1. Chapter 3 Mass Relationships in Chemical Reactions
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2. 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 Reaction yield.

3. 3.1 Atomic mass.

4. 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 = amu
16O = amu
3.1

5. 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 x 7.016
100
= amu
3.1

6. Average atomic mass (6.941)

7. (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.

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

9. The atomic masses of and are 6. 0151 amu and 7. 0160 amu, respectively
The atomic masses of and are amu and amu, respectively. Calculate the natural abundance of these two isotopes. The average atomic mass of Li is amu.
Solution:
Average atomic mass = amu
Isotope II
Atomic mass = amu
Abundance percent = ?
Isotope I
Atomic mass = 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= X X
694.1= X
= X
X= -7.5
-7.5 = X
X= %
100-X= = %

10. 3.2 Avogadro’s number and the molar mass of an element.

11. 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 grams of 12C isotope
1 mol = NA = x 1023
Avogadro’s number (NA)
عدد أفوجادرو هو حلقة الوصل بين عالم الجسيمات غير المرئية (الجزئيات والذرات) وعالم المرئيات كالماء والأملاح
3.2

12. 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 = x 1023 atoms = g
1 12C atom = amu
Atomic mass
1 mole lithium atoms = g Li
1 lithium atom = amu
For any element
atomic mass (amu) = molar mass (g/mol)

13. 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

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

15. 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

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

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

19. 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
= mol He

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

22. Worked Example 3.4

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

24. 3.3 molecular mass.

25. 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 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 = amu
1 mole SO2 = g SO2

27. 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)
= amu

28. Worked Example 3.6

29. n (mol) = m (g) / M (g/mol)
No. of moles ?
CH4
m= 6.09 g
n (mol) = m (g) / M (g/mol)
= 6.07 / ( (1.008))
= mol CH4

30. Worked Example 3.7

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

32. 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 × H atoms
No. of H atoms in
molecule

33. 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

34. 3.5 percent composition of compounds.

35. 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= 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% % % = 100.0%
Check the answer!

37. Molar mass of H3PO4 = 3(1.008) + 1(30.97) + 4(16.00)
= 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%

38. Percent Composition and Empirical Formulas

40. strategy

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

42. ثانياً: تتم قسمة نواتج المولات المحسوبة على أصغر عدد للمولاتC=
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

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

44. 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

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 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

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

49. How to “Read” Chemical Equations
2 Mg + O 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 grams O2 makes 80.6 g MgO √ √
2(24.31)
2(16.0)
2[ ]
reactant =80.6
Product = 80.6
IS NOT X
2 grams Mg + 1 gram O2 makes 2 g MgO

50. 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

51. 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

52. 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
C2H O2
2CO2 + 3H2O 7 2
remove fraction
multiply both sides by 2 2
2C2H6 + 7O2
4CO2 + 6H2O
3.7

53. 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

54. Worked Example 3.12

55. 3.8 amounts of reactants and products.

56. 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

57. Methanol burns in air according to the equation
2CH3OH + 3O CO2 + 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 g of H2O
209 g x g
x = 72×209/64
= 235 g H2O
3.8

58. Worked Example 3.13a

59. Worked Example 3.13b

60. 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 C6H12O 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

62. 2 × Molar mass = 2(6.941)
Molar mass = (2 × 1.008)
g of Li g of H2
x g g
x = ×9.89 / 2.016
= 68.1 g Li

63. 3.9 limiting reagents

64. Excess Reagents
Limiting Reagent

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

66. 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

67. Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
n (mol) = m (g) / M (g/mol)
nAl = 124 / 26.98
= 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

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

69. Worked Example 3.15a

71. 3.10 reaction yield

72. 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

73. 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

76. 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

77. لحساب الحصيلة النظرية لـ Ti نستخدم الكاشف المحدد
189.7 g of TiCl g of Ti
3.54×107 g x g
x = × 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%

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