Unit 4 Stoichiometry Notes PDF

1 Unit 4 pt 2- STOICHIOMETR Y - the study of the quantitative aspects of chemical reactions. © 2009 Brooks/Cole - Cengage  2...

1 Unit 4 pt 2- STOICHIOMETR Y - the study of the quantitative aspects of chemical reactions. © 2009 Brooks/Cole - Cengage  2 Theoretical Yield Stoichiometry involves using balanced chemical equations to determine quantities of reactants and products involved in a chemical reaction. Example: To make a ham and cheese sandwich, you use 2 pieces of bread, 1 slice of cheese, and 1 slice of ham. The reaction would look like: 2 bread slices + 1 slice of cheese + 1 slice of ham → 1 ham & cheese sandwich © 2009 Brooks/Cole - Cengage  3 How many sandwiches could you make if you had 4 pieces of bread, 2 slices of cheese, 3 slices of ham? _____________ This is called the theoretical yield. What limits the amount of product (# of sandwiches) we can make in the reaction above? In a chemical reaction, this would be called the limiting reagent (or reactant). © 2009 Brooks/Cole - Cengage  4 Applying this to a chemical reaction: Example: Combustion of octane: 2 C8H18 + 25 O2 ---------> 16 CO2 + 18 H2O Use dimensional analysis where the stoichiometic coefficients become conversion factors: (a) How many moles of O2 are needed to react fully with 4 moles of octane? © 2009 Brooks/Cole - Cengage  5 Applying this to a chemical reaction: Example: Combustion of octane: 2 C8H18 + 25 O2 ---------> 16 CO2 + 18 H2O (b) If this reaction is to be used to synthesize 8 mole of CO2, how many moles of oxygen are needed? How many moles of octane? © 2009 Brooks/Cole - Cengage  6 PROBLEM: PROBLEM: IfIf454 454ggofofNH NH44NO NO33decomposes, decomposes,how howmuchmuchNN22OO and andHH22OOare areformed? formed? What Whatisisthe thetheoretical theoreticalyield yieldofofproducts? products? STEP 1 Write the balanced chemical equation NH4NO3(s) → N2O(g) + 2 H2O(g) © 2009 Brooks/Cole - Cengage  7 GENERAL GENERAL PLAN PLAN FOR FOR STOICHIOMETR STOICHIOMETRYY CALCULATIONS CALCULATIONS Mass Mass reactant product Stoichiometric Moles factor Moles reactant product © 2009 Brooks/Cole - Cengage  8 454 454 gg of of NH NO33 → NH44NO →NN22O O ++ 22 H H22O O STEP 2 Convert mass of reactant (454 g) to amount (mol) STEP 3 Convert amount of reactant (5.68 mol) to amount (mol) of product. © 2009 Brooks/Cole - Cengage  9 454 454 gg of of NH NO33 → NH44NO →NN22O O ++ 22 H H22O O STEP 3 Convert moles reactant (5.67 mol) to amount (mol) of product = 11.3 mol H2O produced © 2009 Brooks/Cole - Cengage  10 454 454 gg of of NH NO33 → NH44NO →NN22O O ++ 22 H H22O O STEP 4 Convert amount of product (11.3 mol) to mass of product Called the THEORETICAL YIELD © 2009 Brooks/Cole - Cengage  11 454 454 gg of of NH NO33 → NH44NO →NN22O O ++ 22 H H22O O STEP 5 What mass of N2O is formed? Total mass of reactants = total mass of products 454 g NH4NO3 = ___ g N2O + 204 g H2O mass of N2O = 250. g © 2009 Brooks/Cole - Cengage  Theoretical Yield The theoretical yield is the maximum amount of product that can be made. – In other words, it’s the amount of product possible as calculated through the stoichiometry problem. This is different from the actual yield, which is the amount one actually produces and measures in an experiment.  13 454 454 gg of of NH NO33 → NH44NO →NN22O O ++ 22 H H22O O STEP 6 Calculate the percent yield If you isolated only 131 g of N2O, what is the percent yield? This compares the theoretical (250. g) and actual (131 g) yields. © 2009 Brooks/Cole - Cengage  14 PROBLEM: PROBLEM: Using Using 5.00 5.00 gg of of HH22OO22,, what what mass mass of of OO22 and and of of HH22OO can can be be obtained? obtained? If If only only 11 gg of of OO22 was was produced, produced, what what isis the the %% yield yield of of oxygen? oxygen? 2 H2O2(liq) → 2 H2O(g) + O2(g) Step 1: amount (mol) of H2O2 Step 2: use STOICHIOMETRIC FACTOR to calculate amount (mol) of O2 Step 3: mass of O2 and H2O? © 2009 Brooks/Cole - Cengage  15 “Map” for stoichiometric calculations Mass Mass A B Moles Moles liters A stoichiometric B liters A factor B particles particles A B © 2009 Brooks/Cole - Cengage  16 LIMITING REACTANTS Reactant Products 2sNO(g) + O2(g) → 2 NO2(g) In a given reaction, there is not enough of one reagent to use up the other reagent completely. The reagent in short supply LIMITS the quantity of product that can be formed. Limiting reactant = ___________ Excess reactant = ____________ © 2009 Brooks/Cole - Cengage  17 PROBLEM: PROBLEM: Mix Mix 5.40 5.40 gg of of Al Al with with 8.10 8.10 gg of of Cl Cl22.. What What mass mass of of Al Al22Cl Cl66 can can form? form? Mass Mass reactant reactant Stoichiometric Moles factor Moles reactant reactant © 2009 Brooks/Cole - Cengage  18 Using Stoichiometry 1. Pick one reactant (it doesn’t matter which one), and calculate how much of the other reactant you would need for it to completely react. 2. Compare the amount needed (what you calculated in #1) to the amount you have (given in the problem). If you have more than you need, it is the excess reagent. If you don’t have enough, it is the limiting reagent. © 2009 Brooks/Cole - Cengage  19 Step Step 11 of of LR LR problem: problem: How How much much chlorine chlorine isis needed needed to to completely completely react react with with 5.40 5.40 gg Al. Al. 2 Al(s) + 3 Cl2(g) → Al2Cl6(s) © 2009 Brooks/Cole - Cengage  20 Step Step 22 of of LR LR problem: problem: Compare Compare the the amount amount of of chlorine chlorine needed needed with with what what you you have have We have 8.1 g of Cl2 © 2009 Brooks/Cole - Cengage  21 CALCULATIONS: CALCULATIONS: calculatecalculate mass mass of of Al Al22Cl Cl66 expected. expected. Step 1: Calculate moles of Al2Cl6 expected based on LR. Step 2: Calculate mass of Al2Cl6 expected based on LR. © 2009 Brooks/Cole - Cengage  22 How How much much of of which which reactant reactant will will remain remain when when reaction reaction isis complete? complete? Cl2 was the limiting reactant. Therefore, Al was present in excess. But how much? First find how much Al was required. Then find how much Al is in excess. © 2009 Brooks/Cole - Cengage  23 Calculating Calculating Excess Excess Al Al 2 Al + 3 Cl2 products 0.200 0.200mol mol 0.114 0.114 mol mol == LR LR Excess Al = Al available - Al required = 0.200 mol - 0.0760 mol = 0.124 mol Al in excess © 2009 Brooks/Cole - Cengage  24 SOLUTION STOICHIOMETR Y Section 4.7 Zinc reacts with acids to produce H2 gas. Have 10.0 g of Zn What volume of 2.50 M HCl is needed to convert the Zn completely? © 2009 Brooks/Cole - Cengage  Aqueous Reactions © 2015 Pearson Education, Inc.  26 Zinc Zinc reacts reacts with with acids acids to to produce produce H H22 gas. gas. IfIf you you have have 10.0 10.0 gg of of Zn, Zn, what what volume volume of of 2.50 2.50 M M HClHCl isis needed needed toto convert convert the the Zn Zn completely? completely? Step 1: Write the balanced equation Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g) Step 2: Calculate amount of Zn Step 3: Use the stoichiometric factor © 2009 Brooks/Cole - Cengage  27 Zinc Zinc reacts reacts with with acids acids to to produce produce H H22 gas. gas. IfIf you you have have 10.0 10.0 gg of of Zn, Zn, what what volume volume of of 2.50 2.50 M M HClHCl isis needed needed toto convert convert the the Zn Zn completely? completely? Step 3: Use the stoichiometric factor Step 4: Calculate volume of HCl req’d © 2009 Brooks/Cole - Cengage  28 Limiting Reactants with Solutions 200 mL of 0.10 M silver nitrate is mixed with 100mL of 0.15 M barium chloride. How many grams of silver chloride precipitate are formed? (MM AgCl = 143.3 g ) You will first need to determine which reactant is in excess and which is in limited supply. © 2009 Brooks/Cole - Cengage  29 Limiting Reactants with solutions Based on the limited reactant, the amount of product can be calculated. How much excess reactant remains? © 2009 Brooks/Cole - Cengage  Gases and Stoichiometry 2 H2O2(liq) --> 2 H2O(g) + O2(g) If 0.11 g of H2O2 decomposes in a flask with a volume of 2.50 L. What is the pressure of O2 at 25 oC? Of H2O? Solution Strategy: Calculate moles of H2O2 and then moles of O2 and H2O. Finally, calc. P from n, R, T, and V. Gas es © 2015 Pearson Education, Inc.  31 Gases and Stoichiometry 2 H2O2(liq) → 2 H2O(g) + O2(g) Decompose 0.11 g of H2O2 in a flask with a volume of 2.50 L. What is the pressure of O2 at 25 oC? Of H2O? Solution © 2009 Brooks/Cole - Cengage  32 Gases and Stoichiometry 2 H2O2(liq) →2 H2O(g) + O2(g) Decompose 0.11 g of H2O2 in a flask with a volume of 2.50 L. What is the pressure of O2 at 25 oC? Of H2O? Solution P of O2 = 0.016 atm © 2009 Brooks/Cole - Cengage  33 Gases and Stoichiometry 2 H2O2(liq) → 2 H2O(g) + O2(g) What is P of H2O? Could calculate as above. But recall Avogadro’s hypothesis. V ∝ n at same T and P P ∝ n at same T and V There are 2 times as many moles of H2O as moles of O2. P is proportional to n. Therefore, P of H2O is twice that of O2. P of H2O = 0.032 atm © 2009 Brooks/Cole - Cengage  Gases and Stoichiometry If a 0.452 L flask has a pressure of O2 of.0035 atm at 27 ℃, what mass of magnesium will react, according to the balanced equation below. 2 Mg (s) + O2 (g) → 2 MgO (s) Gas es © 2015 Pearson Education, Inc.  Titration A titration is an analytical technique in which one can calculate the concentration of a solute in a solution. Aqueous Reactions © 2015 Pearson Education, Inc.  Titration A solution of known concentration, called a standard solution, is used to determine the unknown concentration of another solution. The reaction is complete at the equivalence point. Aqueous Reactions © 2015 Pearson Education, Inc.  37 ACID-BASE ACID-BASE REACTIONS REACTIONS Titrations Titrations H2C2O4(aq) + 2 NaOH(aq) → acid base Na2C2O4(aq) + 2 H2O(liq) Carry out this reaction using a TITRATION. Oxalic acid, H2C2O4 © 2009 Brooks/Cole - Cengage  38 Titration Titration 1. Add solution from the buret. 2. Reagent (base) reacts with compound (acid) in solution in the flask. 3. Indicator shows when exact stoichiometric reaction has occurred. 4. Net ionic equation H3O+(aq) + OH-(aq) → 2 H2O(liq) 5. At equivalence point moles H3O+ = moles OH- © 2009 Brooks/Cole - Cengage  39 Endpoint Endpoint 1. The experimental determination that equal moles of H+ and OH- are present Often an acid base indicator is used that is one color in excess OH- and another color in excess H+ © 2009 Brooks/Cole - Cengage  40 © 2009 Brooks/Cole - Cengage  41 LAB LAB PROBLEM PROBLEM #1: #1: Standardize Standardize aa solution solution of of NaOH NaOH — — i.e., i.e., accurately accurately determine determine its its concentration. concentration. 1.065 g of H2C2O4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH? © 2009 Brooks/Cole - Cengage  1.065 1.065 gg of of H H22CC22OO44 (oxalic (oxalic acid) acid) requires requires 35.62 35.62 42 mL mL of of NaOH NaOH for for titration titration toto an an equivalence equivalence point. point. What What isis the the concentration concentration of of the the NaOH? NaOH? Step 1: Calculate amount of H2C2O4 Step 2: Calculate amount of NaOH req’d © 2009 Brooks/Cole - Cengage  43 1.065 1.065 gg of of HH22CC22O O44 (oxalic (oxalic acid) acid) requires requires 35.62 35.62 mL mL of of NaOH NaOH forfor titration titration toto an an equivalence equivalence point. point. What What isis the the concentration concentration of of the the NaOH? NaOH? Step 1: Calculate amount of H2C2O4 = 0.0118 mol acid Step 2: Calculate amount of NaOH req’d = 0.0237 mol NaOH Step 3: Calculate concentration of NaOH [NaOH] = 0.6642 M © 2009 Brooks/Cole - Cengage  44 LAB LAB PROBLEM PROBLEM #2: #2: Use Use standardized standardized NaOH NaOH to to determine determine the the amount amount of of an an acid acid in in an an unknown. unknown. Apples contain malic acid, C4H6O5. C4H6O5(aq) + 2 NaOH(aq) → Na2C4H4O5(aq) + 2 H2O(liq) 76.80 g of apple requires 34.56 mL of 0.664 M NaOH for titration. What is weight % of malic acid? © 2009 Brooks/Cole - Cengage  45 76.80 76.80 gg of of apple apple requires requires 34.56 34.56 mL mL ofof 0.664 0.664 M M NaOH NaOH forfor titration. titration. What What isis weight weight % % ofof malic malic acid? acid? Step 1: Calculate amount of NaOH used. C V = (0.664 M)(0.03456 L) = 0.0229 mol NaOH Step 2: Calculate amount of acid titrated. = 0.0115 mol acid © 2009 Brooks/Cole - Cengage  46 76.80 76.80 gg of of apple apple requires requires 34.56 34.56 mL mL ofof 0.664 0.664 M M NaOH NaOH forfor titration. titration. What What isis weight weight % % ofof malic malic acid? acid? Step 1: Calculate amount of NaOH used. = 0.0229 mol NaOH Step 2: Calculate amount of acid titrated = 0.0115 mol acid Step 3: Calculate mass of acid titrated. © 2009 Brooks/Cole - Cengage  47 76.80 76.80 gg of of apple apple requires requires 34.56 34.56 mL mL ofof 0.663 0.663 M M NaOH NaOH forfor titration. titration. What What isis weight weight % % ofof malic malic acid? acid? Step 1: Calculate amount of NaOH used. = 0.0229 mol NaOH Step 2: Calculate amount of acid titrated = 0.0115 mol acid Step 3: Calculate mass of acid titrated. = 1.54 g acid Step 4: Calculate % malic acid. © 2009 Brooks/Cole - Cengage  Combustion Analysis to Determine a Formula Burn 0.115 g of a hydrocarbon, CxHy, and produce 0.379 g of CO2 and 0.1035 g of H2O. CxHy + some oxygen → 0.379 g CO2 + 0.1035 g H2O What is the empirical formula of CxHy? Intermolecular Forces © 2015 Pearson Education, Inc.  Using Stoichiometry to Determine a Formula CxHy + some oxygen → 0.379 g CO2 + 0.1035 g H2O First, recognize that all C in CO2 and all H in H2O is from CxHy. 1. Calculate amount of C in CO2 8.61 x 10-3 mol CO2 → 8.61 x 10-3 mol C 2. Calculate amount of H in H2O 5.744 x 10-3 mol H2O → 1.149 x 10-2 mol H Intermolecular Forces © 2015 Pearson Education, Inc. 104 Using Stoichiometry to Determine a Formula CxHy + some oxygen → 0.379 g CO2 + 0.1035 g H2O Now find ratio of mol H/mol C to find values of x and y in CxHy. 1.149 x 10 -2 mol H/ 8.61 x 10-3 mol C = 1.33 mol H / 1.00 mol C Multiply by 3 to get whole number coefficients. Therefore, we have 4 mol H / 3 mol C Empirical formula = C3H4 Intermolecular Forces © 2015 Pearson Education, Inc.  3.5 Other Types of Quantitative Analysis In quantitative analysis, chemists often want to analyze the components of a mixture of two or more substances. This can be accomplished through mixture analysis, which is a method of determining the percent composition of two or more substances. Each problem is solved a little differently, but most generally require converting from grams to moles of a See Active Figure 4.7 reactant or product and using a balanced equation to determine the relative amounts of each reactant and products. A mixture of potassium chlorate ( KClO3) and potassium chloride (KCl) is analyzed in the example problem. Intermolecular Forces © 2015 Pearson Education, Inc.  Mixture Analysis A mixture of potassium chlorate ( KClO3) and potassium chloride (KCl) weighs 6.55 grams. When the sample is heated the KClO3 decomposes according to the equation below but the KCl does not change. After completion of the reaction, the residue weighed 4.82 grams. What was See Active the4.7 Figure percent KCl in the original mixture? (MM KClO3 = 122.4) Intermolecular 2KClO3(s) → 2 KCl(s) + 3 O2(g) Forces © 2015 Pearson Education, Inc.  53 Chemical Analysis An impure sample of a mineral sample contains Na2SO4. Mass of mineral sample = 0.123 g The Na2SO4 in the sample is converted to insoluble BaSO4. The mass of BaSO4 is 0.177 g What is the mass percent of Na2SO4 in the mineral? © 2009 Brooks/Cole - Cengage  54 Chemical Analysis Na SO (aq) + BaCl (aq) 2 4 2 → 2 NaCl(aq) + BaSO4(s) 0.177 g BaSO (1 mol/233.4 g) 4 = 7.58 x 10-4 mol BaSO4 7.58 x 10-4 mol BaSO (1 mol Na SO /1 mol BaSO ) 4 2 4 4 = 7.58 x 10-4 mol Na2SO4 7.58 x 10-4 mol Na SO (142.0 g/1 mol) 2 4 = 0.108 g Na2SO4 (0.108 g Na SO /0.123 g sample)100% 2 4 = 87.6% Na2SO4 © 2009 Brooks/Cole - Cengage