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All chem-notes

  1. 1. Chemistry: The Study of ChangeTEACHER: QBA MIGUEL ANGEL CASTRO RAMÍREZ
  2. 2. Introduction to Chemistry and theScientific Method
  3. 3. ask a draw a question conclusion do researchanalyze data design an experiment make observations and collect data
  4. 4. Chemistry: A Science for the 21st CenturyHealth and Medicine •Sanitation systems • Surgery with anesthesia • Vaccines and antibiotics Energy and the environment •Fossil fuels • Solar energy • Nuclear energy 1.1
  5. 5. Chemistry: A Science for the 21st Century Materials Technology•Polymers, ceramics, liquid crystals• Room-temperature superconductors?• Molecular computing? Food Technology •Genetically modified crops • “Natural” pesticides • Specialized fertilizers 1.1
  6. 6. The Study of Chemistry Macroscopic MicroscopicChemists study the microscopic properties of matter, which in turnproduce matter’s observable macroscopic properties – thus, weoften switch back and forth between microscopic and macroscopicviews of matter in this course. 1.2
  7. 7. The scientific method is a systematic approachto research. Although it is systematic, it is not arigid series of steps that must be done in a particularorder. ask a question draw a do research conclusion researcher’s form a hidden bias hypothesis analyze design an data experiment make observations and collect data
  8. 8. A hypothesis is a tentative tested modifiedexplanation for a set ofobservations that can be tested. A theory is a unifying principle that explains a body of facts and/or those laws that Atomic Theory are based on them.A law is a concise statement of arelationship between phenomena Force = mass xthat is always the same under the accelerationsame conditions. 1.3
  9. 9. Classification of Matter
  10. 10. SubstancesMatter is anything that has mass and occupiesspace.Matter that has a uniform and unchangingcomposition is called a (pure) substance.examples of pure substances include table salt, purewater, oxygen, gold, etc.
  11. 11. States of MatterMatter normally occupies one of threephases, or states. These are: P Solid P Liquid P Gas* Plasma is a 4th state of matter in which the particles areat extremely high temperatures (over 1,000,000 °C).
  12. 12. States of MatterAs we shall see in more detail later, the phase (orstate) of a substance is determined by the averagekinetic energy of the particles that make up thesubstance, (i.e., temperature) and the strength of theattractive forces holding the substance’s particlestogether. moderate liquid weak gas strong solid
  13. 13. States of MatterSolids Solids have a definite shape and volume. The particles of a solid cannot exchange positions. Solids are incompressible.
  14. 14. States of MatterLiquids Liquids have definite volumes Liquids do not have a fixed shape Like solids, liquids are also incompressible
  15. 15. States of MatterGases Gases take on the shape and volume of their container Unlike solids and liquids,gases are highly compressible
  16. 16. States of MatterTechnically, the word“gas” refers to asubstance that is in thegas phase at roomtemperature.The word “vapor” refersto the gaseous state of asubstance that is normallya solid or liquid at roomtemperature.
  17. 17. Classification of MatterMatter can be classified based on its characteristicsinto the following categories and subcategories: 1. mixtures • homogeneous (solution) • heterogeneous 2. (pure) substances • compounds • elements
  18. 18. Classification of MatterA pure substance is a form of matter that has adefinite composition and distinct properties.examples: gold, salt, iron, pure water, sugarA mixture is a combination of two or moresubstances in which each substance retains itsown distinct identity.examples: salt water, oil & vinegar dressing, granite, air
  19. 19. Classification Summary see page 13
  20. 20. Classification of MatterMixturesMixtures can be heterogeneous or homogeneous.Heterogeneous mixture : the composition is not uniform throughout. You can visibly see the different components. examples: cement, iron filings in sand, granite, milk, oil and water, etc.
  21. 21. Classification of MatterHomogenous mixture (also called a solution): The composition of the mixture is the same throughout.Solutions are made up of two components:(1) the solute which is dissolved in(2) the solvent.If the solvent is water, the solution is called an aqueous solutionwhich is symbolized: (aq).
  22. 22. Classification of MatterWe often think of a solution as being a soliddissolved in a liquid. However…In a solution, both the solvent and solute can be inany phase – solid, liquid or gas. solvent solute example liquid solid salt dissolved in water liquid liquid gasoline (a mix of liquids) gas gas air (O2 dissolved in nitrogen) solid solid alloys (brass, bronze, etc.)
  23. 23. Classification of MatterIf a substance dissolves in another substance, wesay the first substance is soluble in the second. Ifthey do not dissolve, they are said to be insoluble.example: carbon dioxide is soluble in air gold is insoluble in waterIn the case of liquids, we use a special term:If two liquids completely dissolve in each other, theyare said to be miscible. If they do not, they areimmiscible.example: alcohol and water are miscible gasoline and water are immiscible
  24. 24. Classification of MatterA mixture can be separated into its purecomponents by simple physical methods.Filtration is a means of separating asolids from liquids. For example, wecan filter out the sand from a mix ofsand and water. Magnetic substances can be separated using a magnet.
  25. 25. Separation of a MixtureFractional crystallization is a means of separatingtwo solids by adding a solvent that will dissolve oneof the solids but not the other; the mixture is thenfiltered to separate out the insoluble solid. Finally,the solvent is evaporated off to recover theremaining solid.For example, we can separatesalt from sand by adding hotwater to dissolve the salt, thenfilter off the sand. The water isthen evaporated off, leaving thesalt behind.
  26. 26. Separation of a MixtureDistillation is a means of separating two liquidsbased on differences in their boiling points. The substance with the lowest boiling point “boils off” and is then cooled and condensed back into a liquid. The liquid is collected in a receiver flask. This method is only effective for substances that are liquids at room temperature with significant differences in their boiling points. distillation apparatus
  27. 27. Separation of a MixtureChromotography is the separation of a mixturebased on solubility in a “mobile” solvent coupled withan adherence to a “stationary phase” medium, suchas paper or silica gel, etc. column chromotography is a common means of separating components from a mixture Thin Layer Chromotography can be used to separate the components of chlorophyll from a crushed plant leaf.
  28. 28. Separation of a MixtureOther means of separation:Other techniques of separating a mixture includesublimation, extraction, and leaching, etc.If you had a jar containing both nails and marbles,the only way to separate them would be by hand speak to the hand…
  29. 29. Example:You are given a test tube which contains a mixture of water, methanol, aspirin,acetanilide and aluminum oxide. (Acetanilide, aluminum oxide and aspirin areall white, powdery solids at room temperature and are thus visiblyindistinguishable from each other. Water and methanol are both colorlessliquids at room temperature and are also visibly indistinguishable from eachother.) Assume your only source of heat is a Bunsen burner which can producea maximum temperature of 600C. Using the following information, devise amethod to separate this mixture. Be specific and complete in your answer.substance melting point boiling point what it dissolves/does not dissolve inwater 0 C 100 C dissolves in cold or hot methanolmethanol – 97 C 65C dissolves in cold or hot wateraspirin 135 C decomposes at dissolves in methanol or water (if above 140C 10C)aluminum 2072C 2980 C does not dissolve in either methanol oroxide water at any temperatureacetanilide 114C 304C dissolves only in hot (50C) water or warm (25°C) methanol
  30. 30. Pure Substances:Elements and Compounds Mixtures are composed of two or more substances physically combined. Recall that a (pure) substance is matter that has a uniform, unchanging composition Pure substances may be elements orcompounds
  31. 31. Classification Summary see page 13
  32. 32. Classification of MatterElementsAn element is a substance that cannot beseparated into simpler substances by chemicalmeans. carbon sulfur • 114 elements have been identified mercury • 82 elements occur naturally on Earth examples include carbon, sulfur, copper iron copper, iron, and mercury • 32 elements have been synthesized by scientists. examples: technetium, americium, and seaborgium
  33. 33. Symbols for ElementsElements are identified by a one or two-letter symbol. The first letter, which is ALWAYS capitalized, istypically the first letter in the name of the element. eg, C = carbon, H = hydrogenThe second letter (which is only used if other elementshave the same first letter) is NEVER capitalized. eg, Cl = chlorine, He = helium.Some symbols are based on the Latin name eg, iron is Fe (for ferrum) and sodium is Na (for natrium)
  34. 34. Symbols for Elementshttp://www.privatehand.com/flash/elements.html
  35. 35. Classification of MatterCompoundsA compound is a substance composed of atomsof two or more different elements chemicallybonded in fixed proportions.As such, they can be chemically decomposedinto their component elements. table salt (NaCl) sugar Water (H2O) Sucrose (C12H22O11)
  36. 36. Classification of MatterThe properties of a compound are different fromthe properties of its component elementsFor example, table salt is composed of sodium and chlorine.Sodium is a soft, silver colored metal that reacts violentlywith water, and chlorine is a pale-green poisonous gas – yetwhen chemically combined, they form table salt, a whitecrystalline solid you put on your eggs in the morning! + =
  37. 37. CompoundsCompounds can only be separated into their purecomponents (elements) by chemical means.For example:Iron is separated from iron ore (Fe2O3)by heating the ore in a blast furnace andreacting it with carbon monoxide andelemental carbon (in the form of “coke”).Water can be separated into itselements, hydrogen and oxygen, bypassing an electric current through it,a process called electrolysis.
  38. 38. CompoundsThere are TWO kinds of compounds, depending onthe nature of the chemical bond holding the atomstogether.Molecules form when two or more neutral atomsform bonds between them by sharing electronsNote that some elements exist as molecules. For example,the following elements occur innature as molecular diatomic elements: H2 O2 H2, N2, O2, F2, Cl2, Br2 and I2They are molecules, but they are NOT N2compounds, because they have only Cl2one kind of element present.
  39. 39. CompoundsIonic compounds are composed of ions, which areatoms that have a (+) or (-) charge.+ ions are called cations and form + ─ + ─ + ─ + + when an atom loses ─ + ─ ─ ─electrons + ─ + + ─ ions are called anions and form when an atom gains electronsIonic compounds form when cations and anionsform electrostatic attractions between them(opposite charges attract)
  40. 40. Classification MATTER Summary can it be separated YES NO by physical means? MIXTURE PURE SUBSTANCE is the mixture uniform can the substance be throughout? chemically decomposed into simpler substances? YES NO YES NO heterogeneoussolution compound element mixture
  41. 41. Propertiesof Matter
  42. 42. Physical & Chemical PropertiesPhysical Properties are measurable properties • mass $ density • boiling point $ solubility in waterChemical Properties describe how asubstance reacts with other substances • flammability $bonds with oxygen • reacts with water $decomposes when heated
  43. 43. Extensive and Intensive PropertiesPhysical properties can be classified as being eitherextensive or intensive properties.An extensive property of a material depends uponhow much matter is being considered. Extensiveproperties are additive. • mass • length • volume
  44. 44. Extensive and Intensive PropertiesAn intensive property of a material is independentof the amount of matter is being considered, and isnot additive. • density • melting point • temperature •color Note that ALL chemical properties are intensive properties.
  45. 45. Physical & Chemical ChangesA physical change does not alter the compositionor identity of a substance. sugar dissolving ice melting in waterA chemical change (reaction) alters the identityor composition of the substance(s) involved. hydrogen burns in air to form water
  46. 46. Physical & Chemical ChangesEvidence of a chemical reaction include:1. Heat and light (both) produced2. Gas produced (bubbles)3. Solid precipitate forms4. Color changes occur
  47. 47. Measurement
  48. 48. MeasurementThe SI System of MeasurementScientists around the world use a unified system ofmeasurement (Le Systeme Internationale d’Unites,or SI for short).There are seven fundamental “quantities” thatcan be measured: Length Temperature Luminous intensity Mass Electric Current Time Chemical quantity
  49. 49. International System of Units (SI)Each base quantity is given a unit with aspecific name and symbol page 16
  50. 50. International System of Units (SI)The SI units are based on metrics. Each powerof ten change is given a special prefix used withthe base unit. You must know these prefixes see page 17
  51. 51. Measurements with SI UnitsLength (SI unit = meter) The meter is oftendivided into cm and mm. (10 mm = 1 cm ).Your little finger is about 1 cm in width.A dime is about 1 mm thick.English/Metric equivalencies 1 inch = 2.54 cm 1 meter = 39.37 inches
  52. 52. Measurements with SI UnitsVolume (SI unit = m3) Volume is the amount ofspace occupied by something.A more common unit is the dm3 =1 liter.A smaller unit that we will use frequently is the cm 3. 1 cm3 = 1 ml 1000 ml = 1 liter English/Metric equivalencies 1 liter = 1.057 quarts 1 ml ~ 15 drops
  53. 53. Measurements with SI UnitsMeasuring Volumeregular solids: volume = length x width x heightliquids—use a graduated cylinder. To read the scale correctly, read the volume at the lowest part of the meniscus - the curve of the liquid’s surface in a container. Your eye should be level with the meniscus when reading the volume meniscus
  54. 54. Measurements with SI UnitsMeasuring Volume continuedirregular solids: volume is found by displacement.Begin with a known volume of water. Add the solid.The amount of water displaced is the volume of thesolid. volume of solid = 6 6 volume displaced : 6.0 – 4.0 = 2.0 cm3 4 4 2 2
  55. 55. Measurements with SI UnitsMass (SI unit = kilogram): the amount of matter.The mass of a given object is constant.A kilogram is about 2.2 pounds -- this is too large aunit for most chemistry labs, so we will use gramsinstead. Note that mass and weight are two different things…
  56. 56. Measurements with SI UnitsWeight is a measure of the force due to gravityacting on a mass. The weight of an objectchanges, depending on the gravitational forceacting on it.For example, on the moon you would weigh only 1/6th whatyou do on Earth, because the force of gravity on the moonis only 1/6th that of Earth. http://www.exploratorium.edu/ronh/weight/
  57. 57. The Importance of UnitsOn 9/23/99, the Mars Climate Orbiter entered Mar’s atmosphere 100 km (62miles) lower than planned and was destroyed by heat because theengineers that designed the rocket calculated the force provided by theengines in pounds, but NASA engineers thought the force was given in theunits of Newtons (N) when they determined when to fire the rockets… 1 lb = 1 N 1 lb = 4.45 N “This is going to be the cautionary tale that will be embedded into introduction to the metric system in elementary school, high school, and college science courses till the end of time.”
  58. 58. Measurements with SI UnitsMeasuring Mass Triple beam balance Electronic balanceWe still use the term “weighing” even though we are findingthe mass of an object, not its weight…English/Metric equivalencies 1 kg = 2.203 lbs 1 paperclip  1 gram 1 lb = 453.6 grams
  59. 59. Measurements with SI UnitsTemperature (SI unit = kelvin) is a measure of theaverage kinetic energy (energy due to motion) ofthe atoms and molecules that make up a substance.There are three common temperature scalesFahrenheit (oF) – English system, based on the freezingpoint of salt water.Centigrade (oC) – metric system, based on the freezing andboiling points of pure waterKelvin (K) – SI unit, also called the “Absolute” scale; 0 K(Absolute Zero) is defined as the temperature at which allmotion stops (kinetic energy = 0).
  60. 60. Temperature Conversions: K = oC + 273.15 273 K = 0 oC 373 K = 100 oCo C = 5 (oF – 32) 9 9o F= (oC) + 32 5 32 oF = 0 oC 212 oF = 100 oC
  61. 61. TemperatureExamplesA thermometer reads 12o F. What would this be in oC ?The conversion formula from oF to oC is: oC = 5/9(oF – 32)Inserting the values gives: : oC = 5/9(12oF – 32) o C = 5/9(-20) = -11.1oCA thermometer reads 315.3 K. What would this be in oF ?First convert K to oC: 315.3 K – 273.15 = 42.15oCThe conversion formula from oC to oF is: oF = 9/5(oC) +32.Inserting the values gives: : oC = 9/5(42.15 oC) + 32 o C = (75.9) + 32 = 107.9 oF
  62. 62. page 21
  63. 63. Measurements with SI UnitsTime (SI unit = second). This is the only non-metric SI unit. We still use 1 day = 24 hours,1 hour = 60 minutes, 1 minute = 60 seconds We do use metric fractions of time, however, such as milliseconds (1/1000th of a second), etc.Chemical Quantity ( SI unit = mole). Since atomsare so tiny, it takes a LOT of them to make evenone gram. In fact, you would have to put602,200,000,000,000,000,000,000 atoms of carbon(that’s 6.022 X 1023) on a balance to get just 12grams of carbon!
  64. 64. Measurements with SI UnitsThe Mole continuedThat huge number (6.022 X 1023) is given a special name; it iscalled “Avogadro’s Number,” symbolized NA, after the Italianphysicist, Lorenzo Romano Amedeo Avogadro who livedbetween 1776-1856.Just like 1 dozen = 12 things, we define: 1 mole = 6.022 X 1023 things Avogadro
  65. 65. JUST HOW BIG IS AVOGADRO’S NUMBER?? 1 mole of oranges would cover the surface of the earth to a depth of 9 miles! If you stacked 1 mole of notebook paper, it would take you 5,800 years, traveling at the speed of light (186,000,000miles per second) to reach the top of the stack! If you were given 1 mole of dollar bills when the universebegan 13 billion years ago, and you immediately beganspending money at the rate of one million dollars per second, you would still have about 190 billion trillion dollars left ! but 1 mole of Hydrogen atoms would only mass about 1gram!
  66. 66. Working with Numbers: Scientific Notation The number of atoms in 12 g of carbon: 602,200,000,000,000,000,000,000 6.022 x 1023 The mass of a single carbon atom in grams: 0.0000000000000000000000199 1.99 x 10-23N is a number n is a positive orbetween 1 and 10 N x 10 n negative integer
  67. 67. Measurements with SI UnitsDerived UnitsAlthough we need only seven fundamental SI units,we can combine different units to obtain new units,called derived units.For example, speed is distance per unit time, so wemust combine the unit for distance (m) and time(sec) to get the SI unit for speed:speed = meters per second (m/s)We will be working with many different derivedunits in this course. It is important to pay attentionto the individual units that make up derived units!!
  68. 68. Derived UnitsDensity is the mass per unit volume of a substance. It iscalculated using the equation: mass m density = volume d= VSI derived unit for density is kg/m 3 . This is not a convenientunit in chemistry, so we usually use the units g/cm3 or g/mLEvery substance has a unique density. For example: substance density You need to know the density of water. gasoline 0.70 g/cm3 Any object that is more dense than water water 1.00 g/cm 3 will sink in water; if it is less dense, it will aluminum 2.70 g/cm3 float in water lead 11.35 g/cm3
  69. 69. page 18
  70. 70. page 18
  71. 71. Dimensional Analysis: A problem solving technique desired unit given unit x = desired unit given unit
  72. 72. The Mathematics of UnitsIn algebra, we learn that: u x u = u2 and… (2u)3 = 8 u3 u u = 1 (the u’s cancel!) and u x a = a uIf we let “u” = units, then every measured quantity isa number x a unit. We can solve problems bysetting them up so that the unit we do NOT wantgets cancelled out by dividing u/u in the problem.Thus, if a/u is a conversion (say 100 cm/1 m) thenwe can convert cm to meters etc. using thisconversion factor so that the cm cancel…
  73. 73. Dimensional Analysis Method of Solving Problems1. Determine which unit conversion factor(s) are needed2. Carry units through calculation3. If all units cancel except for the desired unit(s), then the problem was solved correctly. given quantity x conversion factor = desired quantity given unit x desired unit = desired unit given unit
  74. 74. Dimensional Analysis Method of Solving ProblemsExample: How many μm are in 0.0063 inches?Begin with what units you have “in hand,” then make a list ofall the conversions you will need.conversion factors needed: 0.0063 in = ? 1 inch = 2.54 cm 106 μm = 1 m 1 m = 100 cm 2.54 cm 1m 106 μm0.0063 inch x x x = 160 μm 1 inch 10 cm 2 1m
  75. 75. Dimensional Analysis Method of Solving ProblemsExample: The speed of sound in air is about 343 m/s. Whatis this speed in miles per hour? (1 mile = 1609 meters) conversion units meters to miles 1 mi = 1609 m seconds to hours 1 min = 60 s 1 hour = 60 min m 1 mi 60 s 60 min mi 343 x x x = 767 s 1609 m 1 min 1 hour hour
  76. 76. page 29
  77. 77. page 30
  78. 78. Uncertainty,Precision & Accuracyin Measurements
  79. 79. Measurements with SI UnitsUncertainty, Precision and Accuracy inMeasurementsWhen you measure length using a meterstick, youoften have to estimate to the nearest fraction of aline.The uncertainty in a measured value is partly dueto how well you can estimate such fractional units.The uncertainty also depends on how accurate themeasuring device, itself, is. http:// www.mhhe.com/physsci/chemistry/chang7/esp/folder_structure/ch/
  80. 80. Precision and AccuracyAccuracy – how close a measurement is to the true oraccepted valueTo determine if a measured value is accurate, you wouldhave to know what the true or accepted value for thatmeasurement is – this is rarely known!Precision – how close a set of measurements are toeach other; the scatter of repeated measurementsabout an average.We may not be able to say if a measured value is accurate,but we can make careful measurements and use goodequipment to obtain good precision, or reproducibility.
  81. 81. Precision and AccuracyA target analogy is often used to compare accuracy andprecision. accurate precise not accurate & but & precise not accurate not precise
  82. 82. Precision and Accuracyexample: which is more accurate: 0.0002 g or 2.0 g?answer: you cannot tell, since you don’t know what theaccepted value is for the mass of whatever object this is thatyou are weighing!example: which is more precise: 0.0002 g or 2.0 g?answer: surprisingly, the most precise value is 2.0 g, not the0.0002 g. The number of places behind the decimal is notwhat determines precision! If that were so, I could increasemy precision by simply converting to a different metric prefixfor the same measurement:Which is more precise: 2 cm or 0.00002 km? They are, infact, identical!
  83. 83. Precision and AccuracyPrecision is a measure of the uncertainty in a measured value.Any measured value is composed of those digits of which youare certain, plus the first estimated digit. 1 2 3 4 1 The length of the object is at least 1.7 cm, and we might estimate the last digit to be half a unit, and say it is 1.75 cm long. Others might say 1.74 or possibly 1.76 – the last digit is an estimate, and so is uncertain.
  84. 84. Precision and AccuracyWe always assume an uncertainty of ±1 in the last digit.The percent error in a measured value is defined as: % error = ± uncertainty x 100 measured valueThe smaller the percent error, the greater the precision – thesmaller the % error, the more likely two measurements will beclose together using that particular measuring instrument.Thus: 2.0 ± 0.1 has a % error of (0.1/2.0) x 100 = ±5%but 0.0002 has a % error of (0.0001/0.0002) x 100 = ± 50%
  85. 85. Percent DifferenceWhen determining the accuracy of an experimentallydetermined value, it must be compared with the “acceptedvalue.” One common method of reporting accuracy is calledthe percent difference ( %) – this gives how far off yourvalue is, as a percent, from the accepted value:Percent difference: experimental value – accepted value % = x 100 accepted value
  86. 86. example: In an experiment, a student determines thedensity of copper to be 8.74 g/cm 3. If the accepted value is8.96 g/cm3, determine the student’s error as a percentdifference.% = experimental value – accepted value x 100 accepted value 8.74 – 8.96 x 100 = − 2.46 % % = 8.96 The (-) sign indicates the experimental value is 2.46% smaller than the accepted value; a (+) % means the experimental value is larger than the accepted value.
  87. 87. Precision and AccuracyWe will be doing math operations involving measurementswith uncertainties, so we need a method of tracking how theuncertainty will affect calculated values – in other words, howmany places behind the decimal do we really get to keep theanswer?The method requires us to keep track of significant digits.Significant digits (or significant figures) are all ofthe known digits, plus the first estimated oruncertain digit in a measured value.
  88. 88. Significant Figures: Rules• Any digit that is not zero is significant 1.234 kg 4 significant figures• Zeros between nonzero digits are significant 606 m 3 significant figures• Zeros to the left of the first nonzero digit are not significant 0.08 L 1 significant figure• If a number is greater than 1, then all zeros to the right ofthe decimal point are significant 2.0 mg 2 significant figures• If a number is less than 1, then only the zeros that are atthe end and in the middle of the number are significant 0.00420 g 3 significant figures
  89. 89. Significant Figures: Rules How many significant figures are in each of the following measurements?24 mL 2 significant figures3001 g 4 significant figures0.0320 m3 3 significant figures6.4 x 104 molecules 2 significant figures560 kg You cannot tell!!
  90. 90. Significant Figures: RulesSuppose you wanted to estimate the number of jellybeans in ajar, and your best guess is around 400.Now – is the uncertainty in your estimate ±1 jellybean, or is it±10 jellybeans, or maybe even ±100 jellybeans (if you weren’tvery good at estimating jellybeans…)We need a way to write 400 and indicate in some waywhether that was 400 ±1 vs 400±10 vs 400±100. The plainnumber “400” is ambiguous as to where the uncertain digit is.Use scientific notation to remove the ambiguity: 400 ± 1 = 4.00 x 102 = 3 sig figs 400 ± 100 = 4 x 102 = 1 sig fig 400 ± 10 = 4.0 x 102 = 2 sig figs
  91. 91. Rounding NumbersGiven the number 6.82 and asked to round to 2 sig digits wewould write 6.8. We write 6.8 because 6.82 is closer to 6.8than it is to 6.9Given the number 6.88 and asked to round to 2 sig digits, wewould write 6.9. We write 6.9 because 6.88 is closer to 6.9than it is to 6.8You were taught this long ago.You were also probably taught that, given the number 6.85,and asked to round this to 2 sig digits, you would write 6.9.My question is, WHY did you round UP? 6.85 is JUST asclose to 6.8 as it is to 6.9! Since it is in the middle, it could berounded either way! And we should round it “either way.”
  92. 92. Rounding NumbersSince the rounding is “arbitrarily” up, this can introducesome round-off errors in chain calculations involving thisnumber – the final value will be too large if you alwaysround up when the next digit is exactly 5.Because rounding is “arbitrary” when the next digit isexactly 5, we introduce the following “odd-even roundingrule: When the next digit is exactly 5, round up or down to make the number an even number. e.g. : 4.65 ≈ 4.6 but 4.75 ≈ 4.8Note however, that 4.651 is closer to 4.7 than 4.6, so weround it to 4.7: only invoke the “odd-even rule” when thenext digit is exactly 5.
  93. 93. Math Operations with Significant DigitsWe need a set of rules to determine how theuncertainty or error will “propagate” or move through aseries of calculations and affect the precision of ourfinal answer.There is one rule for additionand subtraction, and one rulefor multiplication and division.Do not mix them and matchthem and confuse them!
  94. 94. Significant FiguresAddition or Subtraction The answer cannot have more digits to the right of the decimal point than any of the original numbers. 89.332 +1.1 one digit after decimal point 90.432 round off to 90.4 3.70 two digits after decimal point -2.9133 0.7867 round off to 0.79
  95. 95. Significant FiguresAddition or SubtractionWe often encounter two numbers that must be added thatare in scientific notation. We cannot add them anddetermine the number of places “behind the decimal” unlessthey have the same power of 10 – we may have to convert!Example: What is the sum of 2.4 x 102 + 3.77 x 103 ? 3.77 x 103 Always convert the smaller power of 10 to the 0.24 x 103 larger power of 10 4.01 x 103 The answer is good to 2 behind the decimal when written as x103 -- that is, the uncertain digit is in the “tens” place (± 10)
  96. 96. Significant FiguresTo determine the power of 10, visualize a see-saw when youmove the decimal point: 10n Increasing the power of 10 + n means you must move the decimal to the LEFT one n place for each power of 10 increase 10n + n Moving the decimal determines both the magnitude and the +/- n value of 10n Decreasing the power of 10 +n means you must move the 10 n decimal to the RIGHT by one n place for each power of 10 decrease.
  97. 97. Significant FiguresExample: What is the answer to the following, to the correctnumber of significant digits? 3.0268 x 10-2 - - 0.012 x 10-2 0 1.2 x 10-4 3.0148 x 10-2 = 3.015 x 10-2
  98. 98. Significant FiguresMultiplication or DivisionThe number of significant figures in the result is set by theoriginal number that has the smallest number of significantfigures 4.51 x 3.6666 = 16.536366 = 16.5 3 sig figs round to 3 sig figs 6.8 ÷ 112.04 = 0.0606926 = 0.061 round to 2 sig figs 2 sig figs
  99. 99. page 25
  100. 100. Significant FiguresExact Numbers Numbers from definitions or numbers of objects are considered to have an infinite number of significant figuresExample: Find the average of three measured lengths: 6.64, 6.68 and 6.70 cm. These values each have 3 significant figures 6.64 + 6.68 + 6.70 = 6.67333 = 6.67 =7 3Because 3 is an exact number the answer is not rounded to 7, butrather reported to be 6.67 cm (three sig figures). 1.8
  101. 101. Atoms, Molecules and Ions Chapter 2Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
  102. 102. Early IdeasOur understanding of the structure ofmatter has undergone profound changesin the past century.Nonetheless, what we know today didnot arrive on a sudden inspiration. Wecan trace a fairly steady ploddingtowards our current understanding,starting as far back as 400 BCE…
  103. 103. Early IdeasDemocritus (c.a. 400 BCE)All matter was composed of tiny, indivisibleparticles called atoms (atomos = indivisible)Each kind of matter had its own unique kind ofatom – ie., there were water atoms, air atoms, fireatoms, bread atoms, etc.The properties of matter could be explained by theshape and size of its atoms. Fire atoms water atoms “ouch!” rolls & flows
  104. 104. Early IdeasMost importantly, Democritus believed atomsexisted in a vacuum – that is, there was “nothing”in the spaces between atoms… Vacuum??Aristotle, among others, refusedto believe in the existence of“nothingness” that still occupiedspace…As a result, Democritus’ ideas were not very wellreceieved. It would be some 1200 years before theidea of atoms was revisited!
  105. 105. Early IdeasAristotleAristotle was the court philosopher to Alexander theGreat. Because of this, Aristotle’s ideas were given alot of weight . Aristotle believed that all matterwas composed of four elements: earth, air, fire and water.
  106. 106. Early IdeasThese elements could be “inter-converted” into eachother by exchanging the “properties” of hot, cold, dryand wet. FIRE hot dry AIR EARTH wet cold WATERexampleHeating WATER exchanged “hot” for “cold” which created “AIR”(which we see as steam…) WATER (cold, wet) AIR (hot, wet)
  107. 107. Early IdeasThis idea that one kind ofelement could be convertedinto another eventually ledto the belief in Alchemy –that one could turn lead intogold by performing the rightchemical reaction!
  108. 108. Early IdeasThe “scientific method” of inquiry was developedduring the 17th and 18th centuries. The invention ofthe balance and other instruments soon led to a newunderstanding about the nature of matter.The French chemist, Antoine-Laurent Lavoisier (1743-1794),presented two important ideaswhich would later help lead toa new, more developed atomictheory of matter…
  109. 109. Lavoisier1. The Law of Conservation of Matter: matter is not created or destroyed in chemical reactions. Any atomic theory would have to explain why matter is not gained or lost in reactions.2. Lavoisier defined element as any substance that could not be chemically broken down into a simpler substance. Lavoisier was a meticulous experimenter. He also helped develop the metric system of measurement. He is often called the “Father of Modern Chemistry,” in recognition of his pioneering works.
  110. 110. Lavoiser experimenting with respiration
  111. 111. Early IdeasJoseph Proust, another 18th century Frenchscientist, proposed the Law of DefiniteProportion, which states that the mass ratios ofelements present in different samples of the samecompound do not vary.For example, the percent by mass of the elementspresent in sugar are always found to be:53.3% oxygen, 40.0% carbon and 6.7% hydrogen.
  112. 112. John Dalton (1766-1824)Dalton started out as anapothecarys assistant (today,we would call him a pharmacist). He was also interested in bothmeteorology and the study ofgases.Dalton developed a new atomic theory of the natureof matter based on several postulates. His theorydiffered significantly from the early ideas ofDemocritus, but they both agreed that the simplestform of matter was the atom.
  113. 113. Dalton’s Atomic Theory (1808)1. Elements are composed of extremely small particles called atoms.2. All atoms of a given element are identical, having the same size, mass and chemical properties. The atoms of a given element are different from the atoms of all other elements.3. Compounds are composed of atoms of more than one element. In any compound, the ratio of the numbers of atoms of any two of the elements present is either an integer or a simple fraction.4. A chemical reaction involves only the separation, combination, or rearrangement of atoms; it does not result in their creation or destruction.
  114. 114. Dalton’s Atomic TheoryLaw of Conservation of Matter andDefinite Proportion Explained… + = 16 X + 8Y 8 X2Y
  115. 115. Law of Multiple ProportionsIf Dalton’s ideas about atoms were correct, then heproposed that the mass of a compound containingdifferent numbers of a given element (atom) wouldvary by the mass of that one whole atom – that is:If two elements can combine to form more thanone compound, then the masses of one elementthat combine with a fixed mass of the otherelement are in ratios of small, whole numbers.
  116. 116. Dalton’s Atomic Theory Consider the mass ratio of oxygen to carbon in the two compounds: CO and CO2 16 = 1.33 12 2.67 / 1.33 = 2 32 = 2.67 12Note that the mass of oxygen that combines with 12 g ofcarbon in carbon dioxide is 2 x greater than the mass of oxygenthat combines with 12 g of carbon in carbon monoxide.
  117. 117. Modern IdeasIn the late 19th and early 20th centuries, threeimportant experiments that shed light on thenature of matter were conducted: 1. J.J. Thomson’s investigation of cathode rays that led to the discovery of the electron. 2. Robert Millikan’s “Oil drop experiment” that determined the charge and mass of the electron. 3. Ernest Rutherford’s “Gold foil experiment” that finally gave us the current “nuclear” model of the atom.
  118. 118. Cathode rays, discovered by William Crookes,are formed when a current is passed through anevacuated glass tube. Cathode rays areinvisible, but a phosphor coating makes themvisible.
  119. 119. J.J. ThomsonThe Electron is Discovered J.J. Thomson helped show that cathode rays are made up of negatively charged particles (based on their deflection by magnetic and electric fields). Sir Joseph John Thomson 1856-1940 N S
  120. 120. J.J. ThomsonThomson showed that all cathode rays are identical,and are produced regardless of the type of metalsused for the cathode and anode in the cathode raytube.Thomson was unable to determine either the actualelectric charge or the mass of these cathode rayparticles. He was, however, able to determine theratio of the electric charge to the mass of theparticles.
  121. 121. J.J. ThomsonTo do this, he passed cathode rays simultaneously throughelectric and magnetic fields in such a way that the forcesacting on the cathode ray particles (now called electrons)due to the fields cancelled out. The ratio of the electric fieldstrength to the square of the magnetic field strength at thispoint was proportional to the charge to mass ratio of theelectron. Electric field only + Both Magnetic field only _ http://highered.mcgraw-hill.com/classware/ala.do?isbn=0072980605&alaid=ala_729122&showSelfStudyTree=true#
  122. 122. J.J. ThomsonThe value he obtained, −1.76 x 108 C/g*, was alwaysthe same, regardless of the source of the cathoderays.This value was nearly 2000 times larger than thecharge to mass ratio of a hydrogen ion (H+)!This indicated that either the charge of the electronwas very large, or that the mass of the electron wasvery, very small – much smaller than the mass of ahydrogen atom, which was the lightest atom known.*the SI unit of electric charge is the Coulomb (C)
  123. 123. J.J. ThomsonThomson proposed that these electrons werenot just very small particles, but were actually asub-atomic particle present in all atoms.We thus credit Thomson with the “discovery”of the electron because of his work in determiningtheir physical characteristics, and his rather boldhypothesis that they were present in all atoms(which was later shown to be true).
  124. 124. The Plum Pudding ModelSince the atom is neutrally charged, if it has (-)charged electrons, there must also be a (+) part tothe atom to cancel the negative electrons.This showed that Dalton’s idea that atoms wereindivisible is NOT correct – instead, the atom iscomposed of TWO oppositely charged parts.Thomson thought the atom was a diffuse (+)charged object, with electrons stuck in it, likeraisins in pudding (the plum pudding model).
  125. 125. Thomson’s Plum Pudding Model of the Atomhttp://highered.mcgraw-hill.com/classware/ala.do?isbn=0072980605&alaid=ala_729122&showSelfStudyTree=true
  126. 126. Millikan’s Oil Drop ExperimentRobert Millikan (1911) designed an experiment todetermine the actual charge of an electron.He suspended charged oildrops in an electric field.The drops had becomecharged by picking up freeelectrons after passingthrough ionized air.
  127. 127. Millikan’s Oil Drop Experiment FELEC = E x q when the downward force of gravity on the drop was balanced by the upward force FGRAVITY = m x g of the electric field, then: E x q = m x g or q = mg/EKnowing the mass (m) of the oil drop, and thestrength of the electric field (E), he was able to findthe charge (q) on the oil drop.
  128. 128. Millikan’s Oil Drop ExperimentTo find the charge of the electron, he found thesmallest difference between the charges on anytwo oil drops.eg: Suppose you find three oil drops have the following charges: 12.4, 7.6, 10.8. The differences between the charges are: 12.4 – 10.8 = 1.6 10.8 – 7.6 = 3.2 12.4 – 7.6 = 4.8 4.8 – 3.2 = 1.6You would conclude the charge of the electron was 1.6charge units.
  129. 129. Millikan’s Oil Drop ExperimentUsing this technique, Millikan was able to determinethe charge of an electron to be: e = C 1.602 x 10C 19 CUsing Thomson’s charge to mass ratio and thecharge for the electron, Millikan determined themass of the electron to be 9.11 x 10-31 kilogram.For his work, Millikan received the 1923 Nobel Prize inPhysics. http://highered.mcgraw-hill.com/classware/ala.do?isbn=0072980605&alaid=ala_729122&showSelfStudyTree=true
  130. 130. Radioactivity was discovered in 1895 It was found that there are three distinct types of radiation: (+) alpha particles, (-) beta particles, and neutral gamma rays. (Uranium compound) http://highered.mcgraw-hill.com/classware/ala.do?isbn=0072980605&alaid=ala_729122&showSelfStudyTree=true
  131. 131. Rutherford’s Gold Foil Experiment (1908 Nobel Prize in Chemistry)Rutherford designed an experiment using thesenewly discovered alpha-particles to test if Thomson’splum pudding model was correct.He fired (+) alpha particles at the gold foil. If theThomson model was correct, most of the alphaparticles would pass through the foil with littledeflection.
  132. 132. Rutherford’s Experiment Expected Results of Rutherford’s ExperimentThe force of repulsion is directly proportional to the product ofthe charges of the alpha particle and nucleus and inverselyproportional to the square of the distance between the centerof the two charges. F = kQ1Q2/R2A large, diffuse positive charge is not able to repel a (+) alphaparticle very strongly, because the alpha particle cannot makea close approach, so the angle of deflection,θ, would be fairlysmall. θ (+) -particle R
  133. 133. Rutherford’s Gold Foil ExperimentWhen Rutherford performed the experiment,nearly all the alpha particles passed through thefoil without deflection, as expected…However, some particles were deflectedsignificantly, and perhaps one in 2000 wereactually deflected nearly 180 degrees!
  134. 134. Rutherford’s Experiment Rutherford was stunned. This would be like firing a machine gun at an apple, and having most of the bullets pass through -- but every once in a while one of the bullets would bounce off the apple and come back and hit you! Why would this happen??? DUCK, ERNIE!?!? something small and massive must be in there that ? deflects only those bullets that directly hit it…
  135. 135. Rutherford’s ExperimentOnly a positive charge with a very, very smallradius would allow the alpha particle to approachclose enough to experience a significant repulsion. Strong repulsion! -particle θ nucleus RBy carefully measuring the angles of deflection, θ,Rutherford was able to determine the approximatesize of this positive core to the atom.
  136. 136. Rutherford’s ExperimentNext, by measuring the kinetic energy of the alphaparticle before and after the collision, Rutherfordwas able to apply conservation of momentum anddetermine the mass of the atom’s positive core.Putting it all together, he was able to conclude thatall the positive charge -- and about 99.9% of themass -- of an atom was concentrated in a very tinyarea in the middle of the atom, which he called thenucleus.
  137. 137. Rutherford’s ExperimentOnly the very few (+) α-particles that passed verynear this incredibly tiny (+) nucleus were stronglydeflected; most α-particles never came near thenucleus and so were not deflected significantly. *note carefully that the (+) α-particles never actually collide with the (+) nucleus – the repulsive force between the like charges is too great for that to occur!
  138. 138. Rutherford’s Model of the AtomThe estimated size of this nucleus was such a tiny fraction ofthe total volume of the atom, that at first Rutherford doubtedhis own conclusion. atomic radius ~ 100 pm = 1 x 10-10 m nuclear radius ~ 5 x 10-3 pm = 5 x 10-15 m “If the atom is the Houston Astrodome, then the nucleus is a marble on the 50-yard line.”
  139. 139. As another size comparison, if the nucleus were the size of abasketball, placed at PHS, the atom would be over 20 km indiameter, reaching Martin to the North, and just missing theUS 131 Business Loop exit to the South! The basketball- sized nucleus would also mass about 70,000,000,000 tons! This is equivalent to about 100,000 cruise ship ocean liners!
  140. 140. • Rutherford fired (+) charged alpha particles at thin sheets of gold foil and measured the angles at which the alpha particles were deflected.• Rutherford was testing the validity of Thomson’s plum pudding model. If this model were correct, the (+) alpha particles would not be deflected by the diffuse (+) charge of Thomson’s atom.• When Rutherford performed the experiment, he found that the majority of alpha particles did, in fact, pass without significant deflection. However, a small number were significantly deflected, and a very few were strongly deflected nearly 180 degrees.• By measuring the angles of deflection, Rutherford was able to calculate the size and mass of the (+) center that could produce the observed deflections. He found that all the (+) charge and about 99.9% of the atom’s mass was concentrated in a tiny region (about 1/100,000 the volume of the atom).• Only those alpha particles that passed very close to the nucleus experienced a strong enough repulsion to produced significant deflections – most particles never came near the nucleus, and so were not deflected.• AP Extras:• The repulsive force depends on 1/R 2 between the (+) alpha particle and the (+) charge of the nucleus.• He also relied on conservation of momentum to help him determine the mass of the nucleus which was repelling the alpha particles.
  141. 141. Chadwick’s Experiment (1932) (1935 Noble Prize in Physics)Discovery of the Neutron •H atoms have 1 p; He atoms have 2 p?? •ratio of mass He/mass H should be 2/1 = 2 •measured ratio of mass He/mass H = 4 ???James Chadwick discovered that when 9Be was bombardedwith alpha particles, a neutral particle was emitted, which wasnamed the neutron. α + 9Be 1 n + 12C + energy neutron (n) is neutral (charge = 0) n mass ~ p mass = 1.67 x 10-24 gNow the mass ratios can be explained if He has 2 neutronsand 2 protons, and H has one proton with no neutrons
  142. 142. mass n  mass p  1,840 x mass e-
  143. 143. Atomic number, Mass number and IsotopesAtomic number (Z) = number of protons in nucleusMass number (A) = number of protons + number of neutronsalso called the nucleon number = atomic number (Z) + number of neutronsIsotopes are atoms of the same element (X) with differentnumbers of neutrons in their nuclei Mass Number A ZX Element Symbol nuclide Atomic Number 1 2 3examples 1H protium 1H (D) deuterium 1H (T) tritium 14 235 C Carbon-14 U Uranium-235 6 92
  144. 144. Atomic number, Mass number and IsotopesExamples: How many protons, neutrons, and electrons are in 14C ? 6 6 protons, 8 (14 - 6) neutrons, 6 electronsHow many protons, neutrons, and electrons are in 59 Fe ? 26 26 protons, 33 (59 - 26) neutrons, 26 electrons
  145. 145. see page 50
  146. 146. The Periodic Table of the ElementsWe now understand that the number of protons inthe nucleus of the atom is what “defines” theelement and gives each element its uniqueproperties.
  147. 147. The Periodic Table of the Elements Transition metals group period
  148. 148. ElementsProperties of Metals• malleable and ductile• lustrous• good conductors• lose e- to form cationsProperties of Non-metals• brittle• dull• poor conductors• gain e- to form anions
  149. 149. ElementsProperties of Metalloids• properties are intermediate between those of metalsand nonmetals• semi-conductorsNames of Families or Groups 1A = alkali metals 5A = pnictides 2A = alkaline earths 6A = chalcogens 3A = boron family 7A = halogens 4A = carbon family 8A = noble gasesThe chemical properties of elements within aFamily or Group are similar
  150. 150. Elements Natural abundance of elements in the Earth’s crustNatural abundanceof elements in the human body
  151. 151. Moleculesand Ions
  152. 152. Molecules & IonsA molecule is an aggregate of two or more neutralatoms in a definite arrangement held together bychemical forcesNote that some elements exist as molecules. Forexample,the following elements occur in nature as moleculardiatomic elements: H2, N2, O2, F2, Cl2, Br2 and I2 H2 F2They are molecules, but they are NOTcompounds, because they have onlyone kind of element present. O2 N2
  153. 153. Molecules & IonsA polyatomic molecule contains more than two atoms O3, H2O, NH3, C3H6O An allotrope is one of two or more distinct molecular forms of an element, each having unique properties. For example, O 2 and O3 are allotropes of oxygen; diamond, graphite and buckminster fullerene (C60) are all different allotropes of carbon.
  154. 154. Classification of MatterIonic compounds are composed of ions, which areatoms that have a (+) or (-) charge.+ ions are called cations and form when C + C + + C +C + C an atom loses electrons C C + C-ions are called anions andC + when + form + an atom gains electrons C Ionic compounds form when cations and anions form electrostatic attractions between them (opposite charges attract)
  155. 155. Molecules and IonsA monatomic ion contains only one atom Examples: Na+, Cl-, Ca2+, O2-, Al3+, N3-note that the convention is to indicate the magnitude of thecharge first, and then the sign: e.g., Ca2+, not Ca+2A polyatomic ion contains more than one atom Examples: ClO3-, NO2- , CN- , SO42-
  156. 156. Molecules and IonsExamples 27 3+How many electrons are in 13 Al ? 13 protons, so there are 13 – 3 =10 electrons 78How many electrons are in 34 Se2- ? 34 protons, so there are 34 + 2 = 36 electrons
  157. 157. Charges of common monatomic ions see page 54Note that some atoms, especially transition metals, have multiple charge statesNote also that metals typically form (+) charged ions, nonmetals form (-)charged ions.
  158. 158. Also note the relation between the magnitude of the chargeand the group number (1A, 5A, etc) for most elements.The charge of representative metals (group 1A, 2A and 3A)is equal to the group numberThe charge of representative nonmetals (group 4A-7A) isequal to: (the group number – 8)
  159. 159. Chemical Nomenclature Determining the names and formulas of chemical compoundsIUPAC = International Union of Pure and Applied Chemists.This is the group that determines the official rules ofnomenclature for all chemical elements and compounds
  160. 160. Chemical FormulasA chemical formula is a combination of elementsymbols and numbers that represents thecomposition of the compound.Subscripts following an element’s symbolindicate how many of that particular atom arepresent. If no subscripts are given, it isassumed that only one of that atom is present inthe compound. NH3 C3H6S P4O10 1 N + 3 H atoms 3 C + 6 H + 1 S atoms 4 P + 10 O atoms
  161. 161. Chemical FormulasA molecular formula shows the exact number ofatoms of each element in the smallest unit of asubstance
  162. 162. Chemical FormulasAn empirical formula shows the simplest whole-number ratio of the atoms in a substance molecular empirical H2O H2O N2H4 NH2 C2H8O2 CH2O C6H12O6 CH2O note that different molecular compounds may have the same empirical formula
  163. 163. Ionic FormulasFor ionic compounds the formula is always thesame as the empirical formula.The sum of the charges of the cation(s) and anion(s) in each formula unit must equal zero. Thus, the ratioof cations to anions can always be reduced to simple, wholenumber ratios. The ionic compound NaCl Na+500Cl-500 = NaCl
  164. 164. Naming BinaryMolecular Compounds
  165. 165. Naming Molecular CompoundsWe will only consider naming binary molecules.Binary molecular compounds typically formbetween two non-metals, or a non-metal and ametalloid.Naming Molecules: 1st element + root of 2nd element + “-ide” e.g. : HCl = hydrogen chloride
  166. 166. Naming Molecular Compounds See page 62If there is more than one ofa given element, we useprefixes to indicate thenumber of each kind ofatom present.The prefix mono is onlyused for atoms that canform more than onecompound with the secondelement. For this class,oxygen is the mainelement that does this.
  167. 167. Naming Molecular CompoundsExamples of naming molecules HI hydrogen iodide NF3 nitrogen trifluoride SO2 sulfur dioxide N2Cl4 dinitrogen tetrachloride NO2 nitrogen dioxide N2O dinitrogen monoxide (laughing gas)
  168. 168. Naming Molecular CompoundsIf the second element begins with a vowel, theterminal vowel of the prefix is allowed to bedropped.For exampleN2O4 could be called dinitrogen tetroxide, ratherthan dinitrogen tetraoxide.CO would be called carbon monoxide, not carbonmonooxideNote, however, that the official IUPAC rule statesthat the vowel is only dropped for “compellinglinguistic reasons.”
  169. 169. Naming Molecular CompoundsNaming Compounds containing HydrogenCompounds containing hydrogen can be named using theGreek prefixes, but most have common names that areaccepted by IUPAC. The most common examples are: B2H6 diboron hexahydride diborane CH4 carbon tetrahydride methane SiH4 silicon tetrahydride silane NH3 nitrogen trihydride ammonia phosphorus trihydride phosphine PH3 dihydrogen monoxide water H2O dihydrogen sulfide hydrogen sulfide H2S
  170. 170. Naming Molecular CompoundsDetermining the formula of molecules from thenameThe subscripts tell you the number of each typeof element present, so naming molecules fromthe formula is straightforward. e.g. sulfur hexafluoride = SF6 dichlorine heptoxide = Cl2O7The order in which the atoms are listed in molecules is basedon something called electronegativity. For now, we can predictthe order using the chart on the next slide…
  171. 171. Chemical Formulas Order of Elements in Writing Molecular Formulas HB C N O F Si P S Cl Ge As Se Br Sb Te I
  172. 172. Organic chemistry is the branch of chemistry thatdeals with carbon compoundsCarbon is unique among all the elements in itsability to catenate, or form long or branchingchains of carbon atoms.We usually write these chains as “condensed formulas” thatassumes carbons are bonded to each other as follows: H H H = CH3CH2CH3 H C C C H note that we could also H write this as: C3H8 H H
  173. 173. Organic molecules that contain only carbon andhydrogen are called hydrocarbons. The first 10 simple hydrocarbons Hydrocarbon compounds are named based on the number of carbon atoms in the “backbone” or chain of carbon atoms.
  174. 174. Naming Ionic Compounds
  175. 175. Naming Ionic CompoundsIonic CompoundsIonic compounds are typically composedof a metal cation and a non-metal anion$ name of cation = simply the name of the element$ name of anion = root of element’s name + - “ide”
  176. 176. Naming Ionic CompoundsBinary ionic compounds are named:name of metal ion + root of non-metal + “-ide” e.g. BaCl2 barium chloride K2O potassium oxide Na2S sodium sulfide Mg3N2 magnesium nitride Al2O3 aluminum oxide
  177. 177. Formula of Ionic CompoundsDetermining the formula of ionic compounds fromthe name is a little more involved – unlikemolecular compounds, the name does not give usthe subscripts. These must be determined basedon the charges of each ion.Remember that the total number of (+)and (-) charges in any ionic compoundmust sum to zero.
  178. 178. Formula of Ionic Compounds 2 x +3 = +6 3 x -2 = -6aluminum oxide Al2O3 Al3+ O2- 1 x +2 = +2 2 x -1 = -2calcium bromide CaBr2 Ca2+ Br- 1 x +2 = +2 1 x -2 = -2magnesium sulfide MgS Mg2+ S2-
  179. 179. Formula of Ionic CompoundsNote that if you take the magnitude of the charge ofthe cation, and make it the subscript on the anion,and take the magnitude of the anion’s charge andmake it the subscript of the cation, the compoundwill always end up with a net neutral charge. Now, ifpossible, reduce the subscripts to a simpler ratio,and you have the correct formula for the compound! +3 -2 Al O Al2O3 2 3 Al3+ O2-
  180. 180. see page 58
  181. 181. Pb Cu Multivalent ions: WThe Non-Representative Atoms Fe Mn Co
  182. 182. Transition and other multi-valent metal ionsMost elements form only ions with one charge.However, most of the transition metals, as well asPb and Sn, have more than one possible chargestate. We say they are multi-valent.e.g. : copper can exist in either a +1 or +2 chargestate: Cu+ or Cu2+The formula or name of the compound must indicatewhich charge state the metal cation is in.
  183. 183. Transition and other multi-valent metal ionsOlder method gives a common name foreach valence state Cu+ cuprous Fe2+ ferrous Cu2+ cupric Fe3+ ferric Cr2+ chromous Hg22+ mercurous Cr3+ chromic Hg2+ mercurice.g. CuCl = cuprous chloride Hg2I2 = mercurous iodide Fe2O3 = ferric oxide PbO = plumbous oxide
  184. 184. Transition and other multi-valent metal ionsTo determine which charge state the cation is in, youmust look at the anion, and calculate the charge ofthe cation…CuSS is always -2, and there is only one Cu to cancel this out,so copper must be +2. Thus, this is cupric sulfide.Fe2O3Subscript on O is the charge of the iron! Thus, Fe is +3 andthis compound is ferric oxide.
  185. 185. Transition and other multi-valent metal ionsStock System:We indicate charge on metal with Roman numeralsFeCl2 2 Cl- = -2 so Fe is 2+ iron(II) chlorideFeCl3 3 Cl- = -3 so Fe is 3+ iron(III) chlorideCr2S3 3 S-2 = -6 so Cr is 3+ chromium(III) sulfide
  186. 186. 2-NH4 + C2O 4 Polyatomic Ions SO4 2- 2- C2H3O 2
  187. 187. Naming Polyatomic IonsThere are certain groups of neutral atoms that bondtogether, and then gain or lose one or moreelectrons from the group to form what is called apolyatomic ion. Most polyatomic ions arenegatively charged anions.Examples:OH- = hydroxide ion CN- = cyanide ionNO3- = nitrate ion NH4+ = ammonium ionSO42- = sulfate ion SO32- = sulfite ion
  188. 188. See page 60
  189. 189. Naming Polyatomic IonsNaming ionic compounds containing polyatomic ionsis straightforward:Name the cation + name the (polyatomic) anionExamples: NaOH = sodium hydroxide K2SO4 = potassium sulfate Fe(CN)2 = iron (II) cyanide (NH4)2CO3 = ammonium carbonate
  190. 190. page 61
  191. 191. page 62
  192. 192. Compound Summary see page 64
  193. 193. NAMING ACIDS AND BASESThere are a different set ofrules for naming acids. Someof the rules are based on amuch older system ofnomenclature, and so therules are not as simple as theyare for molecular and normalionic compounds.
  194. 194. AcidsAn acid can be defined as a substance that yieldshydrogen ions (H+) when dissolved in water. TheseH+ ions then bond to H2O molecules to form H3O+,called the hydronium ion.Many molecular gases, whendissolved in water, become acids: •HCl (g) = hydrogen chloride •HCl (aq) = HCl dissolved in water which forms (H3O+,Cl-) = hydrochloric acid
  195. 195. AcidsAll acids have hydrogen as the first listedelement in the chemical formula.For nomenclature purposes, there are two majortypes of acids:Oxoacids (also called oxyacids) = acids thatcontain oxygen. eg: H2SO4, HC2H3O2Non-oxo acids = acids that do not contain oxygen.eg: HCl (aq), H2S (aq)
  196. 196. AcidsRules for naming non-oxoacids acid = “hydro-” + root of anion + “-ic acid” see page 65 **note that we add an extra syllable for acids with sulfur and phosphorus:it’s not hydrosulfic acid, but hydrosulfuric acid. Similarly, acids withphosphorus will end in phosphoric, not phosphic acid.
  197. 197. AcidsAn oxoacid is an acid that contains hydrogen,oxygen, and another element –That is, oxoacids are the protonated form ofthose polyatomic ions that have oxygen in theirformulas.examples: HClO3 chloric acid HNO2 nitrous acid H2SO4 sulfuric acid
  198. 198. When naming oxoacids, NO “hydro” prefix is used.Instead, the acid name is the root of the name of theoxoanion + either “-ic” acid or “-ous” acid, as follows:If the name of the polyatomic anion ends in“ate,” drop the -ate and add “ic acid.”eg: SO42- = sulfate anion H2SO4 = sulfuric acid C2H3O2- = acetate anion HC2H3O2 = acetic acidIf the name of the polyatomic anion ends in“ite,” drop the -ite and add “ous acid.”eg: SO32- = sulfite anion H2SO3 = sulfurous acid NO2- = nitrite anion HNO 2 = nitrous acid
  199. 199. AcidsNaming Oxoacids and Oxoanions see page 66
  200. 200. AcidsAs a mnemonic aid, I always use the following:ic goes with ate because….”IC…I ATE it!ite goes with ous like……tonsil-ITE-OUS, senior-ITE-OUS
  201. 201. BasesA base can be defined as a substancethat yields hydroxide ions (OH-) whendissolved in water. NaOH sodium hydroxide KOH potassium hydroxide Ba(OH)2 barium hydroxide
  202. 202. HydratesHydrates are compounds that have a specificnumber of water molecules attached to them. BaCl2•2H2O barium chloride dihydrate LiCl•H2O lithium chloride monohydrate MgSO4•7H2O magnesium sulfate heptahydrate Sr(NO3)2 •4H2O strontium nitrate tetrahydrateCuSO4•5H2O CuSO4cupric sulfate anhydrouspentahydrate cupric sulfate
  203. 203. HydratesOther terms associated with hydratesAnhydrous: without water; this term describeshydrated compounds after “drying.”Hygroscopic: readily absorbs moisture directlyfrom the air.Deliquescent: absorbs moisture from the air soreadily, that these compounds can take on enoughwater to actually start to dissolve.Water of hydration: the water absorbed andincorporated into hygroscopic compounds
  204. 204. see page 68
  205. 205. Mass Relationships in Chemical Reactions Chapter 3Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
  206. 206. Relative Masses of the Elements Micro World Macro World atoms & molecules gramsAtomic mass is the mass of an atom in atomicmass units (amu). This is a relative scalebased on the mass of a 12C atom. By definition: 1 atom 12C “weighs” 12 amu On this scale 1H = 1.008 amu and 16O = 16.00 amu
  207. 207. Relative Masses of the ElementsHow do we find the relative masses of theother elements?Imagine we have 66.00 grams of CO2. The compoundis decomposed and yields 18.00 grams of C and 48grams of O. Since there are two oxygen atoms forevery 1 carbon atom, we can say that 48 g O xygen 2 × 24 gra m s O xyg en 24 g O = so = 1 .3 3 3 18 g C arbon 1 8 g C a rbon 18 g CThis means that the relative mass of each oxygen atom is1.333 x the mass of a carbon atom (12.00 amu) , or…mass of oxygen = 1.333 x 12.00 amu = 16.00 amu
  208. 208. Average Atomic MassThe average atomic mass of an element is theweighted average mass of that element, reflectingthe relative abundances of its isotopes.example: consider lithium (Li), which has twoisotopes with the following relative percentabundances: 7.42% 6Li (6.015 amu) 92.58% 7Li (7.016 amu)The Average atomic mass of lithium would be: 7.42 +  92.58  6.015 amu   7.016 amu = 6.941 amu 100   100 
  209. 209. Average Atomic Mass IA The masses reported at the1 bottom of the “box” for each H 1. 0079 IIA element in the Periodic Table3 Li 4 Be is the average atomic mass 6.941 9. 012 for that element, (in amu).11 12 Na Mg IIIB IVB 22.99 24.30519 20 21 22 K Ca Sc Ti 39.098 40.08 44.956 47.90
  210. 210. Average Atomic Masssee page 79
  211. 211. The Mole & Avogadro’s NumberThe mole (mol) is the SI unit for the amount of asubstance that contains as many “things” as thereare atoms in exactly 12.00 grams of 12C.This number, called Avogadro’s number (NA),has been experimentally determined to beapproximately 6.0221367 X 1023 things. 1 mol = NA = 6.022 x 1023 “things”We can have 1 mole of atoms, or molecules, or even dumptrucks. The mole refers only to a number, like the term“dozen” means 12.
  212. 212. The Mole & Avogadro’s NumberJUST HOW BIG IS AVOGADRO’S NUMBER??• If you stacked 1 mole of notebook paper, it would take you5,800 years, traveling at the speed of light (186,000,000 milesper second) to reach the top of the stack!• If you were given 1 mole of dollar bills when the universebegan 13 billion years ago, and you immediately beganspending money at the rate of one million dollars per second,you would still have about 190 billion trillion dollars left !• 1 mole of oranges would cover the surface of the earth to adepth of 9 miles!• but 1 mole of Hydrogen atoms would only mass about 1gram!
  213. 213. The Mole & Molar MassMolar mass is the mass, in grams, of exactly1 mole of any object (atoms, molecules, etc.)Note that because of the way we defined the mole : 1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g 1 12C atom = 12.00 amuThus, for any element atomic mass (amu) = molar mass (grams) For example: 1 mole 12C atoms = 12.00 g 12C 1 mole lithium atoms = 6.941 g of Li
  214. 214. The Mole & Molar MassOne Mole of: C = 12.01 g S = 32.06 g Hg = 200.6 g Cu = 63.55 g Fe = 55.85 g
  215. 215. The Mole & Molar MassSolving Mole ProblemsWe can now add the definitions of the mole, Avogadro’snumber, and molar mass to our repertoire of conversionfactors we can use in dimensional analysis problems. Thus, given the mass, we can use the molar mass to convert this to moles, and then use Avogadro’s number to convert moles to particles, and vice versa… M = molar mass in g/mol NA = Avogadro’s number
  216. 216. Solving Mole ProblemsHow many atoms are in 0.551 g of potassium (K) ? 1 mol K = 39.10 g K conversion factors 1 mol K = 6.022 x 1023 atoms K 1 mol K 6.022 x 1023 atoms K 0.551 g K x x 39.10 g K 1 mol K = 8.49 x 1021 atoms K
  217. 217. Solving Mole Problemssee page 81
  218. 218. Solving Mole Problemssee page 82
  219. 219. Solving Mole Problems see page 82
  220. 220. Molecular MassMolecular mass (or molecular weight) is the sum ofthe atomic masses of the atoms in a molecule.Example: consider SO2 1S 32.07 amu 2O + 2 x 16.00 amu SO2 64.07 amu SO2 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2As was the case for atoms, for any molecule molecular mass (amu) = molar mass (grams)
  221. 221. Molecular Masssee page 83
  222. 222. Formula MassFormula mass is the sum of the atomic masses(in amu) in a formula unit of an ionic compound. 1Na 22.99 amu NaCl 1Cl + 35.45 amu NaCl 58.44 amu 1 formula unit NaCl = 58.44 amu 1 mole NaCl = 58.44 g NaClFor any ionic compound formula mass (amu) = molar mass (grams)
  223. 223. Formula MassWhat is the formula mass of Ca3(PO4)2 ? 1 formula unit of Ca3(PO4)2 3 Ca 3 x 40.08 2P 2 x 30.97 8O + 8 x 16.00 310.18 amuSince the formula mass, in grams (per mole), isnumerically equal to the molar mass, in amu, we findthat the formula mass of Ca3(PO4)2 = 310.18 gramsper mole of Ca3(PO4)2.
  224. 224. Molecular/Formula MassesUsing Molecular/Formula Masses in DimensionalAnalysis ProblemsWe can now add molecular & formula masses to our list ofconversion factors. They are used similarly to the way weused the molar mass of the elements as conversion factors.Example: How many H atoms are in 72.5 g of C3H8O ? 1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O conversion factors 1 mol C3H8O molecules = 8 mol H atoms 1 mol H = 6.022 x 1023 atoms H 1 mol C3H8O 8 mol H atoms 6.022 x 1023 H atoms72.5 g C3H8O x x x 60 g C3H8O 1 mol C3H8O 1 mol H atoms = 5.82 x 1024 atoms H
  225. 225. Solving Mole Problems see page 84
  226. 226. Solving Mole Problems see page 85
  227. 227. The Mass SpectrometerAtomic and molecular masses of unknown compounds aredetermined using a mass spectrometer.A gaseous sample of the unknown is bombarded withelectrons in an electron beam. This knocks electrons loosefrom the unknown to produce cations. These cations arethen accelerated through perpendicular electric andmagnetic fields. The charge:mass ratio (e/m) of theunknown ions determines the degree to which the particlesare deflected.The greater the charge:mass ratio, the smaller the anglethrough which the beam is deflected.
  228. 228. The Mass SpectrometerWe know the angle that a given e/m produces, so we canidentify the unknown ion when it registers on a special screen. high e/m low e/m Mass Spectrometer
  229. 229. Percent compositionPercent composition of an element in a compound isthe percent, by mass, of that element in the compound.It can be calculated as follows: n x molar mass of element x 100% molar mass of compound where n is the number of moles of the element in 1 mole of the compoundKnowing the percent composition, one can determinethe purity of a substance, (are there contaminantspresent in the sample?) and you can even determinethe empirical formula of an unknown compound.
  230. 230. Percent compositionExample: What is the percent composition ofethanol, which has the formula, C2H6O ?First, we find the molecular mass of ethanol. This is foundto be: 2(12.01) + 6(1.008) + 1(16.00) = 46.07 grams/mole. 2 x (12.01 g)% Composition: %C = x 100% = 52.14% 46.07 g 6 x (1.008 g) %H = x 100% = 13.13% 46.07 g 1 x (16.00 g) %O = x 100% = 34.73% 46.07 g C2H6O check: 52.14% + 13.13% + 34.73% = 100.0%
  231. 231. Percent compositionWe can also determine the % by mass of groupsof atoms present in a compound in the samemanner.Example: what is the percent water in epsom salts, whichhas the formula: MgSO4 • 7 H2O ? mass of water% H2O = x 100 mass of compound this is the molar 7(18.02) mass of water = 24.31 + 32.07 + 4(16.00) + 7(18.02) 126.14 g H2O = x 100 = 51.17% H2O 246.52 g cmpd
  232. 232. Percent compositionExample: How many grams of CaCl2 • 2 H2O must be weighed outto obtain 12.20 grams of CaCl2?There are two ways of solving this problem:Method 1:First determine the % CaCl2 in CaCl2 • 2 H2O: 110.98 g CaCl2 i. % CaCl2 = x 100 = 75.49% 147.02 g CaCl2 • 2 H2O Then we note that the 12.20 g of CaCl2 desired must be 75.49% of the mass of the hydrate used:ii. 75.49% of (X grams) of CaCl2•2 H2O = 12.20 g of CaCl2  0.7549(X) = 12.20 or X = 12.20/0.7549 = 16.16 grams

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