#  is the study of the relative quantities stoichiometry A. Using Mole Ratios Stoichiometry and Quantitative Analysis of reactants and products in a chemical.

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 is the study of the relative quantities stoichiometry A. Using Mole Ratios Stoichiometry and Quantitative Analysis of reactants and products in a chemical reaction – stoich video– 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: W R = W n G R = G n

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

c) What amount, in moles, of O 2 (g) is required to react with 10.2 mol of C 2 H 6 (g)? 2 C 2 H 6 (g) + 7 O 2 (g)  4 CO 2 (g) + 6 H 2 O(g) n = 10.2 mol n = 10.2 mol  = 35.7 mol 7 2 d) What amount, in moles, of H 2 O(g) is formed when 100 mmol of CO 2 (g) is formed? 2 C 2 H 6 (g) + 7 O 2 (g)  4 CO 2 (g) + 6 H 2 O(g) n = 100 mmol n = 100 mmol  = 150 mmol 6 4 = 0.150 mol

 gravimetric =mass measurements B. Gravimetric Stoichiometry Steps 4. Calculate of the wanted species using 3. Find the of the species using 2. Find the of the species using Write a including the states. Write the information balanced equation given. molesgiven n=m M moleswanted mole ratio mass m=nM W R = W n G R = G n

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? CO 2(g) m=? M = 159.70 g/mol Step 1: n = m M = 1000 g 55.85g/mol = 17.90… mol Step 3 : m = nM = (8.95… mol)(159.70 g/mol) = 1429.7225 g = 1430 g Fe 2 O 3(s) CO (g)  Fe (s) ++3231 1 2 Step 2 : n = 17.90… x = 8.95… mol m = 1000 g M = 55.85 g/mol w g

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? H 2 O (g) m=? M = 79.55 g/mol Step 3: m = nM = (0.00904… mol)(79.55 g/mol) = 0.7194862 g = 0.719 g Step 1: n = m M = 1.00 g 221.13 g/mol = 0.00452… mol Cu 2 (CO 3 )(OH) 2(s)  CuO (s) +CO 2(g) +2111 2121 Step 2: n = 0.00452… x m = 1.00 g M = 221.13 g/mol = 0.00904… mol g w

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

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? v=? c= 14.8 mol/L STEP 2: n = 19.35 x 2 1 = 38.70 mol v = 1.50 L c = 12.9 mol/L H 2 O (l) KOH (aq) H 2 SO 4(aq)  K 2 SO 4(aq) ++1122 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 gw

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

Example Consider the following reaction: N 2 (g) + 2 O 2 (g)  N 2 O 4 (g) a) What is the mole ratio for O 2 (g) and N 2 O 4 (g)? b) What is the volume ratio for O 2 (g) and N 2 O 4 (g)? c) If 1 mol of N 2 O 4 (g) is produced, how many moles of O 2 (g) must be consumed? 2:1 1 mol of N 2 O 4 (g)  2 = 2 mol of O 2 (g) 1

d) If 1 L of N 2 O 4 (g) is produced, what volume of O 2 (g) must be consumed? e) If 2.5 L of N 2 (g) is consumed, what volume of O 2 (g) must be consumed? 1 L of N 2 O 4 (g)  2 = 2 L of O 2 (g) 1 2.5 L of N 2 (g)  2 = 5.0 L of O 2 (g) 1

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

Example 1 If 300 g of propane burns in a gas barbeque, what volume of oxygen at SATP is required for the reaction? v=? P = 100.000 kPa T = 298.15 K R = 8.314 kPaL/molK m = 300 g M = 44.11 g/mol n = 6.80… mol x 5 1 = 34.0… mol H 2 O (g) C 3 H 8(g) O 2(g)  CO 2(g) ++5341 PV = nRT (100.000kPa)V = (34.0… mol)(8.314)(298.15K) V = 842.94…L = 843 L n = m M = 300 g 44.11 g/mol = 6.80… mol g w

 limiting reagent = the that determines in a reaction  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   excess reagent = the that is present in than necessary 2 1 2 excess limiting Chapter 8: Applications of Stoichiometry 8.1 Limiting and Excess Reagents double burgers double burger double burgers reactant how much product can be formed reactant larger quantities

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

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? 16 Cu (s) + 1 S 8(s)  8Cu 2 S (s) m=? M = 159.17 g/mol m = 80.0 g M = 63.55 g/mol n = m = 25.0 g M = 256.56 g/mol n/16 = 0.0786…moln/1 = 0.0974… mol  limiting  excess 1.25…mol = 0.629… mol n = 80.0 g 63.55g/mol = 1.25… mol n = 25.0 g 256.56g/mol = 0.0974… mol  8/16 m = (0.629…mol )  (159.17 g/mol) = 100.17… g = 100 g g w

Example 2 You are supplied with 9.00 g of KCl and 6.50 g of AgNO 3. What is the mass of the precipitate formed when these two chemicals react? m = ? M = 143.32 g/mol m = 9.00 g M = 74.55 g/mol n = m = 6.50 g M = 169.88 g/mol n/1 = 0.120…moln/1 = 0.0382… mol  excess  limiting 0.0382… mol n = 9.00 g 74.55g/mol =0.120… mol n = 6.50 g 169.88g/mol = 0.0382… mol  1/1 1 KCl (aq) + 1 AgNO 3(aq)  1 KNO 3(aq)  + 1 AgCl (s) m = (0.0382…mol )  (143.32 g/mol) = 5.48 g

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? m = ? M = 232.66 g/mol C = 0.221 mol/L V = 0.200 L n = C = 0.500 mol/L V = 0.1000 L n/1 = 0.0442…moln/1 = 0.0500… mol  limiting 0.0442…mol m = (0.0442…mol )  (232.66 g/mol) =10.3 g n = CV =(0.221 mol/L)  (0.200 L) =0.0442… mol  1/1 1 HgCl 2(aq) + 1 Na 2 S (aq)  2 NaCl (aq)  + 1 HgS (s) n = CV =(0.500 mol/L)  (0.1000 L) =0.0500… mol

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

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

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

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

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 %

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

 in solution stoichiometry, sometimes you don’t have enough information to solve the problem on paper A. Titrations eg)10 mL of acetic acid reacts with a 0.202 mol/L NaOH solution. What is the concentration of the acetic acid? CH 3 COOH(aq) + NaOH(aq)  H 2 O(l) + NaCH 3 COO(aq) x mol/L v = 0.0100 L c = 0.202 mol/L v = ??? ***** **** you need this volume in order to solve the problem 8.3 Acid-Base Titration

 a is a used to find the of substances so you can calculate  is when a solution of concentration, a, is reacted with a solution of concentration  both need to be since their concentrations will titrationprocedure concentration standardizationknown standard solution unknown strong acids and strong bases standardized change over time volume

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

 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.

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 flasktitrant in burette

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

average volume = NaCl (aq) x mol/L V = 10.00 mL = 0.01000 L n = C = n V = = C = 0.685 mol/L V = mL = L n = = mol ( + + ) 3 = mL HCl (aq) +NaOH (aq) → H 2 O (l) + 1111

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

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

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

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

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

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