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is the study of the

Stoichiometry and Quantitative Analysis . A. Using Mole Ratios . is the study of the. relative quantities. stoichiometry. of reactants and products in a. chemical reaction – stoich video.

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is the study of the

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  1. Stoichiometry and Quantitative Analysis A. Using Mole Ratios • is the study of the relative quantities stoichiometry of reactants and products in a chemical reaction – stoich video • you can use the number of moles for a given reactant or product to find the moles for any other reactant or product • The equation I use is the following: • WR = Wn • GR = Gn

  2. Example Consider the following chemical reaction: 2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g) a) Write the ratio for all components of the reaction. 2:7:4:6 b) What amount, in moles, of CO2(g) is formed if 2.50 mol of C2H6(g) reacts? 2 C2H6(g) + 7 O2(g)4 CO2(g)+ 6 H2O(g) 4 2 n = 2.50 mol n = 2.50 mol  = 5.00 mol

  3. c) What amount, in moles, of O2(g) is required to react with 10.2 mol of C2H6(g)? 2 C2H6(g) + 7 O2(g)4 CO2(g) + 6 H2O(g) 7 2 n = 10.2 mol n = 10.2 mol  = 35.7 mol d) What amount, in moles, of H2O(g) is formed when 100 mmol of CO2(g) is formed? 2 C2H6(g) +7 O2(g) 4 CO2(g) + 6 H2O(g) 6 4 n = 100 mmol n = 100 mmol  = 150 mmol = 0.150 mol

  4. B. Gravimetric Stoichiometry • gravimetric = mass measurements Steps Write a including the states. Write the information balanced equation given. 2. Find the of the species using moles given n=m M 3. Find the of the species using moles wanted mole ratio WR = Wn GR = Gn 4. Calculate of the wanted species using mass m=nM

  5. Example 1 Iron is produced by the reaction of iron(III) oxide with carbon monoxide to produce iron and carbon dioxide. What mass of iron(III) oxide is required to produce 1000 g of iron? g w  1 Fe2O3(s) + 3 CO(g) 2 Fe(s) + 3 CO2(g) m = 1000 g M = 55.85 g/mol m=? M = 159.70 g/mol 1 2 Step 1: n = m M = 1000 g 55.85g/mol = 17.90… mol Step 2 : n = 17.90… x = 8.95… mol Step 3 : m = nM = (8.95… mol)(159.70 g/mol) = 1429.7225 g   = 1430 g

  6. Example 2 The decomposition of the mineral malachite, Cu2(CO3)(OH)2(s), yields copper(II) oxide, carbon dioxide and water vapour. What mass of copper(II) oxide is produced from 1.00 g of malachite? g w  1 Cu2(CO3)(OH)2(s) 2 CuO(s) + 1 CO2(g) + 1 H2O(g) m=? M = 79.55 g/mol m = 1.00 g M = 221.13 g/mol 2 1 Step 2: n = 0.00452… x Step 1: n = m M = 1.00 g 221.13 g/mol = 0.00452… mol = 0.00904… mol Step 3: m = nM = (0.00904… mol)(79.55 g/mol) = 0.7194862 g = 0.719 g

  7. C. Solution Stoichiometry concentration • use to perform calculations c = n v

  8. Example 1 What volume of 14.8 mol/L KOH is needed to react completely with 1.50 L of 12.9 mol/L sulphuric acid? w g  2 KOH(aq) + 1 H2SO4(aq) 1 K2SO4(aq) + 2 H2O(l) v=? c= 14.8 mol/L v = 1.50 L c = 12.9 mol/L STEP 2: n = 19.35 x 2 1 = 38.70 mol Step 1: n = cv = 12.9 mol/L x 1.50 L = 19.35 mol Step 3: v = n c = 38.70 mol 14.8 mol/L = 2.6148649 L = 2.61 L

  9. D. Law of Combining Volumes • you can use the Law of Combining Volumeswhen the pressure and temperature conditions are constant for both the reactants and the products

  10. Example Consider the following reaction: N2(g) + 2 O2(g)  N2O4(g) a) What is the mole ratio for O2(g) and N2O4(g)? 2:1 b) What is the volume ratio for O2(g) and N2O4(g)? 2:1 c) If 1 mol of N2O4(g) is produced, how many moles of O2(g) must be consumed? 1 mol of N2O4(g) 2 = 2 mol of O2(g) 1

  11. d) If 1 L of N2O4(g) is produced, what volume of O2(g) must be consumed? 1 L of N2O4(g) 2 = 2 L of O2(g) 1 e) If 2.5 L of N2(g) is consumed, what volume of O2(g) must be consumed? 2.5 L of N2(g) 2 = 5.0 L of O2(g) 1

  12. E. Gas Stoichiometry • if the other information is given for the chemicals in the reaction (eg. mass), use to perform calculations ideal gas law PV = nRT

  13. Example 1 If 300 g of propane burns in a gas barbeque, what volume of oxygen at SATP is required for the reaction? w g  1 C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) v=? P = 100.000 kPa T = 298.15 K R = 8.314 kPaL/molK m = 300 g M = 44.11 g/mol n = m M = 300 g 44.11 g/mol = 6.80… mol n = 6.80… mol x 5 1 = 34.0… mol PV = nRT (100.000kPa)V = (34.0… mol)(8.314)(298.15K) V = 842.94…L = 843 L

  14. Chapter 8: Applications of Stoichiometry 8.1 Limiting and Excess Reagents • let’s make double burgers… 1 bun + 2 meat patties  1 double burger 2 buns + 4 meat patties  2 buns + 2 meat patties  2 buns + one million meat patties  2 double burgers 1 double burger excess limiting 2 double burgers limiting excess reactant • limiting reagent = the that determinesin a reaction how much product can be formed • excess reagent = the that is present inthan necessary reactant larger quantities

  15. Steps Write the including states. balanced chemical equation, 1. number of moles Calculate the of using each reactant n=m or C = n/v M 2. Do step 2 for both reactants. WR = Wn GR= Gn - Use the lower n for step 3. (it’s the limiting reagent) 3. limiting reagent moles Use to calculate the answer.

  16. Example 1 When 80.0 g copper and 25.0 g of sulphur react, which reactant is limiting and what is the maximum amount of copper(I) sulphide that can be produced? w g 16Cu(s) + 1 S8(s) 8Cu2S(s) m = 80.0 g M = 63.55 g/mol m = 25.0 g M = 256.56 g/mol m=? M = 159.17 g/mol n = 80.0 g 63.55g/mol = 1.25… mol n = 25.0 g 256.56g/mol = 0.0974… mol 8/16 n = 1.25…mol = 0.629… mol n/16 = 0.0786…mol n/1 = 0.0974… mol \ excess \ limiting m = (0.629…mol )  (159.17 g/mol) = 100.17… g = 100 g

  17. Example 2 You are supplied with 9.00 g of KCl and 6.50 g of AgNO3. What is the mass of the precipitate formed when these two chemicals react? 1 KCl(aq) + 1 AgNO3(aq)® 1 KNO3(aq)+ 1 AgCl(s) m = 9.00 g M = 74.55 g/mol m = 6.50 g M = 169.88 g/mol m = ? M = 143.32 g/mol n = 9.00 g 74.55g/mol =0.120… mol n = 6.50 g 169.88g/mol = 0.0382… mol 1/1 n = 0.0382… mol n/1 = 0.120…mol n/1 = 0.0382… mol \ limiting \ excess m = (0.0382…mol )  (143.32 g/mol) = 5.48 g

  18. Example 3 A 200 mL sample of a 0.221 mol/L mercury (II) chloride solution reacts with 100.0 mL of a 0.500 mol/L solution of sodium sulphide. What is the mass of the precipitate formed? 1 HgCl2(aq) + 1 Na2S(aq)® 2 NaCl(aq)+ 1 HgS(s) m = ? M = 232.66 g/mol C = 0.221 mol/L V = 0.200 L C = 0.500 mol/L V = 0.1000 L n = CV =(0.221 mol/L)  (0.200 L) =0.0442… mol 1/1 n = CV =(0.500 mol/L)  (0.1000 L) =0.0500… mol n = 0.0442…mol n/1 = 0.0442…mol n/1 = 0.0500… mol m = (0.0442…mol)  (232.66 g/mol) =10.3 g \limiting

  19. 8.2 Predicted and Experimental Yield • the is called the expected amount of product predicted or theoretical yield • the quantity of the product is called the actually obtained experimental or actual yield • it is for the predicted and experimental yield to be extremely rare the same

  20. factors affecting the experimental yield include: 1. two chemicals can react to give …called different products competing reactions eg) C(s) and O2(g) can react to form CO2(g) or CO(g) 2. reaction is very slow 3. methods collection and transfer 4. reactant or product purity 5. reaction doesn’t go to completion

  21. ideally, percent yield should be as close to as possible 100% • percent yield is calculated as follows: % yield= actual yield  100 predicted yield

  22. % error • we can also calculate our for the experiment, which tells us how far we are from the theoretical yield • the closer to the percent error is, the better the experiment zero % error= actual yield – predicted yield  100 predicted yield

  23. Example 1 Calculate the % error and % yield for the following: predicted mass of ppt = 6.20 g actual mass of ppt = 7.12 g % error = 7.12 g - 6.20 g x 100 6.20 g = 14.8 % % yield = 7.12 g x 100 6.20 g = 115 %

  24. Example 2 Calculate the % error and % yield for the following: predicted mass of ppt = 100 g actual mass of ppt = 93.5 g % error = 93.5 g - 100 g x 100 100 g = -6.50 % % yield = 93.5 g x 100 100 g = 93.5 %

  25. 8.3 Acid-Base Titration A. Titrations • in solution stoichiometry, sometimes you don’t have enough information to solve the problem on paper eg) 10 mL of acetic acid reacts with a 0.202 mol/L NaOH solution. What is the concentration of the acetic acid? CH3COOH(aq)+ NaOH(aq)  H2O(l)+ NaCH3COO(aq) c = 0.202 mol/L v = ??? ***** x mol/L v = 0.0100 L **** you need this volume in order to solve the problem

  26. a is a used to find theof substances so you can calculate titration procedure volume concentration • is when a solution of concentration, a , is reacted with a solution of concentration standardization known standard solution unknown • both need to be since their concentrations will strong acids and strong bases standardized change over time

  27. titrant • a solution called theis transferred from a precisely marked tube called a to a containing the and an burette flask sample indicator • an indicator (eg. methyl orange, bromothymol blue) is used because a sudden change in colour indicates the completion of the reaction

  28. the endpoint is the point where the titrant reacts completely with the sample • equivalence point is the volume needed to reach the endpoint • you need a minimum of 3 trials within of each other to ensure results are accurate 0.20 mL Here is a website to view a virtual titration: Virtual Titration Practice using this simulation.

  29. Example A 10.00 ml sample of HCl(aq) was titrated with a standardized solution of 0.685 mol/L NaOH(aq). Bromothymol blue indicator was used and it changes from yellow to blue at the endpoint. What is the concentration of the HCl(aq)? Note: HCl(aq) “is titrated with” NaOH(aq) sample in flask titrant in burette

  30. Trial 1 overshoot 2 3 4 Final Volume (mL) Initial Volume (mL) Volume NaOH (mL) Endpoint Colour Data Table

  31. average volume = ( + + ) 3 = mL 1 1 1 1 HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) x mol/L V = 10.00 mL = 0.01000 L C = 0.685 mol/L V = mL = L n = = mol n = C = n V = =

  32. B. Titration Curves • a plot of the is called a pH vs. the volume of titrant added pH titration curve • titration curves are S-shaped • when a is titrated with a the willalways have a pH of 7 (at 25C) strong monoprotic acid strong monoprotic base, equivalence point • the on the curve is always the pH of the sample first point

  33. the pH changes at first…this is called the very gradually buffer region • as the endpoint is approached, the pH changes very rapidly • the is (goes to infinity/never ends) to the pH of the overtitration asymptotic titrant

  34. Strong Acid Titrated with Strong Base 14 pH 7  0 volume of titrant added (mL)

  35. Strong Base Titrated with Strong Acid 14  pH 7 0 volume of titrant added (mL)

  36. C. Indicators for Titrations • can be used to carry out a titration but it is much more convenient to use an pH meters indicator • the indicator should change colour immediately after the endpoint is reached • the pH of the endpoint should fall within the pH range of the indicator Review assignment: p. 328 #1-33 (omit 22)

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