1. 1 General Chemistry CHEM 110 Dr. Nuha Wazzan Chapter 3 Mass Relationships in Chemical Reactions

2. 2 Chapter 3 Mass Relationships in Chemical Reactions Chang Pages 80-107 H.W. p110-114 3.5, 3.6, 3.7, 3.8, 3.14, 3.16, 3.18, 3.20, 3.22, 3.24, 3.26, 3.28, 3.40, 3.42, 3.44, 3.46, 3.48, 3.50, 3.52, 3.44, 3.46, 3.48, 3.50, 3.52, 3.60, 3.84, 3.86

3. 3 Chapter 3: Mass Relationships in Chemical Reactions: Outline 3.1 Atomic Mass 3.2 Avogadro’s Number and Molar Mass of an Element 3.3 Molecular Mass 3.5 Percent Composition of Compounds 3.6 Experimental Determination of Empirical Formula 3.9 Limiting Reagents 3.10 Reaction Yield

4. 4 3.1 Atomic Mass Mass of an atom = mass of p + mass of n + mass of e mass of p =mass of n = 1840 mass of e Mass of atom = mass of p + mass of n Atomic mass (or atomic weight): is the mass of the atom in atomic mass unit (amu). Atomic mass unit: is the mass that exactly equal to one-twelfth the mass of one carbon-12 ( 12 C) atom. The atomic mass of 12 C = 12.00 amu

5. 5 12 C is the Standard e.g. Hydrogen atom ) 1 H( is only 8.400% massive as 12 C Thus: the atomic mass of 1 H = 0.084 x 12.0 amu= 1.008 amu and so on for all other elements.

6. 6

7. 7 Average Atomic Mass Atomic mass of carbon = 12.01 not 12.00 The carbon exist in more than one form (isotopes 12 C & 13 C) Carbon is a mixture of Isotopes Thus: Atomic Mass = Average Atomic Mass 6 C 12.01 Natural Abundance الطبيعية الوفرة

8. 8 How we get the Average Atomic Mass? Average atomic mass of natural carbon = ∑ (The natural abundance x Atomic Mass) for each isotope The isotopes of Carbon are 12 C & 13 C The natural abundance of each isotope is: 98.90% & 1.10%, respectively Average Atomic Mass of natural carbon = (0.9890 x 12.00 amu) 12C + (0.0110 x 13.00335 amu) 13C = 12.01 amu  Notes: 98.90% = 98.90/100 = 0.9890; 1.10% = 1.10/100 = 0.0110  Notes: carbon is mainly 12 C → the Average Atomic Mass is 12.01 amu → closer to 12 amu than 13 amu

9. 9 Average atomic mass (12.01)

10. 10 What information would you need to calculate the average atomic mass of an element? (a) The number of neutrons in the element. (b) The atomic number of the element. (c) The mass and abundance of each isotope of the element. (d) The position in the periodic table of the element.

11. 11 Example 3.1 p81 The atomic masses of the Copper isotopes, are 62.93 amu & 64.9278 amu, respectively. Calculate the average atomic mass of copper.

12. 12 Solution: 1.Percentages → Fractions: 69.09% = 69.09/100 = 0.6909; 30.91% = 30.91/100 = 0.3091 2.The AAM will be closer to the mass of the isotope with the largest natural abundance THUS: Average atomic mass of copper = (0.6909 x 62.93) 63Cu + (0.3091 x 64.9278) 65Cu = 63.55 amu H.W. Solve the Practice Exercise p81

13. 13 Iodine has two isotopes 126 I and 127 I, with the equal abundance. Calculate the average atomic mass of Iodine ( 53 I). (a)126.5 amu (b)35.45 amu (c)1.265 amu (d)71.92 amu

14. 14 NOTE: Atomic Mass of an isotope, e.g. 12 C & 13 C Atomic Mass of an element e.g. C = Average Atomic Mass Also: Average Atomic Mass C = Atomic Mass C

15. 15 3.2 Avogadro’s Number and the Molar Mass Mole (mol) is the SI unit of the amount of substance Avogadro’s Number (N A ): one mole of each substance contains 6.022 x 10 23 particles (atoms or molecules or ions etc.), Thus 6.022 x 10 23 is the Avogadro's Number (N A ) N A = 6.022 x 10 23 particles (atoms or molecules or ions etc.)

16. 16 Dozen = 12 1 dozen = 12 Anything 1 mol = 6.022 x10 23 particles

17. 17 THUS: one mole of H atoms has 6.022 x 10 23 atoms & One mole of H 2 molecules has 6.022 x 10 23 molecules

18. 18 One mole of Carbon One mole of Mercury One mole of Copper One mole of Iron One mole of Sulphur

19. 19 Molar Mass Molar mass (M): the mass (in g or kg) of one mole of a substance; M = mass/mol = g/mol For ONE MOLE: 1 amu = 1 g The atomic mass of 12 C is 12.00 amu = 12.00 g 1 mole of 12 C = 12.00 amu = 12.00 g = has N A of atoms = has 6.022 x 10 23 atoms Thus: The Molar Mass (M) of 12 C = 12.00 g/mol

20. 20 Atomic Mass of an isotope, e.g. 12 C & 13 C Atomic Mass of an element e.g. C is Average Atomic Mass Also: Average Atomic Mass C = Atomic Mass C

21. 21 Molar Mass (g/mol) = Atomic Mass (amu) Examples: 1. The atomic mass of Na = 22.99 amu The molar mass of Na = 22.99 g/mol 2. The atomic mass of P = 30.97 amu The molar mass of P = 30.97 g/mol

22. 22 Example 3.2 p83: How many moles of He atoms are in 6.46 g of He?

23. 23 Solution: Number of moles = n The atomic mass of He = The Molar mass of He From the periodic Table: the atomic mass of He = The molar mass of He = 4.003 g/mol Thus: 4.003 g → 1 mole of He 6.46 g of He → ? Moles (?n) Thus: there is 1.61 moles of He atoms in 6.46 g of He H.W. Solve the Practice Exercise p84

24. 24 n = number of moles m = mass (atom or molecule) M = molar mass (atomic mass or molecular mass) What is the relation between them? n = number of moles N = number of atoms or molecules N A = Avogadro's number (atoms (or molecules)/mol) What is the relation between them?

25. 25 How many moles of He atoms are in 6.46 g of He?

26. 26 Example 3.3 p84: How many grams of Zn are in 0.356 mole of Zn?

27. 27 Solution: Number of moles = n The atomic mass of Zn = The Molar mass of Zn From the periodic Table: The Atomic mass of Zn = molar mass of Zn = 65.39 g/mol Thus: 65.39 g → 1 mole of Zn ?g of He → 0.356 mole Thus: there is 23.3 g of Zn atoms in 0.356 moles of Zn H.W. Solve the Practice Exercise p84

28. 28 How many grams of Zn are in 0.356 mole of Zn? حل مختصر

29. 29 Example 3.4 p84: How many S atoms are in 16.3 g of S? Strategy: 1.How many moles in 16.3 g of S = X mol 2.1 mole → 6.022 x10 23 S atoms X moles → ? atoms

30. 30 Solution: From the periodic Table: The atomic mass of S = 32.07 amu The molar mass of S = 32.07 g/mol Thus: 32.07 g → 1 mole of S 16.3 g of S → ? mole We know: 1 mol of S → 6.022 x10 23 S atoms 0.508 mole → ? S atoms There is 3.06 x10 23 atoms of S in 16.3 g of S H.W. Solve the Practice Exercise p85

31. 31 How many S atoms are in 16.3 g of S? حل مختصر

32. 32 Molecular Mass Molecular Mass (molecular weight): is the sum of the atomic masses (in amu) in the molecule. (MOLECULE) Molecular Mass: multiply the atomic mass of each element by the number of atoms of that element present in the molecule and sum over all the elements. e.g. Molecular Mass of H 2 O is: (2 x atomic mass of H) + (1x atomic mass of O) (2 x 1.008 amu) + (1x 16.00 amu) = 18.02 amu

33. 33 Example 3.5 p86: Calculate the molecular masses (in amu) of the following compounds: (a) Sulphur dioxide (SO 2 ) (b) Caffeine (C 8 H 10 N 4 O 2 ) Solution: (a) Molecular mass of SO 2 = (1 x 32.07) + (2 x 16.00) = 64.07 amu (b) Molecular mass of C 8 H 10 N 4 O 2 = (8 x 12.01) + (10 x 1.008) + (4 x 14.01) + (2 x 16.00) =194.20 amu

34. 34 One mole of H atoms has 6.022 x 10 23 H atoms & One mole of H 2 molecules has 6.022 x 10 23 H 2 molecules One mole of H 2 molecules has 2 x 6.022 x 10 23 H atoms

35. 35 One mole of H 2 O has 6.022 x 10 23 H 2 O molecules & One mole of CH 4 molecules has 6.022 x 10 23 CH 4 molecules

36. 36 Example 3.6 p86: How many moles of CH 4 are present in 6.07 g of CH 4 ? Solution: Molecular Mass CH 4 = (1 x 12.01) + (4 x 1.008) = 16.04 amu Molecular Mass (in amu) = Molar Mass (g/mol) 16.04 amu = 16.04 g/mol Thus: 1 mole of CH 4 → 16.04 g CH 4 n? moles → 6.07 g of CH 4 Atomic mass (in amu) = Molar Mass (g/mol) Molecular Mass (in amu) = Molar Mass (g/mol)

37. 37 How many moles of CH 4 are present in 6.07 g of CH 4 ? حل مختصر

38. 38 How many molecules of ethane (C 2 H 6 ) are present in 0.334 g of C 2 H 6 ? (a)2.01 x 10 23 (b)6.69 x 10 21 (c)4.96 x 10 22 (d)8.89 x 10 20

39. 39 Solution: Molecular mass of C 2 H 6 = (2 x 12.01) + (6 x 1.008) = 30.068 amu Molecular mass (amu) = Molar mass (g/mol) = 30.068 g/mol C 2 H 6 30.068 g → 1 mole 0.334g → ?n 1 mole of C 2 H 6 → 6.022 x 10 23 molecules of C 2 H 6 0.011 mole of C 2 H 6 → ? molecules of C 2 H 6

40. 40 How many molecules of ethane (C 2 H 6 ) are present in 0.334 g of C 2 H 6 ? (a)2.01 x 10 23 (b)6.69 x 10 21 (c)4.96 x 10 22 (d)8.89 x 10 20 حل مختصر

41. 41 Example 3.7 p87: How many hydrogen atoms are present in 25.6 g of urea [(NH 2 ) 2 CO]. The molar mass of urea is 60.06 g/mol.

42. 42 H.W. What is the mass, in grams, of one copper atom? (a) 1.055  10 -22 g (b) 63.55 g (c) 1 amu (d) 1.66  10 -24 g (e) 9.476  10 21 g

43. 43 Solution: Atomic mass of Cu = 63.55 amu Molar mass of Cu = 63.55 g/mol 63.55 g of Cu → 1 mol of Cu 1 mol of Cu → 6.022 x 10 23 Cu atoms 63.55g of Cu → 6.022x10 23 Cu atoms ?g of Cu → 1 Cu atom

44. 44 H.W. What is the mass, in grams, of one copper atom? (a) 1.055  10 -22 g (b) 63.55 g (c) 1 amu (d) 1.66  10 -24 g (e) 9.476  10 21 g حل مختصر

45. 45 3.5 Percent Composition of Compounds: outline Percent Composition by Mass  → The Empirical Formula  → The Molecular Formula الصيغة الجزيئية

46. 46 3.5 Percent Composition of Compounds Percent Composition by Mass: is the percent by mass of each element in a compound percent composition of an element in a compound = n x molar mass of element molar mass of compound x 100% n is the number of moles of the element in 1 mole of the compound

47. 47 C2H6OC2H6O %C = 2 x (12.01 g) 46.07 g x 100% = 52.14%H = 6 x (1.008 g) 46.07 g x 100% = 13.13%O = 1 x (16.00 g) 46.07 g x 100% = 34.73% 52.14% + 13.13% + 34.73% = 100.0% Check the answer! PERIODIC TABLE Molar mass of C 2 H 6 O = Molecular mass C 2 H 6 O = (2 x 12.01) + (6 x 1.008) + (1 x16.00) = 46.07 g/mol Molar mass of C = Atomic mass of C = 12.01 g /mol Molar mass of H = Atomic mass of H = 1.008 g/mol Molar mass of O = Atomic mass of O = 16.00 g/mol

48. 48 Example 3.8 p89: Calculate the percent composition by mass of H, P, and O in H 3 PO 4. Solution: Molar mass of H 3 PO 4 = 97.99 Check the answer: 3.08% + 31.61% + 65.31% = 100.01%

49. 49 H.W. Calculate the percent of nitrogen in Ca(NO 3 ) 2 : a) 12.01%. b) 17.10%. c) 18% d) 16%. H.W. All of the substances listed below are fertilizers that contribute nitrogen to the soil. Which of these is the richest source of nitrogen on a mass percentage basis? (a)Urea, (NH 2 ) 2 CO (b)Ammonium nitrate, NH 4 NO 3 (c)Guanidine, HNC(NH 2 ) 2 (d)Ammonia, NH 3 ايا من هذه المواد هواغنى مصدر للنيتروجين على اساس احتوائه على اكبر نسبه وزنيه من النيتروجين؟ (a)%N = 46.6% (b) %N = 58% (c) %N = 71.1% (d) %N = 82.2%

50. 50 Determination of the Empirical Formula from the Percent Composition by Mass Example: Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O 40.51 percent. Solution: 1.Convert to grams & divide by the molar mass → number of moles Thus 24.75g of K, 34.77g of Mn, & 40.51g of O 1. 2. 3.

51. 51 2. Divide by the smallest number of moles: 0.632 mol → mole ratio of the elements 3. All are integers → Finish Thus the Empirical Formula of the compound is: K 1 Mn 1 O 4 → KMnO 4

52. 52 Percent Composition and Empirical Formulas Q: Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O 40.51 percent. KMnO %  100g24.75g34.77g40.51g n=m/M24.75/39.10 =0.633mol 34.77/54.94 =0.6329mol 40.51/16.00 = 2.532mol  on smallest no. of mole0.633/0.632 =1 0.6329/0.632 = 1 2.532/0.632 =4 The empirical formula isK1K1 Mn 1 O4O4 KMnO 4 خطوات الحل 1. ننشأ جدول نضع فيه العناصر المذكورة في السؤال 2. نعتبر أن النسبة المئوية معبر عنها بالجرام فلو كان عندنا 100 جرام من المركب فهذه ال 100 جرام موزعة على العناصر حسب نسبتها. 3. نوجد عدد المولات n لكل عنصر باستخدام القانون n=m/M. 4. نقسم عدد المولات على أصغر مول من العناصر. 5. الأرقام التي نحصل عليها تمثل empirical formula بشرط أن تكون أعداد صحيحة كما في المثال السابق. 6. في حالة ظهور أعداد عشرية نقوم بضرب الأرقام التي في الأسفل الموجودة في الصيغة بأعداد بدأ من 2 ، 3....... حتى نحصل على أعداد صحيحة. Courtesy of Dr. Fawzia Albelwi

53. 53 Determination of the Empirical Formula from the Percent Composition by Mass Example 3.9 p90: Ascorbic acid composed of 40.92% C, 4.58% H, and 54.50% O by mass. Determine its empirical formula. Solution: 1.Thus 40.92g of C, 4.58g of H, & 54.50g of O 1. 2. 3.

54. 54 2. 3. Not all numbers are integers THUS: Convert 1.33 → integer Trail-and error procedure: 1.33 x 2 = 2.66 1.33 x 3 = 3.99 ≈ 4 OK. multiply all the numbers by 3 C = 3 H =4 O = 3 Thus the Empirical Formula of ascorbic acid is: C 3 H 4 O 3

55. 55 CHO %  100g40.92g4.58g54.50g n=m/M40.92/12.01 = 3.407mol 4.58/1.008 =4.54.mol 54.50/16.00 = 3.406 mol  on smallest no. of mole 3.407/3.406 = 1 0.4.54/3.406 = 1.33 3.406/3.406 = 1 Convert into integer x 3 33.99 = 43 The empirical formula is C3C3 H4H4 O3O3 C3H4O3C3H4O3 خطوات الحل 1. ننشأ جدول نضع فيه العناصر المذكورة في السؤال 2. نعتبر أن النسبة المئوية معبر عنها بالجرام فلو كان عندنا 100 جرام من المركب فهذه ال 100 جرام موزعة على العناصر حسب نسبتها. 3. نوجد عدد المولات n لكل عنصر باستخدام القانون n=m/M. 4. نقسم عدد المولات على أصغر مول من العناصر. 5. الأرقام التي نحصل عليها تمثل empirical formula بشرط أن تكون أعداد صحيحة 6. في حالة ظهور أعداد عشرية كما في المثال السابق نقوم بضرب الأرقام التي في الأسفل الموجودة في الصيغة بأعداد بدأ من 2 ، 3....... حتى نحصل على أعداد صحيحة. Example 3.9 p90: Ascorbic acid composed of 40.92% C, 4.58% H, and 54.50% O by mass. Determine its empirical formula.

56. 56 Determination of the Molecular Formula from the Percent Composition by Mass Example 3.11 p93: A sample compound contains 1.52g of N and 3.47g of O. The molar mass of this compound is between 90g and 95g. Determine the molecular formula. Solution: 1. 2. 3. Thus the empirical formula is: NO 2 Present Composition by Mass ↓ Empirical Formula ↓ Molecular Formula

57. 57 4. The molar mass of the empirical formula NO 2 = 14.01 + (2x16.00) = 46.01g 5. The ratio between the empirical formula and the molecular formula: 6. The molecular formula is (NO 2 ) 2 = N 2 O 4 1.956

58. 58 H.W. Which of the following is an empirical formula: a) C 12 H 22 O 10. b) H 2 SO 4. c) Hg 2 Cl 2 d) S 8. H.W. A sample of acid compound contains 40.1 percent of C, 6.6 percent of H, and 53.3 percent of O. The molar mass of this compound is 60 g/mol. What is the molecular formula ? (a) C 5 H 6 O (b) C 2 HO 2 (c) C 2 H 4 O 2 (d) CH 2 O 4 ايا منهم لايمكن تبسيطه اكثر مما هو عليه؟

59. 59 3.7 Chemical Reactions and Chemical Equations  Chemical Reaction: is a process in which one or more substances is changed into one or more new substances  Chemical Equation: uses chemical symbols to show what happens during a chemical reaction reactantsproducts

60. 60  3 ways of representing the reaction of H 2 with O 2 to form H 2 O

61. 61 How to “Read” Chemical Equations? 2 Mg + O 2 2 MgO 2 atoms Mg + 1 molecule O 2 makes 2 formula units MgO 2 moles Mg + 1 mole O 2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O 2 makes 80.6 g MgO 2 grams Mg + 1 gram O 2 makes 2 g MgO X √ √ √ Molar masses Read it!

62. 62 1.Write the correct formula/s for the reactant/s on the left side and the correct formula/s for the product/s on the right side of the equation. Balancing Chemical Equations نكتب الصيغه الصحيحه لكل متفاعل (على الطرف الايسر) ولكل ناتج (على الطرف ا يمن)

63. 63 2.Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 2C 2 H 6 NOT C 4 H 12 وزن المعادله الكيميائيه يكون بتغير الارقام التي بجانب الصيغه وليست التي تحتها بحيث يكون للعنصر نفس العدد على طرفي المعادله.

64. 64 3.Start by balancing those elements that appear in only one reactant and one product. 4. Balance those elements that appear in two or more reactants or products. توزن اولا العناصر ا قل ظهورا ثم توزن العناصر ا كثر ظهورا 5. Check to make sure that you have the same number of each type of atom on both sides of the equation. الخطوه الاخيره هي التأكد من ان لديك نفس العدد من الذرات لكل عنصر على طرفي المعادله

65. 65 Example 1.Write the correct formula/s for the reactant/s on the left side and the correct formula/s for the product/s on the right side of the equation. Nitrogen monoxide reacts with oxygen to form nitrogen dioxide balance this reaction?

66. 66 2. Start by balancing those elements that appear in only one reactant and one product. start with N not O 1 nitrogen on left 1 nitrogen on right N is balanced

67. 67 3 oxygen on left 2 oxygen on right multiply NO by 2 and NO 2 by 2 3.Balance those elements that appear in two or more reactants or products. 2 oxygen + 2 oxygen = 4 on left 2 x 2 = 4 oxygen on right O is balanced

68. 68 4.Check to make sure that you have the same number of each type of atom on both sides of the equation. 2NO + O 2 2NO 2 ReactantsProducts 2 N 4 O 2 N 4 O 2 N 2 O + 2 O = 44 O (2 x 2)

69. 69 Example 1.Write the correct formula/s for the reactant/s on the left side and the correct formula/s for the product/s on the right side of the equation. C 2 H 6 + O 2 CO 2 + H 2 O Ethane reacts with oxygen to form carbon dioxide and water balance this reaction?

70. 70 2. Start by balancing those elements that appear in only one reactant and one product. C 2 H 6 + O 2 CO 2 + H 2 O start with C or H but not O 2 carbon on left 1 carbon on right multiply CO 2 by 2 C 2 H 6 + O 2 2CO 2 + H 2 O 6 hydrogen on left 2 hydrogen on right multiply H 2 O by 3 C 2 H 6 + O 2 2CO 2 + 3H 2 O

71. 71 3.Balance those elements that appear in two or more reactants or products. 2 oxygen on left 4 oxygen (2x2) C 2 H 6 + O 2 2CO 2 + 3H 2 O + 3 oxygen (3x1) multiply O 2 by 7 2 = 7 oxygen on right C 2 H 6 + O 2 2CO 2 + 3H 2 O 7 2 remove fraction multiply both sides by 2 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O

72. 72 4.Check to make sure that you have the same number of each type of atom on both sides of the equation. 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O ReactantsProducts 4 C 12 H 14 O 4 C 12 H 14 O 4 C (2 x 2)4 C 12 H (2 x 6)12 H (6 x 2) 14 O (7 x 2)14 O (4 x 2 + 6)

73. 73 Example 3.12 p98 : Balance the following reaction 3.7 Nitrogen monoxide reacts with oxygen to form nitrogen dioxide and balance this reaction?

74. 74 H.W. What is the coefficient of H 2 O when the equation is balanced: _ Al 4 C 3 + _ H 2 O  _ Al(OH) 3 + 3CH 4 a.13 b.4 c.6 d.12 H.W. What are the coefficients of Al 4 C 3,H 2 O and Al(OH) 3, respectively, when the equation is balanced: _ Al 4 C 3 + _ H 2 O  _ Al(OH) 3 + 3CH 4 a.4,1,5 b.1,12,4 c.1,24, 4 d.4,12,1

75. 75 n = number of moles m = mass (atom or molecule) M = molar mass (atomic mass or molecular mass) What is the relation between them?

76. 76 3.8 Amounts of Reactants and Products Two important questions:  How much product will be formed from specific amount of reactants? e.g. 6.0 g reactant → ? product  How much starting reactants must be used to obtain a specific amount of product? e.g. ? reactant → 6.0 g product

77. 77 1.Write balanced chemical equation 2.Convert quantities of known substances into moles 3.Use coefficients in balanced equation to calculate the number of moles of the sought quantity 4.Convert moles of sought quantity into desired units Mole Method

78. 78 Example 3.13 p101: If 856g of C 6 H 12 O 6 is consumed by a person over a certain period, what is the mass of CO 2 produced?

79. 79 Solution: 1.Write balanced chemical equation 2. Convert quantities of known substances into moles → convert grams of C 6 H 12 O 6 to moles of C 6 H 12 O 6 Balanced!

80. 80 3.Use coefficients in balanced equation to calculate the number of moles of the sought quantity → mole ratio (from the balanced equation): 1 mole C 6 H 12 O 6 → 6 mole of CO 2 4.754 mole → ? mole CO 2

81. 81 4.Convert moles of sought quantity into desired units → convert the moles of CO 2 → grams of CO 2 Summary: grams of C 6 H 12 O 6 → moles of C 6 H 12 O 6 → moles of CO 2 → grams of CO 2

82. 82 Example 3.14 p101: How many grams of Li are needed to produce 9.89g of H 2 ? Strategy: grams of H 2 → moles of H 2 → moles of Li → grams of Li

83. 83 Solution: 1.Write balanced chemical equation 2. convert grams of H 2 to moles of H 2 Balanced!

84. 84 3.Mole ratio (from the balanced equation) → Moles of Li 2 mole of Li → 1 mole H 2 ?mole of Li → 4.927 mole H 2 4. Convert the moles of Li → grams of Li

85. 85 3.9 Limiting Reagent الكاشف المحدد Limiting Reagent: is the reactant used up first in a reaction and thus determine the amount of product Excess Reagent الكاشف الفائض : is the reactant present in quantities greater than necessary to react with the quantity of the limiting reagent (the one that is left at the end of the reaction). → Limiting reagent is in a reaction of more than one reactant!

86. 86 Limiting Reagent: 2NO + O 2 2NO 2 NO is the limiting reagent O 2 is the excess reagent Reactant used up first in the reaction.

87. 87 Questions in Limiting Reagent: First: we have to determine which reactant is the limiting reagent and which is the excess reagent! Second: after we know which one is the limiting reagent, we could determine the amount of the product!! Third: after we know the excess reagent, we could determine how much excess of it is left at the end of the reaction!!!

88. 88 Example 3.15 P102: 637.2g of NH 3 are treated with 1142g of CO 2. (a)Which of the two reactants is the limiting reagent? (b) Calculate the mass of (NH 2 ) 2 CO formed. (c) How much excess reagent (in grams) is left at the end of the reaction?

89. 89 Solution: (a)Which of the two reactants is the limiting reagent? المتفاعل اللذي يعطي مولات اقل من الناتج هو الكاشف المحدد, لانه يحد من كميه الناتج التي يمكن ان تتكون. For the first reactant NH 3 : 1. 637.2 g NH 3 → convert to moles تحول الى مولات

90. 90 Solution: 2. Mole ratio: 2 moles NH 3 → 1 mole of (NH 2 ) 2 CO 37.416moles of NH 3 → ?n (NH 2 ) 2 CO

91. 91 For the second reactant CO 2 : 1. 1142 g CO 2 → convert to moles 2. Mole ratio: 1 mole CO 2 → 1 mole of (NH 2 ) 2 CO 25.949 mol CO 2 → ?n (NH 2 ) 2 CO Thus: the limiting reagent is NH 3 because it produces a smaller amount of (NH 2 ) 2 CO تحول الى مولات

92. 92 (b)Calculate the mass of (NH 2 ) 2 CO formed. We have 18.71 mol of (NH 2 ) 2 CO using NH 3 as the limiting reagent → convert to grams

93. 93 (c)How much excess reagent (in grams) is left at the end of the reaction? Excess reagent is CO 2 : grams of CO 2 left = initial grams – reacted grams 1. moles of CO 2 left = initial moles – reacted moles 2. moles of CO 2 left → grams of CO 2 left 1. Initial moles of CO 2 = 25.95 mol Reacted moles of CO 2 (calculated as follows):

94. 94 Reacted moles of CO 2 (calculated as follows): We know that 18.71 mol (NH 2 ) 2 CO is produced Thus: 1 mol CO 2 → 1 mole (NH 2 ) 2 CO ?n CO 2 → 18.71 mol (NH 2 ) 2 CO THUS: moles of CO 2 reacted is 18.71 mol

95. 95 moles of CO 2 left = initial moles – reacted moles = 25.95 – 18.71 = 7.24 mol 2. moles of CO 2 left → grams of CO 2 left Thus: the mass of CO 2 remaining (left) = 319g

96. 96 H.W. # 1: When 22.0 g NaCl and 21.0 g H 2 SO 4 are mixed and react according to the equation below, which is the limiting reagent? 2NaCl + H 2 SO 4  Na 2 SO 4 + 2HCl (a) NaCl (b) H 2 SO 4 (c) Na 2 SO 4 (d) HCl (e) No reagent is limiting. H.W. # 2: Consider the combustion of carbon monoxide (CO) in oxygen gas: 2CO(g) + O 2 (g) → 2CO 2 (g) Starting with 3.60 moles of CO, calculate the number of moles of CO 2 produced if there is enough oxygen gas to react with all of the CO. (a)7.20 mol (b)44.0 mol (c)3.60 mol (d)1.80 mol The limiting Reagent is CO

97. 97 3.10 Reaction Yield Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. % Yield = Actual Yield Theoretical Yield x 100 Actual Yield is always less.

98. 98 3.10 Reaction Yield Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Limiting regent mass → moles of limiting reagent → moles of product → grams of product (theoretical yield of the product)

99. 99 Example 3.16 p106: 3.54 x10 7 g of TiCl 4 are reacted with 1.13 x 10 7 g of Mg. (a)Calculate the theoretical yield of Ti in grams (b) Calculate the percent yield if 7.91 x 10 6 g of Ti are actually obtained.

100. 100 Solution: (a)Calculate the theoretical yield of Ti in grams. Strategy: 1.Given: Masses of two reactants → limiting reagent problem 2.Masses of reactants → moles of reactant 3.Moles of reactants → used to calculate the moles of product 4.Moles of product is the less number of moles (limiting reagent) 5.Moles of product → grams of product (theoretical yield of the product) أي بمعنى اخر: احسبي وزنه التيتانيوم الناتجه؟

101. 101 grams of TiCl 4 → moles of TiCl 4 → moles of Ti Mole ratio: 1 mole TiCl 4 → 1 mole Ti 1.97 x10 5 mole → ?n Ti n(Ti) =1.87 x10 5 mol

102. 102 grams of Mg → moles of Mg → moles of Ti Mole ratio: 2 mole Mg → 1 mole Ti 4.64 x10 6 mole → ?n Ti

103. 103 n(Ti) is the less number of moles (limiting reagent) = 1.87 x10 5 mol Moles of Ti → grams of Ti Thus: mass of Ti = theoretical yield of Ti = m =nM

104. 104 (b) Calculate the percent yield if 7.91 x 10 6 g of Ti are actually obtained. H.W. Solve the practice exercise p107

105. 105 End of Chapter 3