1. 1 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Stoichiometry: Chemical Calculations Chapter Three

2. 2 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Molecular mass: sum of the masses of the atoms represented in a molecular formula. Simply put: the mass of a molecule. Molecular mass is specifically for molecules. Ionic compounds don’t exist as molecules; for them we use … Formula mass: sum of the masses of the atoms or ions present in a formula unit. Molecular Masses and Formula Masses

3. 3 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Example: water - H 2 O Molecular Mass Molecular mass is the sum of the masses of the atoms represented in a molecular formula. 2 Hydrogen atoms 2(1.0079 u) 1 Oxygen atom + 15.9994 u EOS = 18.0152 u

4. 4 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Formula Mass Formula mass is the sum of the masses of the atoms or ions present in a formula unit. Na + Cl - Na + Cl - Na + Crystal of sodium chloride One Na + and one Cl – make a formula unit for sodium chloride The mass of one formula unit is: = 22.9898 u + 35.4527 u = 58.4425 u EOS

5. 5 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Example 3.1 Calculate the molecular mass of glycerol ( C 3 H 8 O 3 ). Example 3.2 Calculate the formula mass of ammonium sulfate, a fertilizer commonly used by home gardeners.

6. 6 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Determining the Formula Mass of Ammonium Sulfate

7. 7 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Mole (mol): amount of substance that contains as many elementary entities as there are atoms in exactly 12 g of the carbon-12 isotope. Atoms are small, so this is a BIG number … Avogadro’s number (N A ) = 6.022 × 10 23 mol –1 1 mol = 6.022 × 10 23 “things” (atoms, molecules, ions, formula units, oranges, etc.) –A mole of oranges would weigh about as much as the earth! Mole is NOT abbreviated as either M or m. The Mole & Avogadro’s Number

8. 8 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three

9. 9 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three One Mole of Four Elements One mole each of helium, sulfur, copper, and mercury. How many atoms of helium are present? Of sulfur? Of copper? Of mercury?

10. 10 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Molar mass is the mass of one mole of a substance. Molar mass is numerically equal to atomic mass, molecular mass, or formula mass. However … … the units of molar mass are grams (g/mol). Examples: 1 atom Na = 22.99 u 1 mol Na = 22.99 g 1 formula unit KCl = 74.56 u The Mole and Molar Mass 1 mol CO 2 = 44.01 g 1 molecule CO 2 = 44.01 u 1 mol KCl = 74.56 g

11. 11 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three We can use these equalities to construct conversion factors, such as: Note: preliminary and follow-up calculations may be needed. 1 mol Na ––––––––– 22.99 g Na Conversions involving Mass, Moles, and Number of Atoms/Molecules 1 mol Na = 6.022 × 10 23 Na atoms = 22.99 g Na 22.99 g Na ––––––––– 1 mol Na –––––––––––––––––– 6.022 × 10 23 Na atoms

12. 12 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three We can read formulas in terms of moles of atoms or ions.

13. 13 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Hints for Doing Mass/Mole/Atom Problems HINT #1: If you see, ANYWHERE AT ALL IN THE PROBLEM, “grams”, you will NEED to make a SAMT chart! HINT #2: If you see, ANYWHERE AT ALL IN THE PROBLEM, “atoms/molecules/etc”, you will NEED to use Avogadro’s #. HINT #3: If you see, MOLE in the given or find, you will only do a 2-step problem.

14. 14 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Examples Given: 3.50 mol Cu; Find: # g of Cu. Given: 11.5 g of H 2 O; Find: # mol of H 2 O. Given: 0.55 mol of Al; Find: # atoms of Al. Given: 3.73 x 10 23 molecules of HCl; Find: # mol of HCl.

15. 15 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Examples – Part 2 These are more difficult, and are 3-steps: Given = 11.5 g of Ca; Find = # atoms of Ca. Given = 1.22 x 10 23 molecules of NO; Find = # g of NO. Given = 13.5 g of HCl; Find = # particles of HCl.

16. 16 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Examples Calculate the mass in milligrams of 1.34 x 10 -4 mol Ag. Calculate the number of oxygen atoms in 20.5 mol O 2.

17. 17 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Examples Calculate both the (a) # of moles and (b) the # of atoms of Al in a cube of aluminum metal 5.5 cm on an edge. The density of aluminum is known to be 2.70 g/cm 3. Calculate the volume occupied by 4.06 x 10 24 Br atoms present as Br 2 molecules in liquid bromine. (d = 3.12 g/mL)

18. 18 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Example 3.4 Determine (a) the number of NH 4 + ions in a 145-g sample of (NH 4 ) 2 SO 4 and (b) the volume of 1,2,3- propanetriol (glycerol, d = 1.261 g/mL) that contains 1.00 mol O atoms. Example 3.5 An Estimation Example Which of the following is a reasonable value for the number of atoms in 1.00 g of helium? (a) 4.1 × 10 –23 (c) 1.5 × 10 23 (b) 4.0 (d) 1.5 × 10 24

19. 19 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three The mass percent composition of a compound refers to the proportion of the constituent elements, expressed as the number of grams of each element per 100 grams of the compound. In other words … Mass Percent Composition from Chemical Formulas X g element X % element = –––––––––––––– OR … 100 g compound g element % element = ––––––––––– × 100 g compound

20. 20 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Percentage Composition of Butane

21. 21 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Example 3.6 Calculate, to four significant figures, the mass percent of each element in ammonium nitrate. Example 3.7 How many grams of nitrogen are present in 46.34 g ammonium nitrate?

22. 22 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three We can “reverse” the process of finding percentage composition. First we use the percentage or mass of each element to find moles of each element. Then we can obtain the empirical formula by finding the smallest whole-number ratio of moles. –Find the whole-number ratio by dividing each number of moles by the smallest number of moles. Chemical Formulas from Mass Percent Composition

23. 23 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Analysis shows that a compound contains 32.31% sodium, 22.67% sulfur, and 45.02% oxygen. Find the empirical formula for this compound. HINT = Whenever you are given %s, we assume we ALWAYS have a 100-gram sample, so immediately convert you %s to grams. ANS = Na 2 SO 4 Examples

24. 24 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Find the empirical formula of a compound found to contain 26.56% K, 35.41% Cr, and the remainder O. ANS = K 2 Cr 2 O 7 Analysis of a 10.150-g sample of a compound is known to contain only phosphorus and oxygen. There is a phosphorus content of 4.433-g. What is the empirical formula of this compound? ANS = P 2 O 5 Examples (cont.)

25. 25 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Example 3.9 Phenol, a general disinfectant, has the composition 76.57% C, 6.43% H, and 17.00% O by mass. Determine its empirical formula. Example 3.10 Diethylene glycol, used in antifreeze, as a softening agent for textile fibers and some leathers, and as a moistening agent for glues and paper, has the composition 45.27% C, 9.50% H, and 45.23% O by mass. Determine its empirical formula.

26. 26 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Molecular formulas are the actual formulas for the compound. Example: In our last example, we determined that the empirical formula was P 2 O 5. Experimentation shows that the molar mass of this compound is 283.89 g/mol. What is the compound’s molecular formula? Hint = The molar mass given in the problem statement is the numerator!!!! ANS = P 4 O 10 Molecular Formulas

27. 27 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three A molecular formula is a simple integer multiple of the empirical formula. That is, an empirical formula of CH 2 means that the molecular formula is CH 2, or C 2 H 4, or C 3 H 6, or C 4 H 8, etc. So: we find the molecular formula by: = integer (nearly) molecular formula mass empirical formula mass We then multiply each subscript in the empirical formula by the integer. Relating Molecular Formulas to Empirical Formulas

28. 28 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Determine the molecular formula of the compound with an empirical formula of CH and a formula mass of 78.110 amu (g/mol). ANSWER = C 6 H 6 A sample of a compound with a formula mass of 34.00 amu (g/mol) is found to consist of 0.44 gH and 6.92 gO. Determine (a) empirical and (b) molecular formulas for this compound. ANSWER = (a) HO; (b) H 2 O 2 Examples

29. 29 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Example 3.11 The empirical formula of hydroquinone, a chemical used in photography, is C 3 H 3 O, and its molecular mass is 110 u. What is its molecular formula?

30. 30 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three … is one method of determining empirical formulas in the laboratory. This method is used primarily for simple organic compounds (that contain carbon, hydrogen, oxygen). –The organic compound is burned in oxygen. –The products of combustion (usually CO 2 and H 2 O) are weighed. –The amount of each element is determined from the mass of products. Elemental Analysis …

31. 31 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Elemental Analysis (cont’d) The sample is burned in a stream of oxygen gas, producing … … H 2 O, which is absorbed by MgClO 4, and … … CO 2, which is absorbed by NaOH.

32. 32 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Elemental Analysis (cont’d) If our sample were CH 3 OH, every two molecules of CH 3 OH … … would give two molecules of CO 2 … … and four molecules of H 2 O.

33. 33 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Example 3.12 Burning a 0.1000-g sample of a carbon– hydrogen–oxygen compound in oxygen yields 0.1953 g CO 2 and 0.1000 g H 2 O. A separate experiment shows that the molecular mass of the compound is 90 u. Determine (a) the mass percent composition, (b) the empirical formula, and (c) the molecular formula of the compound.

34. 34 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three A chemical equation is a shorthand description of a chemical reaction, using symbols and formulas to represent the elements and compounds involved. Writing Chemical Equations

35. 35 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Sometimes additional information about the reaction is conveyed in the equation. Writing Chemical Equations

36. 36 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Balancing Equations Illustrated The equation is balanced by changing the coefficients … How can we tell that the equation is not balanced? … not by changing the equation … … and not by changing the formulas.

37. 37 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Chemical Reactions What is a chemical reaction? A chemical reaction is any process by which 1 or more substances are changed into one or more different substances. A+B  C+D A and B are referred to as REACTANTS; C and D are referred to as PRODUCTS.

38. 38 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Chemical Equation Chemical reactions are described by chemical equations. Chemical equation = an equation that represents, with symbols and formulas, the identities and relative molecular amounts of the reactants and products in a chemical reaction.

39. 39 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Indications of a Chemical Reaction The following methods are ways we tell a chemical reaction occurs: #1. Evolution of energy as heat and/or light. #2. Production of a gas. #3. Formation of a precipitate. Precipitate = a solid that is produced as a result of a chemical reaction of 2 or more solutions.

40. 40 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Characteristics of a Chemical Equation 1. The equation must represent known facts. All reactants and products MUST be identified. 2. The equation must contain the correct formulas for the reactants and products. 3. The law of conservation of mass must be satisfied. NOTE = we are ONLY allowed to change coefficients, NOT subscripts in a chemical equation!

41. 41 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Coefficients vs. Subscripts

42. 42 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Word and Formula Equations __H 2 O  __H 2 +__O 2 What are the reactant(s)? What are the product(s)? What are the coefficients for each material?

43. 43 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Word Equations A word equation is an equation in which the reactants products in a chemical reaction are represented by words. Remember – what are our diatomic elements? Ex#1: Solid aluminum reacts with gaseous oxygen to produce solid aluminum oxide. Ex#2: Methane, in the presence of oxygen, combusts into water and carbon dioxide.

44. 44 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three A Few More Examples… Example #3: Solid sodium oxide is added to water at room temperature and forms sodium hydroxide. Example #4: Hydrazine, N 2 H 4, is used as rocket fuel. Hydrazine reacts violently with oxygen to produce gaseous nitrogen and water.

45. 45 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three A Few Hints… When balancing chemical equations, here are a few hints: #1. Balance most elements OTHER THAN hydrogen and oxygen 1 st, then go with hydrogen, and save oxygen for LAST. #2. Sometimes, it is easier to call water HOH. #3. Look for polyatomic ions on both sides of the equation. If they are there, you can use a polyatomic as a “pseudo” element.

46. 46 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three What Do I Mean by Hint #3? Example: Magnesium phosphate reacts with lithium chloride to form magnesium chloride and lithium phosphate. Calcium hydroxide and sodium fluoride react to form calcium fluoride and sodium hydroxide.

47. 47 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Pop Quiz…. Try the following….think about what you would get: #1. __Li+__O 2  ??? #2. __NaCl  ??? #3. __H 2 +__F 2  ??? #4. __MgCO 3  ???

48. 48 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three 5 Basic Types of Chemical Reactions There are several ways to classify chemical reactions. We will look at the 5 basic types of reactions: 1. synthesis (composition) 2. decomposition 3. single displacement (replacement) 4. double displacement (replacement) 5. combustion

49. 49 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Synthesis Reactions Synthesis reactions are AKA composition reactions. Synthesis reactions = a reaction in which 2 or more substances combine to form a new compound. A+X  AX

50. 50 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Types of Synthesis Reactions Type #1: Reactions of elements with oxygen and sulfur. Remember, oxygen is diatomic (O 2 ). Sulfur is odd – it can be S 8. __Rb+__S 8  __Rb 2 S __Fe+__O 2  __Fe 2 O 3

51. 51 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Types of Synthesis Rxns (cont.) Type #2: Reactions of metals with halogens. __Na+__Br 2  __NaBr __F 2 +__Mg  __MgF 2 Remember – ALL halogens are diatomic!

52. 52 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Types of Synthesis Rxns. (cont.) Type #3: Synthesis reactions with oxides. __CaO+ __H 2 O  __Ca(OH) 2 __SO 2 +__H 2 O  __H 2 SO 4 Remember, when balancing, call water HOH. Usually, you get a metal hydroxide.

53. 53 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Decomposition Reactions In a decomposition reaction, a single compound undergoes a reaction that produces 2 or more simpler substances. Decomposition reactions are usually the opposite of synthesis reactions. AX  A+X Decomposition reactions usually require heat or electricity in order to occur.

54. 54 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Types of Decomposition Reactions Type #1: Decomposition of a Binary Compound 2 H 2 O  2 H 2 +O 2 (electrolysis) 2 HgO  2 Hg+O 2 (under heat)

55. 55 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Types (cont.) Type #2: Decomposition of Metal Carbonates What IS a metal carbonate? CaCO 3  CaO+CO 2 Breaks down into a metal oxide and carbon dioxide gas.

56. 56 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Types (cont.) Type #3: Decomposition of a Metal Hydroxide What IS a metal hydroxide? Ca(OH) 2  CaO+HOH Breaks down into a metal oxide and water.

57. 57 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Types (cont.) Type #4: Decomposition of a Metal Chlorate What IS a metal chlorate? 2 KClO 3  2 KCl+ 3 O 2 Breaks down into a metal chloride and oxygen.

58. 58 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Types (cont.) Type #5: Decomposition of Acids Think acids that are NOT binary for these! H 2 CO 3  CO 2 +H 2 O H 2 SO 4  SO 3 +H 2 O Breaks down into water and “what’s left”.

59. 59 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Single Displacement Reactions Single displacement reactions are also known as single replacement reactions. We will look at this type of reaction more in the next section…

60. 60 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Double Displacement Reactions In double displacement (replacement) reactions, the ions of 2 compounds exchange places in an aqueous solution to form 2 new compounds. AX+BY  AY+BX __KI+__Pb(NO 3 ) 2  __PbI 2 +__KNO 3

61. 61 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Combustion Reactions In a combustion reaction, a hydrocarbon reacts with oxygen, releasing a LARGE amount of energy, as well as water and carbon dioxide. __CH 4 +__O 2  __CO 2 + __H 2 O __C 3 H 8 +__O 2  __CO 2 + __H 2 O

62. 62 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Example 3.13 Balance the equation Fe + O 2  Fe 2 O 3 (not balanced) Example 3.14 Balance the equation C 2 H 6 + O 2  CO 2 + H 2 O Example 3.15 Balance the equation H 3 PO 4 + NaCN  HCN + Na 3 PO 4 Example 3.16 A Conceptual Example Write a plausible chemical equation for the reaction between water and a liquid molecular chloride of phosphorus to form an aqueous solution of hydrochloric acid and phosphorus acid. The phosphorus-chlorine compound is 77.45% Cl by mass.

63. 63 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three A stoichiometric factor or mole ratio is a conversion factor obtained from the stoichiometric coefficients in a chemical equation. In the equation: CO (g) + 2 H 2 (g)  CH 3 OH (l) –1 mol CO is chemically equivalent to 2 mol H 2 –1 mol CO is chemically equivalent to 1 mol CH 3 OH –2 mol H 2 is chemically equivalent to 1 mol CH 3 OH Stoichiometric Equivalence and Reaction Stoichiometry 1 mol CO ––––––––– 2 mol H 2 1 mol CO ––––––––––––– 1 mol CH 3 OH 2 mol H 2 ––––––––––––– 1 mol CH 3 OH

64. 64 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three One car may be equivalent to either 25 feet or 10 feet, depending on the method of parking. One mole of CO may be equivalent to one mole of CH 3 OH, or to one mole of CO 2, or to two moles of CH 3 OH, depending on the reaction(s). Concept of Stoichiometric Equivalence

65. 65 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Outline of Simple Reaction Stoichiometry Note: preliminary and/or follow-up calculations may be needed.

66. 66 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Example 3.17 When 0.105 mol propane is burned in an excess of oxygen, how many moles of oxygen are consumed?

67. 67 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Outline of Stoichiometry Involving Mass To our simple stoichiometry scheme … … we’ve added a conversion from mass at the beginning … … and a conversion to mass at the end. Substances A and B may be two reactants, two products, or reactant and product. Think: If we are given moles of substance A initially, do we need to convert A to grams?

68. 68 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Examples Magnesium reacts with oxygen to produce magnesium oxide. Given = 2.00 mol Mg Find = # grams of MgO

69. 69 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Examples Tin reacts with hydrofluoric acid to produce tin (II) fluoride and hydrogen gas. Given = 30.00 grams of HF Find = # grams of SnF 2

70. 70 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Example 3.18 The final step in the production of nitric acid involves the reaction of nitrogen dioxide with water; nitrogen monoxide is also produced. How many grams of nitric acid are produced for every 100.0 g of nitrogen dioxide that reacts? Example 3.19 Ammonium sulfate, a common fertilizer used by gardeners, is produced commercially by passing gaseous ammonia into an aqueous solution that is 65% H 2 SO 4 by mass and has a density of 1.55 g/mL. How many milliliters of this sulfuric acid solution are required to convert 1.00 kg NH 3 to (NH 4 ) 2 SO 4 ?

71. 71 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Many reactions are carried out with a limited amount of one reactant and a plentiful amount of the other(s). The reactant that is completely consumed in the reaction limits the amounts of products and is called the limiting reactant, or limiting reagent. The limiting reactant is not necessarily the one present in smallest amount. Limiting Reactants

72. 72 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Limiting Reactant Analogy If we have 10 sandwiches, 18 cookies, and 12 oranges … … how many packaged meals can we make?

73. 73 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Limiting vs. Excess When looking at any chemical reaction, the reaction is rarely carried out as per the balanced equation. We will be looking at reactions with 2 reactants. (A) limiting reactant = the reactant in the chemical reaction that is completely “used up” – none will remain when the reaction is complete. (B) excess reactant = the reactant that is NOT completely used up in a chemical reaction.

74. 74 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three What to Do Silicon dioxide, aka quartz, is usually quite unreactive but reacts readily with hydrogen fluoride to produce silicon tetrafluoride and water. If 6.0g of HF is added to 4.5g of SiO 2, which is the limiting reactant? Step #1: BALANCE THE EQUATION! __SiO 2 +__HF  __SiF 4 +__H 2 O

75. 75 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three What to Do (cont.) SiO 2 +4HF  SiF 4 +2H 2 O Step #2: Pick a product, any product (if the reactant is limiting for 1 product, it will be limiting for the other) = I pick SiF 4. Step #3: Convert BOTH reactants to grams of product. Step #4: The smaller value gives us the reactant that is LIMITING !

76. 76 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Example Fe 2 O 3 is a substance that can be made in the chem lab via the following, UNBALANCED, equation: __Fe+__H 2 O  __Fe 2 O 3 +__H 2 (a) Balance the equation (b) If 36.0 g water is mixed with 67.0 g Fe, determine the limiting reactant. (c) What mass of iron oxide is produced, based on the limiting reactant? (d) What mass of excess reactant remains when the reaction is completed ?

77. 77 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three When 28 g (1.0 mol) ethylene reacts with … … 128 g (0.80 mol) bromine, we get … … 150 g of 1,2- dibromoethane, and leftover ethylene! 1.Why is ethylene left over, when we started with more bromine than ethylene? (Hint: count the molecules.) 2.What mass of ethylene is left over after reaction is complete? (Hint: it’s an easy calculation; why?) Molecular View of the Limiting Reactant Concept

78. 78 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three We recognize limiting reactant problems by the fact that amounts of two (or more) reactants are given. One way to solve them is to perform a normal stoichiometric calculation of the amount of product obtained, starting with each reactant. The reactant that produces the smallest amount of product is the limiting reactant. Recognizing and Solving Limiting Reactant Problems

79. 79 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Example 3.20 Magnesium nitride can be formed by the reaction of magnesium metal with nitrogen gas. (a) How many grams of magnesium nitride can be made in the reaction of 35.00 g of magnesium and 15.00 g of nitrogen? (b) How many grams of the excess reactant remain after the reaction?

80. 80 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three The theoretical yield of a chemical reaction is the calculated quantity of product in the reaction. The actual yield is the amount you actually get when you carry out the reaction. Actual yield will be less than the theoretical yield, for many reasons … can you name some? Yields of Chemical Reactions actual yield Percent yield = ––––––––––––– × 100 theoretical yield

81. 81 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Actual Yield of ZnS Is Less than the Theoretical Yield

82. 82 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Example 3.21 Ethyl acetate is a solvent used as fingernail polish remover. What mass of acetic acid is needed to prepare 252 g ethyl acetate if the expected percent yield is 85.0%? Assume that the other reactant, ethanol, is present in excess. The equation for the reaction, carried out in the presence of H 2 SO 4, is CH 3 COOH + HOCH 2 CH 3  CH 3 COOCH 2 CH 3 + H 2 O Acetic acid Ethanol Ethyl acetate

83. 83 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Solute: the substance being dissolved. Solvent: the substance doing the dissolving. Concentration of a solution: the quantity of a solute in a given quantity of solution (or solvent). –A concentrated solution contains a relatively large amount of solute vs. the solvent (or solution). –A dilute solution contains a relatively small concentration of solute vs. the solvent (or solution). –“Concentrated” and “dilute” aren’t very quantitative … Solutions and Solution Stoichiometry

84. 84 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Molarity (M), or molar concentration, is the amount of solute, in moles, per liter of solution: A solution that is 0.35 M sucrose contains 0.35 moles of sucrose in each liter of solution. Keep in mind that molarity signifies moles of solute per liter of solution, not liters of solvent. Molar Concentration moles of solute Molarity = –––––––––––––– liters of solution

85. 85 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Preparing 0.01000 M KMnO 4 Weigh 0.01000 mol (1.580 g) KMnO 4. Dissolve in water. How much water? Doesn’t matter, as long as we don’t go over a liter. Add more water to reach the 1.000 liter mark.

86. 86 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Example 3.23 What is the molarity of a solution in which 333 g potassium hydrogen carbonate is dissolved in enough water to make 10.0 L of solution? Example 3.24 We want to prepare a 6.68 molar solution of NaOH (6.68 M NaOH). (a) How many moles of NaOH are required to prepare 0.500 L of 6.68 M NaOH? (b) How many liters of 6.68 M NaOH can we prepare with 2.35 kg NaOH? Example 3.25 The label of a stock bottle of aqueous ammonia indicates that the solution is 28.0% NH 3 by mass and has a density of 0.898 g/mL. Calculate the molarity of the solution.

87. 87 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Dilution is the process of preparing a more dilute solution by adding solvent to a more concentrated one. Addition of solvent does not change the amount of solute in a solution but does change the solution concentration. It is very common to prepare a concentrated stock solution of a solute, then dilute it to other concentrations as needed. Dilution of Solutions

88. 88 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Visualizing the Dilution of a Solution We start and end with the same amount of solute. Addition of solvent has decreased the concentration.

89. 89 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Dilution Calculations … … couldn’t be easier. Moles of solute does not change on dilution. Moles of solute = M × V Therefore … M conc × V conc = M dil × V dil

90. 90 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Example 3.26 How many milliliters of a 2.00 M CuSO 4 stock solution are needed to prepare 0.250 L of 0.400 M CuSO 4 ?

91. 91 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Molarity provides an additional tool in stoichiometric calculations based on chemical equations. Molarity provides factors for converting between moles of solute (either reactant or product) and liters of solution. Solutions in Chemical Reactions

92. 92 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three If substance A is a solution of known concentration … If substance B is in solution, then … … we can start with molarity of A times volume (liters) of the solution of A to get here. … we can go from moles of substance B to either volume of B or molarity of B. How? Adding to the previous stoichiometry scheme …

93. 93 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Example 3.27 A chemical reaction familiar to geologists is that used to identify limestone. The reaction of hydrochloric acid with limestone, which is largely calcium carbonate, is seen through an effervescence—a bubbling due to the liberation of gaseous carbon dioxide: CaCO 3 (s) + 2 HCl(aq)  CaCl 2 (aq) + H 2 O(l) + CO 2 (g) How many grams of CaCO 3 (s) are consumed in a reaction with 225 mL of 3.25 M HCl?

94. 94 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Cumulative Example The combustion in oxygen of 1.5250 g of an alkane-derived compound composed of carbon, hydrogen, and oxygen yields 3.047 g CO 2 and 1.247 g H 2 O. The molecular mass of this compound is 88.1 u. Draw a plausible structural formula for this compound. Is there more than one possibility? Explain.