NaOH(aq) → 3 NaCl (aq) + Fe(OH)3(s) nFeCl3 = 0.0250 L x 0.234 mol/L = 0.005850 mol nNaOH = 0.0500 L x 0.453 mol/L = 0.02265 mol Molarity & Reactions in Aqueous Solution FeCl3(aq) + 3 NaOH(aq) → 3 NaCl (aq) + Fe(OH)3(s) 0.005850 mol 0.01925 mol nFe(OH)3 ? 1 Fe(OH)3 0.00585 mol FeCl3 = 0.00585 mol Fe(OH)3 1 FeCl3 1 Fe(OH)3 0.02265 mol NaOH = 0.00755 mol Fe(OH)3 3 NaOH FeCl3 is limiting; 0.00585 mol Fe(OH)3 produced. Stoichiometry in Aqueous Solution A 4.554 g mixture of oxalic acid, H2C2O4 and NaCl was neutralized by 29.58 mL of 0.550M NaOH. What was the weight % of oxalic acid in the mixture? H2C2O4(aq) + 2 NaOH(aq) → Na2C2O4(aq) + 2 H2O(ℓ) nNaOH = 0.02958 L x 0.550 mol/L = 0.01627 mol 1 H2C2O4 ≡ 2 NaOH 1 H2C2O4 nacid = 0.01627 mol NaOH 2 NaOH = 8.135 x10-3 mol Stoichiometry in Aqueous Solution A 4.554 g H2C2O4 / NaCl mixture … Wt % of oxalic acid in the mixture? Mass of acid consumed, massacid = 8.135 x10-3 mol x (90.04 g/mol acid) = 0.7324 g massacid Weight % = sample mass x 100% 0.7324 g Weight % = x 100% = 16.08% 4.554 g Acids Increase the concentration of H+ ions in water. Protons (H+) always combine with water to form H3O+ (hydronium ion) and some larger clusters Sour tasting. Change the color of pigments (indicators) ▪ Litmus, phenolphthalein… H3O+ Most acids are molecular compounds (not ionic), but ionize in water – split into positive and negative ions. Acids Strong acids completely ionize (>99%) in water (strong electrolytes). Hydrochloric acid (HCl) ≈ 100% ionized in H2O Very few HCl molecules exist in solution. HCl(aq) → H+(aq) + Cl-(aq) Weak acids partially ionize (weak electrolytes). Acetic acid (CH3COOH) is 5% ionized in H2O. 95% intact. CH3COOH(aq) H+(aq) + CH3COO-(aq) Bases Increase the concentration of OH- (hydroxide ion) in water. Bases: Counteract an acid (neutralize an acid). Change an indicator’s color (phenolphthalein…). Have a bitter taste. Feel slippery. Bases can be “strong” or “weak”. H2O NaOH(s) Na+(aq) + OH-(aq) strong NH3(aq) + H2O(ℓ) NH4+(aq) + OH-(aq) weak Common Acids & Bases Strong Acids Strong Bases HCl Hydrochloric acid LiOH Lithium hydroxide HBr Hydrobromic acid NaOH Sodium hydroxide HI Hydroiodic acid KOH Potassium hydroxide HNO3 Nitric acid Ca(OH)2 Calcium hydroxide H2SO4 Sulfuric acid Ba(OH)2 Barium hydroxide HClO4 Perchloric acid Sr(OH)2 Strontium hydroxide Weak Acids Weak Bases HF Hydrofluoric acid NH3 Ammonia H3PO4 Phosphoric acid CH3NH2 Methylamine CH3COOH Acetic acid H2CO3 Carbonic acid HCN Hydrocyanic acid HCOOH Formic acid C6H5COOH Benzoic acid Neutralization Reactions acid + base → salt + water salt = ionic compound made from an acid anion and base cation. Salt HX(aq) + MOH(aq) → MX(aq) + H2O(ℓ) HBr(aq) + KOH(aq) → KBr(aq) + H2O(ℓ) H3PO4(aq) + 3 NaOH(aq) → Na3PO4(aq) + 3 H2O(ℓ) Neutralizations are exchange reactions. Net Ionic Equations for Acid-Base Reactions Strong Acid + Strong Base Overall: HX(aq) + MOH(aq) → MX(aq) + H2O(ℓ) Full ionic: H+(aq) + X-(aq) + M+(aq) + OH-(aq) → M+(aq) + X-(aq) + H2O(ℓ) Net ionic: H+(aq) + OH-(aq) → H2O(ℓ) Net Ionic Equations for Acid-Base Reactions Weak Acid + Strong Base Similar: HA(aq) + MOH(aq) → MA(aq) + H2O(ℓ) but the weak-acid remains (mostly) undissociated: Full ionic HA(aq) + M+(aq) + OH-(aq) → M+(aq) + A-(aq) + H2O(ℓ) Net ionic: HA(aq) + OH-(aq) → A-(aq) + H2O(ℓ) Titrations in Aqueous Solution Titration = volume-based method used to determine an unknown concentration. Standard solution (known concentration) is added to a solution of unknown concentration. ▪ Monitor the volume added. ▪ Add until equivalence is reached – stoichiometrically equal moles of reactants added. ▪ A color change (an indicator may be necessary) monitors the end point. Often used to determine acid or base concentrations. Titrations in Aqueous Solution Buret = volumetric glassware used for titrations. Slowly add standard solution until the end point is seen. Vtitrant and the known c(titrant) and Vunknown allow calculation of c(unknown). Titrations in Aqueous Solution 29.5 mL of 0.100 M H2SO4 was required to neutralize 25.0 mL NaOH. What was the NaOH concentration? H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2 H2O(ℓ) 0.100 mol nH2SO4= 0.0295 L = 2.95 x 10-3 mol L 2 NaOH nNaOH = 2.95 x 10-3 mol = 5.90 x 10-3 mol 1 H2SO4 5.90 x 10-3 mol c(NaOH) = = 0.236 M 0.0250L  John W. Moore Conrad L. Stanitski http://academic.cengage.com/chemistry/moore Chapter 4 Energy and Chemical Reactions (Part 1) Stephen C. Foster Mississippi State University The Nature of Energy Energy (E) = the capacity to do work. Work (w) occurs when an object moves against a resisting force: w = −(resisting force) x (distance traveled) w = −F d All energy is either Kinetic or Potential energy. The Nature of Energy Kinetic energy (Ek) - Energy of motion ▪ macroscale = mechanical energy ▪ random nanoscale = thermal energy ▪ periodic nanoscale = acoustic energy Ek = ½mv2 (m = mass, v = velocity of object) Potential energy (Ep) – Energy of position. Stored E. It may arise from: gravity: Ep = m g h (mass x gravity x height). charges held apart. bond energy. Energy Units joule (J) - SI unit (1 J = 1 kg m2s-2) 2.0 kg mass moving at 1.0 m/s (~2 mph): Ek = ½ mv2 = ½ (2.0 kg)(1.0 m/s)2 = 1.0 kg m2 s-2 = 1.0 J 1 J is a relatively small amount of energy. 1 kJ (1000 J) is more common in chemical problems. Energy Units calorie (cal) Originally: “Energy needed to heat of 1g of water from 14.5 to 15.5 °C.” Now: 1 cal = 4.184 J (exactly) Dietary Calorie (Cal) “big C” calorie Used on food products. 1 Cal = 1000 cal = 1 kcal = 4.184 kJ 1 sachet (16 kJ) is… …2 tsps of sugar (140 kJ) Energy Units energy Power = time SI unit: 1 watt (W) = 1 J s-1 A 60-W incandescent bulb is used for 6.0 hours. How much energy does it consume? E = power x time = 60 J s-1 (6.0 h) 60 m 60 s E = 60 J s-1 (6.0 h) 1h 1m E = 1.3 x 106 J = 1.3 MJ Conservation of Energy “Energy can neither be created nor destroyed”. It can only change form. Total E of the universe is constant. Also called the 1st Law of Thermodynamics. Conservation of Energy A diver: a) Has Ep due to macroscale position. b) Converts Ep to macroscale Ek. c) Converts Ek,macro to Ek,nano (motion of water, heat) Energy & Working If an object moves against a force, work is done. Lift a book ▪ you do work against gravity. The book’s Ep increases. Drop the book: ▪ Ep converts into Ek ▪ The book does work pushing the air aside. The book hits the floor ▪ no work is done on the floor (it does not move). ▪ Ek converts to a sound wave and T of the book and floor increase (Ek converts to heat). Energy, Temperature & Heating Temperature is a measure of the thermal energy of a sample. Thermal energy E of motion of atoms, molecules, and ions. Atoms of all materials are always in motion. Higher T = faster motion. Energy, Temperature & Heating Heat Thermal E transfer caused by a T difference. Heat flows from hotter to cooler objects until they reach thermal equilibrium (have equal T ). Keeping Track of Energy Transfers System = the part of the universe under study ▪ chemicals in a flask. ▪ my textbook. Surroundings = rest of the universe (or as much as needed…) ▪ the flask. ▪ perhaps the flask and this classroom. ▪ perhaps the flask and all of the building, etc. Universe = System + Surroundings Keeping Track of Energy Transfers Internal energy = sum of the individual energies of all nanoscale particles in the system. Einternal includes all nanoscale Ek and Ep. A larger sample (larger system) will have larger Einternal Nanoscale Ek = thermal energy Nanoscale Ep: ▪ ion/ion attraction or repulsion ▪ nucleus/electron attraction ▪ proton/proton repulsion ….. Keeping Track of Energy Transfers Internal energy depends on Temperature ▪ higher T = larger Ek for the nanoscale particles. Type of material ▪ nanoscale Ek depends upon the particle mass. ▪ nanoscale Ep depends upon the particle type(s). Amount of material ▪ number of particles. ▪ double sample size, double Einternal, etc. Calculating Thermodynamic Changes Energy change = final E – initial E ΔE = Efinal – Einitial “Change in” Systems can gain or lose E: SURROUNDINGS SURROUNDINGS SYSTEM SYSTEM Efinal Einitial ΔE > 0 E in ΔE < 0 E out Einitial Efinal ΔE positive: internal energy ΔE negative: internal energy increases decreases Conservation of Energy & Reactions No subscript? Refers to the system: E = Esystem E is transferred by heat or by work. Conservation of energy becomes: ΔE = q + w heat work SURROUNDINGS SYSTEM Heat transfer in Heat transfer out q>0 q0 w 0) increasing enthalpy E is absorbed. New bonds are less stable than the old, or products Fewer bonds are formed than broken reactants Measuring Reaction Enthalpies Heat transfers are measured with a calorimeter. Common types: Bomb calorimeter. rigid steel container. filled with O2(g) and a small sample to be burnt. constant V, so qV = ΔrE Flame calorimeter. samples burnt in an open flame. constant P, so qp = ΔrH Coffee-cup calorimeter in lab (constant P). Constant Volume Calorimetry Bomb Calorimeter Measure ΔT of the water. Constant V: q V = Δ rE Conservation of E: qreaction + qbomb + qwater = 0 or −qreaction = qbomb + qwater with qbomb = mcalccalΔT = CcalΔT A constant for a calorimeter Constant Volume Calorimetry Octane (0.600 g) was burned in a bomb calorimeter containing 751 g of water. T increased from 22.15 °C to 29.12 °C. Calculate the heat evolved per mole of octane burned. Ccal = 895 J°C-1. 2 C8H18(ℓ) + 25 O2(g) → 16 CO2(g) + 18 H2O(ℓ) qreaction + qbomb + qwater = 0 qbomb = CcalΔT = 895 J°C-1 (29.12 – 22.15)°C = +6238 J qwater = m c ΔT = 751 g (4.184 J g-1 °C-1)(29.12 – 22.15)°C = +2.190 x 104 J Constant Volume Calorimetry T = 22.15 → 29.12°C. Heat per mol octane (0.600 g) burned. 2 C8H18(ℓ) + 25 O2(g) →16 CO2(g) + 18 H2O(ℓ) −qreaction = qbomb + qwater −qreaction = +6238 J + 2.190 x 104 J = 2.81 x 104 J = 28.1 kJ qreaction = −28.1 kJ Constant Volume Calorimetry Octane (0.600 g)… Calculate the heat evolved per mole of octane burned Molar mass of C8H18 = 114.23 g/mol. nC8H18 = (0.600 g) / (114.23 g/mol) = 0.00525 mol C8H18 Heat evolved /mol octane = -28.1 kJ 0.00525 mol = -5.35 x 103 kJ/mol = -5.35 MJ/mol Constant Volume Calorimetry Coffee-cup calorimeter Nested styrofoam cups prevent heat exchange with the surroundings. Constant P. ΔT measured. q = qp = ΔH Assume the cups do not absorb heat. (Ccal = 0 and qcal = 0) Constant Volume Calorimetry 1.02 g of Mg was reacted with excess 1 M HCl(aq) (255.0 g) in a coffee-cup calorimeter. Tsoln rose from 22.0 to 41.6 °C. cHCl = 3.90 J g-1°C-1. Complete: Mg(s) + 2 HCl(aq) → H2(g) + MgCl2(aq) ΔrH° = ? qsoln = msolnc ΔT (msoln = macid + mMg ) = 256.0 g (3.90 J g-1 °C-1)(41.6 − 22.0)°C = 1.957 x 104 J Conservation of E: qsoln + qrxn = 0 qrxn = -qsoln qrxn = -19.57 kJ = ΔrH° Constant Volume Calorimetry 1.02 g Mg + 255.0 g of acid. T: 22.0 → 42.5 °C. Molar ΔrH° = ? qrxn = -19.57 kJ = ΔrH° for 1.02 g MMg = 24.31 g mol-1 For 1 mol -19.57 kJ 24.31 g ΔrH° = = -476 kJ exothermic 1.02 g 1 mol  John W. Moore Conrad L. Stanitski http://academic.cengage.com/chemistry/moore Chapter 4 Energy and Chemical Reactions (Part 2) Stephen C. Foster Mississippi State University Hess’s Law “If the equation for a reaction is the sum of the equations for two or more other reactions, then ΔrH° for the 1st reaction must be the sum of the ΔrH° values of the other reactions.” Another version: “ΔrH° for a reaction is the same whether it takes place in a single step or several steps.” H is a state function Hess’s Law Multiply a reaction, multiply ΔrH°. Reverse a reaction, change the sign of ΔrH° 2 CO(g) + O2(g) → 2 CO2 (g) ΔrH° = −566.0 kJ/mol Then 2 CO2(g) → 2 CO(g) + O2(g) ΔrH° = -1(-566.0 kJ/mol) = +566.0 kJ/mol 4 CO2(g) → 4 CO(g) + 2 O2(g) ΔrH° = -2(-566.0 kJ/mol) = +1132.0 kJ/mol Hess’s Law Use Hess’s Law to find kJ/mol for unknown reactions. Example It is difficult to measure ΔrH° for: 2 C(graphite) + O2(g) → 2 CO(g) Some CO2 always forms. Calculate ΔrH° given: C(graphite) + O2(g) → CO2(g) ΔrH° = -3