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 3.1 Calculate the molecular mass of glycerol (1,2,3- propanetriol). Example 3.2 Calculate the formula mass of ammonium sulfate, a fertilizer commonly used by home gardeners.

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

5. 5 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

6. 6 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three

7. 7 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?

8. 8 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Example 3.3 Determine (a) the mass of a 0.0750-mol sample of Na, (b) the number of moles of Na in a 62.5- g sample, (c) the mass of a sample of Na containing 1.00 × 10 25 Na atoms, and (d) the mass of a single Na atom.

9. 9 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

10. 10 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

11. 11 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.

12. 12 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

13. 13 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

14. 14 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three Percentage Composition of Butane

15. 15 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? Example 3.8 An Estimation Example Without doing detailed calculations, determine which of these compounds contains the greatest mass of sulfur per gram of compound: barium sulfate, lithium sulfate, sodium sulfate, or lead sulfate.

16. 16 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

17. 17 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.

18. 18 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

19. 19 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?

20. 20 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 …

21. 21 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.

22. 22 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.

23. 23 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.

24. 24 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

25. 25 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

26. 26 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.

27. 27 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Three If an element is present in just one compound on each side of the equation, try balancing that element first. Balance any reactants or products that exist as the free element last. In some reactions, certain groupings of atoms (such as polyatomic ions) remain unchanged. In such cases, treat these groupings as a unit. At times, an equation can be balanced by first using a fractional coefficient(s). The fraction is then cleared by multiplying each coefficient by a common factor. Guidelines for Balancing Chemical Equations

28. 28 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.

29. 29 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

30. 30 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

31. 31 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.

32. 32 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? The reaction is C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O

33. 33 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?

34. 34 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 ?

35. 35 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

36. 36 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?

37. 37 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

38. 38 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

39. 39 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?

40. 40 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

41. 41 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

42. 42 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 Example 3.22 A Conceptual Example What is the maximum yield of CO(g) obtainable from 725 g of C 6 H 14 (l), regardless of the reaction(s) used, assuming no other carbon-containing reactant or product?

43. 43 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

44. 44 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

45. 45 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.

46. 46 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.

47. 47 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

48. 48 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.

49. 49 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

50. 50 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 ?

51. 51 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

52. 52 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 …

53. 53 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?

54. 54 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.