1. C HAPTER 3 M ASS R ELATIONSHIPS IN C HEMICAL R EACTIONS Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. C HAPTER 3 (80-107) 3.1 Atomic mass. 3.2 Avogadro’s number and the molar mass of an element. 3.3 Molecular mass. 3.5 Percent composition of compounds. 3.7 Chemical reactions and chemical equations. 3.8 Amounts of reactants and products. 3.9 Limiting reagents. 3.10 Reaction yield.

3. 3.1 A TOMIC MASS.

4. By definition: 1 atom 12 C “weighs” 12 amu On this scale 1 H = 1.008 amu 16 O = 16.00 amu Atomic mass is the mass of an atom in atomic mass units (amu) One atomic mass unit is defined as a mass exactly equal to one – twelfth the mass of one carbon-12 atom. Micro World atoms & molecules Macro World grams 3.1

5. Natural lithium is: 7.42% 6 Li (6.015 amu) 92.58% 7 Li (7.016 amu) 7.42 x 6.015 + 92.58 x 7.016 100 = 6.941 amu 3.1 Average atomic mass of lithium: Atomic number Average atomic mass Average atomic mass= Atomic mass × abundance percent + Atomic mass × abundance percent 100 Isotope 1Isotope 2

6. Average atomic mass (6.941)

7. 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.

8. The atomic masses of (75.53 percent) and (24.47 percent) are 34.968 amu and 36.956 amu, respectively. Calculate the average atomic mass of chlorine. Average atomic mass= Atomic mass × abundance percent + Atomic mass × abundance percent 100 Isotope 1Isotope 2 Solution: Isotope I Atomic mass = 34.698 amu Abundance percent = 75.53 % Isotope II Atomic mass = 36.956 amu Abundance percent = 24.47 % Average atomic mass= 34.698 × 75.53+ 36.956 × 24.47 100 Isotope 1Isotope 2 35.45 amu =

9. The atomic masses of and are 6.0151 amu and 7.0160 amu, respectively. Calculate the natural abundance of these two isotopes. The average atomic mass of Li is 6.941 amu. Average atomic mass= Atomic mass × abundance percent + Atomic mass × abundance percent 100 Isotope 1Isotope 2 Solution: Isotope I Atomic mass = 6.0151 amu Abundance percent = ? Isotope II Atomic mass = 7.0160amu Abundance percent = ? Abundance percent of all isotopes = 100 % For isotope I: abundance percent = X For isotope II: abundance percent = 100-X 6.941= X (6.0151) + (100-X) (7.0160) 100 Isotope 1Isotope 2 Average atomic mass = 6.941 amu 6.941 × 100= 6.0151X+ 701.60-7.0160X 694.1= -1.0009X+ 701.60694.1-701.60 = -1.0009X -7.5 = -1.0009X X= -7.5 -1.0009 X= 7.493 %100-X= 100-7.493= 92.507%

10. 3.2 A VOGADRO ’ S NUMBER AND THE MOLAR MASS OF AN ELEMENT.

11. The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12 C isotope 3.2 1 mol = N A = 6.0221367 x 10 23 Avogadro’s number (N A ) عدد أفوجادرو هو حلقة الوصل بين عالم الجسيمات غير المرئية (الجزئيات والذرات) وعالم المرئيات كالماء والأملاح يستخدم المول لعد الجزيئات والذرات والأيونات ويستخدمه الكيميائيون لأنه يوفر طريقة ملائمة لمعرفة عدد الجسيمات في العينة

12. Molar mass (M) is the mass of 1 mole of in grams eggs shoes marbles atoms 1 mole 12 C atoms = 6.022 x 10 23 atoms = 12.00 g 1 12 C atom = 12.00 amu 1 mole lithium atoms = 6.941 g Li 1 lithium atom = 6.941 amu For any element atomic mass (amu) = molar mass (g/mol) Molar mass Atomic mass Mass of 1 mole of C = mass of 1 mole Cu M = mass/mol = g/mol

13. One Mole of: C S Cu Fe Hg 3.2 13 32.07 g 12.01 g 200.6 g 55.85 g 63.55 g

14. 14 How many amu are there in 8.4 g? 1 g = 6.022 x 10 23 amu 8.4 g = 5.1 x 10 24 amu What is the mass in grams of 13.2 amu ? 13.2 amu = 2.2 x 10 -23 g 1 amu = 1.66 x 10 -24 g يوضح المثالان أنه يمكن استخدام عدد أفوجادرو لتحويل الكتلة الذرية إلى كتلة بالجرام والعكس صحيح 1 g 6.022 x 10 23 amu 8.4 g ?

15. 3.2 عدد ذرات العنصر N عدد مولات العنصر n كتلة العنصر m nN A N/N A m/ M nMnM عدد أفوجادرو ()الكتلة المولارية () يمكن باستخدام كلا من عدد أفوجادرو (N A ) و الكتلة المولارية ( M ) التحويل بين كتلة العنصر (m) وعدد المولات (n) لنفس العنصر التحويل بين عدد المولات للعنصر (n) وعدد الذرات (N) لنفس العنصر N A = Avogadro’s number M = molar mass in g/mol

16. No. of moles عدد المولات Mass of element كتلة العنصر No. of atoms عدد الذرات Molar mass الكتلة المولارية Avogadro's No. عدد أفوجادرو

17. Do You Understand Molar Mass? How many atoms are in 0.551 g of potassium (K) ? N (atoms) = 0.551 (g)×6.022×10 23 /39.10 (g/mol) = 8.49 x 10 21 atoms K 3.2 N (atoms)=?m =0.551 g From periodic table

19. n (mol) = m (g) / M (g/mol) n = 6.46 / 4.003 = 1.61 mol He From periodic table n (mole)=?m =6.46 g

21. m (g) = n (mol) × M (g/mol) m = 0.356 × 65.39 = 23.28 g Zn n=0.356 molem (g)=?

22. W ORKED E XAMPLE 3.4

23. N (atoms) = m (g) × N A (atoms) / M (g/mol) N = 16.3 × 6.022×10 23 / 32.07 = 3.06 ×10 23 S atoms N (atom)=?m =16.3g

24. 3.3 MOLECULAR MASS.

25. Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. SO 2 1S32.07 amu 2O+ 2 x 16.00 amu SO 2 64.07 amu For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO 2 = 64.07 amu 1 mole SO 2 = 64.07 g SO 2 Molecular mass = (No. of atoms× atomic mass) +( No. of atoms× atomic mass) element Ielement II

27. a- SO 2 = 32.07 amu + 2(16.00 amu) = 64.07 amu b- C 8 H 10 N 4 O 2 = 8(12.01 amu) + 10(1.008 amu) + 4(14.01 amu) + 2(16.00 amu) = 194.20 amu

28. W ORKED E XAMPLE 3.6

29. n (mol) = m (g) / M (g/mol) = 6.07 / (12.01 + 4(1.008)) = 0.378 mol CH 4 No. of moles ?m= 6.09 gCH 4

30. W ORKED E XAMPLE 3.7

31. N (atoms) = m (g) × N A (atoms) / M (g/mol) = 25.6 × 6.022×10 23 × 4 / 60.06 = 1.03 ×10 24 H atoms No. of H atoms ?m= 25.6 g(NH 2 ) 2 CO M = 60.06 g/mol

32. Do You Understand Molecular Mass? How many H atoms are in 72.5 g of C 3 H 8 O ? N (atoms) = m (g) × N A (atoms) / M (g/mol) = 72.5 × 6.022×10 23 × 8 / 60.09 = 5.81 × 10 24 H atoms m= 72.5 gC3H8OC3H8ONo. of H atoms ? M g/mol calculate 60.09 No. of H atoms in molecule

33. Molar mass (g/mol) Atomic mass molecular mass Atoms molecules Al, As, Ni, H HCl, H 2, O 2, NaCl كتلة ذرية (من الجدول الدوري) كتلة جزيئية It must be noted that:

34. 3.5 PERCENT COMPOSITION OF COMPOUNDS.

35. Percent composition of an element in a compound = is the percent by mass of each 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 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% Molar mass= 2(12.01)+ 6(1.008)+ 16.0= 46.07 g Check the answer!

37. %H = 3 x (1.008 g) 97.99 g x 100% = 3.086%P = 30.97 g 97.99 g x 100% = 31.61%O = 4 x (16.00 g) 97.99 g x 100% = 65.31% Molar mass of H 3 PO 4 = 3(1.008) + 1(30.97) + 4(16.00) = 97.99 g

38. Percent Composition and Empirical Formulas

40. strategy

41. n C = 40.92 12.01 g = 3.407 mol Cn = mass percent M of element n H = 4.58 1.008 g = 4.54 mol Hn O = 54.50 16.00 g = 3.406 mol C أولاً: حساب عدد المولات بناء على النسبة المئوية لكل عنصر The smallest No. of moles

42. ثانياً: تتم قسمة نواتج المولات المحسوبة على أصغر عدد للمولات C= 3.407 3.406 ≈ 1 H= 4.54 3.406 = 1.33 O= 3.406 = 1 ثالثاً: الضرب في عدد صحيح للحصول على أرقام صحيحه لعدد المولات 1.33 × 2 = 2.66 1.33 × 3 = 3.99 ≈ 4 1.33 × 4 = 5.32 رابعاً: ضرب مولات الذرات في 3 للحصول على الصيغة الأولية empirical formula وتصبح كالتالي C 3 H 4 O 3 Trail and error procedure

43. 3.5 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. 43 n = percent M n K = 24.75 39.10 g = 0.6330 mol Kn Mn = 34.77 54.94 g = 0.6329 mol Mnn O = 40.51 16.00 g = 2.532 mol O The smallest No. of moles

44. ثانياً: تتم قسمة نواتج المولات المحسوبة على أصغر عدد للمولات K= 0.6330 0.6329 ≈ 1 Mn= 0.6329 = 1 O= 2.532 0.6329 = 4 ثالثاً: بما أننا حصلنا على أرقام صحيحه في الخطوة الثانية فإن الصيغة الأولية empirical formula تصبح كالتالي KMnO 4

45. Determination of Molecular Formulas If we determine the empirical formula and the molar mass of the compound is known we can determine the molecular Formula by using: Molar mass of molecule (compound) Molar mass of empirical formula = n Molecular formula = (empirical formula) n

46. 3.7 CHEMICAL REACTIONS AND CHEMICAL EQUATION.

47. 3.7 3 ways of representing the reaction of H 2 with O 2 to form H 2 O A process in which one or more substances is changed into one or more new substances is a chemical reaction A chemical equation uses chemical symbols to show what happens during a chemical reaction reactantsproducts

48. 1- عدد الذرات أو الجزئيات 2- عدد المولات للذرات أو الجزئيات)) 3- الكتلة المولارية قراءة المعادلة الكيميائية How to “Read” Chemical Equations

49. H OW TO “R EAD ” C HEMICAL E QUATIONS 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 IS NOT 2 grams Mg + 1 gram O 2 makes 2 g MgO 2(24.31)2(16.0)2[24.31+16.0] reactant =80.6Product = 80.6 √ √ √ X

50. Balancing Chemical Equations 1.Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water C 2 H 6 + O 2 CO 2 + H 2 O 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 نكتب الصيغه الصحيحه لكل متفاعل (على الطرف الايسر) ولكل ناتج (على الطرف ا يمن) وزن المعادله الكيميائيه يكون بتغير الارقام التي بجانب الصيغه وليست التي تحتها بحيث يكون للعنصر نفس العدد على طرفي المعادله.

51. Balancing Chemical Equations 3.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 3.7 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 توزن أولا العناصر ا قل ظهورا

52. Balancing Chemical Equations 4.Balance those elements that appear in two or more reactants or products. 3.7 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 2 ثم توزن العناصر الاكثر ظهورا

53. Balancing Chemical Equations 5.Check to make sure that you have the same number of each type of atom on both sides of the equation. 3.7 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) الخطوه الاخيره هي التأكد من ان لديك نفس العدد من الذرات لكل عنصر على طرفي المعادله

54. W ORKED E XAMPLE 3.12

55. 3.8 AMOUNTS OF REACTANTS AND PRODUCTS.

56. 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 Amounts of Reactants and Products 3.8 ReactantProduct Use gram ratio of A and B From balanced equation

57. Methanol burns in air according to the equation 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O If 209 g of methanol are used up in the combustion, what mass of water is produced? 64 g of CH 3 OH 72 g of H 2 O 3.8 209 g x g x = 72 ×209/64 = 235 g H 2 O 2×Molar mass = 2[(1 × 12.01)+(4 × 1.008)+ (1 × 16.00)] 4×Molar mass = 4[(2 × 1.008)+ (1 ×16.00)]

58. W ORKED E XAMPLE 3.13 A

59. W ORKED E XAMPLE 3.13 B

60. n (mol) = m (g) / M (g/mol) n (mol) = 856/180.2 = 4.75 mol 1 mol of C 6 H 12 O 6 6 mol of CO 2 4.75 mol x mol x = 6×4.75 / 1 = 28.5 mol CO 2 m (g) = n (mol) × M (g/mol) = 28.5 × 44.01 = 1.25 × 10 3 g CO 2 Molar mass of Calculate the mole No. of

62. 13.882 g of Li 2.016 g of H 2 x g 9.89 g x = 13.882×9.89 / 2.016 = 68.1 g Li 2 × Molar mass = 2(6.941)Molar mass = (2 × 1.008)

63. 3.9 LIMITING REAGENTS

64. Limiting ReagentExcess Reagents

65. Limiting Reagents is the reactant used up first In a reaction Excess reagents are the Reactants present in quantities Greater than necessary to react With the present quantity of The limiting reagent. 3.9 2NO + 2O 2 2NO 2 NO is the limiting reagent O 2 is the excess reagent

66. .1-حساب عدد المولات للمواد المتفاعلة وذلك بقسمة الوزن بالجرام على الكتلة المولارية 2- تقسم عدد مولات كل مادة متفاعلة على عدد مولاتها الموجودة في المعادلة. 3 - المادة اللي تكون ناتجها أقل هي الكاشف المحدد. 4- يستخدم الكاشف المحدد لحساب كمية المادة الناتجة المطلوبة تبعا للمعادلة الموزونة Identify the Limiting Reactant تحديد الكاشف المحدد Calculate the amount of product obtained from the Limiting Reactant. حساب كمية المادة الناتجة بناء على الكاشف المحدد

67. Do You Understand Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe Calculate the mass of Al 2 O 3 formed. 3.9 n (mol) = m (g) / M (g/mol) n Al = 124 / 26.98 = 4.596 mol n Fe2O3 = 601 / 160 3.756 mol نسبة كل مركب Al = 4.596/2 = 2.298 Fe 2 O 3 = 3.756/1 = 3.756 الأصغر عدديا هو الكاشف المحدد limiting reagent وبناء على النتائج فإن Al هو الكاشف المحدد n (mol) = m (g) / M (g/mol) n Al = 124 / 26.98 = 4.596 mol n Fe2O3 = 601 / 160 3.756 mol نسبة كل مركب Al = 4.596/2 = 2.298 Fe 2 O 3 = 3.756/1 = 3.756 الأصغر عدديا هو الكاشف المحدد limiting reagent وبناء على النتائج فإن Al هو الكاشف المحدد

68. لحساب كمية Fe 2 O 3 نستخدم الكاشف المحدد 53.96 g of Al 101.96 g of Al 2 O 3 124 g x g x = 124 ×101.96/53.96 = 234.3 g Al 2 O 3 لحساب كمية Fe 2 O 3 نستخدم الكاشف المحدد 53.96 g of Al 101.96 g of Al 2 O 3 124 g x g x = 124 ×101.96/53.96 = 234.3 g Al 2 O 3

69. W ORKED E XAMPLE 3.15 A

71. 3.10 REACTION YIELD

72.  Quantities of product calculated represent the maximum amount obtainable (100 % yield)  Most chemical reactions do not give 100 % yield of product because of: 1-Side reactions (unwanted reactions) 2-Reversible reactions ( reactants products) 3-Losses in handling and transferring Reaction Yield

73. 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 3.10 Reaction Yield The percent Yield is the proportion of the actually yield to the theoretical yield which can be obtained from the following relation:

76. أولاً يتم تحديد الكاشف المحدد n (mol) = m (g) / M (g/mol) n TiCl4 = 3.54 ×10 7 / 189.7 = 1.87 ×10 5 mol n Mg = 1.13 ×10 7 / 24.31 4.65 ×10 5 mol نسبة كل مركب TiCl 4 = 1.87 ×10 5 /1 = 1.87 ×10 5 Mg = 2.32 ×10 5 /2 = 2.32 ×10 5 الأصغر عدديا هو الكاشف المحدد limiting reagent وبناء على النتائج فإن TiCl 4 هو الكاشف المحدد أولاً يتم تحديد الكاشف المحدد n (mol) = m (g) / M (g/mol) n TiCl4 = 3.54 ×10 7 / 189.7 = 1.87 ×10 5 mol n Mg = 1.13 ×10 7 / 24.31 4.65 ×10 5 mol نسبة كل مركب TiCl 4 = 1.87 ×10 5 /1 = 1.87 ×10 5 Mg = 2.32 ×10 5 /2 = 2.32 ×10 5 الأصغر عدديا هو الكاشف المحدد limiting reagent وبناء على النتائج فإن TiCl 4 هو الكاشف المحدد

77. لحساب الحصيلة النظرية لـ Ti نستخدم الكاشف المحدد 189.7 g of TiCl4 47.88 g of Ti 3.54 ×10 7 g x g x = 47.88 × 3.54 ×10 7 / 189.7 x (theoretical yield) = 8.95×10 6 g Ti %yield = actual yield/theoretical yield × 100 = (7.91×10 6 g / 8.95×10 6 g) × 100 = 88.4% لحساب الحصيلة النظرية لـ Ti نستخدم الكاشف المحدد 189.7 g of TiCl4 47.88 g of Ti 3.54 ×10 7 g x g x = 47.88 × 3.54 ×10 7 / 189.7 x (theoretical yield) = 8.95×10 6 g Ti %yield = actual yield/theoretical yield × 100 = (7.91×10 6 g / 8.95×10 6 g) × 100 = 88.4%

78. Problems 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 – 3.60 3.84 – 3.86