1. Section 3.4—Counting Molecules So the number of molecules affects pressure of an airbag…how do we “count” molecules?

2. What is a Mole? Ted Ed video http://ed.ted.com/lessons/daniel-dulek- how-big-is-a-mole-not-the-animal-the- other-one http://ed.ted.com/lessons/daniel-dulek- how-big-is-a-mole-not-the-animal-the- other-one

3. What is a mole? Mole – metric unit for counting The only acceptable abbreviation for “mole” is “mol”…not “m”!! We use it just like we use the terms dozen and ream!

4. What is a counting unit? You’re already familiar with one counting unit…a “dozen” “Dozen”12 A dozen doughnuts 12 doughnuts A dozen books A dozen cars 12 books 12 cars A dozen = 12

5. What can’t we count atoms in “dozens”? Atoms and molecules are extremely small We use the MOLE to count particles

6. A mole = 6.02  10 23 particles (called Avogadro’s number) “mole” 6.02  10 23 1 mole of doughnuts 6.02  10 23 doughnuts 1 mole of atoms 1 mole of molecules 6.02  10 23 atoms 6.02  10 23 molecules 6.02  10 23 = 602,000,000,000,000,000,000,000 This number was named after Amadeo Avogadro. He did not calculate it!

7. FUNNY!

8. Representative Particles Remember, matter is broken down into either SUBSTANCES or mixtures Substances are broken down into either ELEMENTS or COMPOUNDS Type of MatterExampleRepresentative Particle ElementFeAtom Ionic CompoundNaClFormula Unit Covalent CompoundCO 2 Molecule

9. Example: Particles & Moles Use the conversion factor (1 mol = 6.02 x 10 23 ) particles to convert Example 1: How many molecules of water are in 1.25 moles?

10. = _______ molecules H 2 O Example: Molecules & Moles 1.25 mol H 2 O mol H 2 O Molecules H 2 O 6.02  10 23 1 7.53  10 23 1 mol = 6.02  10 23 molecules Example 1: How many molecules of water are in 1.25 moles?

11. Let’s Practice #2 Example: How many moles are equal to 2.8 × 10 22 formula units of KBr?

12. = _______ moles Let’s Practice #2 2.8 × 10 22 formula units Formula units mole 1 6.02  10 23 0.047 1 mol = 6.02  10 23 formula units Example: How many moles are equal to 2.8 × 10 22 formula units KBr?

13. Let’s Practice #3 Example: How many atoms are equal to 3.56 moles of Fe?

14. = _______ atoms Let’s Practice #3 3.56 moles Fe moles atoms 6.02 x 10 23 1 2.14 x 10 24 1 mol = 6.02  10 23 molecules Example: How many atoms are equal to 3.56 moles of Fe?

15. Molar Mass Molar Mass – The mass for one mole of an atom or molecule. Other terms commonly used for the same meaning: Molecular Weight Molecular Mass Formula Weight Formula Mass

16. Molar Mass for Elements The average atomic mass = grams for 1 mole ElementMass 1 mole of carbon atoms (C) 12.01 g 1 mole of oxygen atoms (O 2 ) 1 mole of hydrogen atoms (H 2 ) 16.00 g x 2 = 32.00 g O 2 1.01 g x 2 = 2.02 g H 2 Unit for molar mass: g/mole or g/mol Average atomic mass is found on the periodic table

17. Molar Mass for Compounds The molar mass for a molecule = the sum of the molar masses of all the atoms

18. Calculating a Molecule’s Mass Count the number of each type of atom Find the molar mass of each atom on the periodic table Multiply the # of atoms by the molar mass for each atom Find the sum of all the masses 1 2 3 4 To find the molar mass of a molecule:

19. Example: Molar Mass Example: Find the molar mass for CaBr 2

20. Example: Molar Mass Count the number of each type of atom 1 Ca Br 1 2 Example: Find the molar mass for CaBr 2

21. Example: Molar Mass Find the molar mass of each atom on the periodic table 2 Ca Br 1 2 40.08 g/mole 79.90 g/mole Example: Find the molar mass for CaBr 2

22. Example: Molar Mass Multiple the # of atoms  molar mass for each atom 3 Ca Br 1 2 40.08 g/mole 79.90 g/mole   Example: Find the molar mass for CaBr 2 = 40.08 g/mole = 159.80 g/mole

23. Example: Molar Mass Find the sum of all the masses 4 Ca Br 1 2 40.08 g/mole 79.91 g/mole = 40.08 g/mole = 159.80 g/mole + 199.88 g/mole 1 mole of CaBr 2 =199.90 g Example: Find the molar mass for CaBr 2  

24. Example 2: If you see a Parentheses in the Formula Be sure to distribute the subscript outside the parenthesis to each element inside the parenthesis. Example: Find the molar mass for Sr(NO 3 ) 2

25. Example 2: Molar Mass & Parenthesis Be sure to distribute the subscript outside the parenthesis to each element inside the parenthesis. 1 6 87.62 g/mole 16.00 g/mole   = 87.62 g/mole = 96.00 g/mole + 211.64 g/mole 1 mole of Sr(NO 3 ) 2 =211.64 g 214.01 g/mole  = 28.02 g/mole Sr N O Example: Find the molar mass for Sr(NO 3 ) 2

26. Let’s Practice #3 Example: Find the molar mass for Al(OH) 3

27. Let’s Practice #2 Be sure to distribute the subscript outside the parenthesis to each element inside the parenthesis. 1 3 26.98 g/mole 1.01 g/mole   = 26.98 g/mole = 3.03 g/mole + 78.01 g/mole 1 mole of Al(OH) 3 =78.01 g 316.00 g/mole  = 48.00 g/mole Al O H Example: Find the molar mass for Al(OH) 3

28. Using Molar Mass in Conversions

29. Example: Moles to Grams Example: How many grams are in 1.25 moles of water?

30. Example: Moles to Grams 1.25 mol H 2 O = _______ g H 2 O mol H 2 O g H 2 O 18.02 1 22.5 When converting between grams and moles, the molar mass is needed 1 mole H 2 O molecules = 18.02 g H O 2 1 1.01 g/mole 16.00 g/mole  = 2.02 g/mole = 16.00 g/mole + 18.02 g/mole  Example: How many grams are in 1.25 moles of water?

31. Example: Grams to Moles Example: How many moles are in 25.5 g NaCl?

32. 25.5 g NaCl Example: Grams to Moles = ____ moles NaCl g NaCl mol NaCl 1 58.44.436 1 moles NaCl molecules = 58.44 g Na Cl 1 1 22.99 g/mole 35.45 g/mole  = 22.99 g/mole = 35.45 g/mole + 58.44 g/mole  Example: How many moles are in 25.5 g NaCl

33. Example: Grams to Molecules Example: How many formula units are in 25.5 g NaCl?

34. 25.5 g NaCl Example: Grams to Moles = ____ FU’s NaCl g NaCl mol NaCl 1 58.44 2.63 x 10 23 1 moles NaCl formula units = 58.44 g Na Cl 1 1 22.99 g/mole 35.45 g/mole  = 22.99 g/mole = 35.45 g/mole + 58.44 g/mole  Example: How many formula units are in 25.5 g NaCl 1 mol NaCl 6.02 x 10 23 FU’s

35. Percent Composition  Defined as the percent by mass of each element in a compound Steps to Finding Percent Composition 1.Add up the mass of each element within the compound to get the mass of the compound. 2.Divide each element’s mass by the mass of the compound. 3.Multiply by 100 % composition= mass of element x 100 mass of compound

36. Percent Composition by Mass of Air

37. Example: Calculate the % composition of each element in calcium carbonate. CaCO 3 Molar mass = 100.09 g % C = 12.01/100.09 x 100 = 12.00 % %Ca = 40.08/100.09 x 100 = 40.04% %O = 48.00/100.09 x 100 = 47.96%

38. Example : What is the % of each element in a compound that contains 29.00g Ag and 4.30g S only? Total mass of compound = 33.30 g % Ag = 29.00/33.30 x 100 = 87.09 % %S = 4.30/33.30 x 100 = 12.9%

39. Hydrates  A HYDRATE is an ionic compound with water trapped in its crystal. Examples are: CuSO 4 5H 2 O MgSO 4 7 H 2 O CoCl 2 H 2 O  Heating a hydrate removes the water and leaves behind just the salt which is called the anhydrate.

40. Example : What is the % water in the hydrate, CuCl 2  2H 2 O Molar mass of hydrate = 170.48 g % water = 36.04/170.48 x 100 = 21.14%

41. http://group.chem.iastate.edu/Greenb owe/sections/projectfolder/flashfiles/st oichiometry/empirical.html Heating of A Hydrate Animation Calculating the experimental % composition of water in a hydrate.

42. Empirical Formula A chemical formula showing the simplest whole number ratio of moles of elements (subscipts) Hydrogen Peroxide has an actual formula (molecular formula) of H 2 O 2 but an empirical formula of HO

43. How to Calculate Empirical Formula RHYME: Percent to Mass Mass to Mole Divide by Small Multiply til Whole 1.Assume 100 grams of the sample of compound. Switch the percent sign to grams 2.Convert each element’s mass into moles. 3.Divide each element’s mole amount by the smallest mole amount in the entire problem. The answer is the subscript of the element within the compound. 4.OPTIONAL: If mole ratio is not within.1 of a whole number, multiply each amount by the smallest whole number that will produce either a whole number itself or a number within.1 of a whole number.

44. Example : What is the empirical formula for 40.05% S and 59.95% O? 1.Switch the percent sign to grams & convert each element’s mass into moles 40.05 g S / 32.01g = 1.250 mol S 59.95 g O / 16.00 g = 3.747 mol O 2.Divide each element’s mole amount by the smallest mole amount in the entire problem. 1.250 mol S = 13.747 mol O = 2.99 = 3 1.250 mole1.250 mol S 1 O 3  SO 3

45. Example : What is the empirical formula for 43.64% P and 56.36% O? 1.Switch the percent sign to grams & convert each element’s mass into moles 43.64 g P / 30.97g = 1.409 mol S 56.36 g O / 16.00 g = 3.522 mol O 2.Divide each element’s mole amount by the smallest mole amount in the entire problem. 1.409 mol S = 13.522 mol O = 2.49 ≠ 3 1.409 mole1.409 mol 3.If mole ratio is not within.1 of a whole number, multiply each amount by the smallest whole number that will produce either a whole number itself or a number within.1 of a whole number. 1 x 2 = 2 2.49 x 2 = 4.998 = 5 P2O5P2O5

46. Molecular Formula Is the ACTUAL, true formula of the compound. They are usually multiples of their empirical formula N 2 O 4 is the molecular formula; the empirical formula is NO 2 Notice that the molecular formula is 2 times larger than the empirical formula

47. Molecular Formula

48. How to Calculate Molecular Formula 1. You need to find the empirical formula and calculate its molar mass. Call this empirical formula mass EFM. 2. Find the mass of the actual formula which will most likely be given to you in grams. Call this molecular formula mass MFM. 3. Divide the MFM by the EFM to get a factor. 4. Multiply the factor by the empirical formula to get the MOLECULAR FORMULA

49. Example: What is the molecular formula of a compound whose empirical formula is CH 4 N and the molecular mass is 60.12 g/mol? 1. Empirical Formula Mass (EFM) = 12.01 + 4.04 + 14.01 = 30.06 g 2. Molecular Formula Mass (MFM) = 60.12 g 3. 60.12 / 30.06 = 2 4. 2(CH 4 N) = C 2 H 8 N 2

50. Section 3.5—Gas Behavior How does the behavior of gases affect airbags? What is PRESSURE? Force of gas particles running into a surface.

51. Pressure is measured by a Barometer

52. If pressure is molecular collisions with the container… As # of moles increase, pressure increases Think about blowing up a balloon! Pressure and Moles (# of Molecules) As number of molecules increases, there will be more molecules to collide with the wall Collisions between molecules and the wall increase  Pressure increases

53. # of Gas Particles vs. Pressure

54. Pressure & Volume If pressure is molecular collisions with the container… As volume increases, pressure decreases. Think about how your lungs work! http://www.youtube.com/watch?v=q6-oyxnkZC0 As volume increases, molecules can travel farther before hitting the wall Collisions between molecules & the wall decrease  Pressure decreases

55. What is “Temperature”? Temperature – measure of the average kinetic energy of the molecules Energy due to motion (Related to how fast the molecules are moving) As temperature increases, Average Kinetic Energy Increases and Molecular motion increases

56. Pressure and Temperature If temperature is related to molecular motion… and pressure is molecular collisions with the container… As temperature increases, pressure increases As temperature increases, molecular motion increases Collisions between molecules & the wall increase  Pressure increases

57. Volume and Temperature If temperature is related to molecular motion… and volume is the amount of space the gas occupies… As temperature increases, volume increases Think of liquid nitrogen and the balloon. http://www.youtube.com/watch?v=QEpxrGWep4E As temperature increases, molecular motion increases molecules will move farther away from each other  Volume increases

58. Pressure In Versus Out Example: A bag of chips is bagged at sea level. What happens if the bag is then brought up to the top of a mountain. A container will expand or contract until the pressure inside equals atmospheric pressure outside The internal pressure is higher than the external pressure. The bag will expand in order to reduce the internal pressure. The internal pressure of the bag at low altitude is high At high altitude there is lower pressure Higher pressure Lower pressure Lower pressure

59. Can Explodes! When Expansion Isn’t Possible Example: An aerosol can is left in a car trunk in the summer. What happens? Rigid containers cannot expand The internal pressure is higher than the external pressure. The can is rigid—it cannot expand, it explodes! The temperature inside the can begins to rise. As temperature increases, pressure increases. Higher pressure Lower pressure

60. Air Pressure Crushing Cans http://www.csun.edu/scied/4-discrpeant-event/the_can_crush/index.htm

61. http://www.youtube.com/watch?v=Zz95_VvTxZM Another cool video http://www.youtube.com/watch?v=JsoE4F2Pb20 Air Pressure Crushing “Cans”

62. Kinetic Molecular Theory(KMT): explains gas behavior based upon the motion of molecules based on an ideal gas  IDEAL gases are IMAGINARY gases that follow the assumptions of the KMT

63. 1 Assumptions of the KMT All gases are made of atoms or molecules that are in constant, rapid, random motion Gas particles are not attracted nor repelled from one another *** All gas particle collisions are perfectly elastic (no kinetic energy is lost to other forms) The volume of gas particles is so small compared to the space between the particles, that the volume of the particle itself is insignificant*** 2 3 4 5 The temperature of a gas is proportional to the average kinetic energy of the particles

64. So what is a “REAL” gas? Real gases, (like nitrogen), will eventually condense into a liquid when the temperature gets too low or the pressure gets too high BECAUSE: Assumption #3 Assumption #5 Gas particles do have attractions and repulsions towards one another Gas particles do take up space

65. Real Gases Deviate from Ideal Gas Behavior when at high pressure  The gas molecules are compressed making the volume they take up more significant than if they were spread out

67. Real Gases Deviate from Ideal Gas Behavior when at low temperature.  The lower kinetic energy causes the molecules to move slower and ATTRACTIVE FORCES that really exist start to take effect ---------------------------  Polar gases (HCl) deviate more than nonpolar gases (He or H 2

68. At Lower Temperature

69. Gas Movement: Effusion vs Diffusion Effusion –gas escapes from a tiny hole in the container Effusion is why balloons deflate over time!

70. Diffusion –gas moves across a space from high to low concentration Diffusion is the reason we can smell perfume across the room

71. Effusion, Diffusion & Particle Mass How are particle size (mass) and these concepts related? As mass of the particles increases, rate of effusion and diffusion is lowered. As particle size (mass) increases, the particles move slower  it takes them more time to find the hole or to go across the room

72. Rate of Diffusion & Particle Mass Watch as larger particles take longer to get to your nose H2H2 CO 2

73. Section 3.6—Gas Laws How can we calculate Pressure, Volume and Temperature of our airbag?

74. Pressure Units Several units are used when describing pressure UnitSymbol atmospheresatm Pascals, kiloPascals millimeters of mercury pounds per square inch Pa, kPa mm Hg psi 1 atm = 101300 Pa = 101.3 kPa = 760 mm Hg = 14.7 psi

75. Conversions Between Different Pressure Units 1 atm = 760 mmHg = 101.3 kPa Examples 1.Convert 654 mm Hg to atm 1.Convert 879 mm Hg to kPa 1.Convert 15.6 atm to kPa 654 mmHg x 1atm = 760 mmHg.861 atm 879 mmHg x 101.3 Kpa = 760 mmHg 1.16 Kpa 15.6 atm x 101.3 Kpa = 1atm 1580 Kpa

76. Temperature Unit used in Gas Laws Kelvin (K)– temperature scale with an absolute zero Temperatures cannot fall below an absolute zero Examples 1.Convert 15.6 °C into K 2. Convert 234 K into °C 15.6 + 273 = K 288.6  289 K °C + 273 = 234-39 °C

77. Standard Temperature & Pressure (STP) the conditions of:  1 atm (or the equivalent in another unit)  0°C (273 K) Problems often use “STP” to indicate quantities…don’t forget this “hidden” information when making your list!

78. GAS LAWS: “Before” and “After” This section has 5 gas laws which have “before” and “after” conditions. For example: 1= initial amount 2= final amount P= Pressure V= Volume T=Temperature n= moles(molecules)

79. Boyle’s Law Pressure Increases as Volume Decreases

80. Boyles’ Law Volume & Presssure are INVERSELY proportional when temperature and moles are held constant P = pressure V = volume The two pressure units must match and the two volume units must match! Example: A gas sample is 1.05 atm when at 2.5 L. What volume is it if the pressure is changed to 0.980 atm?

81. Boyles’ Law ***The two pressure units must match & the two volume units must match! Example: A gas sample is 1.05 atm when 2.5 L. What volume is it if the pressure is changed to 0.980 atm? P 1 = 1.05 atm V 1 = 2.5 L P 2 = 0.980 atm V 2 = ? L V 2 = 2.7 L

82. Boyles Law: Graph

83. Charles’ Law Volume Increases as Temperature Increases

84. Charles’ Law Volume & Temperature are DIRECTLY proportional when pressure and moles are held constant. V = Volume T = Temperature The two volume units must match & temperature must be in Kelvin! Example: What is the final v olume if a 10.5 L sample of gas is changed from 25  C to 50  C? V 1 = 10.5 L T 1 = 25  C V 2 = ? L T 2 = 50  C Temperature needs to be in Kelvin! 25  C + 273 = 298 K 50  C + 273 = 323 K

85. Charles’ Law ***The two volume units must match & temperature must be in Kelvin! Example: What is the final volume if a 10.5 L sample of gas is changed from 25  C to 50  C? V 1 = 10.5 L T 1 = 25  C V 2 = ? L T 2 = 50  C V 2 = 11.4 L = 298 K = 323 K

86. Charles Law: Graph

87. Gay-Lussac’s Law Temperature decreases as Pressure decreases

88. Gay-Lussac’s Law Pressure & temperature are DIRECTLY proportional when moles and volume are held constant P = Pressure T = Temperature The two pressure units must match and temperature must be in Kelvin! Example: A sample of hydrogen gas at 47  C exerts a pressure of.329 atm. The gas is heated to 77  C at constant volume and moles. What will the new pressure be? P 1 =.329 atm T 1 = 47  C P 2 = ? atm T 2 = 77  C Temperature needs to be in Kelvin! 47  C + 273 = 320 K 77  C + 273 = 350 K

89. Gay-Lussac’ Law Example: A sample of hydrogen gas at 47  C exerts a pressure of.329 atm. The gas is heated to 77  C at constant volume and moles. What will the new pressure be? P 1 =.329 atm T 1 = 47  C P 2 = ? atm T 2 = 77  C P 2 =.360 atm = 320 K = 350 K

90. Gay Lussac Law: Graph

91. Avogadro’s Law Moles and Volume are directly proportional when temp. & pressure are held constant V = Volume n = # of moles of gas Example: A sample with 0.15 moles of gas has a volume of 2.5 L. What is the volume if the sample is increased to 0.55 moles? The two volume units must match!

92. Avogadro’s Law Example: A sample with 0.15 moles of gas has a volume of 2.5 L. What is the volume if the sample is increased to 0.55 moles? The two volume units must match! n 1 = 0.15 moles V 1 = 2.5 L n 2 = 0.55 moles V 2 = ? L V 2 = 9.2 L

93. Combined Gas Law P = Pressure V = Volume n = # of moles T = Temperature Each “pair” of units must match and temperature must be in Kelvin! Example: What is the final volume if a 15.5 L sample of gas at 755 mmHg and 298 K is changed to STP?

94. Combined Gas Law P = Pressure V = Volume T = Temperature Moles is not mentioned so remove it from equation! Example: What is the final volume if a 15.5 L sample of gas at 755 mmHg and 298K is changed to STP? P 1 = 755 mmHg V 1 = 15.5 L T 1 = 298 K P 2 = 760mmHg V 2 = ? L T 2 = 273 K V 2 = 14.1 L STP is standard temperature (273 K) and pressure (1 atm)

95. The combined gas law can be used for all “before” and “after” gas law problems! For example, if volume is held constant, then and the combined gas law becomes: Why you really only need 1 of these

96. Watch as variables are held constant and the combined gas law “becomes” the other 3 laws Hold pressure and temperature constant Avogadro’s Law Hold moles and temperature constant Boyles’ Law Hold pressure and moles constant Charles’ Law Transforming the Combined Law

97. Dalton’s Law

98. Each gas in a mixture exerts its own pressure called a partial pressure = P 1, P 2 …. Total Pressure = P T Example: A gas mixture is made up of oxygen(2.3 atm) and nitrogen(1.7 atm) gases. What is the total pressure? P T = 2.3 atm + 1.7 atm 4.0 atm

99. 2KClO 3 (s) 2KCl (s) + 3O 2 (g) P T = P O + P H O 22 Modified Dalton’s Law When a gas is Collected over water, the total pressure of the mixture collected is a combination of water vapor and the gas you are collecting!

100. Modified Dalton’s Law Example : What is the pressure of the water vapor if the total pressure of the flask is 17.5 atm and the pressure of the oxygen gas is 16.1 atm? 17.5 = 16.1 atm + P H2O 1.4 atm

101. The Ideal Gas Law ( an “ AT NO W” equation ) The volume of a gas varies directly with the number of moles and its Kelvin temperature P = Pressure V = Volume n = moles R = Gas Law Constant T = Temperature There are three possibilities for “R”! Choose the one with units that match your pressure units! Volume must be in Liters when using “R” to allow the unit to cancel!

102. The Ideal Gas Law Example: A sample with 0.55 moles of gas is at 105.7 kPa and 27°C. What volume does it occupy?

103. The Ideal Gas Law Example: A sample with 0.55 moles of gas is at 105.7 kPa and 27°C. What volume does it occupy? n = 0.55 moles P = 105.7 kPa T = 27°C + 273 = 300 K V = ? R = 8.31 L kPa / mole K V 2 = 13 L The Ideal Gas Law does not compare situations—it describes a gas in one situation. Chosen to match the kPa in the “P” above

104. The Ideal Gas Law Example 2: What mass of hydrogen gas in grams is contained in a 10.0 L tank at 27°C and 3.50 atm of pressure? n = ? P = 3.50 atm T = 27°C + 273 = 300 K V = 10.0 L R =.0821 L atm /mole K Chosen to match the atm in the “P” above n = 1.42 mol  1.42 mol x 2.02 g = 2.87 g 1 mol