1. { Topic 1: Stoichiometric Relationships 1.1-1.3 Honors Chemistry Mrs. Peters 2014-15

2. Applications: 1. Deduction of chemical equations when reactants and products are specified. 2. Application of the state symbols (s), (l), (g), and (aq) in equations. *Understandings 1-3, Application 3 were covered in Unit 1 with Topic 11. 1.1: Introduction to the particulate nature of matter and chemical change

3. What is a chemical reaction? Rearrangement of atoms forming new substances Reactants  Products Atoms (matter) are not created or destroyed, they are rearranged! (Law of conservation of matter) 1.1 A1. Deduce chemical equations when all reactants and products are specified

4. Some reactions are desirable… Glucose + oxygen  Carbon dioxide + water …some are not. Iron + oxygen  iron (III) oxide 1.1 A1. Deduce chemical equations when all reactants and products are specified

5. Symbols used in chemical reactions + Used to separate two reactants or products “Yields” separates products from reactants “Yields” separates products from reactants  Used in place of  for reversible reactions 1.1 A.2 Application of the state symbols (s), (l), (g) and (aq)

6. Symbols used in chemical reactions *(s) Designates a solid reactant or product *(l) Designates a liquid reactant or product *(g) Designates a gaseous reactant or product *(aq) Designates an aqueous reactant or product  Indicates that heat is supplied to the reaction  Indicates that heat is supplied to the reaction * Required 1.1 A.2 Application of the state symbols (s), (l), (g) and (aq)

7. Balancing chemical equations The Law of Conservation of Matter: In a chemical (non- nuclear) reaction, atoms are neither created nor destroyed. For an equation to be balanced the number of atoms of each element is the same on both sides of the equation. H 2 (g) + O 2 (g)  H 2 O (l)(unbalanced) 2H 2 (g) + O 2 (g)  2H 2 O (l)(balanced) K(s) + H 2 O(l)  KOH (aq) + H 2 (g)(unbalanced) 2K(s) + 2H 2 O(l)  2KOH (aq) + H 2 (g)(balanced) C 6 H 12 O 6 + O 2  CO 2 + H 2 O(unbalanced) C 6 H 12 O 6 + 6O 2  6CO 2 + 6H 2 O(balanced) 1.1 A1. Deduce chemical equations when all reactants and products are specified.

8. Rules for Balancing Equations 1)Write the correct formulas for the reactants on the left side of the yield sign and products on the right side. 2)Count the number of atoms of each element in the products and the reactants. 3)Balance the elements one at a time by using coefficients. Do not change the subscripts in the chemical formulas. 4)Check each atom or polyatomic ion to make sure the equation is balanced. 5)Make sure all coefficients are in the lowest possible ratio. 1.1 A1. Deduce chemical equations when all reactants and products are specified.

9. Rules for Balancing Equations 1)Write the correct formulas for the reactants on the left side of the yield sign and products on the right side. 2)Count the number of atoms of each element in the products and the reactants. 3)Balance the elements one at a time by using coefficients. Do not change the subscripts in the chemical formulas. 4)Check each atom or polyatomic ion to make sure the equation is balanced. 5)Make sure all coefficients are in the lowest possible ratio. N 2 + H 2  NH 3 1) Formulas given 2) ReactantsProducts 2N 2H  1N 3H 3) Balance N by putting 2 in front of NH 3 N 2 + H 2  2NH 3 Balance H by putting 3 in front of H 2 N 2 + 3H 2  2NH 3 4) 2N 6H  2N 6H 5) Lowest ratio of coefficients

10. 1.1 A1. Deduce chemical equations when all reactants and products are specified. Rules for Balancing Equations 1)Write the correct formulas for the reactants on the left side of the yield sign and products on the right side. 2)Count the number of atoms of each element in the products and the reactants. 3)Balance the elements one at a time by using coefficients. Do not change the subscripts in the chemical formulas. 4)Check each atom or polyatomic ion to make sure the equation is balanced. 5)Make sure all coefficients are in the lowest possible ratio. KClO 3  KCl + O 2 1) Formulas given 2) Reactants Products 1K 1Cl 3O  1K 1Cl 2O 3) Balance O by putting 2 in front of KClO 3 and 3 in front of O 2 2KClO 3  KCl + 3O 2 Balance K & Cl by putting 2 in front of KCl Balance K & Cl by putting 2 in front of KCl 2KClO 3  2KCl + 3O 2 4) 2K 2Cl 6O  2K 2Cl 6O 5) Lowest ratio of coefficients

11. Balance the following equations  Fe(s) + O 2 (g)  Fe 2 O 3 (s)  CaCO 3 (s)  CaO(s) + CO 2 (g)  Al(s) + Fe 2 O 3 (s)  Al 2 O 3 (s) + Fe(s)  H 2 SO 4 (aq) + NaCN(aq)  Na 2 SO 4 (aq) + HCN(g)  C 2 H 5 OH(l) + O 2 (g)  CO 2 (g) + H 2 O(g) 1.1 A1. Deduce chemical equations when all reactants and products are specified.

12. Balance the following equations  4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s)  CaCO 3 (s)  CaO(s) + CO 2 (g)  2Al(s) + Fe 2 O 3 (s)  Al 2 O 3 (s) + 2Fe(s)  H 2 SO 4 (aq) + 2NaCN(aq)  Na 2 SO 4 (aq) + 2HCN(g)  C 2 H 5 OH(l) + 3O 2 (g)  2CO 2 (g) + 3H 2 O(g) 1.1 A1. Deduce chemical equations when all reactants and products are specified.

13. The 7 Diatomics (KNOW THESE) Seven elements are diatomic, meaning they have to have two atoms when alone: H 2, N 2, O 2, F 2, Cl 2, Br 2, I 2 1.1 A1. Deduce chemical equations when all reactants and products are specified.

14. Write the equation, then balance: Aluminum bromide + chlorine yield aluminum chloride + bromine AlBr 3 + Cl 2  AlCl 3 + Br 2

15. 1.1 A1. Deduce chemical equations when all reactants and products are specified. Write the equation, then balance: Aluminum bromide + chlorine yield aluminum chloride + bromine 2AlBr 3 + 3Cl 2  2AlCl 3 + 3Br 2

16. 1.1 A1. Deduce chemical equations when all reactants and products are specified. Write the equation, then balance: Copper + oxygen produces copper(I) oxide Cu + O 2  Cu 2 O

17. 1.1 A1. Deduce chemical equations when all reactants and products are specified. Write the equation, then balance: Copper + oxygen produces copper(I) oxide 4Cu + O 2  2Cu 2 O

18. Write the equations, then balance: 1. Sodium chlorate decomposes to sodium chloride and oxygen gas 2. Aluminum nitrate plus sodium hydroxide yields aluminum hydroxide plus sodium nitrate 3. Ethane (C 2 H 6 ) burns in oxygen to produce carbon dioxide and water vapor 1.1 A1. Deduce chemical equations when all reactants and products are specified.

19. 1. Sodium chlorate decomposes to sodium chloride and oxygen gas NaClO 3  NaCl + O 2 2. Aluminum nitrate plus sodium hydroxide yields aluminum hydroxide plus sodium nitrate Al(NO 3 ) 3 + NaOH  Al(OH) 3 + NaNO 3 3. Ethanol (C 2 H 5 OH) burns in oxygen to produce carbon dioxide and water vapor C 2 H 5 OH + O 2  CO 2 + H 2 O 1.1 A1. Deduce chemical equations when all reactants and products are specified.

20. 1. Sodium chlorate decomposes to sodium chloride and oxygen gas 2NaClO 3  2NaCl + 3O 2 2. Aluminum nitrate plus sodium hydroxide yields aluminum hydroxide plus sodium nitrate Al(NO 3 ) 3 + 3NaOH  Al(OH) 3 + 3NaNO 3 3. Ethanol (C 2 H 5 OH) burns in oxygen to produce carbon dioxide and water vapor C 2 H 5 OH + 3O 2  2CO 2 + 3H 2 O 1.1 A1. Deduce chemical equations when all reactants and products are specified.

21. Two or more substances combine to form a single substance. Synthesis or Combination Reaction: Two or more substances combine to form a single substance. R + S  RS  2Na(s) + Cl 2 (g)  2NaCl(s)  CaO(s) + H 2 O(l)  Ca(OH) 2 (aq) 1.1 A1. Deduce types of reactions

22. A single substance is broken down into two or more substances. Decomposition Reactions: A single substance is broken down into two or more substances. RS  R + S  CaCO 3 (s)  CaO(s) + CO 2 (g)  2H 2 O(l)  2H 2 (g) + O 2 (g) 1.1 A1. Deduce types of reactions

23. Single-Replacement Reactions: One element replaces another in a compound. R + ST  RT + S  Mg(s) + Zn(NO 3 ) 2 (aq)  Mg(NO 3 ) 2 (aq) + Zn(s)  2K(s) + 2H 2 O(l)  2KOH (aq) + H 2 (g) 1.1 A1. Deduce types of reactions

24. Double-Replacement Reactions: Ions of two reacting compounds trade places. RS + TU  RU + TS (cation1 anion1 + cation2 anion2  cation1 anion2 + cation2 anion1)  Na 2 CO 3 (aq) + CaCl 2 (aq)  CaCO 3 (s) + 2NaCl(aq)  Na 2 S(aq) + Cd(NO 3 ) 2 (aq)  CdS(s) + 2NaNO 3 (aq) 1.1 A1. Deduce types of reactions

25. Combustion Reactions: An element or compound reacts with oxygen. Products are usually carbon dioxide and water. C x H y + O 2  CO 2 + H 2 O  CH 4 (g) + O 2 (g)  CO 2 (g) + H 2 O(g)  C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g) 1.1 A1. Deduce types of reactions

26. Combination Reaction: R + S  RS Decomposition Reaction: RS  R + S Single-Replacement Reaction: T + RS  TS + R Double-Replacement Reaction: R + S - + T + U -  R + U - + T + S - Combustion Reaction: C x H y + O 2  CO 2 + H 2 O 1.1 A1. Identify the types of reactions to predict products

27. How do you calculate the quantity of reactants and products in chemical reactions? The calculation of quantities in chemical reactions is called stoichiometry. Use the chemical equation to predict ratio of reactants and products N 2 +3H 2  2NH 3 1 molecule N 2 reacts with 3 molecules H 2 producing 2 molecules NH 3 1 mole N 2 reacts with 3 moles H 2 producing 2 moles NH 3 The balanced equations gives us the ratio in particles or moles (not mass) of the chemicals involved in the reaction. The ratio of N 2 to NH 3 is 1:2 The ratio of NH 3 to H 2 is 2:3 The ratio of H 2 to N 2 is 3:1 1.1 Identify the mole ratio of any two species in a chemical reaction.

28. Coefficients are in particles or in moles 2 NaClO 3  2 NaCl + 3 O 2 We will use moles because we measure moles in grams The coefficients give us mole ratios Mole ratio NaClO 3 :NaCl is 2:2 Mole ratio NaCl:O 2 is 2:3 Mole ratio O 2 :NaCl is 3:2 1.1 Identify the mole ratio of any two species in a chemical reaction.

29. Consider the following equation 2K 2 Cr 2 O 7 + 2H 2 O + 3S  4KOH + 2Cr 2 O 3 + 3SO 2 What is the KOH:S mole ratio? What is the K 2 Cr 2 O 7 :Cr 2 O 3 mole ratio? What is the mole ratio between sulfur dioxide and water? What species will give mole ratios of 4:2? 1.1 Identify the mole ratio of any two species in a chemical reaction

30. Consider the following equation 2K 2 Cr 2 O 7 + 2H 2 O + 3S  4KOH + 2Cr 2 O 3 + 3SO 2 What is the KOH:S mole ratio? 4:3 What is the K 2 Cr 2 O 7 :Cr 2 O 3 2:2 What is the mole ratio between sulfur dioxide and water? 3:2 What species will give mole ratios of 4:2? KOH:K 2 Cr 2 O 7 KOH:H 2 OKOH:Cr 2 O 3 1.1 Identify the mole ratio of any two species in a chemical reaction.

31. Consider the reaction: 2C 8 H 18 + 25O 2  16CO 2 + 18H 2 O What is the ratio of O 2 to CO 2 ? 25:16 What is the ratio of H 2 O to C 8 H 18 ? 18:2 What is the ratio of CO 2 to H 2 O? What is the ratio of O 2 to C 8 H 18 ? Remember these ratios are in particles or moles. We will be using moles. 1.1 Identify the mole ratio of any two species in a chemical reaction.

32. Understandings: 1. The mole is a fixed number of particles and refers to the amount,n, of a substance. 2. Masses of atoms are compared on a scale relative to 12 C and are expressed as relative atomic mass (A r ) and relative formula/molecular mass (M r ) 3. Molar Mass (M) has the units g mol -1 4. The empirical formula and molecular formula of a compound give the simplest ratio and the actual number of atoms present in a molecule respectively. 1.2: The Mole Concept

33. Applications and Skills: 1. Calculation of the molar masses of atoms, ions, molecules and formula units. 2. Solutions of problems involving the relationships between the number of particles, the amount of substance in moles and the mass in grams. 3. Inter conversion of the percentage composition by mass and the empirical formula 4. Determination of the molecular formula of a compound from its empirical formula and molar mass 5. Obtaining and using experimental data for deriving empirical formulas from reactions involving mass changes. 1.2: The Mole Concept

34. 1.2 The Mole 34 1 mole of any substance contains 6.02 x 10 23 particles of that substance  This is called Avogadro’s number (named after a scientist).  Applies to all kinds of particles such as: atoms, molecules, formula units, ions, electrons  How many atoms are in 1 mol of Au?6.02 X 10 23 atoms  How many molecules are in 1 mol of CO?6.02 x 10 23 molecules  How many formula units are in 1 mol of NaCl?6.02 x 10 23 f.u.  How many ions are in 1 mol of S 2- ?6.02 x 10 23 ions  How many electrons are in 1 mol of H atoms?6.02 x 10 23 electrons

35. 1.2 U2. Relative atomic mass (A r ) and relative molecular mass (M r ). 35  The relative atomic mass (A r ) of an element is the weighted mean mass of all of the naturally occurring isotopes of an element relative to one twelfth of the mass of a carbon-12 atom.  This explains why the relative atomic masses given in data books are not exactly whole numbers.  Relative atomic masses have no units.  (This is review from the Unit on Atomic Structure and Relative Atomic Mass calculations)

36. 1.2 U2. Relative atomic mass (A r ) and relative molecular mass (M r ). 36  The relative molecular mass or formula mass (M r ) of a compound is the sum of the relative atomic masses of the elements in the compound.

37. 1.2 U3. Molar Mass 37  Molar Mass = mass of one mole of a substance  Units: g per mole or g mol -1

38. 1.2 Molar Mass (g mol -1 ) 38 Moles 1 mole of a substance contains the number of particles in the formula mass measured in grams (molar mass) 1 particle of oxygen has a relative mass of 16.0 so 1 mole of oxygen particles has a mass of 16.0 g 1 particle of water has a relative mass of 18.0 so 1 mole of water particles has a mass of 18.0 g

39. 1.2 A1. Molar Mass 39 Molar Mass (all masses round to the hundredths place)  The relative atomic mass of an element can be found on the periodic table (e.g. Na = 23.00).  The relative molecular mass or formula mass of a compound is the sum of the atomic masses of the elements in the compound. Examples: CompoundFormula Molar Mass Calculation OxygenO 2 2(16.00) = 32.00 WaterH 2 O 2(1.00) + 16.00 = 18.00 Sodium phosphateNa 3 PO 4 3(23.00) + 31.00 + 4(16.00) = 164.00 Iron(III) nitrateFe(NO 3 ) 3 55.80 + 3(14.00) + 9(16.00) = 241.80

40. 1.2 A1. Molar Mass Calculations 40 Molar Mass (all formula masses round to the hundredth place)  Find the molar mass of Cl 2 O 7  Find the molar mass of Mg(MnO 4 ) 2  Find the molar mass of sulfur trioxide  Find the molar mass of copper (II) carbonate

41. 1.2 A1. Molar Mass Calculations 41 Molar Mass (all formula masses round to the hundredths place) (all formula masses round to the hundredths place)  Find the molar mass of Cl 2 O 7 2(35.50) + 7(16.00) = 183.00  Find the molar mass of Mg(MnO 4 ) 2 1(24.30) + 2(54.90) + 8(16.00) = 262.10  Find the molar mass of sulfur trioxide (SO 3 ) 1(32.10) + 3(16.00) = 80.10  Find the molar mass of copper (II) carbonate (CuCO 3 ) 1(63.50) + 1(12.00) + 3(16.00) = 123.50

42. 1.2 A1. Molar Mass (g mol -1 ) 42 MOLES Scientists need to measure amounts of substances in grams, not amu. 1 mole of a substance contains the number of particles found in the formula mass measured in grams (gram-formula mass) 1 mole of Na 3 PO 4 particles has a mass of 164.0 grams 1 mole of Fe(NO 3 ) 3 particles has a mass of 241.8 grams What is the mass in grams of 1 mole of K 2 CrO 4 particles? What is the mass in grams of 1 mole of CH 3 OH particles?

43. 1.2 A2. Problems involving the relationship between the number of particles, amount of substance in moles, and mass 43 Converting Moles to Grams Multiply the mole amount given by the molar mass to find grams What is the mass in grams of 1 mole of K 2 CrO 4 particles? 194.2 g What is the mass in grams of 1 mole of CH 3 OH particles? 32.0 g What would be the mass of 1.72 moles of K 2 CrO 4 ? 1.72 mol x 194.2 g/mol = 334 g What would be the mass of 0.308 mol CH 3 OH? 0.308 mol x 32.0 g/mol = 9.86 g

44. 1.2 A2. Problems involving the relationship between the number of particles, amount of substance in moles, and mass 44 Converting Moles to Grams Multiply the mole amount given by the molar mass to find grams What is the mass in grams of 1 mole of K 2 CrO 4 particles? 194.2 g What is the mass in grams of 1 mole of CH 3 OH particles? 32.0 g What would be the mass of 1.72 moles of K 2 CrO 4 ? 1.72 mol x 194.2 g/mol = 334 g What would be the mass of 0.308 mol CH 3 OH? 0.308 mol x 32.0 g/mol = 9.86 g

45. 1.2 A2. Problems involving the relationship between the number of particles, amount of substance in moles, and mass 45 Converting Grams to Moles Divide the gram amount given by the molar mass to find moles What is the mass in grams of 1 mole of K 2 CrO 4 particles? 194.2 g What is the mass in grams of 1 mole of CH 3 OH particles? 32.0 g How many moles are in 124 g of K 2 CrO 4 ? 124 g / 194.2 g/mol = 0.639 mol If you had 1.93 g of CH 3 OH, how many moles would you have? 1.93 g / 32.0 g/mol = 0.0603 mol

46. 1.2 A2. Problems involving the relationship between the number of particles, amount of substance in moles, and mass 46 Review  Moles  Grams:mol x molar mass  Grams  Moles:grams / molar mass  Moles  Particles:mol x 6.02 x 10 23  Particles  Moles:particles / 6.02 x 10 23 Moles Grams Particles / Formula mass / 6.02 E23 X Formula mass X 6.02 E23

47. 1.2 A2. Problems involving the relationship between the number of particles, amount of substance in moles, and mass 47 Grams to Particles  How many particles are in 6.91 grams BaSO 4 ?  Convert grams to moles: 6.91 / 233.4 = 0.0296 mol  Convert moles to particles : 0.0296 x 6.02 E23 = 1.78 x 10 23 f.u. Moles Grams Particles / Formula mass / 6.02 E23 X Formula mass X 6.02 E23

48. 1.2 A2. Problems involving the relationship between the number of particles, amount of substance in moles, and mass 48 Particles to Grams  How many grams are in 7.04 x 10 23 molecules N 2 O?  Convert molecules to moles: 7.04 x10 23 / 6.02 x 10 23 = 1.17 mol  Convert moles to grams : 1.17 x 44.0 = 51.5 g Moles Grams Particles / Formula mass / 6.02 E23 X Formula mass X 6.02 E23

49. 1.2 A2. Problems involving the relationship between the number of particles, amount of substance in moles, and mass 49 Practice  How many grams are in 1.91 x 10 24 f.u. Na 2 CO 3  How many molecules are in 154 g PCl 3 ? Moles Grams Particles / Formula mass / 6.02 E23 X Formula mass X 6.02 E23

50. 1.2 A2. Problems involving the relationship between the number of particles, amount of substance in moles, and mass 50 Practice  How many grams are in 1.91 x 10 24 f.u. Na 2 CO 3  336 g  How many molecules are in 154 g PCl 3 ?  6.74 x 10 23 Moles Grams Particles / Formula mass / 6.02 E23 X Formula mass X 6.02 E23

51. 1.2 A3. Percentage Composition 51 Determine the % composition of a compound. 1. Find molar mass (formula mass) 2. Divide the individual atom masses by the molar mass and multiply by 100.

52. 1.2 A3. Percentage Composition 52 Determine the % composition of ammonium nitrate. 1. Find formula mass NH 4 NO 3 = 14.00 + 4(1.00) + 14.00 + 3(16.00) = 80.00 2. Divide the individual masses by the formula mass and multiply by 100% Percent nitrogen= (2(14.00) / 80.00) * 100= 35.00 % N Percent hydrogen = (4(1.00) / 80.0) * 100= 5.00 % H Percent oxygen= (3(16.00) / 80.0) * 100= 60.00 % O

53. 1.2 A3. Percentage Composition 53 Determine the % composition of sodium chloride. 1. Find formula mass Formula mass of NaCl is 23.0 + 35.5 = 58.5 2. Divide the individual masses by the formula mass and multiply by 100% %Na = (23.0 / 58.5) x 100 = 39.3 % % Cl = (35.5 / 58.5) x 100 = 60.7 %

54. 1.2 A3. Percentage Composition 54 Determine the % of nitrogen in urea (NH 2 ) 2 CO. Formula mass of (NH 2 ) 2 CO is 2(14.0) + 4(1.0) + 12.0 + 16.0 = 60.0 % N = 28.0/60.0 * 100% = 46.7% N Determine the mass of nitrogen in 85.8 g urea 85.8 g x 0.467 = 40.1 g  Determine the percentage composition of Mg(ClO 3 ) 2.  Determine the percentage of Fe in FeCl 3.  Determine the mass of Fe found in 51.3 grams FeCl 3.

55. 1.2 A3. Percentage Composition 55  Determine the percentage composition of Mg(ClO 3 ) 2. 24.3 + 2(35.5) + 6(16.0) = 191.3 % Mg = 24.3 / 191.3 * 100% = 12.7% % Cl = 71.0 / 191.3 * 100% = 37.1% % O = 96.0 / 191.3 * 100% = 50.2%  Determine the percentage of Fe in FeCl 3. 55.9 + 3(35.5) = 162.4 %Fe = 55.9 / 162.4 * 100% = 34.4%  Determine the mass of Fe found in 51.3 grams FeCl 3. 51.3g x.344 = 17.6 g

56. 1.2 U4. Empirical formula and Molecular formula. Empirical Formula Smallest whole number ratio of atoms in a compound Molecular Formula Actual ratio of atoms in a compound Sometimes the empirical formula and molecular formula are the same Molecular Formula Empirical Formula N2O4N2O4N2O4N2O4 NO 2 CO 2 C 6 H 12 O 6 CH 2 O C3H6O3C3H6O3C3H6O3C3H6O3 56

57. 1.2 A5. Determine Empirical Formula 57 Find the empirical formula of a compound 1. Assume 100 g of the compound and change % to grams. 2. Convert the grams to moles 3. Divide each mole quantity by the smallest number of moles

58. 1.2 A5. Determine Empirical Formula 58 What is the empirical formula of a compound that contains 40.0% C, 6.7% H, and 53.3% O? 1. Assume 100 g of the compound and change % to grams. 2. Convert the grams to moles 40.0 g C / 12.0 = 3.33 mol C 6.7 g H / 1.0 = 6.7 mol H 53.3 g O / 16.0 = 3.33 mol O 3. Divide each mole quantity by the smallest number of moles 3.33 mol C / 3.33 = 1 C 6.7 mol H / 3.33 = 2 H 3.33 mol O / 3.33 = 1 O 4. Empirical formula is CH 2 O

59. 1.2 A5. Determine Empirical Formula 59 What is the empirical formula of a compound that contains 26.2% N, 7.53% H and 66.4% Cl?  Assume 100 g of the compound and change % to grams.  Convert the grams to moles  Divide each mole quantity by the smaller number of moles  Empirical formula is

60. 1.2 A5. Determine Empirical Formula 60 What is the empirical formula of a compound that contains 26.2% N, 7.53% H, and 66.4% Cl?  Assume 100 g of the compound and change % to grams.  Convert the grams to moles 26.2 g N / 14.0 = 1.87 mol N 7.53 g H / 1.01 = 7.46 mol H 66.4 g Cl / 35.5 = 1.87 mol Cl  Divide each mole quantity by the smallest number of moles 1.87 mol N / 1.87 = 1 mol N 7.46 mol H / 1.87 = 4 mol H 1.87 mol Cl / 1.87 = 1 mol Cl  Empirical formula is NH 4 Cl

61. 1.2 A5. Determine Empirical Formula 61 What is the empirical formula of a compound that contains 1.67 g Ce and 4.54 g I?  Convert the grams to moles 1.67 g Ce / 140.1 =.0119 mol Ce 4.54 g I / 126.9 =.0358 mol I  Divide each mole quantity by the smallest number of moles.0119 mol Ce /.0119 = 1 Ce.0358 mol I /.0119 = 3.01 = 3 I  The empirical formula is CeI 3

62. 1.2 A5. Determine Empirical Formula 62 What is the empirical formula of a compound that contains 2.74 g Na, 0.120 g H, 1.43 g C and 5.71 g O?

63. 1.2 A5. Determine Empirical Formula 63 What is the empirical formula of a compound that contains 2.74 g Na, 0.120 g H, 1.43 g C and 5.71 g O?  Convert the grams to moles 2.74g Na / 23.0 = 0.119 mol Na1.43 g C / 12.0 = 0.119 mol C 0.120 g H / 1.0 = 0.12 mol H5.71 g O / 16.0 = 0.357 mol O  Divide each mole quantity by the smallest number of moles 0.119 mol Na / 0.119 = 1 Na0.119 mol C / 0.119 = 1 C 0.12 mol H / 0.119 = 1 H0.357 mol O / 0.119 = 3 O  The empirical formula is NaHCO 3

64. 1.2 A4. Determine Molecular Formula 64 To find molecular formula you need the empirical formula and the formula (molar) mass A substance has an empirical formula of CH and a molar mass of 78.0. What is the molecular formula?  Find the mass of the empirical formula Mass of CH is 12.0 + 1.0 = 13.0  Divide the molar mass by the empirical formula mass 78.0 / 13.0 = 6  Multiply each subscript in the EF by the number (CH) 6 = C 6 H 6

65. 1.2 A4. Determine Molecular Formula 65 To find molecular formula you need the empirical formula and the formula (molar) mass A substance consists of 42.9% C and 57.1% O with a molar mass of 56.0. What is the molecular formula?

66. 1.2 A4. Determine Molecular Formula 66 To find molecular formula you need the empirical formula and the formula (molar) mass A substance consists of 42.9% C and 57.1% O with a molar mass of 56.0. What is the molecular formula? Find Empirical Formula: 42.9 g C / 12.0 = 3.5857.1 g / 16.0 = 3.57 EF is CO Mass of CO: (12.0 + 16.0) = 28.0. 56.0 / 28.0 = 2 2(CO): MF = C 2 O 2

67. Understandings: 1. Reactants can be either limiting or excess. 2. The experimental yield can be different from the theoretical yield *Understandings 4-6, will be covered in IB Chem 2 1.3: Reacting Masses and Volumes

68. Applications and Skills: 1. Solution of problems relating to reacting quantities, limiting and excess reactants, theoretical, experimental, and percentage yields *Applications and Skills 2-8, will be covered in IB Chem 2 1.3: Reacting Masses and Volumes

69. Stoichiometry: the quantitative method of examining the relative amounts of reactants and products. Reactants can limit the amount of product made or reactants can be left over when a specific amount of product is made 1.3 U1. Limiting Reactants

70. The Four Steps to Stoichiometric Enlightenment Step 1: Write the balanced chemical equation. Step 2: Convert the known chemical amount to moles If amount in grams: Divide by formula weight If amount in particles: Divide by 6.02x10 23 Step 3: Ratio the unknown to the known molar quantities. From ProblemFrom Balanced Equation From ProblemFrom Balanced Equation X moles of known X Coefficient of Unknown = moles unknown Coefficient of Known X moles of known X Coefficient of Unknown = moles unknown Coefficient of Known Step 4: Convert molar quantity from step 3 to amount asked for If amount in grams: Multiply by formula weight If amount in particles: Multiply by 6.02x10 23 1.3 A1. Theoretical Yield Calculations

71. Step 3: Ratio the unknown to the known molar quantities. From ProblemFrom Balanced Equation From ProblemFrom Balanced Equation X moles of known X Coefficient of Unknown = moles unknown Coefficient of Known Coefficient of Known 1.3 A1. Theoretical Yield Calculations

72. Step 4: Convert molar quantity from step 3 to amount asked for If amount in grams: Multiply by formula weight If amount in particles: Multiply by 6.02x10 23 1.3 A1. Theoretical Yield Calculations

73. How many grams of hydrogen are produced from the decomposition of 1,234 g of water? Step 1: Write the balanced equation. Step 2: Convert grams to moles. Step 3:Ratio unknown to known. From Problem From Balanced Equation From Problem From Balanced Equation moles of known X Coefficient of Unknown = Coefficient of known Coefficient of known = moles of unknown Step 4:Convert moles to grams. How many grams of hydrogen are produced from the decomposition of 1,234 g of water?  2H 2 O  2H 2 + O 2  1,234 g / 18.0 g/mol = 68.56 mol H 2 O  From Problem From Equation: 68.56 Moles H 2 O X 2 Moles H 2 2 Moles H 2 O 68.56 Moles H 2 O X 2 Moles H 2 2 Moles H 2 O = 68.56 moles H 2  Mass H 2 = 68.56 mol H 2 (2.02 g/mol) = 138 g (3 s.f.)

74. 1.3 A1. Theoretical Yield Calculations How many grams of sodium carbonate are needed to precipitate the calcium from 254 g of calcium chloride? Step 1: Write the balanced equation. Step 2: Convert grams to moles. Step 3:Ratio unknown to known. From Problem From Balanced Equation From Problem From Balanced Equation moles of known x Coefficient of Unknown = Coefficient of Known Coefficient of Known = moles of unknown Step 4:Convert moles to grams. How many grams of sodium carbonate are needed to precipitate the calcium from 254 g of calcium chloride?  CaCl 2 + Na 2 CO 3  CaCO 3 + 2NaCl  254 g / 111.1 g/mol = 2.29 mol CaCl 2  From Problem From Equation: 2.29 Moles CaCl2 x 1 mol Na 2 CO 3 = 1 mole CaCl 2 1 mole CaCl 2 = 2.29 moles Na 2 CO 3  Mass Na 2 CO 3 = 2.29 mol Na 2 CO 3 (106.0 g/mol) = 243 g (3 s.f.)

75. How much oxygen does it take to burn 100.0 grams of octane? Step 1: 2C 8 H 18 + 25O 2  16CO 2 + 18H 2 O 1.3 A1. Theoretical Yield Calculations

76. How much oxygen does it take to burn 100.00 grams of octane? Step 1: 2C 8 H 18 + 25O 2  16CO 2 + 18H 2 O Step 2: moles of octane = 100.00 g / 114.00 g/mol = 0.8772 mol 1.3 A1. Theoretical Yield Calculations

77. How much oxygen does it take to burn 100.0 grams of octane? Step 1: 2C 8 H 18 + 25O 2  16CO 2 + 18H 2 O Step 2: moles of octane = 100.0 g / 114.0 g/mol = 0.8772 mol Step 3: Moles of O 2 required: 0.8772 mol C 8 H 18 x 25 O 2 = 10.97 mol O 2 2 C 8 H 18 1.3 A1. Theoretical Yield Calculations.

78. How much oxygen does it take to burn 100.0 grams of octane? Step 1:2C 8 H 18 + 25O 2  16CO 2 + 18H 2 O Step 2:moles of octane = 100.0 g / 114.0 g/mol = 0.8772 mol Step 3: Moles of O 2 required: 0.8772 C 8 H 18 x 25 O 2 = 10.97 mol O 2 2 C 8 H 18 0.8772 C 8 H 18 x 25 O 2 = 10.97 mol O 2 2 C 8 H 18 Step 4: Mass of O 2 required: 10.97 mol (32.00 g/mol) = 351.0 g O 2 1.3 A1. Theoretical Yield Calculations

79. In a spectacular reaction called the thermite reaction, iron(III) oxide reacts with aluminum producing iron and aluminum oxide. How many grams of iron will be produced from 43.7 grams of aluminum? 1.4.1 Calculate theoretical yields from chemical equations.

80. In a spectacular reaction called the thermite reaction, iron(III) oxide reacts with aluminum producing iron and aluminum oxide. How many grams of iron will be produced from 43.7 grams of aluminum? Step 1:Fe 2 O 3 + 2Al  2Fe + 2Al 2 O 3 Step 2: The moles of Al = 43.7 g / 27.0 g/mol = 1.62 mol Step 3: Moles of Fe produced: 1.62 Al x 2 mol Fe = 1.62 mol Fe 2 mol Al Step 4: Mass of Fe produced: 1.62 mol (55.8 g/mol) = 90.4 g Fe 1.3 A1. Theoretical Yield Calculations

81. Limiting Reactants You are given amounts for two reactants and one reactant will run out first. This is called the limiting reactant. It limits the amount of product that can be made. The reactant that is left over is called the excess reactant. A strip of zinc metal weighing 2.00 g is placed in a solution containing 2.50 g of silver nitrate causing the following reaction to occur: Zn + 2AgNO 3  2Ag + Zn(NO 3 ) 2 How many grams of Ag will be produced? 1.3 A1. Determining Limiting Reactants

82. 1) Balanced equation: Zn + 2AgNO 3  2Ag + Zn(NO 3 ) 2 2) Moles of starting substances: 2.00 g Zn/ 65.39 g/mol =.03059 mole Zn and 2.50 g AgNO 3 / 169.88 g/mol =.01472 mol AgNO 3 3) For each value above, use mole ratio for substance you are trying to find..03059 Zn X 2 mole Ag =.06118 mol Ag.03059 Zn X 2 mole Ag =.06118 mol Ag 1 mole Zn 1 mole Zn.01472 AgNO 3 x 2 Ag=.01472 mol Ag.01472 AgNO 3 x 2 Ag=.01472 mol Ag 2 AgNO 3 2 AgNO 3 4) Choose the smaller value from step 3 to calculate the grams..01472 mol Ag x 107.87 = 1.58 g Ag (3 sig figs) AgNO 3 is the limiting reactant because it will produce the least amount of Ag. Zn is the reactant in excess. There will be Zn left over after the reaction 1.3 A1. Determining Limiting Reactants

83. How many grams of Cu 2 S will be produced when 80.0 g Cu reacts with 25.0 g S? 1.3 A1. Determining Limiting Reactants

84. How many grams of Cu 2 S will be produced when 80.0 g Cu reacts with 25.0 g S? 1) 2Cu + S  Cu 2 S 1.3 A1. Determining Limiting Reactants

85. How many grams of Cu 2 S will be produced when 80.0 g Cu reacts with 25.0 g S? 1) 2Cu + S  Cu 2 S 2) 80.0 g Cu/63.55 = 1.259 molCu 25.0 g S/32.06 =.7798 mol S 1.3 A1. Determining Limiting Reactants

86. How many grams of Cu 2 S will be produced when 80.0 g Cu reacts with 25.0 g S? 1) 2Cu + S  Cu 2 S 2) 80.0 g Cu/63.55 = 1.259 mol 25.0 g S/32.06 =.7798 mol 1) 1.259 Cu x 1 mole Cu 2 S = 0.6295 mol Cu 2 S 2 moles CU 2 moles CU 0.7798 S x 1mol Cu 2 S = 0.7798 mol Cu 2 S 1 mole S 1 mole S 1.3 A1. Determining Limiting Reactants

87. How many grams of Cu 2 S will be produced when 80.0 g Cu reacts with 25.0 g S? 1) 2Cu + S  Cu 2 S 2) 80.0 g Cu/63.55 = 1.259 mol 25.0 g S/32.06 =.7798 mol 3) 1.259 Cu x 1 mole Cu 2 S = 0.6295 mol Cu 2 S 2 moles CU 2 moles CU 0.7798 S x 1mol Cu 2 S = 0.7798 mol Cu 2 S 1 mole S 1 mole S 4) 0.6295 mol Cu 2 S x 159.16 = 100 g or 1.00 x 10 2 g (3 sig figs) 1.3 A1. Determining Limiting Reactants

88. The theoretical yield from a chemical reaction is the yield calculated by assuming the reaction goes to completion. In practice we often do not obtain as much product from a reaction mixture as theoretically possible due to a number of factors: In practice we often do not obtain as much product from a reaction mixture as theoretically possible due to a number of factors:  Many reactions do not go to completion  Some reactants may undergo two or more reactions simultaneously forming undesired products  Not all of the desired product can be separated from the rest of the products The amount of a specified pure product actually obtained from a given reaction is the experimental or actual yield. 1.3 A1. Solve problems involving theoretical, experimental and percentage yield.

89. Percentage yield indicates how much of a desired product is obtained from a reaction Percentage yield = (experimental yield of a product) (theoretical yield of a product) (theoretical yield of a product) 1.3 A1. Solve problems involving theoretical, experimental and percentage yield. X 100

90. Consider the reaction: P 4 + 5O 2  P 4 O 10 If 272 g of phosphorus reacts, the percentage yield of tetraphosphorus decoxide is 89.5%. What mass of P 4 O 10 is obtained? Theoretical amt:  272 g/124.0 g/mol = 2.194 mol P 4  2.19 P 4 x 1 mole P 4 O 10 = 2.194 mol P 4 O 10 1 mole P 4 1 mole P 4  2.194 mol P 4 O 10 x 284.0 = 623 g Experimental amt:  623 g x.895 = 558 g 1.3 A1. Solve problems involving theoretical, experimental and percentage yield.

91. Consider the reaction: P 4 O 10 + 6H 2 O  4H 3 PO 4 If 558 g of P 4 O 10 reacts, the experimental yield of H 3 PO 4 is 746 g. What is the percentage yield of H 3 PO 4 ? Theoretical amt:  558 g/284 g/mol = 1.965 mol P 4 O 10  1.965 P 4 O 10 x 4 H 3 PO 4 = 7.86 mol H 3 PO 4 1 P 4 O 10 1 P 4 O 10  7.86 mol H 3 PO 4 x 98.0 = 770. g H 3 PO 4 Percentage yield:  746 g x 100% = 96.9% 770 g 1.3 A1. Solve problems involving theoretical, experimental and percentage yield.

92. Consider the reaction: Fe 2 O 3 + 3CO  2Fe + 3CO 2 When 84.8 g iron(III) oxide reacts with an excess of carbon monoxide, 54.3 g of iron is produced. What is the percentage yield of this reaction? 1.3 A1. Solve problems involving theoretical, experimental and percentage yield.

93. Consider the reaction: Fe 2 O 3 + 3CO  2Fe + 3CO 2 When 84.8 g iron(III) oxide reacts with an excess of carbon monoxide, 54.3 g of iron is produced. What is the percentage yield of this reaction? Theoretical amt:  84.8 g/159.2 g/mol = 0.5327 mol Fe 2 O 3  0.5327 mol Fe 2 O 3 x 2 mol Fe= 1.065 mol Fe 1 mol Fe 2 O 3 1 mol Fe 2 O 3  1.065 mol Fe x 55.8 g.mol = 59.4 g Fe Percentage yield:  54.3 g x 100% = 91.4% 59.4 g 1.3 A2. Solve problems involving theoretical, experimental and percentage yield.

94. If 50.0 g silicon dioxide is heated with an excess of carbon, the percentage yield of silicon carbide 76.0%. What mass of silicon carbide will actually be produced? SiO 2 + 3C  SiC + 2CO SiO 2 + 3C  SiC + 2CO 1.3 A2. Solve problems involving theoretical, experimental and percentage yield.

95. If 50.0 g silicon dioxide is heated with an excess of carbon, the percentage yield of silicon carbide 76.0%. What mass of silicon carbide will actually be produced? SiO 2 + 3C  SiC + 2CO Theoretical amt:  50.0 g/60.1 g/mol = 0.8319 mol SiO 2  0.8319 mol SiO 2 x 1 mol SiC = 0.8319 mol SiC 1 mol SiO 2 1 mol SiO 2  0.8319 mol SiC x 40.1 g.mol = 33.4 g SiC Experimental yield:  33.4 g x.760% = 25.4 g SiC 1.3 A1. Solve