CHEMISTRY FOR UNIVERSITY INTRO 1210_STOCHIOMETRY_A.pptx

Chemistry for University students a summarized approach.

1. INTRODUCTION TO CHEMISTRY 1210 STOCHIOMETRY PART A Naming simple compounds; The atom; composition, isotopes, isotopic abundance; Molar mass; The mole concept; Percent composition of compounds; Empirical and molecular formulae; Balancing chemical equations; Concept of Limiting Reagents; Percentage Yield

2. THE ATOM: COMPOSITION

3. INTRODUCTION All matter is made up of atoms. Atoms are the smallest part of elements that can take part in chemical reactions They are the building blocks of all elements The center of the atom is a dense nucleus that contains protons and neutrons . Protons have a positive charge and neutrons are neutral. Negatively charged electrons are found in orbitals that spiral around the nucleus. Electrons, neutrons and protons are referred to as subatomic particles .

4. SUBATOMIC PARTICLES Electrons, protons and neutrons are so small that it is not possible to measure their masses or charges by conventional units such as grams or coulombs We compare them to each other and therefore call them: relative atomic mass and relative atomic charge 1 relative atomic mass is 1/12 the Mass of C 12. C 12 is used as a standard measure

5. Why do orbital electrons remain in orbit? An orbital electron can be compared to a weight being whirled about at the end of a string. If you let go of the string, the weight moves off in a straight line in whatever direction it happened to be going at the time. As you pull on the string, holding it tight, the weight cannot move away from you. In the same way, the electrical force of attraction between the positive nucleus and the negative electron keeps the electron in its orbit. Like charges repel while opposite charges attract!

6. For each proton in the nucleus, the atom has one orbital electron. In other words, the number of orbital electrons in an atom is equal to the number of protons in its nucleus. The total positive charge of the nucleus therefore equals the total negative charge of all the orbital electrons. The net charge of the whole atom is exactly zero

7. Atomic Number (Z) The number of protons in the nucleus, which makes atoms different from one another in their structure and character, is such an important number that it is given a special name, the atomic number The atomic number of carbon is 6 Since the number of protons and electrons are equal, therefore, the number of electrons of a carbon atom is 6

8. THE ATOM: ISOTOPES

9. Isotopes ISOTOPES of an element are atoms that have the same atomic number (Z) as the other atoms of the element, but a different mass number (A) In writing the symbols for atoms, we need a system to identify the different isotopes of the same element. For example, we can identify the uranium isotope that has 92 protons and 146 neutrons in any of the following three versions: U-238

10. Atomic Mass Is the average of all naturally occuring isotopes of an element. Calculated this way: AM= M 1 (%) + M 2 (%) + M 3 (%) + …

11. ISOTOPE 12 C 13 C 14 C Protons 6 6 6 Neutrons 6 7 8 (A) 12 13 14 % 98.89 1.11 0.01 CALCULATE THE ATOMIC MASS OF CARBON BELOW AM= 12(98.89/100)+ 13(1.11/100)+ 14(0.01/100) = 12.0125

12. Example: Boron has two naturally occurring isotopes. In a sample of boron, 20% of the atoms are B-10, which is an isotope of boron with 5 neutrons and a mass of 10 amu . The other 80% of the atoms are B-11, which is an isotope of boron with 6 neutrons and a mass of 11 amu . What is the atomic mass of boron? Solution: Boron has two isotopes. We will use the equation: Atomic mass = (%1)(mass1) + (%2)(mass2) + ... Isotope 1: %1=0.20 mass1=10 Isotope 2: %2=0.80, mass2=11 Substitute these into the equation, and we get: Atomic mass = (0.20)(10) + (0.80)(11) Atomic mass = 10.8 amu The mass of an average boron atom, and thus boron’s atomic mass, is 10.8 amu .

13. THE ATOM: ISOTOPIC ABUNDANCE Bromine has two naturally occurring isotopes. Bromine-79 has a mass of 78.918 amu and is 50.69% abundant. Using the atomic mass reported on the periodic table, determine the mass of bromine-81, the other isotope of bromine. Solution: 79.904 = (78.918 x 0.5069) + (mass2 x 0.4931) → 79.904 = 40.0035 + mass20.4931 → mass2 = 80.9177 amu

14. The element magnesium (relative atomic mass 24.3) has three naturally occurring isotopes, 24Mg, 25Mg and 26Mg. If the percentage of the heaviest isotope is 11.0%, what is the percentage of the lightest isotope present?

15. RELATIVE MOLECULAR MASS The relative molecular mass is the sum of the relative atomic masses of the atoms that make up the molecule. Example, the chemical formula for water is H 2 O Its relative molecular mass = 2 (1.00794) + 15.9994 = 18.0153

16. NAMING SIMPLE COMPOUNDS The formula for a compound uses the symbols to indicate the type of atoms involved and uses subscripts to indicate the number of each atom in the formula. For example, aluminum combines with oxygen to form the compound aluminum oxide. To form aluminum oxide requires two atoms of aluminum and three atoms of oxygen. Therefore, we write the formula for aluminum oxide as Al2O3. The symbol Al tells us that the compound contains aluminum, and the subscript 2 tells us that there are two atoms of aluminum in each molecule. The O tells us that the compound contains oxygen, and the subscript 3 tells us that there are three atoms of oxygen in each molecule.

17. Polyatomic Ions Thus far, we have been dealing with ions made from single atoms. Such ions are called monatomic ions. There also exists a group of polyatomic ions, ions composed of a group of atoms that are covalently bonded and behave as if they were a single ion. Almost all the common polyatomic ions are negative ions. A table of many common polyatomic ions is given. The more familiar you become with polyatomic ions, the better you will be able to write names and formulas of ionic compounds. It is also important to note that there are many polyatomic ions that are not on this chart.

18. Important Polyatomic ions Ammonium, NH 4 + -1 -2 -3 Hypochlorite, ClO - Chlorite, ClO 2 - Chlorate, ClO 3 - Perchlorate, ClO 4 - Sulfite, SO 2- Sulfate, S 3 2- O 4 Phosphate, PO 4 3- Nitrite, NO 2 - Nitrate, NO 3 - Bicarbonate, HCO 3 - Carbonate, CO 3 2- Hydroxide, OH - Peroxide, O 2- 2 Acetate, C 2 H 3 O 2 - Oxalate, C 2 O 2- 4 Silicate, SiO 2- Thiosulfate, 3 O 2- S 2 3 Permanganate, MnO 4 - Chromate, CrO 4 2- Dichromate, Cr 2 O 7 2- Cyanide, CN - Thiocyanate, SCN -

19. Naming Transition Metals Some metals are capable of forming ions with various charges. These include most of the transition metals and many post transition metals. Iron, for example, may form Fe 2+ ions by losing 2 electrons or Fe 3+ ions by losing 3 electrons. The rule for naming these ions is to insert the charge (oxidation number) of the ion with Roman numerals in parentheses after the name. These two ions would be named iron (II) and iron (III). When you see that the compound involves any of the variable oxidation number metals (iron, copper, tin, lead, nickel, and gold), you must determine the charge (oxidation number) of the metal from the formula and insert Roman numerals indicating that charge.

20. Consider FeO and Fe 2 O 3 . These are very different compounds with different properties. When we name these compounds, it is absolutely vital that we clearly distinguish between them. They are both iron oxides but in FeO iron is exhibiting a charge of 2+ and in Fe2O3, it is exhibiting a charge of 3+. The first, FeO , is named iron (II) oxide. The second, Fe2O3, is named iron (III) oxide.

21. To name ionic compounds, we will need to follow these steps: Split the formula into the cation and anion. The first metal listed will be the cation and the remaining element(s) will form the anion. Name the cation . We learned two types of cations : Main group cations in which the name of the ion is the same as that of the element (for example, K + is potassium). Transition metals with variable charges with Roman numerals indicating the charge of the ion (you will have to do a little bit of math to find this charge). Name the anion. There are also two general types of anions: Main group anions in which the name of the anion ends in “-ide” (for example, F - is fluoride) Polyatomic ions (as listed on the polyatomic ion chart)

22. Section Summary When an atom gains one or more extra electrons, it becomes a negative ion, an anion When an atom loses one or more of its electrons, it becomes a positive ion, a cation. Polyatomic ions are ions composed of a group of atoms that are covalently bonded and behave as if they were a single ion. Some transition elements have fixed oxidation numbers and some have variable charges. When naming these charge variable ions, their charges are included in Roman numerals.

23. THE MOLE CONCEPT

24. Because samples of matter typically contain so many atoms, a unit of measure called the mole has been established for use in counting atoms. For our purposes, it is most convenient to define the mole (abbreviated mol) as the number equal to the number of carbon atom in exactly 12 grams of pure 12C. Techniques such as mass spectrometry, which count atom very precisely, have been used to determine this number as 6.02214 x 10 23 (6.022 x 10 23 ) will be sufficient for our purposes. This number is called Avogadro’s number to honour his contributions to chemistry. Just as a dozen eggs is 12 eggs, a mole of eggs is 6.022 x 10 23 eggs.

25. Avogadro’s Number Avogadro’s number: the number of atoms in exactly 12 g of 12 C. N A = 6.022137 x 10 23 Sodium (Na) has a relative atomic mass of 22.98977 Hence a sodium atom is (22.98977) times as heavy as 12 C If N A atoms of 12 C have a mass of 12g then, the mass of N A atoms of sodium must be (22.98977) 12g = 22.98977 g 12 12

26. The mass, in grams, of N A atoms of ANY element is numerically equal to the relative atomic mass of that element. Same applies to molecules. Since the relative molecular mass of water is 18.0152, the mass of N A water molecules is 18.0152g

27. Aluminum (Al) is a metal with a high strength-to-mass ratio and a high resistance to cor rosion ; thus it is often used for structural purposes. Compute both the number of mole of atoms and the number of atoms in a 10.0-g sample of aluminum . Solution The mass of 1 mole (6.022 x 10 23 atoms) of aluminum is 26.98 g. The sample we are considering has a mass of 10.0 g. Since the mass is less than 26.98 g, this sample contains less than 1 mole of aluminum atoms. We can calculate the number of moles of aluminum atoms in 10.0 g as follows: 1 mol Al = 26.98 g Al X = 10.0 g Al 0.371 mol Al atoms The number of atoms in 10.0 g (0.371 mol) of aluminum is 0.371 mol Al x 6.022 x10 23 atoms = 2.23 x10 23 atoms

28. The previous question you have just solved highlights the following relationships:

29. MOLAR MASS Molar mass tells us how much mass is in one mole of substance. Periodic Table masses are in units of (g/mol). Very important – the units are grams per 1 mole, not just grams How many grams in one mole of: carbon? 12.01 g/mol gold? 196.97 g/mol sodium? 22.99 g/mol What do they all have in common? They all have 6.02 x 10 23 atoms!

30. Calculate the molar mass of barium hydroxide. What is the formula? Add up the mass for every atom. Ba: 137.33 g/ mol + O: 2 (16.00) g/ mol + H: 2 (1.01) g/ mol = 171.35 g/ mol

31. PERCENTAGE COMPOSITION OF COMPOUNDS There are two common ways of describing the composition of a compound: in terms of the numbers of its constituent atoms and in terms of the percentages (by mass) of its el - ements . We can obtain the mass percents of the elements from the formula of the com- pound by comparing the mass of each element present in 1 mole of the compound to the total mass of 1 mole of the compound.

32. For example, for ethanol, which has the formula C2H5OH, the mass of each eleme present and the molar mass are obtained as follows: Mass of C = 2 mol x 12.01 g mol = 24.02g Mass of H = 6 mol x 1.008 g mol = 6.048g Mass of O = 1 mol x 16.00 g mol = 16 g Mass of 1 mol C2H5OH = 46.07 g The mass percent (often called the weight percent ) of carbon in ethanol can be compute by comparing the mass of carbon in 1 mole of ethanol to the total mass of 1 mole o ethanol and multiplying the result by 100:

33. Mass percent of C = mass of C in 1 mol C2H5OH / mass of 1 mol C2H5OH x 100% 24.02 g/ 46.07 g x 100% = 52.14% The mass percents of hydrogen and oxygen in ethanol are obtained in a similar manne r Mass percent of H 􏰂 mass of H in 1 mol C2H5OH 􏰀 100% mass of 1 mol C2H5OH 􏰂 6.048 g 􏰀 100% 􏰂 13.13% 46.07 g Mass percent of O 􏰂 mass of O in 1 mol C2H5OH 􏰀 100% mass of 1 mol C2H5OH 􏰂 16.00 g 􏰀 100% 􏰂 34.73% 46.07 g

34. EMPIRICAL & MOLECULAR FORMULAE

35. An empirical formula tells us the relative ratios of different atoms in a compound. The ratios hold true on the molar level as well. Thus, H2O is composed of two atoms of hydrogen and 1 atom of oxygen. Likewise, 1.0 mole of H2O is composed of 2.0 moles of hydrogen and 1.0 mole of oxygen. We can also work backwards from molar ratios since if we know the molar amounts of each element in a compound we can determine the empirical formula.                                                                                                            Learning Objectives To understand the definition and difference between empirical formulas and chemical formulas

36. Example : Mercury Chloride Mercury forms a compound with chlorine that is 73.9% mercury and 26.1% chlorine by mass. What is the empirical formula? Solution Let's say we had a 100 gram sample of this compound. The sample would therefore contain 73.9 grams of mercury and 26.1 grams of chlorine. How many moles of each atom do the individual masses represent?

37. For Mercury: For Chlorine: (73.9 g )x ( 1 mol. ) =0.368 moles 200.59 g (26.1 g )x ( 1 mol ) =0.736 mol              35.45 g        What is the molar ratio between the two elements? 0.736 mol Cl = 2.0 0.368 mol Hg Thus, we have twice as many moles (i.e. atoms) of Cl as Hg . The empirical formula would thus be (remember to list cation first, anion last): HgCl2

38. Chemical Formula from Empirical Formula The chemical formula for a compound obtained by composition analysis is always the empirical formula. We can obtain the chemical formula from the empirical formula if we know the molecular weight of the compound. The chemical formula will always be some integer multiple of the empirical formula (i.e. integer multiples of the subscripts of the empirical formula)

39. Example Vitamin C (ascorbic acid) contains 40.92 % C, 4.58 % H, and 54.50 % O, by mass. The experimentally determined molecular mass is 176 amu. What is the empirical and chemical formula for ascorbic acid? Solution Consider an arbitrary amount of 100 grams of ascorbic acid, so we would have: 40.92 grams C 4.58 grams H 54.50 grams O This would give us how many moles of each element?

40. Carbon Hydrogen Oxygen (40.92 g/ 12.011 g/mol ) =3.407 molC (4.58 g/ 1.008 g/mol )      =4.544 molH (54.50 g/ 15.999 g/mol ) =3.406 molO                

41. Determine the simplest whole number ratio by dividing by the smallest molar amount (3.406 moles in this case - Carbon : 3.407 / 3.406 = 1 Hydrogen : 4.544 / 3.406 = 1.33 0r 1 1/3 Oxygen : 3.406. /3.406 = 1 The relative molar amounts of carbon and oxygen appear to be equal, but the relative molar amount of hydrogen is higher. Since we cannot have "fractional" atoms in a compound, we need to normalize the relative amount of hydrogen to be equal to an integer. 1.333 would appear to be 1 and 1/3, so if we multiply the relative amounts of each atom by '3', we should be able to get integer values for each atom. C = (1.0)*3 = 3 H = (1.333)*3 = 4 O = (1.0)*3 = 3 This is our empirical formula for ascorbic acid. C3 H4 O3

42. What about the chemical formula? We are told that the experimentally determined molecular mass is 176 amu. What is the molecular mass of our empirical formula? (3*12.011) + (4*1.008) + (3*15.999) = 88.062 amu The molecular mass from our empirical formula is significantly lower than the experimentally determined value. What is the ratio between the two values? (176 amu/88.062 amu) = 2.0 Thus, it would appear that our empirical formula is essentially one half the mass of the actual molecular mass. If we multiplied our empirical formula by '2', then the molecular mass would be correct. Thus, the actual chemical formula is: 2* C 3 H 4 O 3 = C 6 H 8 O 6

43. BALANCING CHEMICAL REACTIONS

44. Chemical Reactions & Equations In a chemical reaction elements and/or compounds collide, interact and react to form new compounds. The compounds that come together are called the REACTANTS and the new compounds formed are the PRODUCTS.

45. An important aspect of a chemical reaction is that MASS IS ALWAYS CONSERVED - i.e. the total mass of the reactants must equal the total mass of the products. To ensure that mass is conserved, we have to keep track of the number of atoms of each element in the reactants and number of atoms of each element in the products

46. Writing Chemical Equations reactants H 2 + O 2  H 2 O products Mass has not been conserved - equation is not BALANCED Need to make sure that the number of atoms of a given element is the same on either side.

47. H 2 + O 2  H 2 O H is balanced, but O is not. To balance O, multiply H 2 O by 2 H 2 + O 2  2 H 2 O Now O is balanced but not H (4 H’s on right, 2 on left) Multiply H 2 by 2 2 H 2 + O 2  2 H 2 O

48. Now the equation is balanced. However, for a complete chemical equation the state of all reactants and products must be included 2H 2 + O 2  2 H 2 O 2H 2 (g) + O 2 (g)  2 H 2 O (g) g - gas; l - liquid; s- solid aq - denotes a solution of a solute in water (solvent) Note: the subscripts in the molecular formula must never be changed in balancing equations. Changing the subscript corresponds to a different molecular formula, hence a different molecule, with completely different properties

49. “Rules” to balance chemical equations Consider the reaction of methane, CH 4 , burning in air to produce CO 2 and H 2 O (combustion reaction) CH 4 + O 2  CO 2 + H 2 O It is usually best to first balance those elements that occur in the fewest chemical formulas on each side of the equation. So in this example balance C and H first

50. Ag 2 S(s) + KCN( aq ) + O 2 (g) + H 2 O(l)  KAg (CN) 2 ( aq ) + S(s) + KOH( aq ) Sometimes it is hard to balance a chemical reaction by “inspection” We Can then use algebraic expressions to balance equations

51. a Ag 2 S(s) + b KCN(aq) + c O 2 (g) + d H 2 O(l)  1 KAg(CN) 2 (aq) + e S(s) + f KOH(aq) Element Reactant Product Ag 2 a 1 S a e K b 1 + f C b 2 N b 2 O 2c + d f H 2d f

52. 2 a = 1 ; a = 0.5 b = 2 a = e ; e = 0.5 b = 1 + f; f = 1 2d = f; d = 0.5 2c + d = f; c = 0.25 Divide all by 0.25 a = 2; b = 8; c = 1; d = 2; e = 2; f = 4 Hence 2 Ag 2 S(s) + 8 KCN(aq) + O 2 (g) + 2 H 2 O(l)  4 KAg(CN) 2 (aq) + 2 S(s) + 4 KOH(aq)

53. Chemical Stoichiometry Reactants are consumed and products are formed in definite proportions. These proportions are given by the coefficients in the balanced equations for chemical reactions The calculation of the quantities of reactants and products is called STOICHIOMETRY. Stoichiometry is the use of chemical equations to calculate quantities of substances that take part in chemical reactions.

54. To do a stoichiometric calculation, the chemical equation for the calculation must be balanced. Equations are read in terms of moles of reactants and product 2H 2 (g) + O 2 (g)  2H 2 O(g) 2 moles of H 2 (g) reacts with 1 mole of O 2 (g) to form 2 moles of H 2 O(g) Or: 2N A molecules of H 2 (g) + N A molecules of O 2 (g)  2N A molecules of H 2 O(g)

55. Problem Consider the reaction of 100 g of H 2 (g) with sufficient O 2 (g) to produce the stoichiometric quantity of H 2 O(g). (stoichiometric quantities are the exact amounts of reactants and products predicted by balanced equations). Calculate the mass of H 2 O formed. Need a balanced equation for this reaction 2H 2 (g) + O 2 (g)  2H 2 O(g) (mass of H 2 ) = (100.g H 2 )/(2.02 g/mol) = 49.5 mol H 2 (molar mass of H 2 ) To find the mass of water formed, need to find the number of moles of H 2 that reacted:

56. 2 moles of H 2 reacts with 1 mole of O 2 to form 2 moles of H 2 O => 49.5 moles of H 2 will form 49.5 moles of H 2 O Hence, the mass of H 2 O(g) formed =(49.5 moles H 2 O) x (18.02 g/mol) = 892 g H 2 O

57. Problem Diethyl ether ( C 4 H 10 O ) combusts in air, reacting with O 2 to form H 2 O and CO 2 . How many grams of CO 2 would be produced if 350.0 mL of diethyl ether were combusted in an unlimited amount of oxygen? The density of diethyl ether is 0.713 g/ mL. First write a balanced equation ? C 4 H 10 O(l) + ? O 2 (g)  ? H 2 O(l) + ? CO 2 (g) Then convert the volume of diethyl ether to mass Next, determine the number of moles of diethyl ether reacted Determine the number of moles of CO 2 formed based on the stoichiometry of the reaction Finally determine the mass of CO 2 produced Answer: 592 g CO 2

58. For reactions occurring in solution, instead of using moles as a unit of quantity, define moles/liter = MOLARITY Molarity is defined as the number of moles of a solute per liter of solution. Note: the solute can be a solid, liquid, or gas dissolved in a solvent.

59. Problem For the reaction: 2 PbO 2 (s) + 4 HNO 3 ( aq )  2 PbNO 3 ( aq ) + 2 H 2 O (l) + O 2 (g) What volume of 7.91M solution of nitric acid, HNO 3 , is just sufficient to react with 15.9 g of lead dioxide, PbO 2 ? Answer: 0.0168 L

60. Volume Relationship of Gases in Chemical Reactions Gay Lussac’s law of combining volumes – states that the volumes of gaseous reactants and products stand in ratios of simple integers, as long as those volumes are measured at the SAME temperature and pressure. These integers are the same as the integers used to balance the chemical equation. A balanced chemical equation therefore provides a relationship between the volumes of gases reacting.

61. Problem: Gaseous H 2 and N 2 react according to the equation 3H 2 (g) + N 2 (g)  2 NH 3 (g) If in a given reaction 5.00 L of NH 3 are formed, what volume of H 2 reacted, assuming that the temperature and pressure before and after the reaction are the same. From the balanced reaction above, assuming the same temperature and pressure before and after the reaction, 2 volumes of NH 3 are formed from 3 volumes of H 2 Hence, 5.00 L of NH 3 are formed from (3/2)x5.00 L = 7.5 L of H 2

62. CONCEPT OF LIMITING REAGENT AND THEORATICAL YIELD

63. If we were to mix reactants together in non-stoichiometric amounts then what determines the amount of product formed? The reactant which is first used up determines the amount of product formed. This reactants is called the LIMITING REACTANT. The other reactants are in EXCESS when the reaction stops. If additional amounts of the limiting reactant is added the reaction starts again.

64. Limiting Reactants and Product Yields If everything went perfectly in the reaction : Fe(s) + S (s)  FeS (s) 55.8 g (1 mole) of iron will react with 32.1 g (1 mole) of S to form 87.9 g (1 mole) of FeS . If stoichiometric amounts of reactants were used, and assuming that there are no competing factors to limit the amount of products being formed, then a stoichiometric amount of product is formed

65. Real-World Stoichiometry: Limiting Reactants LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 366 Ideal Stoichiometry Limiting Reactants Fe + S FeS S = Fe = excess

66. Grilled Cheese Sandwich Bread + Cheese  ‘Cheese Melt’ 2 B + C  B 2 C 100 bread 30 slices ? sandwiches Multiple Guess: 130 sandwiches 100 sandwiches 90 sandwiches 60 sandwiches 30 sandwiches Not enough information given 30 sandwiches

67. Limiting Reactants Limiting Reactant used up in a reaction determines the amount of product Excess Reactant added to ensure that the other reactant is completely used up cheaper & easier to recycle Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

68. Limiting Reactants aluminum + chlorine gas  aluminum chloride Al(s) + Cl 2 (g)  AlCl 3 2 Al(s) + 3 Cl 2 (g)  2 AlCl 3 100 g 100 g ? g A. 200 g B. 125 g C. 667 g D. 494 g

69. Limiting Reactants aluminum + chlorine gas  aluminum chloride 2 Al(s) + 3 Cl 2 (g)  2 AlCl 3 100 g 100 g x g Al AlCl 3 x g AlCl 3 = 100 g Al 27 g Al = 494 g AlCl 3 1 mol Al 2 mol AlCl 3 2 mol Al 133.5 g AlCl 3 1 mol AlCl 3 How much product would be made if we begin with 100 g of aluminum? Cl 2 AlCl 3 x g AlCl 3 = 100 g Cl 2 71 g Cl 2 = 125 g AlCl 3 1 mol Cl 2 2 mol AlCl 3 3 mol Cl 2 133.5 g AlCl 3 1 mol AlCl 3 How much product would be made if we begin with 100 g of chlorine gas?

70. Limiting Reactants – Method 1 1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. 3. Smaller answer indicates: limiting reactant amount of product Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

71. Limiting Reactants – Method 2 Begin by writing a correctly balanced chemical equation Write down all quantitative values under equation (include units) Convert ALL reactants to units of moles Divide by the coefficient in front of each reactant The smallest value is the limiting reactant!

72. Theoratical Yield

73. 77 Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO 2 , is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) First copy down the the BALANCED equation! Now place numerical information below the compounds.

74. 78 Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO 2 , is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Two starting amounts? Where do we start? Hide one

75. 79 Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO 2 , is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Hide Ratio Based on: KO 2 0.15 mol KO 2 4 mol KO 2 0.1 mol H 2 O = 0.0375 to 0.5 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol Water is in excess, thus our limiting reactant is KO 2 2 mol H 2 O

76. 80 Potassium superoxide, KO 2 , is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? Determine the limiting reactant. KO 2 (s) Determine the moles of oxygen produced 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol excess ? moles Based on: KO 2 = mol O 2 0.15 mol KO 2 0.1 1 25 Limiting/Excess/ Reactant and Theoretical Yield Problems :

77. Problem: Calcium carbonate CaCO 3 (s) is decomposed by HCl( aq ) to give CaCl 2 ( aq ), CO 2 (g) and H 2 O(l). If 10.0 g of CaCO 3 are treated with 10.0 g of HCl. Which compound is the limiting reagent? how many grams of CO 2 are generated? If the actual yield of carbon dioxide is 2.5g, what is the percentage yield? First write a balanced equation for the reaction: CaCO 3 (s) + 2HCl( aq )  CaCl 2 ( aq ) + H 2 O(l) + CO 2 (g) 1 mole of CaCO 3 reacts with 2 moles of HCl to form 1 mole of CO 2 Moles of CaCO 3 = (10.0 g)/(100.0g/mol) = 0.100 mole CaCO 3 Moles of HCl = (10.0 g)/(36.5 g/mol) = 0.274 mol HCl

78. Hence 0.1 mole of CaCO 3 reacts with 0.2 moles of HCl. Since the amount of HCl present is greater that 0.2, HCl is in excess and CaCO 3 is the limiting reactant. Hence, 0.1 mole of CO 2 formed. Mass of CO 2 formed = 0.1 mol x 44.01 g/mol = 4.40 g CO 2

79. Ideally for an industrial process, the % yield should be large. Performing stoichiometric calculations, determining % yields, are important in analyzing the “success” of a chemical reaction. Sometime, a reaction performed on a lab-scale may have good product yields but when the reaction is scaled up to an industrial process, the yield may be lower.

80. PERCENTAGE YIELD

81. Theoretical yield : the maximum amount of product, which is calculated using the balanced equation • Actual yield: the amount of product obtained when the reaction takes place There are many reasons you might not get the expected amount of product from a reaction • Sources of error, nature of rxn • Percent Yield measures how much of the expected product you actually get • If a package of chocolate chips is supposed to make 48 cookies, but you only make 34 cookies out of it, you didn’t get 100% yield

82. Percent Yield Formula percent yield = actual yield x 100 theoretical yield

83. What is the percent yield of this reaction if 24.8 g of CaCO3 is heated to give 13.1 g of CaO ? CaCO3 à  CaO + CO2 • To find out how much product should be made: • Set the problem up like you would a Mass-Mass Stoichiometry Problem 1. BalanceTheEquation 2. Convert Grams to Moles 3. CompleteMoleRatio 4. Convert Moles to Grams • Then find % yield using the formula

84. What is the percent yield of this reaction if 24.8 g of CaCO 3 is heated to give 13.1 g of CaO? CaCO 3  CaO + CO 2 Balance The Equation: DONE Convert Grams to Moles: Molar Mass of CaCO3 100.09 g 24.8 g CaCO 3 = 0.247 Moles CaCO 3 100.09 g/mol 3. Complete Mole Ratio: CaCO 3 = _1__ = 0.247 M CaCO 3 = 0.247 M CaO               1                      1                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                           CaO 1 x M CaO 4. Convert Moles to Grams 0.247 M CaO × 56.08 g CaO = 13.9 g CaO                       

85. CaCO 3  CaO + CO 2 13.1 g CaO is the ACTUAL YIELD (it’s given in the problem!) 13.9 g CaO is the THEORETICAL YIELD (it’s what you just solved for) Now that you found out the theoretical value, plug your answer into the formula percent yield = actual yield x 100 theoretical yield percent yield = 13.1 g × 100 = 94.2 % 13.9 g                                       

86. THE END. ACKNOWLEDGEMENTS Hudson City Schools Teachlearnchem Oakpark Unified School District A nd other authors