Chemical Equations. Chemical Equations

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1 Page III-4a-1 / Chapter Four Part I Lecture Notes Chemical Reactions Chapter 4 Part 1 Chemistry as Cooking! - the Chemical Reaction "Recipe" and technique leads to successful creations Must know amounts to add, how much will be produced Haphazard additions can be disastrous! Reactants: Zn + I2 Product: ZnI2 Chemistry 221 Professor Michael Russell Chemical Equations Reaction of Phosphorus with Cl2 Depict the kind of reactants and products and their relative amounts in a reaction. 4 Al(s) + 3 O2(g) ---> 2 Al2O3(s) The numbers in the front are called stoichiometric coefficients The letters (s), (g), (aq) and (l) are the physical states of compounds. Notice the stoichiometric coefficients and the physical states of the reactants and products. Reaction of Iron with Cl2 Chemical Equations 4 Al(s) + 3 O2(g) ---> 2 Al2O3(s) This equation means 4 Al atoms + 3 O2 molecules ---give---> 2 molecules of Al2O3 4 moles of Al + 3 moles of O2 ---give---> 2 moles of Al2O3 Evidence of a chemical reaction: heat change, precipitate formation, gas evolution, color change Page III-4a-1 / Chapter Four Part I Lecture Notes

2 Page III-4a-2 / Chapter Four Part I Lecture Notes Chemical Equations Because the same atoms are present in a reaction at the beginning and at the end, the amount of matter in a system does not change. The Law of the Conservation of Matter Also known as the Law of Mass Action Chemical Equations Because of the principle of the conservation of matter, an equation must be balanced. It must have the same number of atoms of the same kind on both sides. Lavoisier, 1788 Balancing Equations 2 _ Al(s) + 3 _ Br 2 (liq) ---> Al _ 2 Al Br 2 Br 6 (s) 6 (s) Balancing Equations C C 3 H 8 (g) 3 H+ 8 (g) 5 O+ 2 (g) ----> O 2 (g) ----> 3 CO CO 2 (g) + H 2 O(g) 2 (g) + 4 H 2 O(g) 2 B B 4 H 10 (g) 4 H 10 + (g) 11 + O 2 (g) ----> O 2 (g) ----> 4 B 2 O 3 (g) B 2 O+ 3 (g) 10 + H 2 O(g) H 2 O(g) Balancing Equations - Hints Balance those atoms which occur in only one compound on each side last (i.e. O2 in previous examples) Balance the remaining atoms first Reduce coefficients to smallest whole integers Check your answer if uncertain Helpful but optional: Check that charges are balanced STOICHIOMETRY Stoichiometry is the study of the quantitative aspects of chemical reactions. Stoichiometry rests on the principle of the conservation of matter. Page III-4a-2 / Chapter Four Part I Lecture Notes

3 Page III-4a-3 / Chapter Four Part I Lecture Notes Stoichiometry The balanced chemical equation 4 Al(s) + 3 O 2 (g) ---> 2 Al 2 O 3 (s) implies all of the following ratios: 4 mol Al 3 mol O 2 4 mol Al 2 mol Al 2 O 3 3 mol O 2 2 mol Al 2 O 3 PROBLEM: If 454 g of NH 4 NO 3 decomposes, how much N 2 O and H 2 O are formed? What is the theoretical yield of products? STEP 1 Write the balanced chemical equation 3 mol O 2 4 mol Al 2 mol Al 2 O 3 4 mol Al 2 mol Al 2 O 3 3 mol O 2 These are nothing more than "conversion units" in dimensional analysis! NH 4 NO 3 ---> N 2 O + 2 H 2 O STEP 2 Convert mass reactant (454 g) --> moles 454 g g = 5.68 mol NH 4 NO 3 STEP 3 Convert moles reactant --> moles product Relate moles NH 4 NO 3 to moles product expected. NH 4 NO 3 --> 2 mol H 2 O Express as a STOICHIOMETRIC FACTOR: 2 mol H 2 O produced NH 4 NO 3 used STEP 3 Convert moles reactant (5.68 mol) --> moles product 5.68 mol NH 4 NO 3 2 mol H 2 O produced NH 4 NO 3 used = 11.4 mol H 2 O produced How many moles of N2O produced? Answer = 5.68 mol N2O STEP 4 Convert moles product (11.4 mol) --> mass product This is called the THEORETICAL YIELD 11.4 mol H 2 O g Page III-4a-3 / Chapter Four Part I Lecture Notes = 204 g H 2 O ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS!

4 Page III-4a-4 / Chapter Four Part I Lecture Notes STEP 5 How much N 2 O is formed? Total mass of reactants = total mass of products 454 g NH 4 NO 3 = g N 2 O g H 2 O mass of N 2 O = 250. g law of mass action! could also turn mol NH4NO3 into mol N2O, then grams of N2O: 5.68 mol N2O * g/mol = 250. g Compound NH 4 NO 3 N 2 O H 2 O Initial (g) 454 g 0 0 Initial (mol) 5.68mol 0 0 Change (mol) (5.68) Final (mol) Final (g) Mass is conserved! STEP 6 Calculate the percent yield We predicted a yield of 250. g of N 2 O. If you isolated only 131 g of N 2 O, what is the percent yield of N 2 O? This compares the theoretical yield (250. g) and actual yield (131 g) of N 2 O. STEP 6 Calculate the percent yield % yield = % yield = 131 g 250. g actual yield theoretical yield 100% 100% = 52.4% GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS PROBLEM: Using 5.00 g of H 2 O 2, what mass of O 2 and of H 2 O can be obtained? 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Reaction is catalyzed by MnO 2 Molarity in next chapter - See Stoichiometry Guide Page III-4a-4 / Chapter Four Part I Lecture Notes

5 Page III-4a-5 / Chapter Four Part I Lecture Notes PROBLEM: Using 5.00 g of H 2 O 2, what mass of O 2 and of H 2 O can be obtained? 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Reaction is catalyzed by MnO 2 Step 1: moles of H 2 O 2 Step 2: use STOICHIOMETRIC FACTOR to calculate moles of O 2 Step 3: mass of O 2 (2.35 g) Step 4: mass of H 2 O (2.65 g) Try this problem yourself! Reactions Involving a LIMITING REACTANT 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 REACTANTS LIMITING REACTANTS Reactants Products 2 NO(g) + O 2 (g) 2 NO 2 (g) Limiting reactant = Excess reactant = LIMITING REACTANTS React solid Zn with mol HCl (aq) Zn (s) + 2 HCl (aq) ---> ZnCl 2(aq) + H 2(g) LIMITING REACTANTS React solid Zn with mol HCl (aq) Zn (s) + 2 HCl (aq) ---> ZnCl 2(aq) + H 2(g) 0.10 mol HCl [ Zn/2 mol HCl] = mol Zn Left: Balloon inflates fully, some Zn left * More than enough Zn to use up the mol HCl Center: Balloon inflates fully, no Zn left * Right amount of each (HCl and Zn) Right: Balloon does not inflate fully, no Zn left. * Not enough Zn to use up mol HCl Page III-4a-5 / Chapter Four Part I Lecture Notes Left Center Right mass Zn (g) mol Zn mol HCl mol HCl/mol Zn Lim Reactant LR = HCl no LR LR = Zn

6 Reaction to be Studied: Page III-4a-6 / Chapter Four Part I Lecture Notes Step 1 of the Limiting Reactant problem: Compare actual mole ratio of reactants to theoretical mole ratio. Reactants must be in the mole ratio PROBLEM: Mix 5.40 g of Al with 8.10 g of Cl 2. How many grams of Al 2 can form? mol Cl 2 mol Al = 3 2 Deciding on the Limiting Reactant Deciding on the Limiting Reactant If mol Cl 2 mol Al then there is not enough Al to use up all the Cl 2, and the limiting reagent is? Al > 3 2 If mol Cl 2 < 3 mol Al 2 then there is not enough Cl 2 to use up all the Al, and the limiting reagent is? Cl 2 Step 2 of the Limiting Reactant problem: Calculate moles of each reactant We have 5.40 g of Al and 8.10 g of Cl 2. How much Al 2 can form? 5.40 g Al 27.0 g = mol Al 8.10 g Cl g = mol Cl 2 Step 3 of the Limiting Reactant problem: Compare moles to find limiting reactant mol Cl 2 mol Al = mol mol = This should be 3/2 or 1.5/1 if reactants are present in the exact stoichiometric ratio. Limiting reagent is Cl 2 Page III-4a-6 / Chapter Four Part I Lecture Notes

7 Page III-4a-7 / Chapter Four Part I Lecture Notes Mix 5.40 g of Al with 8.10 g of Cl 2. What mass of Al 2 can form? moles Cl 2 Limiting reactant = Cl 2 Base all calculations on Cl 2 grams Cl 2 Al 2 3 mol Cl 2 grams Al 2 moles Al 2 CALCULATIONS: calculate mass of Al 2 expected using limiting reactant. Step 1: Calculate moles of Al 2 expected using chlorine: mol Cl 2 Al 2 3 mol Cl 2 = mol Al 2 Step 2: Calculate mass of Al 2 expected based on chlorine: mol Al g Al 2 mol = 10.1 g Al 2 Alternate Limiting Reactant Method Calculate theoretical yield of product based on both reactants. Smaller theoretical yield comes from limiting reactant, greater yield from excess reactant g Cl g Al Cl g 3 mol Cl g Al 27.0 g Al 2 2 mol Al g = 10.1 g Al 2 = 26.6 g Al g < 26.6 g, so: limiting reactant = Cl 2, theoretical yield = 10.1 g, excess reactant = Al Cl 2 was the limiting reactant. Therefore, Al was present in excess. But by how much? First find how much Al was required based on limiting reactant (Cl2). Then find how much Al is in excess. How much of which reactant will remain when reaction is complete? Calculating Excess Al products mol mol = LR 8.10 g Cl g 2 mol Al g = 2.05 g Al 3 mol Cl 2 Using Stoichiometry to Determine a Formula Hydrocarbons, C x H y, can be burned in oxygen to give CO 2 and H 2 O (combustion reaction). The CO 2 and H 2 O can be collected to determine the empirical formula of the hydrocarbon. C x H y + O 2 ---> CO 2 + H 2 O Excess Al = Al available - Al required = 5.40 g g = 3.35 g Al unused in reaction Page III-4a-7 / Chapter Four Part I Lecture Notes

8 Page III-4a-8 / Chapter Four Part I Lecture Notes Using Stoichiometry to Determine a Formula What is the empirical formula of a hydrocarbon, C x H y, if burning g produces g CO 2 and g H 2 O? C x H y + some O2 ---> g CO g H 2 O Puddle of C x H y g +O g CO 2 1 CO 2 molecule forms for each C atom in C x H y +O g H2 O 1 H 2 O molecule forms for each 2 H atoms in C x H y Using Stoichiometry to Determine a Formula C x H y + some oxygen ---> g CO g H 2 O First, recognize that all C in CO 2 and all H in H 2 O comes from C x H y. 1. Calculate amount of C in CO x 10-3 mol CO 2 --> 8.61 x 10-3 mol C C per CO2 2. Calculate amount of H in H 2 O x 10-3 mol H 2 O -- >1.149 x 10-2 mol H 2 mol H per water! Using Stoichiometry to Determine a Formula C x H y + some oxygen ---> g CO g H 2 O Now find ratio of mol H/mol C to find values of x and y in C x H y x 10-2 mol H/ 8.61 x 10-3 mol C = 1.33 mol H / 1.00 mol C = 4 mol H / 3 mol C Empirical formula = C 3 H 4 Formulas with C, H and O Caproic acid, the substance responsible for "dirty gym socks" smell, contains C, H and O. Combustion analysis of g caproic acid gives g H 2 O and g CO 2, and the molar mass was found to be g mol -1. What is the molecular formula of caproic acid? C x H y O z + some oxygen ---> g CO g H 2 O Careful: oxygen comes from caproic acid and O 2, need special technique Formulas with C, H and O Combustion analysis of g caproic acid gives g H 2 O and g CO 2, and the molar mass is g mol -1. What is the molecular formula? Start with "regular" approach for mol H & mol C: g H 2 O * (mol/18.02 g) * (2 mol H/mol H 2 O) = mol H mol H * (1.01 g/mol H) = g H g CO 2 * (mol/44.01 g) * ( C/mol CO 2 ) = mol C mol C * (12.01 g/mol C) = g C Why did we convert to grams? Law of Mass Action! Formulas with C, H and O g caproic acid: g H 2 O ( mol H, g H) and g CO 2 ( mol C, g C), molar mass = g/mol. What is the molecular formula? Realize that g of caproic acid equals all the g C, g H and g O in the complex. Converting mol H and mol C to grams, then subtracting from g, gives g O in caproic acid: g g g = g O caproic acid g of H in acid g of C in acid g of O in acid Page III-4a-8 / Chapter Four Part I Lecture Notes g O * (mol O / g) = mol O

9 You can now find empirical Formulas with formulas C, H based and on combustion O analysis (this chapter) and g caproic acid: g H 2 O ( mol H) and g CO elemental percentages 2 ( mol C), molar mass = g/mol, mol O. What is (previous chapter)! the molecular formula? Now compare moles: C H O gives C 3 H 6 O = empirical formula C 3 H 6 O has a molar mass of 58.1 g/mol, which is half of the g/mol value Molecular Formula = (C 3 H 6 O) 2, or C 6 H 12 O 2 Page III-4a-9 / Chapter Four Part I Lecture Notes End of Chapter 4 Part 1 See also: Chapter Four Part 1 Study Guide Chapter Four Part 1 Concept Guide Page III-4a-9 / Chapter Four Part I Lecture Notes