1. Mass Relationships in Chemical Reactions 3.1 Atomic Mass 3.2 Avogadro’s Number and the Molar Mass of an Element 3.3 Molecular Mass 3.4 The Mass Spectrometer 3.5 Percent Composition of Compounds 3.6 Experimental Determination of Empirical Formulas 3.7 Chemical Reactions and Chemical Equations 3.8 Amounts of Reactants and Products 3.9 Limiting Reagents 3.10 Reaction Yield

2. 3.1 Atomic Mass The mass of an atom depends on the number of electrons, protons, and neutrons it contains. By international agreement, atomic mass (sometimes called atomic weight ) is the mass of the 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. Carbon-12 is the carbon isotope that has six protons and six neutrons. 3.2 Avogadro’s Number (mole) and the Molar Mass of an Element In the SI system the mole (mol) is the amount of a substance that contains a number of atoms, ( molecules, or other particles) as there are atoms in exactly 12 g (or 0.012 kg) of the carbon-12 isotope. The actual number of atoms in 12 g of carbon-12 is determined experimentally. This number is called Avogadro’s number N A = 6.022 x 10 23 1 mole of hydrogen atoms contains 6.022 x 10 23 H atoms. Figure 3.1 shows samples containing 1 mole each of several common elements.

3. 3 Molar mass 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 12 C atoms = 12.00 g 12 C 1 mole lithium atoms = 6.941 g of Li For any element atomic mass (amu) = molar mass (grams)

4. 4 M = molar mass in g/mol N A = Avogadro’s number Figure 3.2 The relationships between mass (m in grams) of an element and number of moles of an element (n) and between number of moles of an element and number of atoms (N) of an element.. m is the molar mass (g/mol) of the element and NA is Avogadro’s number

5. We have seen that 1 mole of carbon-12 atoms has a mass of exactly 12 g and contains 6.022 3 1023 atoms. This mass of carbon-12 is its molar mass ( M), define d as the mass (in grams or kilograms) of 1 mole of units (such as atoms or molecules) of a substance. Note that the molar mass of carbon-12 (in grams) is numerically equal to its mass in amu. Likewise, the atomic mass of sodium (Na) is 22.99 amu and its molar mass is 22.99 g; the atomic mass of phosphorus is 30.97 amu and its molar mass is 30.97 g; and so on. If we know the atomic mass of an element, we also know its molar mass.

6. Do You Understand Molar Mass? How many atoms are in 0.32 g of Sodium (Na) ? N (atoms) = 0.32 (g)×6.022×10 23 /23 (g/mol) = 8.38 x 10 21 atoms Na N (atoms)=?m =0.32 g From periodic table

7. 77 x 6.022 x 10 23 atoms K 1 mol K = How many atoms are in 0.32 g of Sodium (Na) ? 1 mol K = 23.00 g K 1 mol K = 6.022 x 10 23 atoms K 0.32 g K 1 mol K 23.00 g K x 8.38 x 10 21 atoms Na

9. 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

11. m (g) = n (mol) × M (g/mol) m = 0.356 × 65.39 = 23.28 g Zn n=0.356 mole m (g)=? طريقة أخرى Solution: 1 mol Zn = 65.39 g Zn 0.356 mole = ? G Zn mass of Zn = 0.356x65.39 = 23.28 g 1

13. 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

14. 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

16. 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

18. 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

20. 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

21. 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

22. 22 How many H atoms are in 56.8 g of C 3 H 8 O ? 1 mol C 3 H 8 O = (3 x 12) + (8 x 1) + 16 = 60 g C 3 H 8 O 1 mol H = 6.022 x 10 23 atoms H 4.56 x 10 24 atoms H 1 mol C 3 H 8 O molecules = 8 mol H atoms 56.8 g C 3 H 8 O 1 mol C 3 H 8 O 60 g C 3 H 8 O x 8 mol H atoms 1 mol C 3 H 8 O x 6.022 x 10 23 H atoms 1 mol H atoms x =

23. 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:

24. 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!

26. %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

27. 3.3 Molecular Mass The molecular mass (sometimes called molecular weight ) is the sum of the atomic masses (in amu) in the molecule. For example, the molecular mass of H 2 O is 2(atomic mass of H) + 1 atomic mass of O or 2(1.008 amu) + 16.00 amu = 18.02 amu What is the formula mass of Al 2 (SO 4 ) 3 ? 2 Al 2 x 26.98 3 S 3 x 32.06 12 O + 12 x 16.00 342.14 amu

28. 28 Ex. 2 In 1 mole of hydrogen peroxide (H 2 O 2 ) there are 2 moles of H atoms and 2 moles of O atoms. The molar masses of H 2 O 2, H, and O are 34.02 g, 1.008 g, and 16.00 g, respectively. Therefore, the percent composition of H 2 O 2 is calculated as follows: The sum of the percentages is 5.926% + 94.06% = 99.99%.

29. 29 Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound. 1Na22.99 amu 1Cl + 35.45 amu NaCl 58.44 amu For any ionic compound formula mass (amu) = molar mass (grams) 1 formula unit NaCl = 58.44 amu 1 mole NaCl = 58.44 g NaCl NaCl

30. 30 What is the formula mass of Ca 3 (PO 4 ) 2 ? 1 formula unit of Ca 3 (PO 4 ) 2 3 Ca 3 x 40.08 2 P2 x 30.97 8 O + 8 x 16.00 310.18 amu

31. 31 3 ways of representing the reaction of H 2 with O 2 to form H 2 O 3.7 Chemical Reactions and Chemical Equations 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

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

33. 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 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

34. 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.

35. 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 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

36. Balancing Chemical Equations 4.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 2 ثم توزن العناصر الاكثر ظهورا

37. 37 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. 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

39. 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 ReactantProduct Use gram ratio of A and B From balanced equation

40. 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 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)]

41. 41 Methanol burns in air according to the equation 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O If 123 g of methanol are used up in the combustion, what mass of water is produced? grams CH 3 OHmoles CH 3 OHmoles H 2 Ograms H 2 O molar mass CH 3 OH coefficients chemical equation molar mass H 2 O 123 g CH 3 OH 1 mol CH 3 OH 32.0 g CH 3 OH x 4 mol H 2 O 2 mol CH 3 OH x 18.0 g H 2 O 1 mol H 2 O x = 138.38 g H 2 O

42. Limiting ReagentExcess Reagents LIMITING REAGENTS

43. 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

44. 44 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. g Almol Almol Fe 2 O 3 needed g Fe 2 O 3 needed OR g Fe 2 O 3 mol Fe 2 O 3 mol Al needed g Al needed 124 g Al 1 mol Al 27.0 g Al x 1 mol Fe 2 O 3 2 mol Al x 160. g Fe 2 O 3 1 mol Fe 2 O 3 x = 367 g Fe 2 O 3 Start with 124 g Alneed 367 g Fe 2 O 3 Have more Fe 2 O 3 (601 g) so Al is limiting reagent

45. .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

46. 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. 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

47. 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

48. 48 Use limiting reagent (Al) to calculate amount of product that can be formed. g Almol Almol Al 2 O 3 g Al 2 O 3 124 g Al 1 mol Al 27.0 g Al x 1 mol Al 2 O 3 2 mol Al x 102. g Al 2 O 3 1 mol Al 2 O 3 x = 234 g Al 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe At this point, all the Al is consumed and Fe 2 O 3 remains in excess.

49.  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

50. 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: