1. Gases II Dr. Ron Rusay

2. Experimentally Determining The Moles & Molar Mass of Hydrogen Applying Avogadro’s Law Using the Ideal Gas Law & Partial Pressures Dr. Ron Rusay Mg(s) + 2HCl(aq)  MgCl 2 (aq) + H 2 (g) Zn(s) + 2HCl(aq)  ZnCl 2 (aq) + H 2 (g) http://chemconnections.org/general/chem106/Magnesium-Zinc-wo.1.mov

4. What is wrong with this set up? Mg or Zn

5. Mg(s) + 2HCl(aq)  MgCl 2 (aq) + H 2 (g) Zn(s) + 2HCl(aq)  ZnCl 2 (aq) + H 2 (g)

6. Ideal Gas Law PV = n RT  R = “proportionality” constant = 0.08206 L atm   mol   P = pressure in atm  V = volume in liters  n = moles  T = temperature in Kelvins

7. Standard Conditions Temperature, Pressure & Moles  “STP”  For 1 mole of a gas at STP:  P = 1 atmosphere  T =  C (273.15 K)  The molar volume of an ideal gas is 22.42 liters at STP

8. P 1 V 1 = P 2 V 2 P V V 1 / n 1 = V 2 / n 2 V n V 1 / T 1 = V 2 / T 2 V T Isobaric process: pressure constant Isochoric process: volume constant Isothermal process: temperature constant

9. Provide 3 different sets of conditions (Pressure and Temperature) which can account for the volume of the gas decreasing by ½. Applying a Gas Law

10. Applications If the molecules of a trapped gas maintain the same average velocity when their volume is changed, the pressure will be inversely related to the volume change. Spore Dispersal / Boyle’s Law Spore Dispersal / Boyle’s Law “Spud Guns” and a Voyage to Near Outer Space “Spud Guns” and a Voyage to Near Outer Space

11. Hydrogen & the Ideal Gas Law n H 2 (g) = PV / RT  n = moles H 2 (g)  P H 2 (g) = pressure of H 2 (g) in atm (mm Hg  atm)  V = experimental volume (mL  L)  T = experimental temperature ( o C  K) Mg(s) + 2HCl(aq)  MgCl 2 (aq) + H 2 (g) Zn(s) + 2HCl(aq)  ZnCl 2 (aq) + H 2 (g)

12. Total Pressure: Sum of the Partial Pressures  For a mixture of gases, the total pressure is the sum of the pressures of each gas in the mixture. P Total = P 1 + P 2 + P 3 +... P Total  n Total n Total = n 1 + n 2 + n 3 +. n Total = n 1 + n 2 + n 3 +.

13. P H 2 (g) = P Total ( barometric) - P H 2 O (g) [TABLE] - P HCl (g)P H 2 (g) = P Total ( barometric) - P H 2 O (g) [TABLE] - P HCl (g) P HCl (g) = HCl Height (mm) ÷ 12.95 __________ Density Hg is 12.95 times > density HCl(aq) P HCl (g) = HCl Height (mm) x 0.0772 __________ Density Hg is 12.95 times > density HCl(aq) 0.772 mm Hg/cm of acid solution

14. Ideal Gas Law: Moles & Avogadro’s Law n H 2 (g) = PV / RT  n = moles H 2 (g)  P H 2 (g) = pressure of H 2 (g) in atm (mm Hg  atm)  P H 2 (g) = P Total ( barometric) - P H 2 O (g) [TABLE] - P HCl (g)  V = experimental volume (mL  L)  T = experimental temperature ( o C  K) Mg(s) + 2HCl(aq)  MgCl 2 (aq) + H 2 (g) Zn(s) + 2HCl(aq)  ZnCl 2 (aq) + H 2 (g)

15. Do you have enough oxygen to climb Mt. Everest? http://chemconnections.org/chemwiki/everest/everest.htm

16. QUESTION A typical total capacity for human lungs is approximately 5,800 mL. At a temperature of 37°C (average body temperature) and pressure of 0.98 atm, how many moles of air do we carry inside our lungs when inflated? (R = 0.08206 L atm/ K mol) A)1.9 mol B) B)0.22 mol C) C)230 mol D) D)2.20 mol E) E)0 mol: Moles can harm a person’s lungs.

17. ANSWER B) is correct. The units for temperature must be in K, pressure in atm, and volume in L. Then using the universal constant 0.08206 L atm/ K mol in the PV = nRT equation moles may be solved. n O 2 (g) = PV / RT

18. An average pair of human lungs actually contains only about 3.5 L of air after inhalation and about 3.0 L after exhalation. Assuming that air in your lungs is at 37°C and 1.0 atm a) How many moles of O 2 are actually in a typical breath?. b) What is the mass of O 2 in a typical breath?. c) How much of the O 2 is essential biochemically?

19. QUESTION An average pair of human lungs actually contains only about 3.5 L of air after inhalation and about 3.0 L after exhalation. Assuming that air in your lungs is at 37°C and 1.0 atm: How many moles of air are actually in a typical breath? 0.0020 mol A) 0.0020 mol B) 0.020 mol C) 0.030 mol D) 0.025 mol E) 0.0042 mol

20. An average pair of human lungs actually contains only about 3.5 L of air after inhalation and about 3.0 L after exhalation. Assuming that air in your lungs is at 37°C and 1.0 atm a) How many moles of O 2 are actually in a typical breath?. b) What is the mass of O 2 in a typical breath?. c) How much of the O 2 is essential biochemically? n O 2 (g) = (20.9%) * PV / RT n O 2 (g) = (0.209 mol O 2 (g) / mol air) (1.0 atm x (3.5 L-3.0 L) x mol air * K / 0.0821 L * atm x 300 K) n O 2 (g) = 0.0042 mol (E)

21. ANSWER An average pair of human lungs actually contains only about 3.5 L of air after inhalation and about 3.0 L after exhalation. Assuming that air in your lungs is at 37°C and 1.0 atm: How many moles of air are actually in a typical breath? 0.0020 mol A) 0.0020 mol B) 0.020 mol C) 0.030 mol D) 0.025 mol E) 0.0042 mol

22. QUESTION An average pair of human lungs actually contains only about 3.5 L of air after inhalation and about 3.0 L after exhalation. Assuming that air in your lungs is at 37°C and 1.0 atm: What is the mass of O 2 in a typical breath? A) 0.0042 mol x 16 g/mol B) 0.020 mol x 16 g/mol C) 0.0042 mol x 32 g/mol D) 0.020 mol x 32 g/mol

23. An average pair of human lungs actually contains only about 3.5 L of air after inhalation and about 3.0 L after exhalation. Assuming that air in your lungs is at 37°C and 1.0 atm a) How many moles of O 2 are actually in a typical breath?. b) What is the mass of O 2 in a typical breath?. c) How much of the O 2 is essential biochemically? n O 2 (g) = (20.9%) * PV / RT n O 2 (g) = (0.209 mol O 2 (g) / mol air) (1.0 atm x (3.5 L-3.0 L) x mol air * K / 0.0821 L * atm x 300 K) n O 2 (g) = 0.0042 mol (E) g O 2 (g) = 0.0042 mol x 32.0 g/mol g O 2 (g) = 0.13 g

24. An average pair of human lungs actually contains only about 3.5 L of air after inhalation and about 3.0 L after exhalation. Assuming that air in your lungs is at 37°C and 1.0 atm c) How much of the O 2 is essential biochemically? Two estimates for a person with normal physical activity range from 0.67 - 0.84 kg of O 2 being used per day (NASA provided the higher value). How many breaths do you take in one day? ~ 5 mol % of the O 2 is actually used per breath. Hard exercise Hard exercise increases this oxygen demand (intake) about 10 fold.

25. QUESTION The primary source of exhaled CO 2 is from the combustion of glucose, C 6 H 12 O 6 (molar mass = 180. g/mol.). The balanced equation is shown here: C 6 H 12 O 6 (aq) + 6 O 2 (g)  6 CO 2 (g) + 6 H 2 O (l) If you oxidized 5.42 grams of C 6 H 12 O 6 while tying your boots to climb Mt. Everest, how many liters of O 2 @ STP conditions did you use? A)0.737 L B)0.672 L C)4.05 L D)22.4 L

26. ANSWER C), assuming you were not holding your breath, is correct. The number of moles of glucose must first be determined (5.42/180. = 0.0301 moles), then this is multiplied by 6 to account for the stoichiometric ratio between glucose and oxygen. From this, PV = nRT may be used with the appropriate substitutions.

27. Molar Mass of a Gas (Hydrogen)  PV = n RT  n = g of gas/ MM gas [MM gas = g/mol]  PV = (g of gas/ MM gas )RT  MM gas = g of gas/V (RT/P) Density of gas [separate experiment]  MM gas = g of gas/V (RT/P)  MM gas = density of gas (RT/P)

28. QUESTION Freon-12, CF 2 Cl 2, a “safe” compressible gas, was widely used from 1935-1994 as a refrigerant in refrigerators, freezers, and air conditioning systems. However, it had been shown to be a greenhouse gas and to catalytically destroy the ozone layer. It was phased out and banned. Freon-12, CF 2 Cl 2, a “safe” compressible gas, was widely used from 1935-1994 as a refrigerant in refrigerators, freezers, and air conditioning systems. However, it had been shown to be a greenhouse gas and to catalytically destroy the ozone layer. It was phased out and banned. 200 ml of Freon-12 was collected by syringe. It weighed 0.927 grams, had a temperature of 30.0°C, and a pressure of 730 mm of Hg. What is the experimental molar mass of Freon-12? A. 12.1 g/mol B. 84 g/mol C. 92.7 g/mol D. 115 g/mol E. 121 g/mol

29. ANSWER E) is the molar mass of Freon-12. The pressure must be corrected for the presence of water by subtracting 31.8 mmHg from the total pressure. This should also be converted to atm. The temperature must be converted to K. Then PV = nRT can be used if n is written as g/mm and solved for mm. ……. Or use the Periodic Table.

30. The density of an unknown atmospheric gas pollutant was experimentally determined to be 1.964 g/ L @ 0 o C and 760 torr. What is the molar mass of the gas?What is the molar mass of the gas? What might the gas be?What might the gas be? QUESTION A) COB) SO 2 C) H 2 OD) CO 2

31. ANSWER MM gas = density of gas (RT/P) MM gas = 1.964 g/ L x 0.08206 L atm   mol  x 273K/ 760 torr x 760 torr/ 1atm x 273K/ 760 torr x 760 torr/ 1atm 1.964 g/ L @ 0 o C and 760 torr. R = 0.08206 L atm   mol  o C  K torr  atm MM gas = MM gas = 44.0 g/mol D) CO 2

32. QUESTION 0.0820 grams of the volatile, gaseous phase, of a compound, which smells like fresh raspberries, was trapped in a syringe. It had a volume of 12.2 mL at 1.00 atmosphere of pressure and 25.0°C. What is the molar mass of this pleasant smelling compound ? A)13.8 g/mol B)164 g/mol C)40.9 g/mol D)224 g/mol

33. ANSWER B)164 g/mol Using PV = nRT : 0.0122 L for V, 298 K for T, 0.08206 for R and solving for n the number of moles represented by 0.0820 grams can be obtained. Then the MOLAR MASS (grams in one mole) can be determined. MM gas = density of gas (RT/P) MM gas = 0.0820 g/ L x 0.00122 L x 0.0821 atm   mol  x 298K/ 1atm x 298K/ 1atm

34. QUESTION For the compound that smells like fresh raspberries, the following structure matches its molecular formula. A)TRUE B)FALSE

35. ANSWER Based on your answers for the compound, which smells like fresh raspberries, in the previous two questions, the following structure matches its molecular formula. A)TRUE B)FALSE 164 g/mol = C 10 H 12 O 2

36. Which sequence represents the gases in order of increasing density at STP? A) Fluorine < Carbon monoxide < Chlorine < Argon B) Carbon monoxide < Fluorine < Argon < Chlorine C) Argon < Carbon monoxide < Chlorine < Fluorine D) Fluorine < Chlorine < Carbon monoxide < Argon QUESTION

37. Which sequence represents the gases in order of increasing density at STP? A) Fluorine < Carbon monoxide < Chlorine < Argon B) Carbon monoxide < Fluorine < Argon < Chlorine C) Argon < Carbon monoxide < Chlorine < Fluorine D) Fluorine < Chlorine < Carbon monoxide < Argon ANSWER

38. Gases & Airbags Use of Chemical Reactions and Physical Properties