1. Regents Chemistry: Mrs. Ingersoll The Study of Matter

2. Unit 1 – Matter and Energy 2

3. Matter has mass............and volume can be divided into substances + mixtures Substancematter with identical properties and composition, regardless of its source can be divided into elements and compounds 3

4. Matter SubstanceMixture ElementCompound ElementA substance that can not be broken down to simpler substances Made of only one kind of atom Examples: gold carbon oxygen 4

5. Atoms consist of: 1. a heavy, dense nucleus 2. a thin, mostly empty electron cloud atoms are made of even smaller, sub-atomic particles Particle Where found Mass Electrical charge proton nucleus 1 u+1 neutron nucleus 1 u 0 electroncloud 1/1836 u (almost zero) 5

6. Describing Atoms: 1. Atomic Number= number of protons determines what element the atom is 2. Atomic Mass (or mass number) = number of protons + neutrons B 10.81 5 atomic mass atomic number number of neutrons = atomic mass – atomic number 3. Electrical Charge= protons(+) + electrons(-) a neutral atom has a charge of 0 (protons = electrons) 6

7. Examples: = proton = neutron e - atomic number = _________ atomic mass = ________ charge = _______ element = ______ How many neutrons are in an atom of F? _______ How many electrons are in a neutral atom of boron? ______ Suppose an atom has 14 protons, 16 neutrons, and 16 electrons -- what element is it? __________ -- atomic number = _______ -- atomic mass = ______ -- charge = _______ 3 7 +1 Li 10 5 Si 14 30 -2 7

8. Compounds: --have two or more kinds of atoms held together by chemical bonds -- properties of the original elements are no longer apparent -- can be decomposed by a chemical reaction -- formed by a chemical change Examples water carbon dioxide sodium chloride glucose H 2 O CO 2 NaCl C 6 H 12 O 6 8

9. Mixtures --variable properties and composition --form without a chemical change --properties of ingredients still apparent can be: homogeneous -- evenly mixed Or heterogeneous -- not evenly mixed --can be separated without a chemical reaction 9

10. Mixtures: Separation Techniques 1.Filtration – based on differences in particle size, or solubility 2.Distillation – based on differences in boiling point 10

11. 3.Chromatography – based on differences in solubility and attraction to another substance, ex. paper Differences in density, freezing point, and other properties can also be used to separate mixtures 11

12. SUMMARY Element Compound Mixture Substances element + element + compound + compound 12

13. Changes in Matter 1. Physical change-- a rearrangement of existing particles ex. change in shape; phase change 2. Chemical change-- creation of new particles ex. burning of wood; rusting of iron Changes in matter are accompanied by changes in energy 13

14. Energy can be changed from one form to another.... but can not be created or destroyed this is the Law of Conservation of Energy Kinetic Energy: energy of motion includes heat (thermal) energy...motion of molecules Potential Energy: stored energy includes chemical energy 14

15. Imagine a change from “A” to “B” A  B 1. If B has less energy than A, --the extra energy is released as heat, light, etc --changes that release energy are exothermic 2. If B has more energy than A, -- the extra energy would be absorbed from the surroundings --changes that absorb energy are endothermic 15

16. Measuring Energy 1. Temperature: the average kinetic energy of a substance’s molecules measured in: degrees Celsius degrees Fahrenheit Kelvins ºC ºF K _ water boils 100 212 373 water freezes032 273 absolute zero -273-459 0 Temperature determines the direction of heat flow – from high temp. to low temp. K = ºC + 273 (ref table T) 16

17. 2. Total Heat Energy --depends on the amount of matter --measured in joules (J) or kilojoules (kJ) q = m · C · ΔT heat mass specific temp energy heat change specific heat capacity is the number of joules required to raise the temp of 1 gram of matter by 1ºC for water, C = 4.18 J/g-ºC (ref table B) it is lower for almost all other substances (ref table T) Examples: if 5.0 grams of water is heated from 50.0ºC to 54.0ºC, how much heat energy was used? if 10.0 grams of water at 30.0ºC has 1500.0 Joules of energy added. What’s the water’s new temperature? 17

18. Sample Heat Energy Problems 1.50.0g of water are heated from 22.0ºC to 31.0ºC. How many joules were added? 2.Some water at 75.0ºC cools to 40.0ºC. If the mass of the water is 7.00g, how many joules were released? 3.200.0g of water has a temp. of 9.0ºC, then 2000.0 J are added. What’s the new temperature? 4.A quantity of water has a temp. of 29.0ºC. 1200.0J are added, and the temp. increases to 79.0ºC. What is the water’s mass? 5.If adding 209J of heat to 5.00 grams of water can make the temperature change to 53.0ºC, what was the water’s initial temperature? 6.How much energy is required to heat 5.00g of silver from 20.0ºC to 200.0ºC? (silver’s specific heat capacity is 0.24J/gºC) 18

19. Phase changes absorb or release energy: Gas Liquid Solid melting boiling condensation freezing The energy change when 1 gram of a substance freezes or melts (at constant temp.) is called its heat of fusion (H f ) For water, H f = 334 J/g When 1 gram boils or condenses (at constant temp.), the energy change is called its heat of vaporization (H v ) For water, H v = 2260 J/g (ref table B) 19

20. Sample Problems 1.How much energy is released when 25.0g of 0ºC water solidifies? solution: q = m · H f q = 25.0 · 334 q = 8350 J 2.7.00g water is at 45.0ºC. How much energy is required to change it to water vapor at 100.0ºC ? solution: here we have a temp change and a phase change q = m C Δt q = 7.00 · 4.18 · 55.0 q = 1610 J to raise temp q = m H v q = 7.00 · 2260 q = 15820 J to vaporize total energy = 1610 + 15820 = 17430 J 20

21. Phase Change Diagram a graph of time vs temperature of a substance being heated or cooled at a constant rate time (min) temp (ºC) solid melting liquid boiling gas melting pt boiling pt HfHf HvHv 21

22. The diagonal parts of the graph show increasing kinetic energy -- temp change The horizontal parts of the graph show increasing potential energy -- phase change 22

23. Gases -- take the shape and volume of their container --molecules far apart; compressible -- elements: H 2 N 2 O 2 F 2 Cl 2 He Ne Ar Kr Xe Rn -- many compounds: ex. CO 2 NH 3 CH 4 etc. Gases exert pressure molecules colliding with container wall weight of air above pushing downwards – atmospheric pressure 23

24. On average, at sea level, atmospheric pressure will support a column of mercury 29.92 inches, or 760 mm, high. The metric unit of pressure is the kilopascal (kPa) 29.92” of mercury = 101.3 kPa For high pressures, we use the atmosphere (atm) 1 atm = 101.3 kPa = standard pressure standard temperature = 0ºC = 273K STP = 273K, 101.3kPa 24

25. Behavior of Gases-- The Gas Laws Robert Boyle, England, 1600’s: Boyle’s Law The volume of a gas is inversely proportional to the pressure exerted on it (at constant temp) as pressure, volume V  1P1P if the pressure is doubled (x2), the volume is halved (x½) Example: A gas has a volume of 20.0 L when the pressure is 100 kPa. What will the volume become if the pressure drops to 50 kPa? 25

26. Example: Suppose the 20.0 L of gas has its pressure change from 101.3kPa to 68.4 kPa. New volume = ? P x V = k P 1 x V 1 = P 2 x V 2 Jacques Charles, France, 1700’s: Charles’ Law the volume of a gas is directly proportional to its Kelvin temperature (at constant pressure) V  T if temp (K) is doubled, volume is doubled Example: A gas has a volume of 50.0mL at a temp. of 200K. What will the volume become if the temp drops to 100K ? as temperature, volume 26

27. Example: Suppose the 50.0 mL of gas had its temperature change from 27ºC to -13ºC. New Volume = ? VTVT = k V1T1V1T1 = V2T2V2T2 If pressure and temperature change at the same time, use the Combined Gas Law -- P 1 V 1 T1T1 = P 2 V 2 T2T2 Example: A 10.0L sample of nitrogen is stored at 25ºC and 120 kPa What is its volume at STP? Example: A scuba diver fills her 8.0L tank to a pressure of 10.0 atm at 30.0ºC. During her dive, the tank cools to 5.0ºC. What is the resulting pressure in the tank? 27

28. John Dalton, England, 1800’s: Law of Partial Pressures In a mixture of gases, the partial pressure of each gas is proportional to the number of molecules of that gas The total pressure in a mixture is equal to the sum of the partial pressures 60 kPa + 100 kPa = 160 kPa 28

29. GAS LAW PROBLEMS --Answer Key 1.3. 5.6.8.1.3. 5.6.8. 120 X 200 800 =X = 480 mL 2. 50  510 = 150  X X = 170 mL 100 X 300 250 = X = 83 mL 1  50 0.25  X 300 150 = X = 100 L 1 st part 45 X 260 280 = X = 48.5 psi, NO 45 50 260 X = X = 289 K, or 16  C 2 nd part 4.4. 25  10 101.3  X 546 273 =X = 1.2 L 7.7. 25  10 1  X 819 273 = X = 83 L 70 + 150 = 220 kPa 9.9. Volume is x2, so pressure is x ½ 110 kPa 10. 15% of 120 120  0.15 = 18 atm 29

30. Chemists have searched for a theory to explain these gas laws The kinetic-molecular theory (KMT) explains them Assumptions of the Theory 1.gas molecules are in constant, random, straight-line motion 2.molecules collide with each other and with the container, but lose no energy 3. molecules don’t attract one another 4. molecules take up no space (compared to the size of the container) a gas which followed all these assumptions would be an ideal gas for real gases, the gas laws are good approximations 30

31. Problems with the Theory: 1. all gas molecules are attracted to each other 2.gas molecules do have volume, and the volume becomes important when the molecules are close-packed... (high pressure or low temperature) SUMMARY: Real gases are closest to ideal when their molecules are: 1. not very attracted to each other................. noble gases 2. small......................................................... low formula mass 3. spread out................................................ high temp, low pressure 31

32. Amedeo Avogadro, Italy, 1800’s at the same temp and pressure, equal volumes of different gases have the same number of molecules Example: 1.0 L of H 2 has the same number of molecules as 1.0 L of CO 2. The CO 2 weighs more because each CO 2 molecule is heavier (formula mass H 2 = 2, CO 2 = 44) Since it’s difficult to work with single molecules, chemists usually work with groups of them: Avogadro’s number = 6.02 x 10 23 6.02 x 10 23 molecules = 1 mole Why 6.02 x 10 23 ? --because 6.02 x 10 23 molecules of any substance weigh the formula mass of that substance, in grams 32

33. Examples: 6.02 x 10 23 Na atoms = ______moles 1 = _______ grams 23 6.02 x 10 23 H 2 O molecules = ______ moles1= _______ grams18 3.01 x 10 23 B atoms = _____ moles 0.5 = _______ grams 5.5 80 g calcium = _____ moles 2 = _______________atoms 12.04 x 10 23 48 g carbon = _____ moles 4 = ______________ atoms 24.08 x 10 23 16 g aluminum = ______ moles moles = grams. formula mass 0.59 = _____________ atoms 3.55 x 10 23 33

34. By Avogadro’s law, 1 mole of any gas must have the same volume The molar volume of a gas at STP = 22.4 L Examples: 2. 44.8 L of hydrogen at STP = ____ moles 2 = ___________molecules 12.04 x 10 23 1. What volume would 4 moles NO 2 occupy, at STP? __________ 89.6 L 3. What is the mass of 22.4 L of O 2 at STP? _________ 32 g 4. a. What volume does 1 mole NO occupy at STP? _________ 22.4 L b. What is the mass of 1 mole of NO? _________ 30 g c. What is the density of NO? __________ density = mass volume 1.3 g/L 5. A gas has a density of 1.78 g/L at STP a. what is the mass of 1 L of this gas? _________1.78 g b. how much does 22.4 L of this gas weigh? ________39.9 g c. what is the formula mass of this gas? ________ 39.9 g 34

35. Liquids Take the shape of the container, but have a definite volume Kinetic Molecular Theory as temp or pressure --molecules are forced closer together --forces of attraction between molecules increase --molecules clump together, but can still move freely 35

36. A few molecules have enough kinetic energy to escape from the surface of the liquid This is evaporation In a closed container, evaporated molecules eventually return to the liquid Equilibrium is reached when the rate of evaporation = the rate of condensation Liquid ⇔ Gas 36

37. Particles of evaporated liquid (vapor) cause pressure, just as any gas molecules would The partial pressure due to evaporated liquid is called the vapor pressure Vapor pressure is a measure of how easily evaporated a liquid is VP depends on: 1.Kinetic energy of molecules (temperature) 37

38. weak forces strong forces high vp low vp 2.Strength of attraction between molecules See reference table H 38

39. Boiling Occurs when vapor pressure= atmospheric pressure Example: a 10-minute boiled egg at sea level pressure = 101.3 kPa bp = 100ºC at 18,000 ft pressure = 50 kPa bp = 82ºC The Normal Boiling Point is the bp at standard pressure (101.3 kPa) So, lower pressure results in a lower boiling point 39

40. SUMMARY The strength of the intermolecular forces can be estimated from: 1. vapor pressure: high vp = weak forces 2. normal boiling point:high bp = strong forces (table H) (table S) 40

41. Solids Have a definite shape and volume Kinetic Molecular Theory at lower temp. & higher pressure: -- molecules very close together -- intermolecular forces very strong -- molecules lock each other in place -- kinetic energy in the form of vibration -- KE can still be quite large 41

42. In most solids, molecules line up in a regular, geometric pattern NaCl crystal latticeNaCl crystals Melting point:temperature at which solid and liquid coexist at equilibrium solid ⇔ liquid ice crystal 42

43. Some solids also undergo: sublimation: solid  gas deposition: gas  solid ex. dry ice (CO 2 ) iodine (I 2 ) Sublimation occurs in solids with weak intermolecular forces and can produce high vapor pressures 43