Chapter 3 Mass Relationships in Chemical Reactions Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

Chapter 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.

3.1 Atomic mass.

Atomic mass is the mass of an atom in atomic mass units (amu) Micro World atoms & molecules Macro World grams 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. By definition: 1 atom 12C “weighs” 12 amu On this scale 1H = 1.008 amu 16O = 16.00 amu 3.1

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

Average atomic mass (6.941)

(a) The number of neutrons in the element. 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.

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. Solution: Isotope II Atomic mass = 36.956 amu Abundance percent = 24.47 % Isotope I Atomic mass = 34.698 amu Abundance percent = 75.53 % Average atomic mass= Atomic mass × abundance percent + Atomic mass × abundance percent 100 Isotope 1 Isotope 2 Average atomic mass= 34.698 × 75.53+ 36.956 × 24.47 100 Isotope 1 Isotope 2 35.45 amu =

The atomic masses of and are 6. 0151 amu and 7. 0160 amu, respectively 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. Solution: Average atomic mass = 6.941 amu Isotope II Atomic mass = 7.0160amu Abundance percent = ? Isotope I Atomic mass = 6.0151 amu Abundance percent = ? Average atomic mass= Atomic mass × abundance percent + Atomic mass × abundance percent 100 Isotope 1 Isotope 2 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 1 Isotope 2 6.941 × 100= 6.0151X+ 701.60-7.0160X 694.1= -1.0009X+ 701.60 694.1-701.60 = -1.0009X X= -7.5 -1.0009 -7.5 = -1.0009X X= 7.493 % 100-X= 100-7.493= 92.507%

3.2 Avogadro’s number and the molar mass of an element.

Dozen = 12 Pair = 2 The mole (mol) is the amount of a substance that يستخدم المول لعد الجزيئات والذرات والأيونات ويستخدمه الكيميائيون لأنه يوفر طريقة ملائمة لمعرفة عدد الجسيمات في العينة 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 12C isotope 1 mol = NA = 6.0221367 x 1023 Avogadro’s number (NA) عدد أفوجادرو هو حلقة الوصل بين عالم الجسيمات غير المرئية (الجزئيات والذرات) وعالم المرئيات كالماء والأملاح 3.2

1 mole lithium atoms = 6.941 g Li 1 lithium atom = 6.941 amu Molar mass (M) is the mass of 1 mole of in grams eggs shoes marbles atoms Mass of 1 mole of C = mass of 1 mole Cu M = mass/mol = g/mol Molar mass 1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g 1 12C atom = 12.00 amu Atomic mass 1 mole lithium atoms = 6.941 g Li 1 lithium atom = 6.941 amu For any element atomic mass (amu) = molar mass (g/mol)

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

How many amu are there in 8.4 g? 1 g = 6.022 x 1023 amu 8.4 g = 5.1 x 1024 amu 1 g 6.022 x 1023 amu 8.4 g ? What is the mass in grams of 13.2 amu ? 1 amu = 1.66 x 10-24 g 13.2 amu = 2.2 x 10-23 g يوضح المثالان أنه يمكن استخدام عدد أفوجادرو لتحويل الكتلة الذرية إلى كتلة بالجرام والعكس صحيح

NA = Avogadro’s number = molar mass in g/mol M يمكن باستخدام كلا من عدد أفوجادرو (NA) 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 M = molar mass in g/mol NA = Avogadro’s number m/M nNA كتلة العنصر m عدد مولات العنصر n عدد ذرات العنصر N nM N/NA 3.2

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

Do You Understand Molar Mass? How many atoms are in 0.551 g of potassium (K) ? N (atoms)=? m =0.551 g From periodic table N (atoms) = 0.551 (g)×6.022×1023/39.10 (g/mol) = 8.49 x 1021 atoms K 3.2

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

m (g) = n (mol) × M (g/mol) n=0.356 mole m (g)=? m (g) = n (mol) × M (g/mol) m = 0.356 × 65.39 = 23.28 g Zn

Worked Example 3.4

N (atoms) = m (g) × NA (atoms) / M (g/mol) = 3.06×1023 S atoms

3.3 molecular mass.

molecular mass (amu) = molar mass (grams) Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. SO2 1S 32.07 amu 2O + 2 x 16.00 amu SO2 64.07 amu element I element II Molecular mass = (No. of atoms× atomic mass) +( No. of atoms× atomic mass) For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2

a- SO2 = 32.07 amu + 2(16.00 amu) = 64.07 amu b- C8H10N4O2 = 8(12.01 amu) + 10(1.008 amu) + 4(14.01 amu) + 2(16.00 amu) = 194.20 amu

Worked Example 3.6

n (mol) = m (g) / M (g/mol) No. of moles ? CH4 m= 6.09 g n (mol) = m (g) / M (g/mol) = 6.07 / (12.01 + 4(1.008)) = 0.378 mol CH4

Worked Example 3.7

N (atoms) = m (g) × NA (atoms) / M (g/mol) No. of H atoms ? m= 25.6 g (NH2)2CO M = 60.06 g/mol N (atoms) = m (g) × NA (atoms) / M (g/mol) = 25.6 × 6.022×1023 × 4 / 60.06 = 1.03 ×1024 H atoms

Do You Understand Molecular Mass? How many H atoms are in 72.5 g of C3H8O ? No. of H atoms ? m= 72.5 g C3H8O M g/mol calculate 60.09 N (atoms) = m (g) × NA (atoms) / M (g/mol) = 72.5 × 6.022×1023 × 8 / 60.09 = 5.81 × 1024 H atoms No. of H atoms in molecule

It must be noted that: Molar mass (g/mol) Atomic mass molecular mass Atoms molecules Al, As, Ni, H HCl, H2, O2, NaCl كتلة ذرية (من الجدول الدوري) كتلة جزيئية

3.5 percent composition of compounds.

n x molar mass of element molar mass of compound x 100% 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 Molar mass= 2(12.01)+ 6(1.008)+ 16.0= 46.07 g %C = 2 x (12.01 g) 46.07 g x 100% = 52.14% C2H6O %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!

Molar mass of H3PO4 = 3(1.008) + 1(30.97) + 4(16.00) = 97.99 g %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%

Percent Composition and Empirical Formulas

strategy

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

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 Trail and error procedure 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 C3H4O3

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

KMnO4 ثانياً: تتم قسمة نواتج المولات المحسوبة على أصغر عدد للمولات K= 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 KMnO4

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

3.7 chemical reactions and chemical equation.

3 ways of representing the reaction of H2 with O2 to form H2O 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 3 ways of representing the reaction of H2 with O2 to form H2O reactants products 3.7

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

How to “Read” Chemical Equations 2 Mg + O2 2 MgO √ 2 atoms Mg + 1 molecule O2 makes 2 formula units MgO 2 moles Mg + 1 mole O2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO √ √ 2(24.31) 2(16.0) 2[24.31+16.0] reactant =80.6 Product = 80.6 IS NOT X 2 grams Mg + 1 gram O2 makes 2 g MgO

Balancing Chemical Equations نكتب الصيغه الصحيحه لكل متفاعل (على الطرف الايسر) ولكل ناتج (على الطرف الايمن) 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 C2H6 + O2 CO2 + H2O وزن المعادله الكيميائيه يكون بتغير الارقام التي بجانب الصيغه وليست التي تحتها بحيث يكون للعنصر نفس العدد على طرفي المعادله. 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. 2C2H6 NOT C4H12

Balancing Chemical Equations توزن أولا العناصر الاقل ظهورا Start by balancing those elements that appear in only one reactant and one product. C2H6 + O2 CO2 + H2O start with C or H but not O 1 carbon on right 2 carbon on left multiply CO2 by 2 C2H6 + O2 2CO2 + H2O 6 hydrogen on left 2 hydrogen on right multiply H2O by 3 C2H6 + O2 2CO2 + 3H2O 3.7

Balancing Chemical Equations ثم توزن العناصر الاكثر ظهورا Balance those elements that appear in two or more reactants or products. multiply O2 by 7 2 C2H6 + O2 2CO2 + 3H2O 2 oxygen on left 4 oxygen (2x2) + 3 oxygen (3x1) = 7 oxygen on right C2H6 + O2 2CO2 + 3H2O 7 2 remove fraction multiply both sides by 2 2 2C2H6 + 7O2 4CO2 + 6H2O 3.7

Balancing Chemical Equations الخطوه الاخيره هي التأكد من ان لديك نفس العدد من الذرات لكل عنصر على طرفي المعادله Check to make sure that you have the same number of each type of atom on both sides of the equation. 2C2H6 + 7O2 4CO2 + 6H2O 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) Reactants Products 4 C 12 H 14 O 3.7

Worked Example 3.12

3.8 amounts of reactants and products.

Amounts of Reactants and Products Use gram ratio of A and B From balanced equation Write balanced chemical equation Convert quantities of known substances into moles Use coefficients in balanced equation to calculate the number of moles of the sought quantity Convert moles of sought quantity into desired units 3.8

Methanol burns in air according to the equation 2CH3OH + 3O2 2CO2 + 4H2O If 209 g of methanol are used up in the combustion, what mass of water is produced? 4×Molar mass = 4[(2 × 1.008)+ (1 ×16.00)] 2×Molar mass = 2[(1 × 12.01)+(4 × 1.008)+ (1 × 16.00)] 64 g of CH3OH 72 g of H2O 209 g x g x = 72×209/64 = 235 g H2O 3.8

Worked Example 3.13a

Worked Example 3.13b

n (mol) = m (g) / M (g/mol) n (mol) = 856/180.2 = 4.75 mol Calculate the mole No. of n (mol) = m (g) / M (g/mol) n (mol) = 856/180.2 = 4.75 mol Molar mass of 1 mol of C6H12O6 6 mol of CO2 4.75 mol x mol x = 6×4.75 / 1 = 28.5 mol CO2 m (g) = n (mol) × M (g/mol) = 28.5 × 44.01 = 1.25 × 103 g CO2

2 × Molar mass = 2(6.941) Molar mass = (2 × 1.008) 13.882 g of Li 2.016 g of H2 x g 9.89 g x = 13.882×9.89 / 2.016 = 68.1 g Li

3.9 limiting reagents

Excess Reagents Limiting Reagent

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. 2NO + 2O2 2NO2 NO is the limiting reagent O2 is the excess reagent 3.9

Identify the Limiting Reactant تحديد الكاشف المحدد .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. Calculate the amount of product obtained from the Limiting Reactant . حساب كمية المادة الناتجة بناء على الكاشف المحدد 4- يستخدم الكاشف المحدد لحساب كمية المادة الناتجة المطلوبة تبعا للمعادلة الموزونة

Do You Understand Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe2O3 2Al + Fe2O3 Al2O3 + 2Fe Calculate the mass of Al2O3 formed. n (mol) = m (g) / M (g/mol) nAl = 124 / 26.98 = 4.596 mol nFe2O3 = 601 / 160 3.756 mol نسبة كل مركب Al = 4.596/2 = 2.298 Fe2O3 = 3.756/1 = 3.756 الأصغر عدديا هو الكاشف المحدد limiting reagent وبناء على النتائج فإن Al هو الكاشف المحدد 3.9

لحساب كمية Fe2O3 نستخدم الكاشف المحدد 53.96 g of Al 101.96 g of Al2O3 124 g x g x = 124×101.96/53.96 = 234.3 g Al2O3

Worked Example 3.15a

3.10 reaction yield

Reaction Yield 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 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. The percent Yield is the proportion of the actually yield to the theoretical yield which can be obtained from the following relation: % Yield = Actual Yield Theoretical Yield x 100 3.10

أولاً يتم تحديد الكاشف المحدد n (mol) = m (g) / M (g/mol) nTiCl4 = 3.54×107 / 189.7 = 1.87×105 mol nMg = 1.13×107 / 24.31 4.65×105 mol نسبة كل مركب TiCl4= 1.87×105 /1 = 1.87×105 Mg = 2.32×105 /2 = 2.32×105 الأصغر عدديا هو الكاشف المحدد limiting reagent وبناء على النتائج فإن TiCl4 هو الكاشف المحدد

لحساب الحصيلة النظرية لـ Ti نستخدم الكاشف المحدد 189.7 g of TiCl4 47.88 g of Ti 3.54×107 g x g x = 47.88 × 3.54×107 / 189.7 x (theoretical yield) = 8.95×106 g Ti %yield = actual yield/theoretical yield × 100 = (7.91×106 g / 8.95×106 g) × 100 = 88.4%

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