1. Unit 1 Matter and Measurements Honors Chemistry

2. Wolpa/Advanced Placement Chemistry 2 How to be Successful in Chemistry Memorize strategies not equations! Study a lot! Work ALL the problems. Self-evaluate after test results. Make use of Tutorial. Start a Study Group

3. Wolpa/Advanced Placement Chemistry 3 Chapter 1: Matter and Measurement Overview: The Study of Chemistry Classifications of Matter Properties of Matter Units of Measurement Uncertainty in Measurement Dimensional Analysis Basic Math Concepts

4. Wolpa/Advanced Placement Chemistry 4 Chemistry The study of matter and the changes it undergoes

5. Wolpa/Advanced Placement Chemistry 5 Matter Anything that has mass and occupies space Characterized by physical and chemical properties Law of the Conservation of Mass - matter is neither created nor destroyed in chemical reactions

6. Wolpa/Advanced Placement Chemistry 6 Element An element is a pure substance composed of one type of atom. An atom is the smallest particle of an element that retains the chemical properties of the element. An element is the most basic form of matter under ordinary circumstances Simplest chemical substance Only a few elements are found in their free state (nitrogen, oxygen, gold, etc.)

7. Wolpa/Advanced Placement Chemistry 7 Elements and the Periodic Table Each element is represented by a name and a symbol. (Periods/groups - alkali metals, alkaline earth metals, halogens, noble gases) The first letter is always capitalized the second (and third) are never capitalized.

8. Wolpa/Advanced Placement Chemistry 8 Compound A unique substance composed of two or more elements that are chemically combined (i.e. joined intimately, not just mixed together) Pure compounds have definite compositions and properties Require complex chemical procedures to separate into simpler substances (elements) Compounds include water, table salt, sugar, etc

9. Wolpa/Advanced Placement Chemistry 9 Properties of Substances Elements and Compounds are pure substances. Properties describe the particular characteristics of a substance Pure substances have definite composition and definite, unchanging properties Physical properties - can be observed without changing the substance Chemical properties - require that the substance change into another

10. Wolpa/Advanced Placement Chemistry 10 Physical States The three physical states are solid, liquid and gas  solids - have a definite shape and volume  liquid - have a definite volume but not a definite shape  gas - neither a definite volume or shape A substance exists in a particular physical state under defined conditions

11. Wolpa/Advanced Placement Chemistry 11

12. Wolpa/Advanced Placement Chemistry 12 Phase Changes Melting point or freezing point  temperature at which a substance changes from solid to liquid Boiling point or condensation point  temperature at which a substance changes from liquid to gas

14. More on Evaporation…. Liquid evaporates faster when heated. This is because heating a liquid increases the average kinetic energy of its particles. As evaporation occurs, the particles with the highest kinetic energy escape first. Evaporation is a cooling process!!

15. Boiling Point vapor pressure = external pressure When a liquid is heated to a temperature at which particles throughout the liquid have enough kinetic energy to vaporize, the liquid begins to boil. Because atmospheric pressure is lower at higher altitudes, boiling points decrease at higher altitudes.

17. Phase Diagram Cont… The Triple Point describes the only set of conditions at which all three phases can exist in equilibrium with one another. A decrease in pressure lowers the boiling point and raises the melting point An increase in pressure will raise the boiling point and lower the melting point.

18. Wolpa/Advanced Placement Chemistry 18 Density ratio of the mass of a substance to the volume of that mass usually measure in g/mL for solids and liquids; g/L for gases also a conversion factor relating the mass of a substance to it’s volume Specific gravity is the ratio of the mass of a substance to the mass of an equal volume of water

19. Wolpa/Advanced Placement Chemistry 19 What’s happening?

20. Wolpa/Advanced Placement Chemistry 20 Density Calculation Equation d=m/V Example: If an object has a mass of 15.0 g and a volume of 10cm 3 what’s the objects density? d = 15.0 g/ 10.0 cm 3 = 1.50 g/cm 3

21. Wolpa/Advanced Placement Chemistry 21 Temperature and its Measurement Temperature - measure of the intensity of the heat of a substance Thermometer - device to measure temperature Kelvin - K - SI unit of temperature Celsius - °C - commonly used unit Fahrenheit - °F - only used in USA

22. Wolpa/Advanced Placement Chemistry 22 Relationships between temperature scales

23. Wolpa/Advanced Placement Chemistry 23 The Kelvin scale The idea of negative temperatures is a problem for any mathematical treatment of temperature dependent properties. It was found that a practical minimum temperature did exist (absolute zero) which has a value of -273.15°C This is defined as 0 K (no degree sign) The Kelvin degree is the same size as the Celsius degree (K = °C + 273.15)

24. Wolpa/Advanced Placement Chemistry 24 Temperature Scale Comparison

25. Wolpa/Advanced Placement Chemistry 25 Chemical Properties Chemical properties - involve how a substance changes into another Sometimes quite difficult to determine Some examples are burning (as opposed to boiling) and color changes

26. Wolpa/Advanced Placement Chemistry 26 A mixture is a combination of two or more substances in which the substances retain their distinct identities. 1. Homogenous mixture – composition of the mixture is the same throughout. 2. Heterogeneous mixture – composition is not uniform throughout. ?

27. Wolpa/Advanced Placement Chemistry 27 Mixtures Combinations of two or more substances Can be separated by exploiting different physical properties (filtration, distillation, crystallization, chromatography) Have chemical and physical properties that are different from the substances that make them up The percentages by mass of the components of a mixture can be varied continuously

28. Wolpa/Advanced Placement Chemistry 28 Heterogeneous Vs. Homogeneous Mixtures

29. Wolpa/Advanced Placement Chemistry 29 Physical means can be used to separate a mixture into its pure components.

30. Wolpa/Advanced Placement Chemistry 30 Physical means can be used to separate a mixture into its pure components. (Mechanical process) magnet filtration distillation

31. Wolpa/Advanced Placement Chemistry 31 Physical Change: the composition of the substance remains the same but the state changes.

32. Wolpa/Advanced Placement Chemistry 32 Chemical Change: a new substance is formed.

33. Wolpa/Advanced Placement Chemistry 33 Physical Properties: Identifying properties of a substance. Density Solubility Color Melting/Boiling Point Crystalline Shape Malleability, Ductility, Conductivity, Luster Etc.

34. Wolpa/Advanced Placement Chemistry 34 An extensive property of a material depends upon how much matter is is being considered. An intensive property of a material does not depend upon how much matter is is being considered. mass length volume density malleability color Extensive and Intensive Properties 1.6

35. Wolpa/Advanced Placement Chemistry 35 Solutions A type of homogeneous mixture Usually involves a liquid phase, but can be solid-solid, liquid-liquid, solid-liquid, etc. The pure substances can be in different phases but form a homogeneous mixture (table salt and water, for example)

36. Wolpa/Advanced Placement Chemistry 36 Alloys important solid solutions of two or more metals dental fillings (silver and mercury) stainless steel (iron, chromium and nickel)

37. Wolpa/Advanced Placement Chemistry 37 Putting it All Together

38. Wolpa/Advanced Placement Chemistry 38 Measurements and Units Measurement - determines the quantity, dimensions or extent of something 1.Consist of two parts a. a numerical quantity (1.23) b. a specific unit (meters) Unit - a definite quantity adapted to as a standard of measurement

39. Wolpa/Advanced Placement Chemistry 39 Features of Measured Quantities When we measure a number, there are physical constraints to the measurement Instruments and scientists are not perfect, so the measurement is not perfect (i. e., it has error) The error in the measurement is related to the accuracy and the precision of the measurement

40. Wolpa/Advanced Placement Chemistry 40 Accuracy and Precision Accuracy – how close the measurement is to the “true” value (of course we have to know what the “true” value is) Precision - the degree to which the measurement is reproducible 1. expressed through how we write the number – significant figures

41. Wolpa/Advanced Placement Chemistry 41 Example: Accuracy and Precision

42. Wolpa/Advanced Placement Chemistry 42 Equations for Precision and Accuracy 1. Precision Percent Relative error % RE = Avg. Dev / Avg Value X 100 2. Accuracy Absolute Error % AE = (True value-Avg Value) X 100 True Value

43. Wolpa/Advanced Placement Chemistry 43 Significant Figures Any digit that is not zero is significant 1.234 kg 4 significant figures Zeros between nonzero digits are significant 6006 m 4 significant figures Zeros to the left of the first nonzero digit are not significant 0.08 L 1 significant figure One or more final zeros to the right of the decimal point are significant 2.00 mg 3 significant figures 0.00420 g 3 significant figures 10.006000 8 sig figs

44. Wolpa/Advanced Placement Chemistry 44 Counting Significant Figures Atlantic / Pacific Method a. Absent Decimal- Start on “atlantic” side of number & cross out all zeroes until 1 st nonzero digit is reached, remaining digits are significant b. Present decimal- start on the “pacific” side of the number & cross out all zeros until the 1 st nonzero digit Is reached, remaining digits are significant

45. Wolpa/Advanced Placement Chemistry 45 2. Examples:

46. Wolpa/Advanced Placement Chemistry 46 How many significant figures are in each of the following measurements? 24 mL2 significant figures 3001 g 4 significant figures 0.0320 m 3 3 significant figures 6.400 x 10 4 molecules 4 significant figures 560 kg2 significant figures

47. Wolpa/Advanced Placement Chemistry 47 Significant Figures Addition or Subtraction The answer cannot be more accurate than any of the original numbers. 89.332 1.1+ 90.432 round off to 90.4 one significant figure after decimal point 3.70 -2.9133 0.7867 two significant figures after decimal point round off to 0.79 370 -291.33 78.67 Number is rounded to “tens” place round off to 80

48. Wolpa/Advanced Placement Chemistry 48 Significant Figures Multiplication or Division The number of significant figures in the result is set by the original number that has the smallest number of significant figures 4.51 x 3.6666 = 16.536366= 16.5 3 sig figsround to 3 sig figs 6.8 ÷ 112.04 = 0.0606926 2 sig figsround to 2 sig figs = 0.061

49. Wolpa/Advanced Placement Chemistry 49 Significant Figures Exact Numbers Numbers from definitions or numbers of objects are considered to have an infinite number of significant figures The average of three measured lengths; 6.64, 6.68 and 6.70? 6.64 + 6.68 + 6.70 3 = 6.67333 = 6.67 Because 3 is an exact number = 7

50. Wolpa/Advanced Placement Chemistry 50 Scientific notation -Expressing very large and small numbers Using scientific Notation - Numbers are expressed with one nonzero digit to the left of the decimal point multiplied by 10 raised to an appropriate power 1. The base is the number with all of the appropriate significant digits 2. The exponent is the power of ten the base is multiplied by

51. Wolpa/Advanced Placement Chemistry 51 Scientific notation format

52. Wolpa/Advanced Placement Chemistry 52 Scientific notation and calculators 1. Calculators handle scientific notation by only inputting the exponent, using an EXP or EE key a. enter the base as you would a regular number, then press EXP or EE, then enter the exponent

53. Wolpa/Advanced Placement Chemistry 53 Scientific notation and significant figures 1. When using scientific notation the base must be written with the correct number of significant digits 2. All zeroes are significant when using scientific notation

54. Wolpa/Advanced Placement Chemistry 54 Measurement of mass, length and volume In the United States, we use a fairly awkward system of measurement for most things - the English system Scientists use the metric and SI systems of units for the measurement of physical quantities This system using standard units based on very precisely known properties of matter and light Prefixes are used in from of the units to indicate powers of ten

55. Wolpa/Advanced Placement Chemistry 55 SI Units MeasurementUnitSymbol MassKilogramkg LengthMeterM TimeSeconds TemperatureKelvinK QuantityMolemol EnergyJouleJ PressurePascalPa

56. Wolpa/Advanced Placement Chemistry 56 SI Prefixes PrefixSymbolPowerPrefixSymbolPower tera-T10 12 deci-d10 -1 giga-G10 9 centi-c10 -2 mega-M10 6 milli-m10 -3 kilo-k10 3 micro-  10 -6 hecto-h10 2 nano-n10 -9 deca-da10 1 pico-p10 -12

57. Wolpa/Advanced Placement Chemistry 57. Base Units Mass - the quantity of matter that a sample contains Note that weight is a measure of the attraction of gravity for a sample and it varies depending on the distance of the mass to a planet or moon Scientists often speak imprecisely of the “weight” of an amount of substance. They really mean mass.

58. Wolpa/Advanced Placement Chemistry 58 Basic SI units/Derived units Used to generate new Units Volume - space a given quantity of matter occupies Volume - expressed in terms of length - m 3 m 3 - an inconveniently large volume, so we use liter (L; one cubic decimeter) We often use a mL (1 cubic centimeter) for more manageable amounts of matter

59. Wolpa/Advanced Placement Chemistry 59 Converting between units The standard method to convert between two different units is the factor-label or dimensional analysis method Dimensional analysis converts a measurement in one unit to another by the use of a conversion factor Conversion factors are developed from relationships between units

60. Wolpa/Advanced Placement Chemistry 60 Dimensional Analysis Method of Solving Problems 1.Determine which unit conversion factor(s) are needed 2.Carry units through calculation; treat them like x and y. 3.If all units cancel except for the desired unit(s), then the problem was solved correctly. m is 10 -3 How many mg are in 1.63 kg? 1kg 1x10 6 mg 1.63 kg x = 1.63 x 10 6 mg 1kg 1x10 6 mg 1.63 kg x = 1.63 x10 -6 kg 2 mg k is 10 3 1kg = 1x10 6 mg

61. Wolpa/Advanced Placement Chemistry 61 The speed of sound in air is about 343 m/s. What is this speed in miles per hour? 1 mi = 1609 m1 min = 60 s1 hour = 60 min 343 m s x 1 mi 1609 m 60 s 1 min x 60 min 1 hour x = 767 mi hour meters to miles seconds to hours

62. Wolpa/Advanced Placement Chemistry 62 Solubility is a conversion factor between solute and solvent amounts How many grams of water would be needed to just dissolve 42 grams potassium chlorate at 70 o C? If 200 grams of water are used, is the sol’n saturated, unsaturated or super saturated?