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Chapter 12

Chapter 12. Molecular Composition of Gases. First, Let’s Review Chapter 9-11. Stoichiometric Calculations. The coefficients in the balanced equation give the ratio of moles of reactants and products. Stoichiometric Calculations.

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Chapter 12

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  1. Chapter 12 Molecular Composition of Gases

  2. First, Let’s Review Chapter 9-11 Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products

  3. Stoichiometric Calculations • From the mass of Substance A you can use the ratio of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)

  4. Stoichiometric Calculations C6H12O6 + 6 O2 6 CO2 + 6 H2O Starting with 1.00 g of C6H12O6… we calculate the moles of C6H12O6… use the coefficients to find the moles of H2O… and then turn the moles of water to grams

  5. Limiting Reactants • The limiting reactant is the reactant present in the smallest stoichiometric amount • In other words, it’s the reactant you’ll run out of first (in this case, the H2)

  6. Limiting Reactants • In the example below, the O2 would be the excess reagent

  7. 12.1 Volume - Mass Relationships of Gases • Last chapter, we related the volume and the mass of a gas: • Reaction Stoichiometry • Note: a “stoich” problem can be recognized by the fact that information from one substances is given and the problem is asking about a different substance. • The Combined Gas Law • Let’s compare the volumes of gases in two example problems…

  8. 1 mole H2 1 mole O2 22.4 L O2 x x x = 25.0 L O2 22.4 L H2 2 moles H2 1 mole O2 1 2 3 …Reaction Stoichiometry at standard conditions • Example 1 - How many L of Oxygen are needed to react with 50.0L of Hydrogen at STP? 2H2(g) + O2(g) ---> 2 H2O(g) 50.0 L H2 • But...we can skip steps 1 and 3. • Why? All gases take up the same amount of space at STP.

  9. P1V1 = P2V2 T1 T2 (760)(22.4) = (1000)(V2) (293) (273) V2 = 18.3 L H2 …The Combined Gas Law at nonstandard conditions • Example 2 - Calculate the volume of one mole of H2 at 20ºC and 1000 torr. P1 = 760 torr P2 = 1000 torr T1 = 273K T2 = 293K V1 = 22.4 L V2 = ?

  10. PV = nRT Pressure n = # of moles R = ideal gas constant Temp. Volume 1 atm in Liters Kelvin at STP, R = (1atm)(22.4L / 1mol)(273K) = .0821 L · atm / mol· K 1 atm = 760 torr 12.2 The Ideal Gas Law • We solved example 2 using the combined gas law – we could have also used • A relationship between pressure, volume, temperature and the # of moles of a gas • A new formula that can help us solve all kinds of gas law problems more easily.

  11. Deriving the Ideal Gas Law • Let’s derive the ideal gas law and gas constant... • Volume is proportional to 1/P (as P is reduced, the V increases) • V is proportional to T (as T increases, the V increases) • V is proportional to n (as more moles are added, the V increases)

  12. In general… • The combined gas law, P1V1 = P2V2 T1 T2 is used for changing conditions. • But, a new gas law, the Ideal Gas Law, can be introduced when you have problems containing: • one set of conditions • solving for grams • solving for moles • calculating molecular weight (molar mass) • calculating density • involving stoichiometry and non-STP conditions

  13. One Set of Conditions Example: Calculate the volume of 1.00 mole of Hydrogen at 20.0 ˚C and 1000 torr Step 1: Convert your pressure to atm. 1000 torr x 1 atm / 760 torr = 1.316 atm Step 2: Write the ideal gas law and derive Volume. PV = nRT V = nRT/P Step3: Using V = nRT/P, we have V = (1 mole)(.0821 L atm/mol K)(293K)/1.316 atm V = 18.3 L H2

  14. Solving for Grams Example: Calculate the number of grams of helium in a 6.0 liter cylinder at 27˚C and 800 torr. Step 1: Convert your pressure to atm and your temperature to Kelvin. 800 torr x 1 atm / 760 torr = 1.05 atm 27˚C + 273 = 300K Step 2: Write the ideal gas law and derive for moles. PV = nRT n = PV/RT Step 3: Calculate the moles. n = (1.05 atm)(6.0 L)/(.0821 L•atm/mol•K)(300K) n = 0.26 moles Step 4: Convert to grams. .26 moles x 4.0 g/mole = 1.04 g He

  15. Solving for Moles Example: A sample of CO2 in a 10.0 L container at 293K exerts a pressure of 50,000 torr. How many moles of CO2 are in your sample? Step 1: Convert your pressure to atm. 50,000 torr x 1 atm / 760 torr = 65.8 atm Step 2: Write the ideal gas law and derive for moles. PV = nRT n = PV/RT Step 3: Calculate the moles. n = (65.8 atm)(10.0 L)/(.0821 L•atm/mol•K)(293K) n = 27.3 moles CO2

  16. Calculating M.W. Example: If 18.0 grams of a gas at 380 torr and 546.0 K occupies 44.8 L, what is the molecular weight of the gas? Step 1: Convert your pressure to atm. 380 torr x 1 atm / 760 torr = 0.50 atm Step 2: Identify the m.w. formula. So, plug in 18.0 grams into the formula. m.w. = g/mol m.w. = 18.0 g/ ? mol Step 3: Now, use PV = nRT to solve for n (the number of moles). n = PV/RT n = (0.50 atm)(44.8 L)/(.0821 L•atm/mol•K)(546 K) n = .499 moles Step 4: Plug this value into the molecular weight (m.w.) equation. m.w. = 18.0 g/ .499 moles = 36.0 g/moles

  17. Calculating Density at STP Example: Find the density of carbon dioxide at STP. Step 1: Identify the density formula. D = m/V Step 2: If it is assumed that one mole of CO2 is present, then the mass can be calculated from the periodic table and the volume at STP is 22.4 L.

  18. Step 1: Identify the density formula. D = m/V Step 2:Assume one mole of CO2. Thus, CO2 weighs 44.0 grams. Plug this into the Density formula. Step 3: Because the conditions are non-STP values, 22.4 L can’t be used. So, use PV=nRT and solve for V (assume one mole). V = nRT/P V = (1.00 mole)(.0821 L•atm/mol•K)(546K)/(4.00 atm) V =11.206 L Step 4: Plug this value into the density equation. D = 44.0 g/ 11.206 L = 3.93 g/ L Calculating Density (not at STP) Example: Find the density of carbon dioxide at 546.0 K and 4.00 atm.

  19. Deviations from Ideal Behavior • Real gases do not behave according to the KMT - Why? • Real gases have molecules that occupy space • Real gases have attractive and repulsive forces • Ideal gases conform exactly to the KMT • no such gas exists • gases only behave close to ideally at low P and high T. • At low T and high P, gases deviate greatly from ideal behavior. • Some gases are close to ideal (if they are small and nonpolar): • H2, He, Ne (these are small and nonpolar!!!) • O2 and N2 are not too bad • NH3, H2O are not even close to ideal

  20. 12.3 Stoichiometry of Gases • At STP, there is nothing new here. • The only thing new is that 1 mole = 22.4 L must be adjusted if not at STP. • Problems for Stoichiometry of Gases include converting: • Grams to Liters • Liters to Grams • Liter to Liter

  21. PV = nRT V = nRT/P V = (1 mole)(.0821 L atm/mol K)(293)/.950 atm V = 25.321 L O2 Grams to Liters Example: How many liters of O2 are generated when 50.0 grams of sodium chlorate decomposes at 0.950 atm and 20.0˚C? • 2NaClO3(s) + heat ---> 2NaCl(s) + 3O2(g) Step 1: The conditions are non-STP values, 22.4 L can’t be used so, use PV=nRT and solve for V (assume one mole). Step 2: Plug this value into the appropriate step of the stoichiometry problem.

  22. V = nRT/P V = (1 mole)(.0821 L atm/mol K)(310K)/.996 atm V = 25.551 L CO2 Liter to Grams Example: If a lawn mower engine generates 555.0 L CO2 on a lovely Sunday afternoon (.996 atm and 37.00 C0 )- how many grams of octane were consumed? • 2C8H18(l) + 25O2(g) ---> 16CO2(g) + 18H2O(g) Step 1: The conditions are non-STP values, 22.4 L can’t be used so, use PV=nRT and solve for V (assume one mole). Step 2: Plug this value into the appropriate step of the stoichiometry problem.

  23. V = nRT/P 1 mole C3H8 13.7 L CO2 V = (1 mole)(.0821 L atm/mol K)(500K)/3.00 atm V = 13.7 L CO2 x x = 8 L CO2 1 2 3 13.7 L C3H8 1 mole CO2 16 moles CO2 2.00 L C3H8 x 4 moles C3H8 Liter to Liter (not at STP) Example: How many L of carbon dioxide can be made from the combustion of 2.00 L of propane (C3H8) at 500. K and 3.00 atm? • 4C8H3(l) + 18O2(g) ---> 16CO2(g) + 6H2O(g) Step 1: The conditions are non-STP values, 22.4 L can’t be used so, use PV=nRT and solve for V (assume one mole). Step 2 : Plug this value into the appropriate step of the stoichiometry problem. • Again...we can skip steps 1 and 3.

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