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Chapter 12: Chemical Quantities

Learn about Avogadro's Principle, STP, molar volume, and the ideal gas law. Practice problems and calculations for moles of gas, volume of gas, and percent yield.

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Chapter 12: Chemical Quantities

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  1. Chapter 12: Chemical Quantities Section 12.2: Using Moles (part 2)

  2. Remember… Avogadro’s Principle In terms of moles, states that equal volumes of gases at the same temperature and pressure contain equal numbers of moles of gases. What is STP?

  3. MOLAR VOLUME For a gas – the volume that a mole of a gas occupies at a pressure of one atmosphere (equal to 101 kPa) and at a temperature of 0.00ºC = STP **At STP, the volume of 1 mol of any gas is 22.4 L**

  4. Steps Vol A ↔ Mol A ↔ Mol B ↔ grams B 22.4 L Coeff. Mass OR Grams A ↔ Mol A ↔ Mol B ↔ Vol A mass Coeff. 22.4 L

  5. Practice Problems 1) What mass of glucose (C6H12O6) must be broken down in your body to produce 2.5 L of CO2 at STP? C6H12O6 + 6O2 → 6H2O + 6 CO2 2.5 L CO2 x 1 mol CO2 x 1 molC6H12O6 x 180 g C6H12O6 22.4 L CO2 6 mol CO2 1 mol C6H12O6 = 3.35 g C6H12O6

  6. Practice Problems (Cont) 2) What volume of oxygen is required to react with 100 g of iron at STP? 4 Fe + 3O2 → 2 Fe2O3 100 g Fe x 1 mol Fe x 3 mol O2 x 22.4 L O2 55.847 g Fe 4 mol Fe 1 mol O2 = 30.08 L O2

  7. IDEAL GAS LAW Pressure P, volume V, temperature T, and the number of particles n of gas are related by: PV = nRT

  8. IDEAL GAS LAW (Cont) R can be determined using the definition of molar volume at STP: P = 101.3 kPa V = 22.4 L n = 1 mol T = 273.15 K (101.3 kPa)(22.4L) = (1 mol)(R)(273.15 K) R = (101.3 kPa)(22.4L) = 8.31 kPa•L (1 mol) (273.15 K) mol•K

  9. R is a constant = 8.31 kPa. L mol. K ** If the given pressure is in kPa, use the value for R above. If the given pressure is in atm, then use the value R = .08205 atm. L mol. K ** If the given pressure is in mm Hg, then use the value R = 62.36 mm Hg . L mol. K ** YOU WILL NOT HAVE TO MEMORIZE THESE!**

  10. Do NOT let the algebraic formula get you! These are all “plug & chug” problems. 

  11. Practice Problem 1) How many moles of gas are contained in a 10 L tank at 300 KPa and 500C? PV= nRT → n= PV/ RT 50+273 = 323 K (300 kPa)(10L) = n (8.31kPa•L)(323K) mol•K n = (300 kPa)(10L) = 1.12 mol (8.31kPa•L)(323K) mol•K

  12. Practice Problems (Cont) 2) What volume of gas is contained if 2 moles of gas are at 20oC and 2 atm? 2 atm x 101 KPa = 202 KPa 1 atm PV = nRT → V= nRT/ P 20+273 = 293 K V = 2 mol (8.31kPa•L)(293K) mol•K 202 kPa V= 24.1 L

  13. THEORETICAL YIELD The amount of product predicted to form is called the theoretical yield The actual yield is usually less than the predicted (theoretical yield) - Theoretical yield is determined through calculation. - Actual yield may be affected by the collection techniques, apparatus used, time, and chemist skills

  14. Efficiency of a reaction can be expressed as percent yield: PERCENT YIELD = actual yield x 100% theoretical yield Manufacturers want to produce chemicals as efficiently and inexpensively as possible

  15. Practice Problems 1) What is the percent yield if the theoretical yield is 4.5 g and the actual is 3.8 g? % yield = 3.8 x 100 = 4.5 =84.4%

  16. 2) 10 g of H2 react with excess O2. When the reaction is over 85 g of H2O are recovered, what is the percent yield? 2H2 + O2 → 2H2O Actual yield= 85g 10 g H2 x 1 mol H2 x 2 mol H2O x 18.01 g H2O = 2.0158 g H2 2 mol H2 1 mol H2O Theoretical yield= 89.34 g H2O % yield = 85 g / 89.34 g x 100 = 95.14%

  17. Mass Percent Steps: 1) Calculate mass of each element 2) Calculate total mass 3) Divide mass of element/ mass of compound

  18. Mass Percent (Cont) Ex #1: Calculate the mass % of C and H in C2H6 2 mol C x 12 g C = 24 g C 6 mol H x 1 g H = 6 g H Total mass = C2H6 = 30 g % C = 24 g C x 100% = 80% 30 g C2H6 % H = 6 g H x 100% = 20% 30 g C2H6 * Should add up to 100

  19. Mass Percent (Cont) Ex #2: Calculate the mass % of C, H, Br in C6H5Br 6 mol C x 12.011 = 72.066 5 mol H x 1.0079 = 5.0395 1 mol Br x 79.904 = 79.904 = 157.0095 g % C = 72.066 g C x 100% = 45.9% 157.0095 g C2H5 Br % H = 5.0395 g H x 100% = 3.2% 157.0095 g C2H5 Br % Br = 79.904 g Br x 100% = 50.9% 157.0095 g C2H5 Br

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