Chem 101 week 2


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Chem 101 week 2

  1. 1. CHEM 101PPts from week of1/25/20121/27/2012
  2. 2. Significant Figures
  3. 3. Significant Digits• Any non-zero digit• A zero between two nonzero digits• Zeroes to the right of a nonzero digit and tothe right of the decimal point
  4. 4. 461All nonzero numbers are significant.Significant Figures
  5. 5. 461All nonzero numbers are significant.Significant Figures
  6. 6. 461All nonzero numbers are significant.Significant Figures
  7. 7. 4613 Significant FiguresAll nonzero numbers are significant.Significant Figures
  8. 8. A zero is significant when it is between nonzerodigits.5 Significant Figures600.39Significant Figures
  9. 9. 3 Significant Figures30.9A zero is significant when it is between nonzerodigits.Significant Figures
  10. 10. A zero is significant at the end of a number thatincludes a decimal point.5 Significant Figures000.55Significant Figures
  11. 11. A zero is significant at the end of a number thatincludes a decimal point.5 Significant Figures0391.2Significant Figures
  12. 12. Non-Significant Digits• Zeroes to the left of the leftmost nonzero digit• Zeroes to the left of an implied decimal point– 567 has an implied decimal point– 450 has an implied decimal point we do not knowif the zero is significant therefore it is non-significant
  13. 13. A zero is not significant when it is before thefirst nonzero digit.1 Significant Figure600.0Significant Figures
  14. 14. A zero is not significant when it is before thefirst nonzero digit.3 Significant Figures907.0Significant Figures
  15. 15. A zero is not significant when it is at the end of anumber without a decimal point.1 Significant Figure00005Significant Figures
  16. 16. A zero is not significant when it is at the end of anumber without a decimal point.4 Significant Figures01786Significant Figures
  17. 17. 12 inches = 1 foot100 centimeters = 1 meter• Exact numbers have an infinite number ofsignificant figures.• Exact numbers occur in simple countingoperationsExact Numbers• Defined numbers are exact.12345
  18. 18. Rounding Off Numbers
  19. 19. • Often when calculations are performed on acalculator extra digits are present in the results.• It is necessary to drop these extra digits so as toexpress the answer to the correct number ofsignificant figures.• When digits are dropped, the value of the lastdigit retained is determined by a process knownas rounding off numbers.
  20. 20. 80.873Rule 1. When the first digit after those you want to retainis 4 or less, that digit and all others to its right aredropped. The last digit retained is not changed.4 or lessRules for Rounding Off
  21. 21. 1.875377Rule 1. When the first digit after those you want to retainis 4 or less, that digit and all others to its right aredropped. The last digit retained is not changed.4 or lessRounding Off Numbers
  22. 22. 5 or greater5.459672Rule 2. When the first digit after those you want to retainis 5 or greater, that digit and all others to its right aredropped. The last digit retained is increased by 1.drop these figuresincrease by 16Rounding Off Numbers
  23. 23. Significant Figures inCalculations
  24. 24. The results of a calculation based onmeasurements cannot be moreprecise than the least precisemeasurement. It is importantto remember that:
  25. 25. Multiplication or Division
  26. 26. • In multiplication or division, the answer must containthe same number of significant figures as in themeasurement that has the least number ofsignificant figures.
  27. 27. (190.6)(2.3) = 438.38438.38Answer given bycalculator.2.3 has two significant figures.190.6 has four significantfigures.The answer should have two significantfigures because 2.3 is the number withthe fewest significant figures.Drop these threedigits.Round off this digit tofour.The correct answer is 440 or 4.4 x 102
  28. 28. Addition or Subtraction
  29. 29. The results of an addition or asubtraction must be expressed to thesame precision as the least precisemeasurement.
  30. 30. The result must be rounded to the samenumber of decimal places as the valuewith the fewest decimal places.
  31. 31. 1.039 - 1.020Calculate1.0391.039 - 1.020= 0.0182868141.039Answer given bycalculator.1.039 - 1.020 = 0.0190.019= 0.0182868141.039The answer should have two significantfigures because 0.019 is the numberwith the fewest significant figures.280.018 6814Two significantfigures.Drop these6 digits.0.018286814Correct answer.
  32. 32. Units of Measurement• Recall that the unit part of a measurementtells us the scale or standard used torepresent the results of a measurement• The two most widely used systems ofmeasurement are the English System and theMetric System
  33. 33. Measurement of mass, length andvolume• In the United States, we use a fairly awkward systemof measurement for most things - the English system• Scientists use the metric or SI system of units for themeasurement of physical quantities• This system uses standard units based on veryprecisely known properties of matter and light• Metric Prefixes are used to change the size of thefundamental or standard units.
  34. 34. The Metric System
  35. 35. The Metric or International System (SI,Systeme International)• Is a decimal system of units.• It is built around standard units.• It uses prefixes representing powers of 10to express quantities that are larger orsmaller than the standard units.
  36. 36. International System’sStandard Units of MeasurementQuantity Name of Unit AbbreviationLength meter mMass kilogram kgTemperature Kelvin KTime second sAmount of substance mole molElectric Current ampere ALuminous Intensity candela cd
  37. 37. Because these fundamental units are notalways convenient size, the SI System usesmetric prefixes to change the size of the unit
  38. 38. Common Prefixes and Numerical Values for SI UnitsPower of 10Prefix Symbol Numerical Value Equivalentgiga G 1,000,000,000 109mega M 1,000,000 106kilo k 1,000 103hecto h 100 102deca da 10 101— — 1 100
  39. 39. Prefixes and Numerical Values for SI Unitsdeci d 0.1 10-1centi c 0.01 10-2milli m 0.001 10-3micro µ 0.000001 10-6nano n 0.000000001 10-9pico p 0.000000000001 10-12femto f 0.00000000000001 10-15Power of 10Prefix Symbol Numerical Value Equivalent
  40. 40. Measurement of Length
  41. 41. The standard unit of length in the SI system isthe meter. 1 meter is the distance that light travelsin a vacuum during of a second.1299,792,458
  42. 42. The Meter• 1 meter = 39.37 inches• 1 meter is a little longer than a yard
  43. 43. Metric Units of LengthExponentialUnit Abbreviation Metric Equivalent Equivalentkilometer km 1,000 m 103mmeter m 1 m 100mdecimeter dm 0.1 m 10-1mcentimeter cm 0.01 m 10-2mmillimeter mm 0.001 m 10-3mmicrometer µm 0.000001 m 10-6mnanometer nm 0.000000001 m 10-9mangstrom Å 0.0000000001 m 10-10m
  44. 44. Converting BetweenUnits
  45. 45. Converting Between Units• The standard method to convert between twodifferent units is the factor-label ordimensional analysis method• Dimensional analysis converts ameasurement in one unit to another by theuse of a conversion factor• Conversion factors are developed fromrelationships between two units
  46. 46. Dimensional Analysis• Dimensional analysis converts one unit to anotherby using conversion factors.• The conversion factor must accomplish two things:1. It must cancel unit12. It must introduce unit2unit1 x conversion factor = unit2
  47. 47. Conversion factors• Unit factors - factors that relate a quantity in acertain unit to one of another unite.g. 103m = 1 km• The conversion factor is created by dividing bothsides by the same quantity103m = 1 = 1 km103m 103m or103m = 1 km = 11km 1 kmEach unit factor gives 2 possible conversionfactors.
  48. 48. Dimensional analysis• Multiplying a quantity in one unit by an appropriate conversionfactor converts the number into the new unit• Note that conversion factors are exact relationships• Exact relationships have unlimited precision, so they can beignored for the purposes of deciding the number of significantdigits in a calculationkm1068.4m10km1m468.0 43−×=×
  49. 49. Some Problems• How many millimeters are there in 2.5 meters?The conversion factor mustaccomplish two things:m x conversion factor = mm It must cancelmeters. It must introducemillimeters
  50. 50. The conversion factor takes afractional form.mmm x = mmm
  51. 51. conversionfactorconversionfactorThe conversion factor isderived from theequality.1 m = 1000 mmDivide both sides by 1000 mmDivide both sides by 1 m1 m 1000 mm= 11 m 1 m=1 m 1000 mm= 11000m 1000 mm=
  52. 52. Use the conversion factor with millimeters in thenumerator and meters in the denominator.1000 mmx1 m2.5 m = 2500 mm32.5 x 10 mmHow many millimeters are there in 2.5meters?1000 mm1 m
  53. 53. How many millimeters are there in 2.5meters?Use the conversion factor with millimeters inthe numerator and meters in the denominator.1000 mm1 m2.5 m1000 mmx1 m= 2500 mm32.5 x 10 mm
  54. 54. 16.0 in2.54 cmx1 in= 40.6 cm2.54 cm1 inUse thisconversionfactorConvert 16.0 inches to centimeters.
  55. 55. Centimeters can be converted to micrometers bya series of two conversion → m → µmeters33.7 x 10 cm1 mx100 cm1= 3.7 x 10 m610 μmx1 m7= 3.7 x 10 μm13.7 x 10 mConvert 3.7 x 103cm to micrometers.
  56. 56. Convert 3.7 x 103cm to micrometers.33.7 x 10 cm1 mx100 cm610 μmx1 m7= 3.7 x 10 μmCentimeters can be converted to micrometers bywriting down conversion factors in → m → µmeters
  57. 57. MeasuringMass and Volume
  58. 58. Mass
  59. 59. The standard unit of mass in the SI systemis the kilogram. 1 kilogram is equal to the mass ofa platinum-iridium cylinder kept in a vault atSevres, France.1 kg = 2.205 pounds
  60. 60. Metric Units of massExponentialUnit Abbreviation Gram Equivalent Equivalentkilogram kg 1,000 g 103ggram g 1 g 100gdecigram dg 0.1 g 10-1gcentigram cg 0.01 g 10-2gmilligram mg 0.001 g 10-3gmicrogram µg 0.000001 g 10-6g
  61. 61. Convert 45 decigrams to grams.45 dg1 gx10 dg= 4.5 g1 g = 10 dg
  62. 62. An atom of hydrogen weighs 1.674 x 10-24g. Howmany ounces does the atom weigh?1 lbx454 g-241.674 x 10 g -273.69 x 10 lb=16 ozx1 lb-265.90 x 10 oz=-273.69 x 10 lb1 lb = 454 g16 oz = 1 lbGrams can be converted to ounces using a seriesof two conversion factors.
  63. 63. An atom of hydrogen weighs 1.674 x 10-24g. Howmany ounces does the atom weigh?-241.674 x 10 g1 lbx454 g16 ozx1 lb-265.90 x 10 oz=Grams can be converted to ounces using asingle linear expression by writing downconversion factors in succession.
  64. 64. Volume
  65. 65. Volume• The amount of 3-dimensional space occupiedby a substance• Fundamental unit of volume in the SI Systemfor volume is based on the volume of a cubemeasuring1m x 1m x 1m = (1m)3= 1m3= one cubicmeter
  66. 66. The liter (L) and milliliter (mL) are thestandard units of volume used in mostchemical laboratories.• 1m3is split into 1000 smallercubes ….each has a volume of1dm3• 1dm3= 1L (liter)• A cube with a volume of 1dm3can be split into 1000 smallercubes each with a volume of1cm3= 1mL(milliliter)1L = 1000mL
  67. 67. Convert 4.61 x 102microliters tomilliliters.Microliters can be converted to millilitersusing a series of two conversion factors.µL → L → mL61 Lx10 μL24.61x10 μL-44.61x10 L=-1= 4.61 x 10 mL-44.61x10 L1000 mLx1 L
  68. 68. Microliters can be converted to millilitersusing a linear expression by writing downconversion factors in succession.µL → L → mL24.61x10 μL 61 Lx10 μL1000 mLx1 L-1= 4.61 x 10 mLConvert 4.61 x 102microliters tomilliliters.
  69. 69. An extensive property of a material depends uponhow much matter is is being considered.An intensive property of a material does notdepend upon how much matter is is beingconsidered.• mass• length• volume• density• temperature• colorExtensive and Intensive Properties
  70. 70. Measurement ofTemperature
  71. 71. Heat• A form of energy that is associated with themotion of small particles of matter.• Heat refers to the quantity of this energyassociated with the system.• The system is the entity that is being heatedor cooled.
  72. 72. Temperature• A measure of the intensity of heat.• It does not depend on the size of the system.• Heat always flows from a region of highertemperature to a region of lowertemperature.
  73. 73. Temperature Measurement• The SI unit of temperature is the Kelvin.• There are three temperature scales: Kelvin,Celsius and Fahrenheit.• In the laboratory, temperature is commonlymeasured with a thermometer.
  74. 74. 3 Temperature Scales1. Celsius Scale – used in Europe and in physical and life sciences• Unit is oC• Freezing point of H2O is 0oC• Boiling point of H2O is 100oC2. Fahrenheit Scale – used in US and Britain• Unit is oF• Freezing point of H2O is 32oF• Boiling point of H2O is 212oF3. Absolute or Kelvin ScaleUnit is K• Freezing point of H2O is 273K• Boiling point of H2O is 373K
  75. 75. Temperature Scales• The size of each temperature unit (eachdegree) is the same for the Celsius andKelvin Scale- the difference between the freezingpoint and boiling point of H2O is 100units on both scales• The Fahrenheit degree is smaller thanthe C or K degree- On the F scale there are 180 F degreesbetween the boiling point and freezingpoint of H2O compared to 100 in theCelsius and Kelvin scales• The zero points are different on all 3scales
  76. 76. Converting Between Scales• Conversions Between K and C Scales• simple because the size of the units is the sametoC + 273 = tKConvert 70oC to K:70 + 273 = tKtK= 343KConvert 77K to oC:toC= 77 -273toC= -196 oC
  77. 77. Conversions Between F and C Scales• Requires 2 adjustments1. Adjustment for different size of unit2. Adjustment for different zero points• To convert Coto Fo:toF= 1.80(toC) + 32Factor 1.8 is due to the fact that there are:100C divisions/180F divisions =1C division/1.8 F divisions or 1.8F div./1C div• To convert Foto Co:Subtract 32 from the F temperature (so both scales start atthe same point) then multiply by the proper conversion unittoC= (toF – 32) (1oC/1.8oF)
  78. 78. o o oF - 32 = 1.8 x CTo convert between the scales, use thefollowing relationships:o o oF = 1.8 x C + 32oK = C + 273.15oo F - 32C =1.8
  79. 79. It is not uncommon for temperatures in the Canadianplains to reach –60oF and below during the winter.What is this temperature in oC and K?oo F - 32C =1.8o o60. - 32C = = -51 C1.8
  80. 80. Density
  81. 81. Density is the ratioof the mass of asubstance to thevolume occupied bythat substance.massd =volume
  82. 82. Mass is usuallyexpressed in gramsand volume in mLor =mL3gd =cmThe density of gases isexpressed in grams =L
  83. 83. Density varies with temperatureo24 CH O1.0000 g gd = = 1.00001.0000 mL mLo280 CH O1.0000 g gd = = 0.971821.0290 mL mL
  84. 84. Examples
  85. 85. A 13.5 mL sample of an unknown liquidhas a mass of 12.4 g. What is the density of the liquid?MDV= 0.919 g/mL=12.4g13.5mL=
  86. 86. 46.0 mL98.1 gA graduated cylinder is filled to the 35.0 mL mark with water.A copper nugget weighing 98.1 grams is immersed into thecylinder and the water level rises to the 46.0 mL. What is thevolume of the copper nugget? What is the density of copper?35.0 mLcopper nugget final initialV = V -V = 46.0mL - 35.0mL = 11.0mLg/mL8.92mL11.0g98.1VMD ===
  87. 87. The density of ether is 0.714 g/mL. What is themass of 25.0 milliliters of ether?Method 1(a) Solve the density equation for mass.massd =volume(b) Substitute the data and calculate.mass = density x volume0.714 g25.0 mL x = 17.9 gmL
  88. 88. The density of ether is 0.714 g/mL. What is themass of 25.0 milliliters of ether?Method 2 Dimensional Analysis. Use density as aconversion factor. Convert:0.714 g25.0 ml x = 17.9 gmLmL → ggmL x = gmLThe conversion of units is
  89. 89. The density of oxygen at 0oC is 1.429 g/L. What isthe volume of 32.00 grams of oxygen at thistemperature?Method 1(a) Solve the density equation for volume.massd =volume(b) Substitute the data and calculate.massvolume =density2232.00 g Ovolume = = 22.40 L1.429 g O /L
  90. 90. The density of oxygen at 0oC is 1.429 g/L. What is thevolume of 32.00 grams of oxygen at this temperature?Method 2 Dimensional Analysis. Use density as aconversion factor. Convert:2 221 L32.00 g O x = 22.40 L O1.429 g Og → LLg x = LgThe conversion of units is
  91. 91. Solving Problems1. Read the problem carefully. Determinewhat is known and what is to be solved forand write it down.– It is important to label all factors and units withthe proper labels.
  92. 92. 2. Determine which principles are involved andwhich unit relationships are needed to solvethe problem.– You may need to refer to tables for neededdata.3. Set up the problem in a neat, organized andlogical fashion.– Make sure unwanted units cancel.– Use sample problems in the text as guides forsetting up the problem.Basic Steps
  93. 93. 4. Proceed with the necessary mathematicaloperations.– Make certain that your answer contains theproper number of significant figures.5. Check the answer to make sure it isreasonable.Basic Steps
  94. 94. Chapter 2Atoms, Molecules and IonsPart 1
  95. 95.  In studying the materials of theEarth (and other parts of the Universe)scientists have found that all matter can bebroken down chemically into about 118different elements It is quite amazing that millions of knownsubstances are composed of so fewfundamental elements ….a phenomena not unlike the hundreds ofthousands of known words composed fromonly 26 letters of the alphabet
  96. 96. Compounds Are made by combining atoms of the variouselements…..just as words are constructed from the 26 letters ofthe alphabetand you had to learn the letters of the alphabetbefore you learned to read and write You now must learn the names andsymbols of the Chemical Elementsbefore you can read and write chemistry!
  97. 97. The Elements There are presently 118 known elements 88 occur naturally the remainder were made in the lab and decomposespontaneously into others by radioactive decay the elements vary tremendously in abundance-only 10 constitute over 99% of the Earth’s crust-about 93% of the mass of our bodies is composed of only 3 elements: C, H, and O-the list of elements found in living matter differs greatly from that of the earth’scrust*** Elements are fundamental to understandingChemistry
  98. 98. How Chemists use the word ElementElement-single atomof that element-sample ofthe element largeenough to weigh ona balance-some elementsthat containmolecules ratherthan individualatoms
  99. 99. Single Atoms of an Element – microscopicform of an elementSample of an Element Large Enough toWeigh on a Balance –Contains many, many atoms of the element and arethe macroscopic form of an elementSome Elements that Contain MoleculesRather than Individual Atoms – macroscopicform of elements like oxygen, hydrogen Nitrogen,chlorine, bromine and iodine exist as O2 , H2 , N2 , Cl2 ,Br2,F2 and I2
  100. 100. Names for Elements The names of chemical elements come from a variety ofsources- they are often derived from Greek, Latin or German wordsthat describe some property of the elementie. Gold- originally called aurum (Latin for “shinning dawn”)Lead- called plumbum (heavy)Names of Cl2(chlorine) and I2(iodine) comefrom Greek words describing their colorsBromine- derived from a Greek word for “stench” Some elements are named for the place they werediscovered: Francium, Germanium, Californium andAmericium
  101. 101. Element SymbolsElement symbols are abbreviations forelement names. Consist of the first letter or first two lettersof the elements name The first letter is always capitalized! The second letter (if present) is nevercapitalized.
  102. 102. Element Symbols• Fluorine – F• Oxygen – O• Neon – Ne• Silicon – Si Some have two letterswhich are not the firstand second letters inthe name• Zinc – Zn• Chlorine – Cl• Cadmium – Cd• Platinum - Pt
  103. 103. More Element Symbols Some other elementsare based on the Latinor Greek name.• Gold (aurum) - Au• Lead (plumbum)- Pb• Sodium (natrium) - Na• Iron (ferrum) - Fe• All 118 elements andtheir symbols are foundin the inside front coverof your text….as well asin the Periodic Table• We will look at thePeriodic Table in detailin chapters to come
  104. 104. Atomic Theory As scientists of the 18thCentury studied the natureof materials, several things became clear:1. Most natural materials are mixtures of pure substances2. Pure substances are either elements or combinations of elementscalled compounds3. A given compound always contains the same proportions by mass of itselementsie. H2O always contains 8g of oxygen for every 1g of hydrogen andCO2always contains 2.7g of oxygen for every 1g of carbon- this is known as the Law of Constant Composition- this law says that a given compound has the same compositionregardless of where it comes from
  105. 105. John Dalton• An English scientist and teacher• He knew of these observations and offered anexplanation for them• His explanation is known as:
  106. 106. DALTON’S ATOMIC THEORYThe main ideas of his theory include:1. Elements are made of tiny particles called atoms.2. All atoms of a given element are identical. Atoms of a givenelement are different from those of any other element.3. Atoms of one element can combine with atoms of otherelements to form compounds. A given compound alwayshas the same relative number and types of atoms4. Atoms are indivisible in chemical processes.** Atoms are not created or destroyed in chemical reactions.A chemical reaction simply changes the way atoms aregrouped together.
  107. 107. * Dalton’s Theory offered simple explainations for some basiclaws of chemistry such as:The Law of Conservation of Mass Mass is neither created or destroyed.If atoms are conserved in a reaction then mass must also beconservedThe Law of Constant Composition Tells us that a cmpd regardless of its origin or method ofpreparation always contains the same elements in the sameproportions by weightLaw of Multiple Proportions When 2 elements combine to form more than 1 cmpd themasses of one element which combines with a fixed mass ofthe other elelment are in a ratio of small whole numberssuch as 2:1