1. Chapter 3 Matter and Energy

2. Tro's "Introductory Chemistry", Chapter 3 2 Matter Matter is anything that occupies space and has mass. Even though it appears to be smooth and continuous, matter is actually composed of a lot of tiny little pieces we call atoms and molecules.

3. Tro's "Introductory Chemistry", Chapter 3 3 Atoms and Molecules Atoms are the tiny particles that make up all matter. In most substances, the atoms are joined together in units called molecules.

4. Tro's "Introductory Chemistry", Chapter 3 4 Classification of Matter Matter can be classified in two ways: 1- according to its physical state 2- according to its composition 4

5. Tro's "Introductory Chemistry", Chapter 3 5 Classification of Matter According to its Physical State (STATES OF MATTER) Matter can be classified as solid, liquid, or gas SOLID e.g. stone, charcoal, diamond LIQUID e.g. water, wine, vinegar GAS e.g. oxygen, carbon dioxide MATTER

6. Tro's "Introductory Chemistry", Chapter 3 6 Solids The particles in a solid are packed close together and are fixed in position. Due to this close parking solids are incompressible. The inability of the particles to move around results in solids retaining their shape and volume when placed in a new container and prevents the particles from flowing. Solids may be further classified as: Crystalline Solids Amorphous Solids

7. Tro's "Introductory Chemistry", Chapter 3 7 Crystalline vs Amorphous Some solids have their particles arranged in an orderly geometric pattern—we call these crystalline solids. Salt and diamonds. Other solids have particles that do not show a regular geometric pattern over a long range—we call these amorphous solids. Plastic and glass.

8. Tro's "Introductory Chemistry", Chapter 3 8 Liquids The particles in a liquid are closely packed, but they have some ability to move around. The close packing results in liquids being incompressible. The ability of the particles to move allows liquids to take the shape of their container and to flow. However, they don’t have enough freedom to escape and expand to fill the container.

9. Tro's "Introductory Chemistry", Chapter 3 9 Gases In the gas state, the particles have complete freedom from each other. The particles are constantly flying around, bumping into each other and the container. In the gas state, there is a lot of empty space between the particles. On average.

10. Tro's "Introductory Chemistry", Chapter 3 10 Gases, Continued Because there is a lot of empty space, the particles can be squeezed closer together. Therefore, gases are compressible. Because the particles are not held in close contact and are moving freely, gases expand to fill and take the shape of their container, and will flow.

11. Tro's "Introductory Chemistry", Chapter 3 11 Summary: Properties of Solids, Liquids and Gases VolumeShapeCompressible? Solidsfixed NO Liquidsfixednot fixedNO Gasesnot fixed YES

12. Tro's "Introductory Chemistry", Chapter 3 12 Classification of Matter by Composition Pure Substance Constant Composition Mixture Variable Composition Matter

13. Tro's "Introductory Chemistry", Chapter 3 13 Classification of Matter as a pure substance A pure substance is made of a single type of particle (i.e., atom or molecule). The composition of a pure substance does not change from one sample to another and because of this, all samples have the same characteristics (or properties)

14. Tro's "Introductory Chemistry", Chapter 3 14 Classification of Matter as a Pure Substance Pure substances can further be sub-divided into two groups:  Elements  Compounds Elements: Pure substances that cannot be broken down into simpler substances e.g., copper, helium, sodium Compounds: Pure substances that are made of two or more elements in definite proportions e.g., sodium chloride NaCl, Carbon dioxide CO 2

15. Tro's "Introductory Chemistry", Chapter 3 15 Classification of Matter as a Mixture A mixture is a combination of two or more substances in which each substance retain their distinct characteristics (or properties). e.g., a mixture of rice and sand Mixtures can also be further sub-divided into two groups: Homogeneous mixture Heterogeneous mixture Homogeneous mixture: Has uniform composition through out the sample e.g., solutions such as tea with sugar, salt solutions Heterogeneous mixture: Does not have a uniform composition through out the sample. e.g., sand and water Note: HOMO means the same while HETERO means different

16. Tro's "Introductory Chemistry", Chapter 3 16 Example—Classify the Following as Homogeneous or Heterogeneous Mixtures A cup of coffee A mixture of table sugar and black pepper A mixture of sugar dissolved in water Sand and water - homogeneous - heterogeneous

17. 17 Classifying Matter

18. Tro's "Introductory Chemistry", Chapter 3 18 How Do We Distinguish Matter? Looking at both flasks, it is had to tell the difference between the two substances How can we tell if one flask contain water and the other contains an acid? We can only tell by studying their properties water acid

19. Tro's "Introductory Chemistry", Chapter 3 19 Properties of Matter Each sample of matter is distinguished by its characteristics. The characteristics of a substance are called its properties.

20. Tro's "Introductory Chemistry", Chapter 3 20 Types of Properties of Matter Physical Properties Properties of matter that can be observed without changing its composition. e.g., coke is dark brown Chemical Properties Properties of matter that can be observed only when matter changes its composition. e.g., gasoline is a very flammable liquid

21. Tro's "Introductory Chemistry", Chapter 3 21 Some Physical Properties

22. Tro's "Introductory Chemistry", Chapter 3 22 Example: Some Physical Properties of Iron Iron melts at 1538 °C Iron boils at 4428 °C. Iron’s density is 7.87 g/cm 3. Iron conducts electricity, but not as well as most other common metals.

23. Tro's "Introductory Chemistry", Chapter 3 23 Some Chemical Properties

24. Tro's "Introductory Chemistry", Chapter 3 24 Example: Some Chemical Properties of Iron Iron is easily oxidized in moist air to form rust. When iron is added to hydrochloric acid, it produces a solution of ferric chloride and hydrogen gas. Iron is more reactive than silver, but less reactive than magnesium.

25. Tro's "Introductory Chemistry", Chapter 3 25 Example—Decide Whether Each of the Observations About Table Salt Is a Physical or Chemical Property Salt is a white, granular solid. Salt melts at 801 °C. A liquid burns with a blue flame. 36 g of salt will dissolve in 100 g of water. Salt solutions conduct electricity. When a clear, colorless solution of silver nitrate is added to a salt solution, a white solid forms. When electricity is passed through molten salt, a gray metal forms at one terminal and a yellow-green gas at the other. physical chemical physical chemical

26. Tro's "Introductory Chemistry", Chapter 3 26 Changes in Matter Changes that alter the state or appearance of matter without altering its composition are called physical changes. Changes that alter the composition of matter are called chemical changes.

27. Tro's "Introductory Chemistry", Chapter 3 27 Physical Changes in Matter Physical Changes—Changes that do not affect composition of matter. Heating water.  Raises its temperature, but it is still water. Evaporating butane from a lighter. Dissolving sugar in water.  Even though the sugar seems to disappear, it can easily be separated back into sugar and water by evaporation.

28. Tro's "Introductory Chemistry", Chapter 3 28 Changes in Matter, Continued Chemical Changes— involve a change in composition. A chemical reaction.  Silver combines with sulfur in the air to make tarnish. Rusting is iron combining with oxygen to make iron(III) oxide. Burning results in butane from a lighter to be changed into carbon dioxide and water.

29. Tro's "Introductory Chemistry", Chapter 3 29 Is it a Physical or Chemical Change? A physical change results in a different form of the same substance. The kinds of molecules don’t change. i.e., composition stays the same. A chemical change results in one or more completely new substances. The new substances have different molecules than the original substances. i.e., composition changes. Appearance may or may not change. You will observe different physical properties because the new substances have their own physical properties.

30. Tro's "Introductory Chemistry", Chapter 3 30 Phase Changes Are Physical Changes Vaporizing = liquid to gas. Melting = solid to liquid. Subliming = solid to gas. Freezing = liquid to solid. Condensing = gas to liquid. Deposition = gas to solid. Changes in the state of matter require heating or cooling the substance.

31. Tro's "Introductory Chemistry", Chapter 3 31 Example—Classify Each Change as Physical or Chemical, Continued Evaporation of rubbing alcohol. Sugar turning black when heated. An egg splitting open and spilling out. Physical change Chemical change Physical change

32. Tro's "Introductory Chemistry", Chapter 3 32 Practice—Classify Each Change as Physical or Chemical Sugar fermenting. Bubbles escaping from soda. Bubbles that form when hydrogen peroxide is mixed with blood.

33. Tro's "Introductory Chemistry", Chapter 3 33 Separation of Mixtures Separating mixtures based on different physical properties of their components. DecantingDensity EvaporationVolatility ChromatographyAdherence to a surface FiltrationState of matter (solid/liquid/gas) DistillationBoiling point TechniqueDifferent Physical Property DecantingDensity EvaporationVolatility ChromatographyAdherence to a surface FiltrationState of matter (solid/liquid/gas) DistillationBoiling point TechniqueDifferent Physical Property

34. Tro's "Introductory Chemistry", Chapter 3 34 Distillation

35. Tro's "Introductory Chemistry", Chapter 3 35 Filtration

36. Tro's "Introductory Chemistry", Chapter 3 36 Law of Conservation of Mass The law of conservation of mass states: “Matter is neither created nor destroyed in a chemical reaction.” The total amount of matter present before a chemical reaction is always the same as the total amount after. Example: 58 grams of butane burns in 208 grams of oxygen to form 176 grams of carbon dioxide and 90 grams of water. butane + oxygen  carbon dioxide + water 58 grams + 208 grams  176 grams + 90 grams 266 grams = 266 grams

37. Tro's "Introductory Chemistry", Chapter 3 37 Practice—A Student Places Table Sugar and Sulfuric Acid into a Beaker and Gets a Total Mass of 144.0 g. Shortly, a Reaction Starts that Produces a “Snake” of Carbon Extending from the Beaker and Steam Is Seen Escaping. If the Carbon Snake and Beaker at the End Have a Total Mass of 129.6 g, How Much Steam Was Produced?

38. Tro's "Introductory Chemistry", Chapter 3 38 Energy Energy is anything that has the capacity to do work. Unlike matter, energy does not have mass and does not occupy any space. Although chemistry is the study of matter, matter is effected by energy. It can cause physical and/or chemical changes in matter.

39. Tro's "Introductory Chemistry", Chapter 3 39 Law of Conservation of Energy “Energy can neither be created nor destroyed.” The total amount of energy in the universe is constant. There is no process that can increase or decrease that amount. However, we can transfer energy from one place in the universe to another, and we can also change energy from one form to another.

40. Tro's "Introductory Chemistry", Chapter 3 40 Kinds of Energy Kinetic and Potential Potential energy is energy that is stored. Water flows because gravity pulls it downstream. However, the dam won’t allow it to move, so it has to store that energy. Kinetic energy is energy of motion, or energy that is being transferred from one object to another. When the water flows over the dam, some of its potential energy is converted to kinetic energy of motion.

41. Tro's "Introductory Chemistry", Chapter 3 41 Some Forms of Energy Electrical Energy Kinetic energy associated with the flow of electrical charge. Heat or Thermal Energy Kinetic energy associated with molecular motion. Light or Radiant Energy Kinetic energy associated with energy transitions in an atom. Nuclear Energy Potential energy stored in the nucleus of atoms. Chemical Energy Potential energy in compounds.

42. Tro's "Introductory Chemistry", Chapter 3 42 Converting Forms of Energy When water flows over the dam, some of its potential energy is converted to kinetic energy. Some of the energy is stored in the water because it is at a higher elevation than the surroundings. The movement of the water is kinetic energy. Along the way, some of that energy can be used to push a turbine to generate electricity. Electricity is one form of kinetic energy. The electricity can then be used in your home. For example, you can use it to heat cake batter you mixed, causing it to change chemically and storing some of the energy in the new molecules that are made.

43. Tro's "Introductory Chemistry", Chapter 3 43 How do We Measure Energy? Units of Energy The SI unit for energy is the joule (J) Other units such as calorie, Calorie, KWh are widely used calorie (cal) is the amount of energy needed to raise one gram of water by 1 °C. 1 kcal = energy needed to raise 1000 g of water 1 °C. 1 food calories (Cal) = 1 kcals. Energy Conversion Factors 1 calorie (cal)=4.184 joules (J) 1 Calorie (Cal)=1000 calories (cal) 1 kilowatt-hour (kWh)=3.60 x 10 6 joules (J)

44. Tro's "Introductory Chemistry", Chapter 3 44 Some Common Uses of Energy Use Unit Energy Required to Raise Temperature of 1 g of Water by 1°C Energy Required to Light 100-W Bulb for 1 Hour Energy Used by Average U.S. Citizen in 1 Day joule (J)4.183.6 x 10 5 9.0 x 10 8 calorie (cal)1.008.60 x 10 4 2.2 x 10 8 Calorie (Cal)1.00 x 10 -3 86.02.2 x 10 5 kWh1.1 x 10 -6 0.1002.50 x 10 2

45. Example 3.5—Convert 225 Cal to Joules 225 Cal = 9.41 x 10 5 J Round:6.Significant figures and round. Solution:5.Follow the solution map to Solve the problem. Solution Map: 4.Write a Solution Map. 1 Cal = 1000 cal 1 cal = 4.184 J Conversion Factors: 3.Write down the appropriate Conversion Factors. ? J Find:2.Write down the quantity you want to Find and unit. 225 Cal Given:1.Write down the Given quantity and its unit. CalcalJ 3 sig figs 3 significant figures

46. 46 Practice 1: The complete combustion of a wooden match produces 512 cal of heat. How many kilojoules are produced? Practice 2: An energy bill indicates that the customer used 955 KWh. How many calories did the customer use? Answer: 8.22 x exp(8) cal. Answer: 2.14 kJ

47. Tro's "Introductory Chemistry", Chapter 3 47 Practice 1: The complete combustion of a wooden match produces 512 cal of heat. How many kilojoules are produced? Answer: 2.14 kJ

48. 48 Practice 2: An energy bill indicates that the customer used 955 KWh. How many calories did the customer use? Answer: 8.22 x exp(8) cal.

49. Tro's "Introductory Chemistry", Chapter 3 49 Energy Changes Processes that involve energy changes can be: Exothermic Endothermic

50. Tro's "Introductory Chemistry", Chapter 3 50 Exothermic Processes When a change results in the release of energy it is called an exothermic process. An exothermic chemical reaction occurs when the reactants have more chemical potential energy than the products. The excess energy is released into the surrounding materials, adding energy to them. Often the surrounding materials get hotter from the energy released by the reaction.

51. Tro's "Introductory Chemistry", Chapter 3 51 An Exothermic Reaction Potential energy Reactants Products Surroundings reaction Amount of energy released

52. Tro's "Introductory Chemistry", Chapter 3 52 Endothermic Processes When a change requires the absorption of energy it is called an endothermic process. An endothermic chemical reaction occurs when the products have more chemical potential energy than the reactants. The required energy is absorbed from the surrounding materials, taking energy from them. Often the surrounding materials get colder due to the energy being removed by the reaction.

53. Tro's "Introductory Chemistry", Chapter 3 53 An Endothermic Reaction Potential energy Products Reactants Surroundings reaction Amount of energy absorbed

54. Tro's "Introductory Chemistry", Chapter 3 54 Temperature Scales Fahrenheit scale, °F. Used in the U.S. Celsius scale, °C. Used in all other countries. A Celsius degree is 1.8 times larger than a Fahrenheit degree. Kelvin scale, K. Absolute scale.

55. Temperature Scales CelsiusKelvinFahrenheit -273°C -269°C -183°C -38.9°C 0°C 100°C 0 K 4 K 90 K 234.1 K 273 K 373 K -459 °F -452°F -297°F -38°F 32°F 212°F Absolute zero BP helium Boiling point oxygen Boiling point mercury Melting point ice Boiling point water 0 R 7 R 162 R 421 R 459 R 671 R Rankine Room temp 25°C 298 K75°F534 R

56. Tro's "Introductory Chemistry", Chapter 3 56 The Fahrenheit Temperature Scale The Fahrenheit temperature scale was setup by assigning 0 °F to the freezing point of concentrated saltwater and 96 °F for normal body temperature. Room temperature is about 72 °F.

57. Tro's "Introductory Chemistry", Chapter 3 57 The Celsius Temperature Scale Was setup by assigning 0 °C to the freezing point of distilled water and 100 °C to the boiling point of distilled water. Most commonly used in science. More reproducible standards. Room temperature is about 22 °C.

58. Tro's "Introductory Chemistry", Chapter 3 58 The Kelvin Temperature Scale Both the Celsius and Fahrenheit scales have negative numbers. The Kelvin scale avoids negative numbers and therefore the lowest temperature is 0 K. 0 K is called absolute zero. It is too cold for matter to exist because all molecular motion would stop. 0 K = -273 ° C = -459 °F. Absolute zero is a theoretical value obtained by following patterns mathematically.

59. Tro's "Introductory Chemistry", Chapter 3 59 Celsius to Kelvin K = ° C + 273 Fahrenheit to Celsius Converting from one temperature scale to another

60. Example 3.7—Convert –25 °C to Kelvins 248 K Round:6.Significant figures and round. Solution: 5.Follow the solution map to Solve the problem. Solution Map: 4.Write a Solution Map. Equation:3.Write down the appropriate Equations. K ? Find: 2.Write down the quantity you want to Find and unit. -25 °C Given:1.Write down the Given quantity and its unit. ° CK K = ° C + 273

61. Example 3.8—Convert 55° F to Celsius Units and magnitude are correct. Check:7.Check. 12.778 °C = 13 °C Round:6.Significant figures and round. Solution: 5.Follow the solution map to Solve the problem. Solution Map: 4.Write a Solution Map. Equation:3.Write down the appropriate Equations. ° C ? Find: 2.Write down the quantity you want to Find and unit. 55 °F Given:1.Write down the Given quantity and its unit. units place and 2 sig figs units place and 2 sig figs ° F° C

62. Example 3.9—Convert 310. K to Fahrenheit Units and magnitude are correct. Check:7.Check. 98.6 °F = 99 °F Round:6.Significant figures and round. Solution: 5.Follow the solution map to Solve the problem. Solution Map: 4.Write a Solution Map. Equation:3.Write down the appropriate Equations. °F ? Find: 2.Write down the quantity you want to Find and unit. 310 K Given:1.Write down the Given quantity and its unit. units place and 3 sig figs units place and 2 sig figs K = °C + 273 °F°CK °C = K - 273

63. Tro's "Introductory Chemistry", Chapter 3 63 Practice 1- A sick child has a body temperature of 40.00 °C. What is the child’s temperature in Kelvins (K) ? Practice 2- During one summer day in Gettysburg, a record temperature of 92.1 ° F was reported. What is the temperature in K ?

64. Tro's "Introductory Chemistry", Chapter 3 64 Practice 1- A sick child has a body temperature of 40.00 °C. What is the child’s temperature in Kelvins (K) ?

65. Tro's "Introductory Chemistry", Chapter 3 65 Practice 2- During one summer day in Gettysburg, a record temperature of 92.1 ° F was reported. What is the temperature in K ?

66. Tro's "Introductory Chemistry", Chapter 3 66 Energy and the Temperature of Matter The increase in temperature of an object depends on the amount of heat energy added (q) and the mass of the object. If you double the added heat energy the temperature will increase twice as much. If you double the mass, it will take twice as much heat energy to raise the temperature the same amount.

67. 67 Heat Capacity Heat capacity is the amount of heat a substance must absorb to raise its temperature by 1 °C. cal/°C or J/°C. Metals have low heat capacities; insulators have high heat capacities. Specific heat = heat capacity of 1 gram of the substance. cal/g°C or J/g°C. Water’s specific heat = 4.184 J/g°C for liquid.  Or 1.000 cal/g°C.

68. Tro's "Introductory Chemistry", Chapter 3 68 Specific Heat Capacity Specific heat is the amount of energy required to raise the temperature of one gram of a substance by 1 °C. The larger a material’s specific heat is, the more energy it takes to raise its temperature a given amount. Like density, specific heat is a property of the type of matter. It doesn’t matter how much material you have. It can be used to identify the type of matter. Water’s high specific heat is the reason it is such a good cooling agent. It absorbs a lot of heat for a relatively small mass.

69. Tro's "Introductory Chemistry", Chapter 3 69 Specific Heat Capacities

70. Tro's "Introductory Chemistry", Chapter 3 70 Heat Gain or Loss by an Object The amount of heat energy gained or lost by an object depends on 3 factors: how much material there is what the material is how much the temperature changed Amount of Heat = Mass x Specific Heat Capacity x Temperature Change q = m x C x  T

71. Example 3.10—Calculate Amount of Heat Needed to Raise Temperature of 2.5 g Ga from 25.0 to 29.9 °C Units and magnitude are correct. Check:7.Check. 4.557 J = 4.6 J Round:6.Significant figures and round. Solution: 5.Follow the solution map to Solve the problem. Solution Map: 4.Write a Solution Map. Equation:3.Write down the appropriate Equations. q, J Find: 2.Write down the quantity you want to Find and unit. m = 2.5 g, T 1 = 25.0 °C, T 2 = 29.9 °C, C = 0.372 J/g°C Given:1.Write down the Given quantity and its unit. 2 significant figures m, C,  T q

72. Example—Calculate the Amount of Heat Released When 7.40 g of Water Cools from 49° to 29 °C q = m ∙ C s ∙  T C s = 4.18 J/g  C (Table 3.4) The unit and sign are correct. T 1 = 49 °C, T 2 = 29 °C, m = 7.40 g q, Jq, J Check: Check. Solution: Follow the concept plan to solve the problem. Solution Map: Relationships: Strategize Given: Find: Sort Information C s m,  T q

73. Tro's "Introductory Chemistry", Chapter 3 73 Practice 1- If you hold gallium in your hand, it melts from body heat. How much heat must 3.5 g of gallium absorb from your hand to raise its temperature from 25.0 °C to 30.5 °C? The heat capacity of gallium is 0.372 J/g°C. Practice 2- A bottle containing 24.1 g of an alcohol was removed from a refrigerator and placed on top of a counter in a kitchen. The amount of heat absorbed by the alcohol was determined to be 1.17 kJ. Given that the specific heat of the alcohol is 2.42 J/g°C 1-Calculate the change in temperature 2-If the initial temperature of the alcohol was 5 °C, find the final temperature

74. Tro's "Introductory Chemistry", Chapter 3 74 Practice 1- If you hold gallium in your hand, it melts from body heat. How much heat must 3.5 g of gallium absorb from your hand to raise its temperature from 25.0 °C to 30.5 °C? The heat capacity of gallium is 0.372 J/g°C.

75. Tro's "Introductory Chemistry", Chapter 3 75 Practice 2- A bottle containing 24.1 g of an alcohol was removed from a refrigerator and placed on top of a counter in a kitchen. The amount of heat absorbed by the alcohol was determined to be 1.17 kJ. Given that the specific heat of the alcohol is 2.42 J/g°C 1-Calculate the change in temperature 2-If the initial temperature of the alcohol was 5 °C, find the final temperature

76. Tro's "Introductory Chemistry", Chapter 3 76 Practice 3- 25.3 g of iron was heated to 53.2 °C and let to cool slowly. If the amount of heat lost by iron during the cooling process is 353.5 J, find the final temperature of the iron. The specific heat of iron is 0.499 J/g°C.

77. Tro's "Introductory Chemistry", Chapter 3 77 Recommended Study Problems Chapter 3 NB: Study problems are used to check the student’s understanding of the lecture material. Students are EXPECTED TO BE ABLE TO SOLVE ALL THE SUGGESTED STUDY PROBLEMS. If you encounter any problems, please talk to your professor or seek help at the HACC-Gettysburg learning center. Questions from text book Chapter 3, p 81 1, 3, 5, 15, 17, 19, 21, 23, 27, 28, 29, 31, 35-37, 39, 41, 44, 45, 47, 51, 52, 55, 56, 57, 60, 63-65, 67, 71, 73, 75, 77, 79, 81, 83, 87, 93, 97, 99, 107, 109, 110 ANSWERS -The answers to the odd-numbered study problems are found at the back of your textbook