Chapter 3 Mass Relationships in Chemical Reactions 國防醫學院生化學科王明芳老師

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1 Chapter 3 Mass Relationships in Chemical Reactions 國防醫學院生化學科王明芳老師

2 Balancing Chemical Equations A balanced chemical equation shows that the law of conservation of mass is adhered to. In a balanced chemical equation, the numbers and kinds of atoms on both sides of the reaction arrow are identical. 2Na(s) + Cl 2 (g) 2NaCl(s) left side: right side: 2 Na 2 Cl 2 Na 2 Cl Chapter 3/2

3 Balancing Chemical Equations Hg(NO 3 ) 2 (aq) + 2KI(aq) left side: 1 Hg 2 N 6 O 2 K 2 I HgI 2 (s) + 2KNO 3 (aq) right side: 1 Hg 2 I 2 K 2 N 6 O Chapter 3/3

4 Balancing Chemical Equations 1. Write the unbalanced equation using the correct chemical formula for each reactant and product. H 2 (g) + O 2 (g) H 2 O(l) 2. Find suitable coefficients the numbers placed before formulas to indicate how many formula units of each substance are required to balance the equation. 2H 2 (g) + O 2 (g) 2H 2 O(l) Chapter 3/4

5 Balancing Chemical Equations 3. Reduce the coefficients to their smallest wholenumber values, if necessary, by dividing them all by a common divisor. 4H 2 (g) + 2O 2 (g) 4H 2 O(l) divide all by 2 2H 2 (g) + O 2 (g) 2H 2 O(l) Chapter 3/5

6 Balancing Chemical Equations 4. Check your answer by making sure that the numbers and kinds of atoms are the same on both sides of the equation. 2H 2 (g) + O 2 (g) left side: 2H 2 O(l) right side: 4 H 2 O 4 H 2 O Chapter 3/6

7 Example 3.1 Balancing a Chemical Equation Propane, C 3 H 8, is a colorless, odorless gas often used as a heating and cooking fuel in campers and rural homes. Write a balanced equation for the combustion reaction of propane with oxygen to yield carbon dioxide and water. Solution Step 1. Write the unbalanced equation using correct chemical formulas for all substances: C 3 H 8 + O 2 CO 2 + H 2 O Unbalanced Step 2. Find coefficients to balance the equation. It s usually best to begin with the most complex substance in this case C 3 H 8 and to deal with one element at a time. C 3 H 8 + O 2 3 CO 2 + H 2 O Balanced for C Next, look at the number of hydrogen atoms. C 3 H 8 + O 2 3 CO H 2 O Finally, look at the number of oxygen atoms. C 3 H O 2 3 CO H 2 O Balanced for C and H Balanced for C, H, and O

8 Example 3.2 Balancing a Chemical Equation The major ingredient in ordinary safety matches is potassium chlorate, KClO 3, a substance that can act as a source of oxygen in combustion reactions. Its reaction with ordinary table sugar (sucrose, C 12 H 22 O 11 ), for example, occurs violently to yield potassium chloride, carbon dioxide, and water. Write a balanced equation for the reaction. Solution Write the unbalanced equation, making sure the formulas for all substances are correct: KClO 3 + C 12 H 22 O 11 KCl + CO 2 + H 2 O Unbalanced Find coefficients to balance the equation by starting with the most complex substance (sucrose) and considering one element at a time. KClO 3 + C 12 H 22 O 11 KCl + 12 CO 2 + H 2 O KClO 3 + C 12 H 22 O 11 KCl + 12 CO H 2 O Balanced for C Balanced for C and H 8 KClO 3 + C 12 H 22 O 11 KCl + 12 CO H 2 O Balanced for C, H, and O 8 KClO 3 + C 12 H 22 O 11 8 KCl + 12 CO H 2 O Balanced for C, H, O, K, and Cl

9 Example 3.3 Balancing a Chemical Equation Write a balanced equation for the reaction of element A (red spheres) with element B (blue spheres) as represented below: Solution 3 A B 2 6 AB 3 or A B 2 2 AB 3

10 Representing Chemistry on Different Levels 2H 2 (g) + O 2 (g) 2H 2 O(l) microscopic: 2 molecules of hydrogen gas react with 1 molecule of oxygen gas to yield 2 molecules of liquid water. Chapter 3/10

11 Representing Chemistry on Different Levels 2H 2 (g) + O 2 (g) 2H 2 O(l) microscopic: macroscopic: 2 molecules of hydrogen gas react with 1 molecule of oxygen gas to yield 2 molecules of liquid water. 2 moles of hydrogen gas react with 1 mole of oxygen gas to yield 2 moles of liquid water. Chapter 3/11

12 Chemical Arithmetic: Stoichiometry Molecular Mass: Sum of atomic masses of all atoms in a molecule (also called molecular weights) Formula Mass: Sum of atomic masses of all atoms in a formula unit of any compound, molecular or ionic HCl: C 2 H 4 : 1.0 amu amu = 36.5 amu 2(12.0 amu) + 4(1.0 amu) = 28.0 amu Chapter 3/12

13 Chemical Arithmetic: Stoichiometry HCl: 1 mole = 36.5 g x molecules = 36.5 g C 2 H 4 : 1 mole = 28.0 g x molecules = 28.0 g Chapter 3/13

14 Chemical Arithmetic: Stoichiometry Stoichiometry: The chemical arithmetic needed for mole-mass conversions aa + bb cc + dd Grams of A Moles of A Moles of B Grams of B Molar Mass of A Mole Ratio Between A and B (Coefficients) Molar Mass of B Chapter 3/14

15 Chemical Arithmetic: Stoichiometry 2NaOH(aq) + Cl 2 (g) NaOCl(aq) + NaCl(aq) + H 2 O(l) How many grams of NaOH are needed to react with 25.0 g Cl 2? Grams of Cl 2 Moles of Cl 2 Moles of NaOH Grams of NaOH Molar Mass of Cl 2 Mole Ratio Between Cl 2 and NaOH Molar Mass of NaOH Chapter 3/15

16 Chemical Arithmetic: Stoichiometry 2NaOH(aq) + Cl 2 (g) NaOCl(aq) + NaCl(aq) + H 2 O(l) How many grams of NaOH are needed to react with 25.0 g Cl 2? 25.0 g Cl 2 x 1 mol Cl g Cl 2 x 2 mol NaOH 1 mol Cl 2 x 40.0 g NaOH 1 mol NaOH = 28.2 g NaOH Chapter 3/16

17 Example 3.4 Calculating a Molecular Mass What is the molecular mass of table sugar (sucrose, C 12 H 22 O 11 ), and what is its molar mass in g/mol? Solution The molecular mass of a substance is the sum of the atomic masses of the constituent atoms. C (12.0 amu); H (1.0 amu); O (16.0 amu) Because one molecule of sucrose has a mass of amu, 1 mol of sucrose has a mass of g. Thus, the molar mass of sucrose is g/mol.

18 Example 3.5 Converting Mass To Moles How many moles of sucrose are in a tablespoon of sugar containing 2.85 g? (the molar mass of sucrose, C 12 H 22 O 11, was calculated in Example 3.4) Solution mass-to-mole conversion

19 Example 3.6 Converting Moles To Mass How many grams are in mol of NaHCO 3? Solution mole-to-mass conversion Formula mass of NaHCO 3 = 23.0 amu amu amu + ( amu) = 84.0 amu Molar mass of NaHCO 3 = 84.0 g/mol

20 Example 3.7 Finding the Mass of One Reactant, Given the Mass of Another Aqueous solutions of sodium hypochlorite (NaOCl), best known as household bleach, are prepared by reaction of sodium hydroxide with chlorine. How many grams of NaOH are needed to react with 25.0 g of Cl 2? 2 NaOH(aq) + Cl 2 (g) NaOCl(aq) + NaCl(aq) + H 2 O(l) Solution Finding the relationship between numbers of reactant formula units. First, find out how many moles of Cl 2 are in 25.0 g of Cl 2. Next, look at the coefficients in the balanced equation carry out a mole-to-gram conversion combining the steps

21 Yields of Chemical Reactions Actual Yield: The amount actually formed in a reaction Theoretical Yield: The amount predicted by calculations Actual yield of product Percent Yield = Theoretical yield of product x 100% Chapter 3/21

22 Example 3.8 Calculating a Percent Yield Methyl tert-butyl ether (MTBE, C 5 H 12 O), a gasoline additive now being phased out in many places because of health concerns, can be made by reaction of isobutylene (C 4 H 8 ) with methanol (CH 4 O). What is the percent yield of the reaction if 32.8 g of methyl tert-butyl ether is obtained from reaction of 26.3 g of isobutylene with sufficient methanol? Solution Isobutylene, C 4 H 8 : Molecular mass = ( amu) + (8 1.0 amu) = 56.0 amu MTBE, C 5 H 12 O: Molecular mass = ( amu) + ( amu) amu = 88.0 amu To calculate the amount of MTBE that could theoretically be produced from 26.3 g of isobutylene, According to the balanced equation, 1 mol of product is produced per mol of reactant a mole-to-mass conversion Dividing the actual amount by the theoretical amount and multiplying by 100% gives the percent yield:

23 Example 3.9 Calculating a Yield In Grams, Given a Percent Yield Diethyl ether (C 4 H 10 O), the ether used medically as an anesthetic, is prepared commercially by treatment of ethyl alcohol (C 2 H 6 O) with an acid. How many grams of diethyl ether would you obtain from 40.0 g of ethyl alcohol if the percent yield of the reaction is 87%? Solution First, calculate the molar masses of the reactant and product: Ethyl alcohol, C 2 H 6 O: Molecular mass = ( amu) + (6 1.0 amu) amu = 46.0 amu Diethyl ether, C 4 H 10 O: Molecular mass = ( amu) + ( amu) amu = 74.0 amu Next, find how many moles of ethyl alcohol are in 40.0 g Finally, we have to multiply the theoretical amount of product by the observed yield (87% = 0.87 ) to find how much diethyl ether is actually formed: 32.2 g diethyl ether 0.87 = 28 g diethyl ether

24 Reactions with Limiting Amounts of Reactants Limiting Reactant: The reactant that is present in limiting amount. The extent to which a chemical reaction takes place depends on the limiting reactant. Excess Reactant: Any of the other reactants still present after determination of the limiting reactant Chapter 3/24

25 Reactions with Limiting Amounts of Reactants At a high temperature, ethylene oxide reacts with water to form ethylene glycol which is an automobile antifreeze and a starting material in the preparation of polyester polymers: C 2 H 4 O(aq) + H 2 O(l) C 2 H 6 O 2 (l) Because water is so cheap and abundant, it is used in excess when compared to ethylene oxide. This ensures that all of the relatively expensive ethylene oxide is entirely consumed. Chapter 3/25

26 Reactions with Limiting Amounts of Reactants At a high temperature, ethylene oxide reacts with water to form ethylene glycol which is an automobile antifreeze and a starting material in the preparation of polyester polymers: C 2 H 4 O(aq) + H 2 O(l) C 2 H 6 O 2 (l) If 3 moles of ethylene oxide react with 5 moles of water, which reactant is limiting and which reactant is present in excess? Chapter 3/26

27 Reactions with Limiting Amounts of Reactants At a high temperature, ethylene oxide reacts with water to form ethylene glycol which is an automobile antifreeze and a starting material in the preparation of polyester polymers: C 2 H 4 O(aq) + H 2 O(l) C 2 H 6 O 2 (l) Chapter 3/27

28 Reactions with Limiting Amounts of Reactants Lithium oxide is used aboard the space shuttle to remove water from the air supply according to the equation: Li 2 O(s) + H 2 O(g) 2LiOH(s) If 80.0 g of water are to be removed and 65.0 g of Li 2 O are available, which reactant is limiting? How many grams of excess reactant remain? How many grams of LiOH are produced? Chapter 3/28

29 Reactions with Limiting Amounts of Reactants Li 2 O(s) + H 2 O(g) 2LiOH(s) Which reactant is limiting? Amount of H 2 O that will react with 65.0 g Li 2 O: 65.0 g Li 2 O x 1 mol Li 2 O 29.9 g Li 2 O x 1 mol H 2 O 1 mol Li 2 O = 2.17 moles H 2 O Amount of H 2 O given: 1 mol H 2 O 80.0 g H 2 O x = 4.44 moles H 2 O 18.0 g H 2 O Li 2 O is limiting Chapter 3/29

30 Reactions with Limiting Amounts of Reactants Li 2 O(s) + H 2 O(g) 2LiOH(s) How many grams of excess H 2 O remain? 2.17 mol H 2 O x 18.0 g H 2 O = 39.1 g H2 O (consumed) 1 mol H 2 O 80.0 g H 2 O g H 2 O = 40.9 g H 2 O initial consumed remaining Chapter 3/30

31 Reactions with Limiting Amounts of Reactants Li 2 O(s) + H 2 O(g) 2LiOH(s) How many grams of LiOH are produced? 2.17 mol H 2 O 2 mol LiOH x x 1 mol H 2 O 23.9 g LiOH = 104 g LiOH 1 mol LiOH Chapter 3/31

32 Example 3.10 Calculating the Amount of an Excess Reactant Cisplatin, an anticancer agent used for the treatment of solid tumors, is prepared by the reaction of ammonia with potassium tetrachloroplatinate. Assume that 10.0 g of K 2 PtCl 4 and 10.0 g of NH 3 are allowed to react. (a) Which reactant is limiting, and which is in excess? (b) How many grams of the excess reactant are consumed, and how many grams remain? (c) How many grams of cisplatin are formed?

33 Worked Example 3.10 Calculating the Amount of an Excess Reactant Solution (a) Finding the molar amounts of reactants always begins by calculating formula masses and using molar masses as conversion factors: Thus, a large excess of NH 3 is present, and K 2 PtCl 4 is the limiting reactant.

34 Worked Example 3.10 Calculating the Amount of an Excess Reactant (b) (c)

35 Concentrations of Reactants in Solution: Molarity Molarity: The number of moles of a substance dissolved in each liter of solution Solution: A homogeneous mixture Solute: The dissolved substance in a solution Solvent: The major component in a solution Chapter 3/35

36 Concentrations of Reactants in Solution: Molarity Chapter 3/36

37 Concentrations of Reactants in Solution: Molarity Molarity converts between mole of solute and liters of solution: Moles of solute Molarity = Volume of solution (L) 1.00 mol of sodium chloride placed in enough water to make 1.00 L of solution would have a concentration equal to: 1.00 mol = L mol or 1.00 M L Chapter 3/37

38 Concentrations of Reactants in Solution: Molarity How many grams of solute would you use to prepare 1.50 L of M glucose, C 6 H 12 O 6? Molar mass C 6 H 12 O 6 = g/mol 1.50 L x mol = mol 1 L mol x g = 49.5 g 1 mol Chapter 3/38

39 Example 3.11 Calculating the Molarity of a Solution What is the molarity of a solution made by dissolving g of sulfuric acid (H 2 SO 4 ) in water and diluting to a final volume of 50.0 ml? Solution The solution has a sulfuric acid concentration of M.

40 Example 3.12 Calculating the Number of Moles of Solute in a Solution Hydrochloric acid is sold commercially as a 12.0 M aqueous solution. How many moles of HCl are in ml of 12.0 M solution?. Solution There are 3.60 mol of HCl in ml of 12.0 M solution

41 Diluting Concentrated Solutions concentrated solution + solvent dilute solution initial final M i V i = M f V f Since the number of moles of solute remains constant, all that changes is the volume of solution by adding more solvent. Chapter 3/41

42 Diluting Concentrated Solutions Chapter 3/42

43 Diluting Concentrated Solutions Sulfuric acid is normally purchased at a concentration of 18.0 M. How would you prepare ml of M aqueous H 2 SO 4? M i = 18.0 M V i =? ml M f = M V f = ml V i = M f V f M i = M 18.0 M x ml = 6.94 ml Add 6.94 ml 18.0 M sulfuric acid to enough water to make ml of M solution. Chapter 3/43

44 Example 3.13 Diluting a Solution How would you prepare ml of M NaOH solution starting from a concentration of M? Solution We need to place ml of M NaOH solution in a ml volumetric flask and fill to the calibration mark with water.

45 Solution Stoichiometry aa + bb cc + dd Volume of Solution of A Moles of A Moles of B Volume of Solution of B Molarity of A Mole Ratio Between A and B (Coefficients) Molarity of B Chapter 3/45

46 Solution Stoichiometry What volume of M H 2 SO 4 is needed to react with 50.0 ml of M NaOH? H 2 SO 4 (aq) + 2NaOH(aq) Na 2 SO 4 (aq) + 2H 2 O(l) Volume of Solution of H 2 SO 4 Moles of H 2 SO 4 Moles of NaOH Volume of Solution of NaOH Molarity of H 2 SO 4 Mole Ratio Between H 2 SO 4 and NaOH Molarity of NaOH Chapter 3/46

47 Solution Stoichiometry H 2 SO 4 (aq) + 2NaOH(aq) Na 2 SO 4 (aq) + 2H 2 O(l) Moles of NaOH available: 50.0 ml NaOH x mol 1 L x 1 L 1000 ml = mol NaOH Volume of H 2 SO 4 needed: 1 mol H 2 SO 4 1 L solution 1000 ml mol NaOH x x x 2 mol NaOH mol H 2 SO 4 1 L 10.0 ml solution (0.250 M H 2 SO 4 ) Chapter 3/47

48 Example 3.14 Reaction Stoichiometry in Solution Stomach acid, a dilute solution of HCl in water, can be neutralized by reaction with sodium hydrogen carbonate, NaHCO 3, according to the equation Solution HCl(aq) + NaHCO 3 (aq) NaCl(aq) + H 2 O(l) + CO 2 (g) How many milliliters of M NaHCO 3 solution are needed to neutralize 18.0 ml of M HCl? Thus, 14.4 ml of the M NaHCO 3 solution is needed to neutralize 18.0 ml of the M HCl solution.

49 Titration Titration: A procedure for determining the concentration of a solution by allowing a measured volume of that solution to react with a second solution of another substance (the standard solution) whose concentration is known. HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) Once the reaction is complete you can calculate the concentration of the unknown solution. How can you tell when the reaction is complete? Chapter 3/49

50 Titration Chapter 3/50

51 Titration 48.6 ml of a M NaOH solution is needed to react with 20.0 ml of an unknown HCl concentration. What is the concentration of the HCl solution? HCl(aq) + NaOH(aq) NaCl(aq) + 2H 2 O(l) Volume of Solution of NaOH Moles of NaOH Moles of HCl Volume of Solution of HCl Molarity of NaOH Mole Ratio Between NaOH and HCl Molarity of HCl Chapter 3/51

52 Titration HCl(aq) + NaOH(aq) NaCl(aq) + 2H 2 O(l) Moles of NaOH available: mol 1 L 48.6 ml NaOH x x = mol NaOH 1 L 1000 ml Moles of HCl reacted: 1 mol HCl mol NaOH x = mol HCl 1 mol NaOH Concentration of HCl solution: mol HCl 20.0 ml solution 1000 ml x 1 L = M HCl Chapter 3/52

53 Percent Composition and Empirical Formulas Percent Composition: Expressed by identifying the elements present and giving the mass percent of each Empirical Formula: It tells the smallest whole-number ratios of the atoms in the compound. Molecular Formula: It tells the actual numbers of atoms in a compound. It can be either the empirical formula or a multiple of it. Multiple = Molecular mass Empirical formula mass Chapter 3/53

54 Percent Composition and Empirical Formulas A colorless liquid has a composition of 84.1 % carbon and 15.9 % hydrogen by mass. Determine the empirical formula. Also, assuming the molar mass of this compound is g/mol, determine the molecular formula of this compound. Mass percents Moles Mole ratios Subscripts Molar masses Relative mole ratios Chapter 3/54

55 Percent Composition and Empirical Formulas Assume g of the substance: Mole of carbon: 1 mol C 84.1 g C x = 7.01 mol C 12.0 g C Mole of hydrogen: 15.9 g H x 1 mol H = 15.9 mol H 1.0 g H Chapter 3/55

56 Percent Composition and Empirical Formulas Empirical formula: C 7.01 H 15.9 C 7.01 H 15.9 = C 1 H 2.27 smallest value for the ratio C 1 H 2.27 C 1x4 H 2.27x4 = C 4 H 9 need whole numbers Molecular formula: multiple = = 2 C 4x2 H 9x2 = C 8 H Chapter 3/56

57 Example 3.15 Calculating an Empirical Formula from a Percent Composition Vitamin C (ascorbic acid) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula of ascorbic acid? Solution 1. Assume that you have g of ascorbic acid. 2. Dividing each of the three numbers by the smallest (3.41 mol) gives a C:H:O mole ratio of 1:1.33:1 and a temporary formula of C 1 H 1.33 O Multiplying the subscripts by small integers in a trial-and-error procedure until whole numbers are found then gives the empirical formula: C (3 1) H (3 1.33) O (3 1) = C 3 H 4 O 3

58 Example 3.16 Calculating a Percent Composition from a Formula Glucose, or blood sugar, has the molecular formula C 6 H 12 O 6. What is the empirical formula, and what is the percent composition of glucose? Solution 1. dividing the subscripts by 6 reduces C 6 H 12 O 6 to CH 2 O. 2. converted into a mass ratio by assuming that we have 1 mol of compound and carrying out mole-togram conversions: 3. Dividing the mass of each element by the total mass, and multiplying by 100%, gives the percent composition. Note that the sum of the mass percentages is 100%. 4. Total mass of 1 mol glucose = 72.0 g g g = g

59 Determining Empirical Formulas: Elemental Analysis Combustion Analysis: A compound of unknown composition is burned with oxygen to produce the volatile combustion products CO 2 and H 2 O, which are separated and have their amounts determined by an automated instrument. hydrocarbon + O 2 (g) xco 2 (g) + yh 2 O(g) carbon hydrogen Chapter 3/59

60 Worked Example 3.17 Calculating an Empirical Formula and a Molecular Formula from a Combustion Analysis Caproic acid, the substance responsible for the aroma of goats, dirty socks, and oldshoes, contains carbon, hydrogen, and oxygen. On combustion analysis, a g sample of caproic acid gives g of H 2 O and g of CO 2. What is the empirical formula of caproic acid? If the molecular mass of caproic acid is amu, what is the molecular formula? Solution 1. First, find the molar amounts of C and H in the sample: 2. Next, find the number of grams of C and H in the sample: 3. Subtracting the masses of C and H from the mass of the starting sample indicates that g is unaccounted for: g ( g g) = g

61 Worked Example 3.17 Calculating an Empirical Formula and a Molecular Formula from a Combustion Analysis Continued 5. Because we are told that oxygen is also present in the sample, the missing mass must be due to oxygen, which can t be detected by combustion. We therefore need to find the number of moles of oxygen in the sample: 6. Knowing the relative numbers of moles of all three elements, C, H, and O, we divide the three numbers of moles by the smallest number ( mol of oxygen) to arrive at a C:H:O ratio of 3:6:1. 7. The empirical formula of caproic acid is therefore C 3 H 6 O, and the empirical formula mass is 58.1 amu. Because the molecular mass of caproic acid is 116.2, or twice the empirical formula mass, the molecular formula of caproic acid must be C (2 3) H (2 6) O (2 1) = C 6 H 12 O 2

62 Determining Molecular Masses: Mass Spectrometry Chapter 3/62