1. 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:
Natural lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
Average atomic mass of lithium:
7.42 x x 7.016
100
= amu
3.1

6. Average atomic mass (6.941)

7. 3.2 Avogadro’s number and the molar mass of an element.

8. Dozen = 12 Pair = 2 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)
3.2

9. Molar mass M is the mass of 1 mole of in grams marbles atoms
eggs
shoes
Molar mass M is the mass of 1 mole of in grams
marbles
atoms
1 mole 12C atoms = x 1023 atoms = g
1 12C atom = amu
1 mole 12C atoms = g 12C
1 mole lithium atoms = g of Li
For any element
atomic mass (amu) = molar mass (grams)
3.2

10. 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

11. How many amu are there in 8.4 g?
1 g = x 1023 amu
8.4 g = 5.1 x 1024 amu
What is the mass in grams of 13.2 amu ?
1 amu = 1.66 x g
13.2 amu = 2.2 x g

12. = molar mass in g/mol M NA = Avogadro’s number m/M nNA كتلة العنصر mnM
N/NA
3.2

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

16. n (mol) = m (g) / M (g/mol)
= mol He

18. m (g) = n (mol) × M (g/mol)
= g Zn

19. Worked Example 3.4

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

21. 3.3 molecular mass.

22. 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
For any molecule
molecular mass (amu) = molar mass (grams)
1 molecule SO2 = amu
1 mole SO2 = g SO2
3.3

24. 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

25. Worked Example 3.6

26. n (mol) = m (g) / M (g/mol)
= 6.07 / ( (1.008))
= mol CH4

27. Worked Example 3.7

28. N (atoms) = m (g) × NA (atoms) / M (g/mol)
= 25.6 × 6.022×1023 × 4 / 60.06
= 1.03 ×1024 H atoms

29. Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ?
N (atoms) = m (g) × NA (atoms) / M (g/mol)
= 72.5 × 6.022×1023 × 8 / 60
= 5.82 × H atoms
3.3

30. 3.5 percent composition of compounds.

31. 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
%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%
3.5

33. 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%

34. Percent Composition and Empirical Formulas

36. أولاً: حساب عدد المولات بناء على النسبة المئوية لكل عنصر
n =
percent M
nC =
40.92
12.01 g
= mol C
nH =
4.58
1.008 g
= mol H
nO =
54.50
16.00 g
= mol C

37. 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
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

38. 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
nO =
40.51
16.00 g
= mol O
3.5

39. ثانياً: تتم قسمة نواتج المولات المحسوبة على أصغر عدد للمولاتK=
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

40. 3.7 chemical reactions and chemical equation.

41. 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

42. 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
IS NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
3.7

43. Balancing Chemical Equations
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
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
3.7

44. Balancing Chemical Equations
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

45. Balancing Chemical Equations
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
2C2H6 + 7O2
4CO2 + 6H2O
3.7

46. Balancing Chemical Equations
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

47. Worked Example 3.12

48. 3.8 amounts of reactants and products.

49. Amounts of Reactants and Products
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

50. 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?
64 g of 2CH3OH g of H2O
209 g x g
x = 72×209/64
= 235 g H2O
3.8

51. Worked Example 3.13a

52. Worked Example 3.13b

53. n (mol) = m (g) / M (g/mol) n (mol) = 856/180.2 = 4.75 mol
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

55. g of Li g of H2
x g g
x = ×9.89 / 2.016
= 68.1 g Li

56. 3.9 limiting reagents

57. 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

58. 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

59. 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

60. Worked Example 3.15a

62. 3.10 reaction yield

63. 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

66. 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

67. لحساب الكمية 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%

68. 1.
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. 2.
The atomic masses of Cl (75.53 percent) and Cl (24.47 percent) are amu and amu, respectively. Calculate the average atomic mass of chlorine. The percentages in parentheses denote the relative abundances. A)
35.96 amu B)
35.45 amu C)
36.47 amu D)
71.92 amu 3.
How many amu are there in 8.4 g? A)
8.4 x1023 amu B)
1.4 x amu C)
8.4 amu D)
5.1 x 1024 amu 4.
How many atoms are there in 5.10 moles of sulfur (S)? A)
3.07 x 1024 B)
9.59 x 1022 C)
6.02 x 1023 D)
9.82 x 1025

69. 5.
What is the mass in grams of a single atom of As? A)
1.244 x g B)
2.217 x g C)
8.039 x 1021 g D)
4.510 x1025 g 6.
How many atoms are present in 3.14 g of copper (Cu)? A)
2.98 x 1022 B)
1.92 x 1023 C)
1.89 x 1024 D)
6.02 x 1023 7.
Calculate the molar mass of Li2CO3. A)
73.89 g B)
66.95 g C)
41.89 g D)
96.02 g 8.
How many molecules of ethane (C2H6) are present in g of C2H6? A)
2.01 x 1023 B)
6.69 x 1021 C)
4.96 x 1022 D)
8.89 x 1020

70. 9.
Allicin is the compound responsible for the characteristic smell of garlic. An analysis of the compound gives the following percent composition by mass: C: 44.4 percent; H: 6.21 percent; S: 39.5 percent; O: 9.86 percent. What is its molecular formula given that its molar mass is about 162 g? A)
C12H20S4O2 B)
C7H14SO C)
C6H10S2O D)
C5H12S2O2
10.
The formula for rust can be represented by Fe2O3. How many moles of Fe are present in 24.6 g of the compound? A)
2.13 mol B)
0.456 mol C)
0.154 mol D)
0.308 mol

71. 11.
How many grams of sulfur (S) are needed to react completely with 246 g of mercury (Hg) to form HgS? A)
39.3 g B)
24.6 g C)
9.66  103 g D)
201 g
12.
Tin(II) fluoride (SnF2) is often added to toothpaste as an ingredient to prevent tooth decay. What is the mass of F in grams in 24.6 g of the compound? A)
18.6 g B)
24.3 g C)
5.97 g D)
75.7 g
13.
What is the empirical formula of the compound with the following composition?
2.1 percent H, 65.3 percent O, 32.6 percent S. A)
H2SO4 B)
H2SO3 C)
H2S2O3 D)
HSO3
14.
Which of the following equations is balanced? A)
2C + O CO B)
2CO + O CO2 C)
H2 + Br HBr D)
2K + H2O KOH + H2

72. 15.
Consider the combustion of carbon monoxide (CO) in oxygen gas:
2CO(g) + O2(g)  2CO2(g)
Starting with 3.60 moles of CO, calculate the number of moles of CO2 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
16.
Nitrous oxide (N2O) is also called “laughing gas.” It can be prepared by the thermal decomposition of ammonium nitrate (NH4NO3). The other product is H2O. The balanced equation for this reaction is:
NH4NO3  N2O + 2H2O
How many grams of N2O are formed if 0.46 mole of NH4NO3 is used in the reaction? A)
2.0 g B)
3.7  101 g C)
2.0  101 g D)
4.6  10-1 g
17.
The fertilizer ammonium sulfate [(NH4)2SO4] is prepared by the reaction between ammonia (NH3) and sulfuric acid:
2NH3(g) + H2SO4(aq)  (NH4)2SO4(aq)
How many kilograms of NH3 are needed to produce 1.00  105 kg of (NH4)2SO4? A)
1.70  104 kg B)
3.22  103 kg C)
2.58  104 kg D)
7.42  104 kg

73. 18.
Nitric oxide (NO) reacts with oxygen gas to form nitrogen dioxide (NO2), a dark-brown gas:
2NO(g) + O2(g)  2NO2(g)
In one experiment mole of NO is mixed with mole of O2. Calculate the number of moles of NO2 produced (note: first determine which is the limiting reagent). A)
0.886 mol B)
0.503 mol C)
1.01 mol D)
1.77 mol
19.
Hydrogen fluoride is used in the manufacture of Freons (which destroy ozone in the stratosphere) and in the production of aluminum metal. It is prepared by the reaction
CaF2 + H2SO4  CaSO4 + 2HF
In one process 6.00 kg of CaF2 are treated with an excess of H2SO4 and yield 2.86 kg of HF. Calculate the percent yield of HF. A)
93.0 % B)
95.3 % C)
47.6 % D)
62.5 %

74. Answer Key 1-C 2-B 3-D 4-A 5-A 6-A 7-A 8-B 9-C 10-D 11-A 12-C 13-A 14-B 15-C 16-C 17-C 18-A 19-A
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