Chemical Equations & Reaction Stoichiometry

Transcription

1 Chemical Equations & Reaction Mr. Matthew Totaro Legacy High School AP Chemistry

2 Law of Conservation of Mass We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends. --Antoine Lavoisier, 1789

3 Chemical Equations Concise representations of chemical reactions

4 Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

5 Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) Reactants appear on the left side of the equation.

6 Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) Products appear on the right side of the equation.

7 Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) The states of the reactants and products are written in parentheses to the right of each compound.

8 Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) Coefficients are inserted to balance the equation.

9 Subscripts and Coefficients Give Different Information Subscripts tell the number of atoms of each element in a molecule

10 Subscripts and Coefficients Give Different Information Subscripts tell the number of atoms of each element in a molecule Coefficients tell the number of molecules

11 Symbols Used in Equations Symbols used to indicate state after chemical (g) = gas; (l) = liquid; (s) = solid (aq) = aqueous = dissolved in water Energy symbols used above the arrow for decomposition reactions = heat hν = light shock = mechanical elec = electrical

12 Reaction Types

13 Synthesis Reactions Two or more substances react to form one product Examples: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) C 3 H 6 (g) + Br 2 (l) C 3 H 6 Br 2 (l) 2 Mg (s) + O 2 (g) 2 MgO (s)

14 2 Mg (s) + O 2 (g) 2 MgO (s)

15 Decomposition Reactions One substance breaks down into two or more substances Examples: CaCO 3 (s) CaO (s) + CO 2 (g) 2 KClO 3 (s) 2 KCl (s) + O 2 (g) 2 NaN 3 (s) 2 Na (s) + 3 N 2 (g)

16 Single Replacement Reactions one element replaces another in a compound metal replaces metal (+) nonmetal replaces nonmetal (-) A + BC B + AC

17 Single Replacement Reactions Cu(s) + 2AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2Ag(s)

18 Double Replacement Reactions ions in two compounds change partners cation of one compound combines with anion of the other AB + CD AD + CB

19 Double Replacement Reactions Pb(NO 3 ) 2 (aq) + K 2 CrO 4 (aq) PbCrO 4 (s) + 2KNO 3 (aq)

20 Combustion Reactions Rapid reactions that produce a flame Most often involve hydrocarbons reacting with oxygen in the air Examples: CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g)

21 Quantities in Chemical Reactions The amount of every substance used and made in a chemical reaction is related to the amounts of all the other substances in the reaction Law of Conservation of Mass Balancing equations by balancing atoms The study of the numerical relationship between chemical quantities in a chemical reaction is called reaction stoichiometry

22 Reaction The coefficients in a balanced chemical equation specify the relative amounts in moles of each of the substances involved in the reaction 2 C 8 H 18 (l) + 25 O 2 (g) 16 CO 2 (g) + 18 H 2 O(g) 2 molecules of C 8 H 18 react with 25 molecules of O 2 to form 16 molecules of CO 2 and 18 molecules of H 2 O 2 moles of C 8 H 18 react with 25 moles of O 2 to form 16 moles of CO 2 and 18 moles of H 2 O 2 mol C 8 H 18 : 25 mol O 2 : 16 mol CO 2 : 18 mol H 2 O

23 Making Pizza The number of pizzas you can make depends on the amount of the ingredients you use 1 crust + 5 oz. tomato sauce + 2 cu cheese 1 pizza This relationship can be expressed mathematically 1 crust : 5 oz. sauce : 2 cu cheese : 1 pizza If you want to make more or less than one pizza, you can use the amount of cheese you have to determine the number of pizzas you can make assuming you have enough crusts and tomato sauce

24 Predicting Amounts from The amounts of any other substance in a chemical reaction can be determined from the amount of just one substance How much CO 2 can be made from 22.0 moles of C 8 H 18 in the combustion of C 8 H 18? 2 C 8 H 18 (l) + 25 O 2 (g) 16 CO 2 (g) + 18 H 2 O(g) 2 moles C 8 H 18 : 16 moles CO 2

25 Reaction Road Map

26 Practice According to the following equation, how many moles of water are made in the combustion of 0.10 moles of glucose? C 6 H 12 O O 2 6 CO H 2 O

27 Practice How many moles of water are made in the combustion of 0.10 moles of glucose? Given: Find: 0.10 moles C 6 H 12 O 6 moles H 2 O Conceptual Plan: mol C 6 H 12 O 6 mol H 2 O Relationships: C 6 H 12 O O 2 6 CO H 2 O 1 mol C 6 H 12 O 6 : 6 mol H 2 O Solution: Check: 0.6 mol H 2 O = 0.60 mol H 2 O because 6x moles of H 2 O as C 6 H 12 O 6, the number makes sense Tro: Chemistry: A Molecular Approach, 2/e

28 Example: Estimate the mass of CO 2 produced in 2007 by the combustion of 3.5 x g gasolne Assuming that gasoline is octane, C 8 H 18, the equation for the reaction is 2 C 8 H 18 (l) + 25 O 2 (g) 16 CO 2 (g) + 18 H 2 O(g) The equation for the reaction gives the mole relationship between amount of C 8 H 18 and CO 2, but we need to know the mass relationship, so the conceptual plan will be g C 8 H 18 mol C 8 H 18 mol CO 2 g CO 2

29 Example: Estimate the mass of CO 2 produced in 2007 by the combustion of 3.5 x g gasoline Given: Find: Conceptual Plan: 3.4 x g C 8 H 18 g CO 2 g C 8 H 18 mol C 8 H 18 mol CO 2 g CO 2 Relationships: Solution: 1 mol C 8 H 18 = g, 1 mol CO 2 = 44.01g, 2 mol C 8 H 18 :16 mol CO 2 Check: because 8x moles of CO 2 as C 8 H 18, but the molar mass of C 8 H 18 is 3x CO 2, the number makes sense

30 Which Produces More CO 2 ; Volcanoes or Fossil Fuel Combustion? Our calculation just showed that the world produced 1.1 x g of CO 2 just from petroleum combustion in x kg CO 2 Estimates of volcanic CO 2 production are 2 x kg/year This means that volcanoes produce less than 2% of the CO 2 added to the air annually 11 kg yr 13 kg yr 100% = 1. 8%

31 Example: How many grams of glucose can be synthesized from 37.8 g of CO 2 in photosynthesis? Given: Find: Conceptual Plan: Relationships: 37.8 g CO 2, 6 CO H 2 O C 6 H 12 O O 2 g C 6 H 12 O 6 g CO 2 mol CO 2 mol C 6 H 12 O 6 1 mol g 1 mol C 6 H 12 O 6 mol CO g 1 mol g C 6 H 12 O 6 1 mol C 6 H 12 O 6 = 180.2g, 1 mol CO 2 = 44.01g, 1 mol C 6 H 12 O 6 : 6 mol CO 2 Solution: 1 mol CO g CO g CO = 25.8 g C H O Check: mol C 6 H 12 O 6 mol CO because 6x moles of CO 2 as C 6 H 12 O 6, but the molar mass of C 6 H 12 O 6 is 4x CO 2, the number makes sense g C 6 H 12 O 1 mol C H O

32 Practice How many grams of O 2 can be made from the decomposition of g of PbO 2? 2 PbO 2 (s) 2 PbO(s) + O 2 (g) (PbO 2 = 239.2, O 2 = 32.00)

33 Practice How many grams of O 2 can be made from the decomposition of g of PbO 2? 2 PbO 2 (s) 2 PbO(s) + O 2 (g) Given: Find: Conceptual Plan: g PbO 2, 2 PbO 2 2 PbO + O 2 g O 2 g PbO 2 mol PbO 2 mol O 2 g O 2 Relationships: 1 mol O 2 = 32.00g, 1 mol PbO 2 = 239.2g, 1 mol O 2 : 2 mol PbO 2 Solution: Check: because ½ moles of O 2 as PbO 2, and the molar mass of PbO 2 is 7x O 2, the number makes sense

34 More Making Pizzas We know that 1 crust + 5 oz. tomato sauce + 2 cu cheese 1 pizza But what would happen if we had 4 crusts, 15 oz. tomato sauce, and 10 cu cheese?

35 More Making Pizzas, Continued Each ingredient could potentially make a different number of pizzas But all the ingredients have to work together! We only have enough tomato sauce to make three pizzas, so once we make three pizzas, the tomato sauce runs out no matter how much of the other ingredients we have.

36 More Making Pizzas, Continued The tomato sauce limits the amount of pizzas we can make. In chemical reactions we call this the limiting reactant. also known as the limiting reagent The maximum number of pizzas we can make depends on this ingredient. In chemical reactions, we call this the theoretical yield. it also determines the amounts of the other ingredients we will use!

37 The Limiting Reactant For reactions with multiple reactants, it is likely that one of the reactants will be completely used before the others When this reactant is used up, the reaction stops and no more product is made The reactant that limits the amount of product is called the limiting reactant sometimes called the limiting reagent the limiting reactant gets completely consumed Reactants not completely consumed are called excess reactants The amount of product that can be made from the limiting reactant is called the theoretical yield

38 Limiting and Excess Reactants in the Combustion of Methane CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) Our balanced equation for the combustion of methane implies that every one molecule of CH 4 reacts with two molecules of O 2

39 Limiting and Excess Reactants in the Combustion of Methane CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) If we have five molecules of CH 4 and eight molecules of O 2, which is the limiting reactant? since less CO 2 can be made from the O 2 than the CH 4, so the O 2 is the limiting reactant

40 Practice How many moles of Si 3 N 4 can be made from 1.20 moles of Si and 1.00 moles of N 2 in the reaction 3 Si + 2 N 2 Si 3 N 4?

41 Practice How many moles of Si 3 N 4 can be made from 1.20 moles of Si and 1.00 moles of N 2 in the reaction 3 Si + 2 N 2 Si 3 N 4? Given: Find: 1.20 mol Si, 1.00 mol N 2 mol Si 3 N 4 Conceptual Plan: Relationships: mol Si mol Si 3 N 4 Pick least amount mol N 2 mol Si 3 N 4 Limiting reactant and theoretical yield 2 mol N 2 : 1 Si 3 N 4 ; 3 mol Si : 1 Si 3 N 4 Solution: Limiting reactant Theoretical yield

42 More Making Pizzas Let s now assume that as we are making pizzas, we burn a pizza, drop one on the floor, or other uncontrollable events happen so that we only make two pizzas. The actual amount of product made in a chemical reaction is called the actual yield. We can determine the efficiency of making pizzas by calculating the percentage of the maximum number of pizzas we actually make. In chemical reactions, we call this the percent yield.

43 Theoretical and Actual Yield As we did with the pizzas, in order to determine the theoretical yield, we should use reaction stoichiometry to determine the amount of product each of our reactants could make The theoretical yield will always be the least possible amount of product the theoretical yield will always come from the limiting reactant Because of both controllable and uncontrollable factors, the actual yield of product will always be less than the theoretical yield

44 Example 4.4: Finding limiting reactant, theoretical yield, and percent yield

45 Example: When 28.6 g of C are allowed to react with 88.2 g of TiO 2 in the reaction below, 42.8 g of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield. TiO 2 (s) + 2 C(s) Ti(s) + 2 CO(g)

46 Example: When 28.6 kg of C reacts with 88.2 kg of TiO 2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield TiO 2 (s) + 2 C(s) Ti(s) + 2 CO(g) Write down the given quantity and its units Given: 28.6 kg C 88.2 kg TiO kg Ti produced

47 Example: Find the limiting reactant, theoretical yield, and percent yield TiO 2 (s) + 2 C(s) Ti(s) + 2 CO(g) Information Given: 28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti Write down the quantity to find and/or its units Find: limiting reactant theoretical yield percent yield

48 Example: Find the limiting reactant, theoretical yield, and percent yield TiO 2 (s) + 2 C(s) Ti(s) + 2 CO(g) Information Given: 28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. Write a conceptual plan kg C kg TiO 2 } smallest amount is from limiting reactant smallest mol Ti

49 Example: Find the limiting reactant, theoretical yield, and percent yield TiO 2 (s) + 2 C(s) Ti(s) + 2 CO(g) Information Given: 28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct g rct mol rct mol Ti pick smallest mol Ti TY kg Ti %Y Ti Collect needed relationships 1000 g = 1 kg Molar Mass TiO 2 = g/mol Molar Mass Ti = g/mol Molar Mass C = g/mol 1 mole TiO 2 : 1 mol Ti (from the chem. equation) 2 mole C : 1 mol Ti (from the chem. equation)

50 Combustion Analysis Mr. Matthew Totaro Legacy High School AP Chemistry

51 Combustion Analysis A common technique for analyzing compounds is to burn a known mass of compound and weigh the amounts of product made generally used for organic compounds containing C, H, O By knowing the mass of the product and composition of constituent element in the product, the original amount of constituent element can be determined all the original C forms CO 2, the original H forms H 2 O, the original mass of O is found by subtraction Once the masses of all the constituent elements in the original compound have been determined, the empirical formula can be found

52 Combustion Analysis

53 Example of a Combustion Analysis Combustion of a g sample of a compound containing only carbon, hydrogen, and oxygen produced the following: CO 2 = g H 2 O = g Determine the empirical formula of the compound

54 Example: Find the empirical formula of compound with the given amounts of combustion products Write down the given quantity and its units Given: compound = g CO 2 = g H 2 O = g

55 Example: Find the empirical formula of compound with the given amounts of combustion products Information Given: g compound, g CO 2, g H Write down the quantity to find and/or its units Find: empirical formula, C x H y O z

56 Example: Find the empirical formula of compound with the given amounts of combustion products Information Given: g compound, g CO 2, g H Find: empirical formula, C x H y O z Write a conceptual plan g CO 2, H 2 O mol CO 2, H 2 O mol C, H g C, H g O mol O mol C, H, O pseudo formula mol ratio empirical formula

57 Example: Find the empirical formula of compound with the given amounts of combustion products Information Given: g compound, g CO 2, g H 2 O Find: empirical formula, C x H y O z CP: g CO 2 & H 2 O mol CO 2 & H 2 O mol C & H g C & H g O mol O mol ratio empirical formula Collect needed relationships 1 mole CO 2 = g CO 2 1 mole H 2 O = g H 2 O 1 mole C = g C 1 mole H = g H 1 mole O = g O 1 mole CO 2 = 1 mole C 1 mole H 2 O = 2 mole H

58 Example: Find the empirical formula of compound with the given amounts of combustion products Apply the conceptual plan calculate the moles of C and H Information Given: g compound, g CO 2, g H 2 O Find: empirical formula, C x H y O z CP: g CO 2 & H 2 O mol CO 2 & H 2 O mol C & H g C & H g O mol O mol ratio empirical formula Rel: MM of CO 2, H 2 O, C, H, O; mol element : 1 mol compound

59 Example: Find the empirical formula of compound with the given amounts of combustion products Apply the conceptual plan calculate the grams of C and H Information Given: g compound, g CO 2, g H 2 O Find: empirical formula, C x H y O z CP: g CO 2 & H 2 O mol CO 2 & H 2 O mol C & H g C & H g O mol O mol ratio empirical formula Rel: MM of CO 2, H 2 O, C, H, O; mol element : 1 mol compound

60 Example: Find the empirical formula of compound with the given amounts of combustion products Apply the conceptual plan Information Given: g compound, g CO 2, g H 2 O Find: empirical formula, C x H y O z CP: g CO 2 & H 2 O mol CO 2 & H 2 O mol C & H g C & H g O mol O mol ratio empirical formula Rel: MM of CO 2, H 2 O, C, H, O; mol element : 1 mol compound calculate the grams and moles of O

61 Example: Find the empirical formula of compound with the given amounts of combustion products Information Given: g compound, g CO 2, g H 2 O mol C, g C, mol H, g H, g O, mol O Find: empirical formula, C x H y O z CP: g CO 2 & H 2 O mol CO 2 & H 2 O mol C & H g C & H g O mol O mol ratio empirical formula Rel: MM of CO 2, H 2 O, C, H, O; mol element : 1 mol compound Apply the conceptual plan write a pseudoformula C H O

62 Example: Find the empirical formula of compound with the given amounts of combustion products Information Given: g compound, g CO 2, g H 2 O mol C, g C, mol H, g H, g O, mol O Find: empirical formula, C x H y O z CP: g CO 2 & H 2 O mol CO 2 & H 2 O mol C & H g C & H g O mol O mol ratio empirical formula Rel: MM of CO 2, H 2 O, C, H, O; mol element : 1 mol compound Apply the conceptual plan find the mole ratio by dividing by the smallest number of moles

63 Example: Find the empirical formula of compound with the given amounts of combustion products Information Given: g compound, g CO 2, g H 2 O mol C, g C, mol H, g H, g O, mol O Find: empirical formula, C x H y O z CP: g CO 2 & H 2 O mol CO 2 & H 2 O mol C & H g C & H g O mol O mol ratio empirical formula Rel: MM of CO 2, H 2 O, C, H, O; mol element : 1 mol compound Apply the conceptual plan multiply subscripts by factor to give whole number, if necessary write the empirical formula C 10 H 12 O 1

64 Practice The smell of dirty gym socks is caused by the compound caproic acid. Combustion of g of caproic acid produced g of H 2 O and 1.92 g of CO 2. If the molar mass of caproic acid is g/mol, what is the molecular formula of caproic acid? (MM C = 12.01, H = 1.008, O = 16.00)

65 Practice Combustion of g of caproic acid produced g of H 2 O and 1.92 g of CO 2. If the molar mass of caproic acid is g/mol, what is the molecular formula of caproic acid?

66 C H O g moles C H O

67 Molecular formula = {C 3 H 6 O} x 2 = C 6 H 12 O 2