1. Chapter 3 Stoichiometry AP Chemistry

2. Finding atomic, molecular and formula masses. Atomic mass is the mass of an atom of an element These are the “red numbers” on the “buck-o-five” card. Atomic mass is the mass of an atom of an element These are the “red numbers” on the “buck-o-five” card. Molecular mass is the sum of the atomic masses of the atoms present in a covalent molecule. Molecular mass is the sum of the atomic masses of the atoms present in a covalent molecule. Formula mass is the sum of the atomic masses of the atoms present in a formula unit of an ionic compound. Formula mass is the sum of the atomic masses of the atoms present in a formula unit of an ionic compound. All three are expressed in atomic mass units (amu). All three are expressed in atomic mass units (amu).

3. Example #1 What is the atomic mass of hydrogen? What is the atomic mass of hydrogen? Look on the periodic table and the red number for hydrogen is 1.0079 amu Look on the periodic table and the red number for hydrogen is 1.0079 amu We’ll always round to 4 digits for mass, We’ll always round to 4 digits for mass, so the answer is 1.008 amu.

4. Example #2 What is the molecular mass of water? What is the molecular mass of water? The formula for water is H 2 O, this represents two hydrogens and one oxygen. The formula for water is H 2 O, this represents two hydrogens and one oxygen. hydrogen’s mass is 1.008 x 2 = 2.016 hydrogen’s mass is 1.008 x 2 = 2.016 oxygen’s mass is 16.00 x 1 = 16.00 oxygen’s mass is 16.00 x 1 = 16.00 molecular mass of water 18.016 molecular mass of water 18.016

5. Example #3 What is the formula mass of sodium hydroxide? What is the formula mass of sodium hydroxide? The formula is NaOH. One sodium, one oxygen and one hydrogen. The formula is NaOH. One sodium, one oxygen and one hydrogen. sodium’s mass is 22.99 x 1 = 22.99 sodium’s mass is 22.99 x 1 = 22.99 hydrogen’s mass is 1.008 x 1 = 1.008 hydrogen’s mass is 1.008 x 1 = 1.008 oxygen’s mass is 16.00 x 1 = 16.00 oxygen’s mass is 16.00 x 1 = 16.00 formula mass of NaOH 39.998 formula mass of NaOH 39.998

6. Percent composition Percent composition is the % mass of each element in a compound. Percent composition is the % mass of each element in a compound. The formula is: The formula is: mass you’re interested in x 100% mass you’re interested in x 100% total mass

7. Example 4 What is the percent composition of water? What is the percent composition of water? First determine the molecular mass. First determine the molecular mass. hydrogen’s mass is 1.008 x 2 = 2.016 hydrogen’s mass is 1.008 x 2 = 2.016 oxygen’s mass is 16.00 x 1 = 16.00 molecular mass of water 18.016 Then divide each by the total and change it to a percentage. Then divide each by the total and change it to a percentage.

8. Example 4 cont’d % hydrogen is 2.016 x 100% = 11.2 % % hydrogen is 2.016 x 100% = 11.2 % 18.016 18.016 % oxygen is 16.00 x 100% = 88.8 % % oxygen is 16.00 x 100% = 88.8 % 18.016 18.016 Therefore, the percent composition of water is 11.2 % H and 88.8% O.

9. The Mole Chemists rarely work with single atoms or molecules and can seldom measure anything as small as an amu in the lab. A larger unit was needed to make the masses more “user-friendly”. Chemists rarely work with single atoms or molecules and can seldom measure anything as small as an amu in the lab. A larger unit was needed to make the masses more “user-friendly”. The mole was developed to be that unit. It was based upon Amedeo Avogadro’s theory about equal volumes of gases having the same number of molecules. The mole was developed to be that unit. It was based upon Amedeo Avogadro’s theory about equal volumes of gases having the same number of molecules.

10. The Mole cont’d Volumes of gases that had the same mass in grams as their atomic mass in amu were found to occupy 22.4 dm 3. Volumes of gases that had the same mass in grams as their atomic mass in amu were found to occupy 22.4 dm 3. The number of molecules of gas in that volume was calculated to be 6.02 x 10 23. The number of molecules of gas in that volume was calculated to be 6.02 x 10 23. If a mole is present, then the mass in amu is equal to the mass in grams. If a mole is present, then the mass in amu is equal to the mass in grams. If the mass in grams is equal to the mass in amu, then a mole is present. If the mass in grams is equal to the mass in amu, then a mole is present.

11. The Mole cont’d An atom of sodium has the mass of 22.99 amu and a mole of sodium atoms has the mass of 22.99 g. Same number, different unit. An atom of sodium has the mass of 22.99 amu and a mole of sodium atoms has the mass of 22.99 g. Same number, different unit. A water molecule has the mass of 18.016 amu and 6.02 x 10 23 water molecules have a mass of 18.016 g. A water molecule has the mass of 18.016 amu and 6.02 x 10 23 water molecules have a mass of 18.016 g. Masses in grams are called the gram atomic, gram formula or gram molecular masses. Masses in grams are called the gram atomic, gram formula or gram molecular masses.

12. Example 5 What is the mass in grams of 1 mole of carbon dioxide gas molecules? What is the mass in grams of 1 mole of carbon dioxide gas molecules? The formula is CO 2. The formula is CO 2. The mass is found by adding the masses. The mass is found by adding the masses. Carbon 12.01 x 1 = 12.01 Carbon 12.01 x 1 = 12.01 Oxygen 16.00 x 2 = 32.00 Oxygen 16.00 x 2 = 32.00 44.01 amu 44.01 amu One mole is therefore 44.01 g.

13. mass to moles & moles to mass One mole is the same number as amu, just expressed in grams. One mole is the same number as amu, just expressed in grams. One molecule of water is 18.016 amu. One molecule of water is 18.016 amu. One mole of water molecules is 18.016g. One mole of water molecules is 18.016g. If you have 18.016g of water, you have a mole of water. If you have 18.016g of water, you have a mole of water. MASS (g) MOLES X ÷ By the red numbers Moles to different, multiply. Different to moles, divide.

14. Example 6 How many grams of carbon dioxide are in 2.0 moles of carbon dioxide? How many grams of carbon dioxide are in 2.0 moles of carbon dioxide? The molecular mass of CO 2 is 44.01 amu. Since we’re working in moles, 1 mole of CO 2 is 44.01 g. 2.0 moles CO 2 x 44.01g CO 2 = 88.02g CO 2 2.0 moles CO 2 x 44.01g CO 2 = 88.02g CO 2 1mole CO 2 1mole CO 2

15. Example 7 How many moles of oxygen molecules are in 96.0 g of O 2 ? How many moles of oxygen molecules are in 96.0 g of O 2 ? The molecular mass of O 2 is 32.00 amu. Since we’re working in moles, 1 mole of O 2 is 32.0 g. 96.0 g O 2 x 1 mole O 2 = 3.00 moles O 2 96.0 g O 2 x 1 mole O 2 = 3.00 moles O 2 32.00 g O 2 32.00 g O 2

16. particles to moles & moles to particles One mole is One mole is 6.02 x 10 23 anything (Avogadro’s Number). One mole of water is 6.02 x 10 23 water molecules One mole of water is 6.02 x 10 23 water molecules If you have If you have 6.02 x 10 23 molecules of water, you have a mole of water. particlesMOLES X ÷ By Avogadro’s number Moles to different, multiply. Different to moles, divide.

17. Example 8 How many sodium atoms are in 2.0 moles of sodium? How many sodium atoms are in 2.0 moles of sodium? 2.0 moles Na x 6.02 x 10 23 Na atoms = 1.2 x 10 24 Na atoms 1 mole Na atoms 2.0 moles Na x 6.02 x 10 23 Na atoms = 1.2 x 10 24 Na atoms 1 mole Na atoms How many moles of hydrogen molecules are in 3.01 x 10 23 hydrogen molecules? How many moles of hydrogen molecules are in 3.01 x 10 23 hydrogen molecules? 3.01 x 10 23 H 2 x 1 mole H 2 molecules = 0.5 mole H 2 6.02 x 10 23 H 2 molecules molecules 3.01 x 10 23 H 2 x 1 mole H 2 molecules = 0.5 mole H 2 6.02 x 10 23 H 2 molecules molecules

18. volumes to moles & moles to volumes One mole of gas is 22.4 dm 3. One mole of gas is 22.4 dm 3. One mole of water is 22.4 dm 3 water vapor. One mole of water is 22.4 dm 3 water vapor. If you have 22.4 dm 3 water vapor, you have a mole of water vapor. If you have 22.4 dm 3 water vapor, you have a mole of water vapor. volumesMOLES X ÷ By 22.4 dm 3 Moles to different, multiply. Different to moles, divide.

19. Example 9 How many dm 3 of nitrogen gas are in 2.0 moles of nitrogen? How many dm 3 of nitrogen gas are in 2.0 moles of nitrogen? 2.0 moles N 2 x 22.4 dm 3 N 2 = 44.8 dm 3 N 2 2.0 moles N 2 x 22.4 dm 3 N 2 = 44.8 dm 3 N 2 1 mole N 2 1 mole N 2 How many moles of hydrogen molecules are in 67.2 dm 3 hydrogen molecules? How many moles of hydrogen molecules are in 67.2 dm 3 hydrogen molecules? 67.2 dm 3 H 2 x 1 mole H 2 = 3.0 moles H 2 67.2 dm 3 H 2 x 1 mole H 2 = 3.0 moles H 2 22.4 dm 3 H 2 22.4 dm 3 H 2

20. Empirical formula and molecular formula Empirical formula is the smallest whole number ratio of atoms of each element in a formula. Ionic compounds are always empirical. Molecular compounds can be empirical. If the real formula isn’t empirical then it is called the molecular formula. Empirical formula is the smallest whole number ratio of atoms of each element in a formula. Ionic compounds are always empirical. Molecular compounds can be empirical. If the real formula isn’t empirical then it is called the molecular formula. More info is needed to determine a molecular formula, it will be given. More info is needed to determine a molecular formula, it will be given.

21. How to determine empirical formula If the analysis is given in percent, assume a 100 g sample and change to grams. If the analysis is given in percent, assume a 100 g sample and change to grams. If the analysis is given in grams continue. If the analysis is given in grams continue. Divide the mass of each element by its gram atomic mass to get moles. Divide the mass of each element by its gram atomic mass to get moles. Divide each by the smallest molar value to get the smallest ratio. Divide each by the smallest molar value to get the smallest ratio. If necessary, multiply to get whole numbers for the ratios. If necessary, multiply to get whole numbers for the ratios.

22. E.F. Example What is the empirical formula for a compound that is 88.2% oxygen and 11.2% hydrogen? What is the empirical formula for a compound that is 88.2% oxygen and 11.2% hydrogen? Change % to grams (just change % to g). Change % to grams (just change % to g). 88.2 g O x 1 mole O = 5.51 moles O 88.2 g O x 1 mole O = 5.51 moles O 16.00 g O 16.00 g O 11.2 g H x 1 mole H = 11.1 moles H 11.2 g H x 1 mole H = 11.1 moles H 1.008 g H 1.008 g H

23. E.F. Example cont’d Divide each by the smallest: Divide each by the smallest: 5.51 moles O = 1.00 mole O 5.51 moles O = 1.00 mole O5.51 11.1 moles H = 2.01 moles H 11.1 moles H = 2.01 moles H5.51 The ratio is 2 H : 1 O, the formula is H 2 0. The ratio is 2 H : 1 O, the formula is H 2 0.

24. AP Level E.F. Example A 3.489 g sample of a compound containing C, H, and O yields 7.832 g of carbon dioxide, and 1.922 grams of water upon combustion. A 3.489 g sample of a compound containing C, H, and O yields 7.832 g of carbon dioxide, and 1.922 grams of water upon combustion. (a) Calculate the number of moles of carbon produced. (b) Calculate the number of moles of hydrogen produced. (c) Calculate the number of moles of oxygen in the compound. (d) Calculate the empirical formula of the compound

25. AP Level E.F. Example, cont'd (a) Calculate the number of moles of carbon produced: C x H y O z + O 2 ----> CO 2 + H 2 O C x H y O z + O 2 ----> CO 2 + H 2 O 7.832 g CO x 1 mol CO 2 x 1 mol C 44.01 g CO 2 1 mol CO 2 44.01 g CO 2 1 mol CO 2 = 0.1780 mol C

26. AP Level E.F. Example, cont'd (b) Calculate the number of moles of hydrogen produced. 1.922 g H O x 1 mol H 2 O x 2 mol H 18.02 g H 2 O 1 mol H 2 O 18.02 g H 2 O 1 mol H 2 O = 0.2134 mol H

27. AP Level Example, cont'd (c) Calculate the number of moles of oxygen in the compound. Mass oxygen = mass compound – mass hydrogen – mass carbon 0.1780 mol C x 12.01 g C = 2.138 g C 1 mol C 1 mol C 0.2134 mol H x 1.008 g H = 0.2151 g H 1 mol H 1 mol H

28. AP Level Example, cont'd (c) Mass oxygen = 3.489 g –2.138 g-0.2151 g = 1.136 g O 1.136 g O x 1 mol O = 0.07100 mol O 16.00 g O 16.00 g O

29. AP Level Example, cont'd (d) Calculate the empirical formula of the compound. 0.1780 mol C = 2.5 x 2 = 5 0.07100 mol O 0.2134 mol H = 3 x 2 = 6 0.07100 mol O 0.07100 mol O = 1 x 2 = 2 0.07100 mol O

30. AP Level E.F. Example, cont'd (d) C 5 H 6 O 2

31. Writing a chemical equation. FOR A CHEMICAL EQUATION TO BE VALID: FOR A CHEMICAL EQUATION TO BE VALID: THE EQUATION MUST REPRESENT THE KNOWN FACTS. THE EQUATION MUST REPRESENT THE KNOWN FACTS. THE EQUATION MUST CONTAIN THE CORRECT FORMULAS OF THE REACTANTS AND PRODUCTS. THE EQUATION MUST CONTAIN THE CORRECT FORMULAS OF THE REACTANTS AND PRODUCTS. THE LAW OF CONSERVATION OF MATTER MUST BE SATISFIED. THE LAW OF CONSERVATION OF MATTER MUST BE SATISFIED.

32. When balancing: BALANCE ATOMS THAT APPEAR ONLY ONCE PER SIDE FIRST. BALANCE ATOMS THAT APPEAR ONLY ONCE PER SIDE FIRST. TREAT POLYATOMIC IONS AS UNITS IF THEY APPEAR ON BOTH SIDES OF THE EQUATION. TREAT POLYATOMIC IONS AS UNITS IF THEY APPEAR ON BOTH SIDES OF THE EQUATION. BALANCE ANY REMAINING NON-H OR O ATOMS NEXT. BALANCE ANY REMAINING NON-H OR O ATOMS NEXT. BALANCE H NEXT. BALANCE H NEXT. BALANCE O LAST. BALANCE O LAST.

33. AVOID MISTAKES-USE THE CORRECT CHEMICAL FORMULA! DON’T CHANGE SUBSCRIPTS!!!!

34. Three Types of Equations The first type of Chemical Equation is the word equation The first type of Chemical Equation is the word equation Word equations are expressed as you would speak a sentence. Word equations are expressed as you would speak a sentence. For example: For example: HYDROGEN PLUS OXYGEN REACT TO FORM WATER. HYDROGEN PLUS OXYGEN REACT TO FORM WATER.

35. Second type: The skeleton or unbalanced chemical equation Correct chemical formulas and symbols are substituted for the words. Correct chemical formulas and symbols are substituted for the words. For the word equation: For the word equation: HYDROGEN PLUS OXYGEN REACT TO FORM WATER HYDROGEN PLUS OXYGEN REACT TO FORM WATER We write H 2 + 0 2 → H 2 O We write H 2 + 0 2 → H 2 O

36. The third type is the balanced chemical reaction The law of conservation of matter must be obeyed. For every atom of oxygen on the reactant side, an atom of oxygen must be represented on the product side. To achieve this you add coefficients in front of the chemical formulas you need more of. The law of conservation of matter must be obeyed. For every atom of oxygen on the reactant side, an atom of oxygen must be represented on the product side. To achieve this you add coefficients in front of the chemical formulas you need more of. DO NOT CHANGE SUBSCRIPTS DO NOT CHANGE SUBSCRIPTS

37. H 2 + O 2 → H 2 0 is not balanced H 2 + O 2 → H 2 0 is not balanced Both sides have two H, but the product side only has one 0. We can't change the subscripts but we can have more than one H 2 O. Add another H 2 O then the 0 are balanced but we have two H on the reactant side and four on the product side. We balance the H by adding another H 2. This gives the balanced chemical equation… Both sides have two H, but the product side only has one 0. We can't change the subscripts but we can have more than one H 2 O. Add another H 2 O then the 0 are balanced but we have two H on the reactant side and four on the product side. We balance the H by adding another H 2. This gives the balanced chemical equation…

38. 2H 2 + O 2 → 2H 2 O This is a balanced chemical equation!

39. COMMON SYMBOLS USED IN CHEMICAL EQUATIONS It is often useful to know what state of matter or phase a reactant or product is in. They are not to be balanced. It is often useful to know what state of matter or phase a reactant or product is in. They are not to be balanced. (aq) aqueous dissolved or made up in a water solution (aq) aqueous dissolved or made up in a water solution (I) exists as a liquid during this rxn (I) exists as a liquid during this rxn (s) exists as a solid during this rxn (s) exists as a solid during this rxn (cr) exists as a crystal solid during the rxn (cr) exists as a crystal solid during the rxn (g) exists as a gas during the rxn (g) exists as a gas during the rxn (↓), precipitates falls out of the solution as a solid product (↓), precipitates falls out of the solution as a solid product (↑) gas product leaves the solution as a gas product (↑) gas product leaves the solution as a gas product → yields separates the reactants from the products indicates the direction of the rxn → yields separates the reactants from the products indicates the direction of the rxn

40. ↔ reversible rxn can go forwards or backwards ↔ reversible rxn can go forwards or backwards

41. Five types of chemical reactions. Synthesis, composition or combination Synthesis, composition or combination Analysis or decomposition Analysis or decomposition Single replacement or displacement Single replacement or displacement Double replacement or displacement (metathesis) Double replacement or displacement (metathesis) Combustion Combustion

42. Synthesis, composition or combination A + X → AX A + X → AX RXN OF ELEMENT WITH OXYGEN → OXIDE RXN OF ELEMENT WITH OXYGEN → OXIDE RXN OF 2 NONMETALS → COVALENT COMPOUND RXN OF 2 NONMETALS → COVALENT COMPOUND RXN OF METAL WITH NONMETALS, NOT OXYGEN → SALTS RXN OF METAL WITH NONMETALS, NOT OXYGEN → SALTS RXN OF OXIDES AND WATER → HYDROXIDES RXN OF OXIDES AND WATER → HYDROXIDES

43. Analysis or decomposition AX → A + X AX → A + X DECOMP OF BINARY → ELEMENTS DECOMP OF BINARY → ELEMENTS DECOMP OF METAL CARBONATE → METAL OXIDE AND CO 2 DECOMP OF METAL CARBONATE → METAL OXIDE AND CO 2 DECOMP OF METAL HYDROXIDE → METAL OXIDE AND H 2 O DECOMP OF METAL HYDROXIDE → METAL OXIDE AND H 2 O DECOMP OF METAL CHLORATES → METAL CHLORIDE AND 0 2 DECOMP OF METAL CHLORATES → METAL CHLORIDE AND 0 2 DECOMP OF ACIDS → NONMETAL OXIDES AND H 2 O DECOMP OF ACIDS → NONMETAL OXIDES AND H 2 O

44. Single replacement or displacement A + BX → AX + B or Y + BX → BY + X A + BX → AX + B or Y + BX → BY + X REPLACE. OF A METAL BY A MORE REACTIVE METAL REPLACE. OF A METAL BY A MORE REACTIVE METAL REPLACE. OF H 2 IN H 2 0 BY A METAL → METAL HYDROXIDES AND H 2 REPLACE. OF H 2 IN H 2 0 BY A METAL → METAL HYDROXIDES AND H 2 REPLACE. OF H 2 IN AN ACID BY A METAL → SALT AND H 2. REPLACE. OF H 2 IN AN ACID BY A METAL → SALT AND H 2. REPLACE. OF HALOGENS REPLACE. OF HALOGENS

45. Double replacement or displacement AY + BX → AX + BY AY + BX → AX + BY FORMATION OF PRECIP.-ANIONS OF ONE FORMATION OF PRECIP.-ANIONS OF ONE REACTANT COMBINE WITH CATIONS OF THE OTHER REACTANT TO FORM AN INSOLUBLE OR SLIGHTLY INSOLUBLE COMPOUND. p 920 FORMATION OF GAS-WHEN THE PRODUCT PRODUCES A GAS WHICH IS INSOLUBLE IN THE NEW SOLUTION. THIS GAS WILL BUBBLE OUT OF SOLUTION. FORMATION OF GAS-WHEN THE PRODUCT PRODUCES A GAS WHICH IS INSOLUBLE IN THE NEW SOLUTION. THIS GAS WILL BUBBLE OUT OF SOLUTION.

46. Combustion C x H x + O 2 → CO 2 + H 2 O C x H x + O 2 → CO 2 + H 2 O Hydrocarbon or carbohydrate and oxygen yields carbon dioxide and water Hydrocarbon or carbohydrate and oxygen yields carbon dioxide and water

47. Reactions that form precipitates Complete ionic equation - break the chemical formulas into the ions they are formed from. Be sure to include symbols. Complete ionic equation - break the chemical formulas into the ions they are formed from. Be sure to include symbols. Start with: Start with: AgNO 3 (aq) + NaCl(aq) → NaNO 3 (aq) + AgCl(s) Break it down into the aqueous ions: Break it down into the aqueous ions: Ag +1 (aq) + NO 3 -1 (aq) +Na +1 (aq) + Cl -1 (aq) → Na +1 (aq) + NO 3 -1 (aq) + AgCl(s) Ag +1 (aq) + NO 3 -1 (aq) +Na +1 (aq) + Cl -1 (aq) → Na +1 (aq) + NO 3 -1 (aq) + AgCl(s)

48. Coefficients in a balanced chemical equation can be used to represent: 1. Moles 1. Moles 2. Particles (Molecules, Atoms or Formula Units) 2. Particles (Molecules, Atoms or Formula Units) 3. Volumes of Gases 3. Volumes of Gases BUT NEVER MASS!!! BUT NEVER MASS!!!

49. If the problem involves mass, CONVERT TO MOLES FIRST! CONVERT TO MOLES FIRST! If you are given atoms and the answer needs to be in volumes, CONVERT TO MOLES FIRST! If you are given atoms and the answer needs to be in volumes, CONVERT TO MOLES FIRST! If you are given mass and asked for molecules, CONVERT TO MOLES FIRST! If you are given mass and asked for molecules, CONVERT TO MOLES FIRST!

50. For the rxn: 2H 2 + 0 2 → 2H 2 O The coefficient 2 in front of the H 2 can represent any of these. The coefficient 2 in front of the H 2 can represent any of these. 2 Moles of Hydrogen Molecules 2 Moles of Hydrogen Molecules 2 Molecules of Hydrogen 2 Molecules of Hydrogen 2 Volumes of Hydrogen Gas 2 Volumes of Hydrogen Gas

51. Molar Ratio The molar ratio is the ratio of coefficients in a BALANCED chemical rxn of the reactants to the products. The molar ratio is the ratio of coefficients in a BALANCED chemical rxn of the reactants to the products.

52. For the reaction: 4 Fe + 3 0 2 → 2 Fe 2 0 3 the ratio of the coefficients for iron and oxygen is 4 : 3. Therefore, 4 moles of iron react with 3 moles of oxygen to produce 2 moles of iron (III) oxide. Also, 4 atoms of iron react with 3 molecules of oxygen to produce 2 formula units of iron (III) oxide. the ratio of the coefficients for iron and oxygen is 4 : 3. Therefore, 4 moles of iron react with 3 moles of oxygen to produce 2 moles of iron (III) oxide. Also, 4 atoms of iron react with 3 molecules of oxygen to produce 2 formula units of iron (III) oxide.

53. BUT 4 grams of iron do not react with 3 grams of oxygen to yield 2 grams of iron (III) oxide!!! BUT 4 grams of iron do not react with 3 grams of oxygen to yield 2 grams of iron (III) oxide!!! For the reaction: 4 Fe + 3 0 2 → 2 Fe 2 0 3

54. REMEMBER: Convert to Moles 1 st ! Convert to Moles 1 st ! Coefficients never represent mass. Coefficients never represent mass.

55. Rules for Stoichiometry 1. Balance the equation. 1. Balance the equation. 2. Convert given to moles 2. Convert given to moles 3. Use molar ratio. 3. Use molar ratio. 4. Convert from moles to what you want. 4. Convert from moles to what you want.

56. Example #1: How many grams of iron (III) oxide are produced by 111.7 g of iron, when it reacts with oxygen? How many grams of iron (III) oxide are produced by 111.7 g of iron, when it reacts with oxygen? Fe + 0 2 → Fe 2 0 3 Fe + 0 2 → Fe 2 0 3 Balance: 4 Fe + 3 0 2 → 2 Fe 2 0 3 Balance: 4 Fe + 3 0 2 → 2 Fe 2 0 3 Convert given to moles : Convert given to moles : 111.7 g Fe x 1 mole Fe = 2 moles Fe 55.85 g Fe 55.85 g Fe

57. Example #1 cont’d Find molar ratio: Find molar ratio: 4:2 reduces to 2:1 Fe: Fe 2 0 3 4:2 reduces to 2:1 Fe: Fe 2 0 3 2 moles Fe :1 mole Fe 2 0 3. Convert moles to grams: Convert moles to grams: 1 mole Fe 2 0 3 x 159.7 g Fe 2 0 3 = 159.7 g Fe 2 0 3 1 mole Fe 2 0 3 x 159.7 g Fe 2 0 3 = 159.7 g Fe 2 0 3 1 mole Fe 2 0 3 1 mole Fe 2 0 3

58. Percent Yield In a perfect world all reactions use up all reactants and none are lost or spilled, etc. However, in the real world the reactions often produce less than anticipated. The amount produced compared to amount that would be produced in a "perfect world" is percent yield. In a perfect world all reactions use up all reactants and none are lost or spilled, etc. However, in the real world the reactions often produce less than anticipated. The amount produced compared to amount that would be produced in a "perfect world" is percent yield.

59. Percent Yield Formula % yield = Actual Yield x 100% % yield = Actual Yield x 100% Calculated Yield Calculated Yield

60. Example #2: 140.0g of iron (III) oxide are produced by 111.7 g of iron, when it reacts with oxygen, what is the percent yield? 140.0g of iron (III) oxide are produced by 111.7 g of iron, when it reacts with oxygen, what is the percent yield? Fe + 0 2 → Fe 2 0 3 Fe + 0 2 → Fe 2 0 3 Balance: 4 Fe + 30 2 →2 Fe 2 0 3 Balance: 4 Fe + 30 2 →2 Fe 2 0 3 Convert given to moles: Convert given to moles: 111.7 g Fe x 1 mole Fe= 2 moles Fe 55.85 g Fe 55.85 g Fe

61. Example #2 cont’d Find molar ratio: Find molar ratio: 4:2 reduces to 2:1 Fe: Fe 2 0 3 4:2 reduces to 2:1 Fe: Fe 2 0 3 2 moles Fe : 1 mole Fe 2 0 3 2 moles Fe : 1 mole Fe 2 0 3 Convert moles to grams : Convert moles to grams : 1 mole Fe 2 0 3 x 159.7 g Fe 2 0 3 = 159.7 g Fe 2 0 3 1 mole Fe 2 0 3 x 159.7 g Fe 2 0 3 = 159.7 g Fe 2 0 3 1 mole Fe 2 0 3 1 mole Fe 2 0 3 This is the calculated yield

62. Example #2 cont’d % yield = Actual Yield x 100% % yield = Actual Yield x 100% Calculated Yield Calculated Yield % yield = 140.0 g Fe 2 0 3 x 100% = 87.66 % % yield = 140.0 g Fe 2 0 3 x 100% = 87.66 % 159.7 g Fe 2 0 3 159.7 g Fe 2 0 3

63. Limiting Reactant (LR) and Excess Reactant (XS). It is seldom that all of the reactants in a chemical rxn are used up at exactly the same time. It is seldom that all of the reactants in a chemical rxn are used up at exactly the same time. When one of the reactants is used up the reaction stops. When one of the reactants is used up the reaction stops.

64. LR and XS The reactant that is used up first is the limiting reactant (LR). The reactant that is left over is the excess reactant (XS). Since the amount of product produced depends upon the amount of limiting reactant, it is very important that you know how to determine which reactant runs out first. Use the limiting reactant to calculate yield. The reactant that is used up first is the limiting reactant (LR). The reactant that is left over is the excess reactant (XS). Since the amount of product produced depends upon the amount of limiting reactant, it is very important that you know how to determine which reactant runs out first. Use the limiting reactant to calculate yield.

65. Example #3 How many grams of iron (III) oxide are produced by 111.7 g of iron, when it reacts with 96.0 g of oxygen? What is the LR? How many grams of iron (III) oxide are produced by 111.7 g of iron, when it reacts with 96.0 g of oxygen? What is the LR? Fe + 0 2 → Fe 2 0 3 Fe + 0 2 → Fe 2 0 3 Balance: 4 Fe + 30 2 → 2 Fe 2 0 3 Balance: 4 Fe + 30 2 → 2 Fe 2 0 3

66. Example #3 cont’d Convert both givens to moles: Convert both givens to moles: 111.7 g Fe x 1 mole Fe = 2 moles Fe 55.85 g Fe 55.85 g Fe 96.0 g O 2 x 1 mole O 2 = 3 moles O 2 32.00 g O 2 32.00 g O 2

67. Example #3 cont’d Compare moles of the reactants: From the eq., if all 3 moles of 0 2 are used it would require 4 moles of Fe. Since we only have 2 moles of Fe, it will be the limiting reactant (LR). Compare moles of the reactants: From the eq., if all 3 moles of 0 2 are used it would require 4 moles of Fe. Since we only have 2 moles of Fe, it will be the limiting reactant (LR). Fe is the LR, 0 2 is XS Fe is the LR, 0 2 is XS

68. Example #3 cont’d Find molar ratio using the LR (always use the LR) Find molar ratio using the LR (always use the LR) 4:2 reduces to 2:1 Fe: Fe 2 0 3 4:2 reduces to 2:1 Fe: Fe 2 0 3 2 moles Fe : 1 mole Fe 2 0 3 2 moles Fe : 1 mole Fe 2 0 3 Convert moles to grams: Convert moles to grams: 1 mole Fe 2 0 3 x 159.7 g Fe 2 0 3 = 159.7 g Fe 2 0 3 1 mole Fe 2 0 3 x 159.7 g Fe 2 0 3 = 159.7 g Fe 2 0 3 1 mole Fe 2 0 3 1 mole Fe 2 0 3

69. Example #4 In the above problem, how much oxygen is left over after the reaction stops? In the above problem, how much oxygen is left over after the reaction stops?

70. Example #4 2 moles of Fe were used. 2 moles of Fe were used. From the ratio for Fe:0 2, 4:3, From the ratio for Fe:0 2, 4:3, 2 moles of Fe require 1.5 moles of 0 2. 2 moles of Fe require 1.5 moles of 0 2. 1.5 mole 0 2 x 32.00 g 0 2 = 48.00 g 0 2 1 mole 0 2 1.5 mole 0 2 x 32.00 g 0 2 = 48.00 g 0 2 1 mole 0 2