Download General Chemistry CHEM 110

General Chemistry
CHEM 110
Dr. Nuha Wazzan
Chapter 3
Mass Relationships in
Chemical Reactions
1
Chapter 3
Mass Relationships in Chemical
Reactions
Chang Pages 80-107
H.W. p110-114
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, 3.44, 3.46, 3.48, 3.50,
3.52, 3.60, 3.84, 3.86
2
Chapter 3: Mass Relationships in
Chemical Reactions: Outline
3.1 Atomic Mass
3.2 Avogadro’s Number and Molar Mass of an
Element
3.3 Molecular Mass
3.5 Percent Composition of Compounds
3.6 Experimental Determination of Empirical
Formula
3.9 Limiting Reagents
3.10 Reaction Yield
3
3.1 Atomic Mass
 Mass of an atom = mass of p + mass of n + mass of e
mass of p =mass of n = 1840 mass of e
Mass of atom = mass of p + mass of n
Atomic mass (or atomic weight): is the
mass of the atom in atomic mass unit (amu).
 Atomic mass unit: is the mass that exactly
equal to one-twelfth the mass of one
carbon-12 (12C) atom.
The atomic mass of 12C = 12.00 amu

4
The atomic mass of 12C = 12.00 amu
 12C is the Standard
e.g. Hydrogen atom )1H( is only 8.400%
massive as 12C
Thus: the atomic mass of 1H = 0.084 x
12.0 amu= 1.008 amu and so on for all
other elements.
5
6
Average Atomic Mass
6
C
Natural Abundance
‫الوفرة الطبيعية‬
12.01




Atomic mass of carbon = 12.01 not 12.00
The carbon exist in more than one form
(isotopes 12C & 13C)
Carbon is a mixture of Isotopes
Thus: Atomic Mass = Average Atomic Mass
7
How we get the Average Atomic Mass?
Average atomic mass of natural carbon =

∑ (The natural abundance x Atomic Mass)


for each isotope
The isotopes of Carbon are 12C & 13C
The natural abundance of each isotope is:
98.90% & 1.10%, respectively
Average Atomic Mass of natural carbon =
(0.9890 x 12.00 amu)12C + (0.0110 x 13.00335 amu)13C =
12.01 amu
Notes: 98.90% = 98.90/100 = 0.9890; 1.10% =
1.10/100 = 0.0110
 Notes: carbon is mainly 12C → the Average Atomic
Mass is 12.01 amu → closer to 12 amu than 13 amu 8

Average atomic
mass (12.01)
9
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.
10
Example 3.1 p81
The atomic masses of the Copper isotopes,
63
29
Cu(69.09%) &
65
29
Cu(30.91%)
are 62.93 amu & 64.9278 amu, respectively.
Calculate the average atomic mass of
copper.
11
Solution:
1. Percentages → Fractions:
69.09% = 69.09/100 = 0.6909; 30.91% =
30.91/100 = 0.3091
2.The AAM will be closer to the mass of the
isotope with the largest natural abundance
THUS: Average atomic mass of copper
= (0.6909 x 62.93)63Cu
+ (0.3091 x 64.9278)65Cu = 63.55 amu
12
H.W. Solve the Practice Exercise p81
Iodine has two isotopes 126I and 127I, with the equal
abundance.
Calculate the average atomic mass of Iodine (53I).
(a) 126.5 amu
(b) 35.45 amu
(c) 1.265 amu
(d) 71.92 amu
13
NOTE: Atomic Mass of an isotope,
e.g. 12C & 13C
Atomic Mass of an element
e.g. C
= Average Atomic Mass
Also:
Average Atomic Mass C = Atomic
Mass C
14
3.2 Avogadro’s Number and the
Molar Mass
Mole (mol) is the SI unit of the amount
of substance
 Avogadro’s Number (NA): one mole of
each substance contains 6.022 x 1023
particles (atoms or molecules or ions etc.),
 Thus 6.022 x 1023 is the Avogadro's
Number (NA)
NA = 6.022 x 1023 particles (atoms or
molecules or ions etc.)

15
Dozen = 12
1 dozen = 12 Anything
1 mol = 6.022 x1023 particles
16
THUS:
one mole of H atoms has
6.022 x 1023 atoms
&
One mole of H2 molecules has
6.022 x 1023 molecules
17
One mole of
Carbon
One mole of
Copper
One mole of
Sulphur
One mole of
Mercury
One mole of
Iron
18
Molar Mass




Molar mass (M): the mass (in g or kg) of one mole
of a substance;
M = mass/mol = g/mol
For ONE MOLE: 1 amu = 1 g
The atomic mass of 12C is 12.00 amu = 12.00 g
1 mole of 12C = 12.00 amu = 12.00 g = has NA of
atoms = has 6.022 x 1023 atoms
Thus:
The Molar Mass (M) of 12C = 12.00 g/mol
19
Atomic Mass of an isotope,
e.g. 12C & 13C
Atomic Mass of an element
e.g. C
is Average Atomic Mass
Also:
Average Atomic Mass C = Atomic
Mass C
20
Molar Mass (g/mol)
=
Atomic Mass (amu)
Examples:
1. The atomic mass of Na = 22.99 amu
The molar mass of Na = 22.99 g/mol
2. The atomic mass of P = 30.97 amu
The molar mass of P = 30.97 g/mol
21
Example 3.2 p83:
How many moles of He atoms are in 6.46 g
of He?
22
Solution:
Number of moles = n
The atomic mass of He = The Molar mass of He
From the periodic Table: the atomic mass of He =
The molar mass of He = 4.003 g/mol
Thus: 4.003 g → 1 mole of He
6.46 g of He → ? Moles (?n)
1 mol x 6.46 g
n
 1.61 mol
4.003 g
Thus: there is 1.61 moles of He atoms in 6.46 g of
He
H.W. Solve the Practice Exercise p84
23
n = number of moles
m = mass (atom or molecule)
M = molar mass (atomic mass or molecular
mass)
What is the relation between them?
m
g
n

 mol
M g / mol
n = number of moles
N = number of atoms or molecules
NA = Avogadro's number (atoms (or
molecules)/mol)
What is the relation between them?
n
N
atoms (or molecules)

 mol
N A atoms (or molecules) / mol
24
How many moles of He atoms are in 6.46 g
of He?
m
6.46 g
n( He ) 

 1.61mol
M 4.003g / mol
25
Example 3.3 p84:
How many grams of Zn are in 0.356 mole
of Zn?
26
Solution:
Number of moles = n
The atomic mass of Zn = The Molar mass of Zn
From the periodic Table: The Atomic mass of Zn
= molar mass of Zn = 65.39 g/mol
Thus: 65.39 g → 1 mole of Zn
?g of He → 0.356 mole
65.39 g x 0.356 mol
mass 
 23.3 g of Zn
1 mol
Thus: there is 23.3 g of Zn atoms in 0.356 moles of
Zn
H.W. Solve the Practice Exercise p84
27
‫حل مختصر‬
How many grams of Zn are in 0.356 mole
of Zn?
m
n( Zn) 
 m  nM
M
m  0.356 mol x65.39 g/mol  23.3 g
28
Example 3.4 p84:
How many S atoms are in 16.3 g of S?
Strategy:
1. How many moles in 16.3 g of S = X mol
2. 1 mole → 6.022 x1023 S atoms
X moles → ? atoms
29
Solution:
From the periodic Table: The atomic mass of S =
32.07 amu
The molar mass of S = 32.07 g/mol
Thus: 32.07 g → 1 mole of S
16.3 g of S → ? mole
1 mol x 16.3 g
n
 0.508 mol
32.07 g
We know: 1 mol of S → 6.022 x1023 S atoms
0.508 mole → ? S atoms
6.022 x1023 atoms x 0.508 mol
number of S atoms 
1 mol
 3.06 x1023 S atoms
There is 3.06 x1023 atoms of S in 16.3 g of S
H.W. Solve the Practice Exercise p85
30
‫حل مختصر‬
How many S atoms are in 16.3 g of S?
m
16.3 g
n( S ) 

 0.508 mol
M 32.07 g/mol
N
n( S ) 
 N  nxN A
NA
 0.508 mol x6.022 x10 atoms/mol
23
 3.06 x10 atoms
23
31
Molecular Mass
Molecular Mass (molecular weight): is
the sum of the atomic masses (in amu)
in the molecule. (MOLECULE)
 Molecular Mass: multiply the atomic
mass of each element by the number of
atoms of that element present in the
molecule and sum over all the elements.
 e.g. Molecular Mass of H2O is:

(2 x atomic mass of H) + (1x atomic mass of O)
(2 x 1.008 amu) + (1x 16.00 amu) = 18.02 amu
32
Example 3.5 p86:
Calculate the molecular masses (in amu) of
the following compounds:
(a) Sulphur dioxide (SO2)
(b) Caffeine (C8H10N4O2)
Solution:
(a) Molecular mass of SO2 =
(1 x 32.07) + (2 x 16.00) = 64.07 amu
(b) Molecular mass of C8H10N4O2 =
(8 x 12.01) + (10 x 1.008) +
(4 x 14.01) + (2 x 16.00) =194.20 amu
33
One mole of H atoms has
6.022 x 1023 H atoms
&
One mole of H2 molecules has
6.022 x 1023 H2 molecules
One mole of H2 molecules has
2 x 6.022 x 1023
H atoms
34
One mole of H2O has
6.022 x 1023 H2O molecules
&
One mole of CH4 molecules has
6.022 x 1023 CH4 molecules
35
Atomic mass (in amu) = Molar Mass (g/mol)
Molecular Mass (in amu) = Molar Mass (g/mol)
Example 3.6 p86:
How many moles of CH4 are present in 6.07 g of
CH4?
Solution:
Molecular Mass CH4 = (1 x 12 .01) + (4 x 1.008) = 16.04 amu
Molecular Mass (in amu) = Molar Mass (g/mol)
16.04 amu = 16.04 g/mol
Thus: 1 mole of CH4 → 16.04 g CH4
n? moles → 6.07 g of CH4
1 mole x 6.07 g
number of moles of CH4 
 0.378 mol
16.04 g
36
‫حل مختصر‬
How many moles of CH4 are present in 6.07 g
of CH4?
m
6.07 g
n(CH 4 ) 

 0.378 mol
M 16.04 g / mol
37
How many molecules of ethane (C2H6) are present in
0.334 g of C2H6?
(a) 2.01 x 1023
(b) 6.69 x 1021
(c) 4.96 x 1022
(d) 8.89 x 1020
38
Solution:
Molecular mass of C2H6 = (2 x 12.01) + (6 x 1.008) = 30.068 amu
Molecular mass (amu) = Molar mass (g/mol) = 30.068 g/mol
C2H6 30.068 g → 1 mole
0.334g → ?n
1 mole x 0.334 g
number of moles of C2 H 6 
 0.011 mol
30.068 g
1 mole of C2H6 → 6.022 x 1023 molecules of C2H6
0.011 mole of C2H6 → ? molecules of C2H6
0.011 mol x 6.022x10 23 molecules
number of molecules of C2 H 6 
1 mole
 6.624 x1021 molecules
39
‫حل مختصر‬
How many molecules of ethane (C2H6) are present in
0.334 g of C2H6?
(a) 2.01 x 1023
(b) 6.69 x 1021
(c) 4.96 x 1022
(d) 8.89 x 1020
m
0.334 g
n(C2 H 6 ) 

 0.011 mol
M 30.07 g / mol
N
n(C2 H 6 ) 
 N  nxN A
NA
 0.011 mol x6.022 x10 23 atoms/mol
 6.624 x10 atoms
21
40
Example 3.7 p87:
How many hydrogen atoms are present in 25.6 g of urea
[(NH2)2CO]. The molar mass of urea is 60.06 g/mol.
m
25.6 g
n[( NH 2 ) 2 CO] 

 0.426 mol
M 60.06 g / mol
N
n[( NH 2 ) 2 CO] 
NA
 N  nxN A  0.426 mol x6.022 x1023 molecules/ mol
N  2.567 x1023 molecules
1 molecule [(NH 2 ) 2 CO]  4 H atoms
2.567x10 23 [(NH 2 ) 2 CO] molecules  ?H atoms
4 atomx2.567 x10 23 molecule
number of H atoms 
 1.03 x1024 atoms
1 molecule
41
H.W. What is the mass, in grams, of one copper
atom?
(a) 1.055  10-22 g
(b) 63.55 g
(c) 1 amu
(d) 1.66  10-24 g
(e) 9.476  1021 g
42
Solution:
Atomic mass of Cu = 63.55 amu
Molar mass of Cu = 63.55 g/mol
63.55 g of Cu → 1 mol of Cu
1 mol of Cu → 6.022 x 1023 Cu atoms
63.55g of Cu → 6.022x1023 Cu atoms
?g of Cu → 1 Cu atom
1 atom x 63.55g
 22
grams of Cu 
 1.055 x10 g
23
6.022x10 atom
43
‫حل مختصر‬
H.W. What is the mass, in grams, of one copper
atom?
(a) 1.055  10-22 g
(b) 63.55 g
(c) 1 amu
(d) 1.66  10-24 g
(e) 9.476  1021 g
N
1 atom
 24
n(Cu ) 


1
.
661
x
10
mol
23
N A 6.022x10 atoms/mol
m
n(Cu ) 
 m  nxM
M
 1.661x10  24 mol x 63.55 g/mol
 1.055 x10  22 g
44
3.5 Percent Composition of
Compounds: outline

Percent Composition by Mass
→ The Empirical Formula
 → The Molecular Formula ‫الصيغة الجزيئية‬

45
3.5 Percent Composition of
Compounds
Percent Composition by Mass: is the
percent by mass of each element in a
compound
 percent composition of an element in a
compound =

n x molar mass of element
x 100%
molar mass of compound
n is the number of moles of the element in 1 mole
of the compound
46
PERIODIC TABLE
Molar mass of C2H6O = Molecular mass C2H6O =
(2 x 12.01) + (6 x 1.008) + (1 x16.00) = 46.07 g/mol
Molar mass of C = Atomic mass of C = 12.01 g /mol
Molar mass of H = Atomic mass of H = 1.008 g/mol
C2H6O
Molar mass of O = Atomic mass of O = 16.00 g/mol
2 x (12.01 g)
%C =
x 100% = 52.14%
46.07 g
6 x (1.008 g)
%H =
x 100% = 13.13%
46.07 g
1 x (16.00 g)
%O =
x 100% = 34.73%
46.07 g
Check the
answer!
52.14% + 13.13% + 34.73% = 100.0%
47
Example 3.8 p89:
Calculate the percent composition by mass
of H, P, and O in H3PO4.
Solution:
Molar mass of H3PO4 = 97.99
3x1.008
%H 
x100  3.086%
97.99
1x30.97
%P
x100  31.61%
97.99
4 x16.00
Check the answer:
%O 
x100  65.31%
3.08% + 31.61% + 65.31% = 100.01%
97.99
48
H.W. Calculate the percent of nitrogen in Ca(NO3)2:
a) 12.01%.
b) 17.10%.
c) 18%
d) 16%.
H.W. All of the substances listed below are
fertilizers that contribute nitrogen to the soil.
Which of these is the richest source of nitrogen
on a mass percentage basis?
(a) Urea, (NH2)2CO
(b) Ammonium nitrate, NH4NO3
(a) %N = 46.6%
(c) Guanidine, HNC(NH2)2
(b) %N = 58%
(d) Ammonia, NH3
‫( ايا من هذه المواد هواغنى مصدر للنيتروجين على اساس‬c) %N = 71.1%
‫( احتوائه على اكبر نسبه وزنيه من النيتروجين؟‬d) %N = 82.2% 49
Determination of the Empirical Formula
from the Percent Composition by Mass
Example: 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.
Solution:
1. Convert to grams & divide by the molar mass →
number of moles
Thus 24.75g of K, 34.77g of Mn, & 40.51g of O
24.75
 0.632 mol of K
39.10
34.77
nMn 
 0.632mol of Mn
54.94
40.51
nO 
 2.532 mol of O
16.00
1.
2.
nK 
3.
50
2. Divide by the smallest number of moles:
0.632 mol → mole ratio of the elements
0.632
0.632
2.532
K:
1
Mn 
1
O:
4
0.632
0.632
0.632
3. All are integers → Finish
Thus the Empirical Formula of the compound is:
→
K1Mn1O4
KMnO4
51
‫‪Percent Composition and Empirical Formulas‬‬
‫‪Q: 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.‬‬
‫‪O‬‬
‫‪Mn‬‬
‫‪K‬‬
‫‪40.51g‬‬
‫‪34.77g‬‬
‫‪24.75g‬‬
‫‪% 100g‬‬
‫‪40.51/16.00‬‬
‫‪= 2.532mol‬‬
‫‪34.77/54.94‬‬
‫‪=0.6329mol‬‬
‫‪24.75/39.10‬‬
‫‪=0.633mol‬‬
‫‪n=m/M‬‬
‫‪2.532/0.632‬‬
‫‪=4‬‬
‫‪0.6329/0.632‬‬
‫‪=1‬‬
‫‪0.633/0.632‬‬
‫‪=1‬‬
‫‪ on smallest no. of mole‬‬
‫‪O4‬‬
‫‪Mn1‬‬
‫‪K1‬‬
‫‪The empirical formula is‬‬
‫‪KMnO4‬‬
‫‪52‬‬
‫خطوات الحل‬
‫‪ .1‬ننشأ جدول نضع فيه العناصر المذكورة في السؤال‬
‫‪ .2‬نعتبر أن النسبة المئوية معبر عنها بالجرام فلو كان عندنا ‪ 100‬جرام من المركب فهذه ال ‪ 100‬جرام موزعة على‬
‫العناصر حسب نسبتها‪.‬‬
‫‪ .3‬نوجد عدد الموالت ‪n‬لكل عنصر باستخدام القانون ‪. n=m/M‬‬
‫‪ .4‬نقسم عدد الموالت على أصغر مول من العناصر‪.‬‬
‫‪ .5‬األرقام التي نحصل عليها تمثل ‪empirical formula‬بشرط أن تكون أعداد صحيحة كما في المثال السابق‪.‬‬
‫‪ .6‬في حالة ظهور أعداد عشرية نقوم بضرب األرقام التي في األسفل الموجودة في الصيغة بأعداد بدأ من ‪.......3 ،2‬‬
‫حتى نحصل على أعداد صحيحة ‪.‬‬
‫‪Courtesy of Dr. Fawzia Albelwi‬‬
Determination of the Empirical Formula
from the Percent Composition by Mass
Example 3.9 p90:
Ascorbic acid composed of 40.92% C, 4.58% H, and 54.50%
O by mass. Determine its empirical formula.
Solution:
1. Thus 40.92g of C, 4.58g of H, & 54.50g of O
40.92
 3.407 mol of C
12.01
4.58
nH 
 4.54 mol of H
1.008
54.50
nO 
 3.406 mol of O
16.00
nC 
1.
2.
3.
53
2.
3.407
C:
1
3.406
4.54
H
 1.33
3.406
3.406
O:
1
3.406
3. Not all numbers are integers
THUS: Convert 1.33 → integer
Trail-and error procedure:
1.33 x 2 = 2.66
1.33 x 3 = 3.99 ≈ 4
OK. multiply all the numbers by 3
C=3
H =4
O=3
Thus the Empirical Formula of ascorbic acid is:
C3H4O3
54
‫‪Example 3.9 p90:‬‬
‫‪Ascorbic acid composed of 40.92% C, 4.58% H, and 54.50%‬‬
‫‪O by mass. Determine its empirical formula.‬‬
‫‪O‬‬
‫‪H‬‬
‫‪C‬‬
‫‪54.50g‬‬
‫‪4.58g‬‬
‫‪40.92g‬‬
‫‪% 100g‬‬
‫‪54.50/16.00‬‬
‫‪= 3.406 mol‬‬
‫‪4.58/1.008‬‬
‫‪=4.54.mol‬‬
‫‪40.92/12.01‬‬
‫‪= 3.407mol‬‬
‫‪n=m/M‬‬
‫‪3.406/3.406‬‬
‫‪=1‬‬
‫‪0.4.54/3.406‬‬
‫‪= 1.33‬‬
‫‪3.407/3.406‬‬
‫‪=1‬‬
‫‪ on smallest no. of‬‬
‫‪mole‬‬
‫‪3‬‬
‫‪3.99 = 4‬‬
‫‪3‬‬
‫‪Convert into integer x‬‬
‫‪3‬‬
‫‪O3‬‬
‫‪H4‬‬
‫‪C3‬‬
‫‪The empirical formula‬‬
‫‪is‬‬
‫‪C3H4O3‬‬
‫خطوات الحل‬
‫‪ .1‬ننشأ جدول نضع فيه العناصر المذكورة في السؤال‬
‫‪ .2‬نعتبر أن النسبة المئوية معبر عنها بالجرام فلو كان عندنا ‪ 100‬جرام من المركب فهذه ال ‪ 100‬جرام موزعة‬
‫على العناصر حسب نسبتها‪.‬‬
‫‪ .3‬نوجد عدد الموالت ‪n‬لكل عنصر باستخدام القانون ‪. n=m/M‬‬
‫‪ .4‬نقسم عدد الموالت على أصغر مول من العناصر‪.‬‬
‫‪ .5‬األرقام التي نحصل عليها تمثل ‪empirical formula‬بشرط أن تكون أعداد صحيحة‬
‫‪ 55‬في حالة ظهور أعداد عشرية كما في المثال السابق نقوم بضرب األرقام التي في األسفل الموجودة في الصيغة‬
‫‪.6‬‬
‫بأعداد بدأ من ‪ .......3 ،2‬حتى نحصل على أعداد صحيحة ‪.‬‬
Determination of the Molecular Formula
from the Percent Composition by Mass
Example 3.11 p93:
A sample compound contains 1.52g of N and 3.47g of
O. The molar mass of this compound is between 90g
and 95g. Determine the molecular formula.
Present Composition
Solution:
by Mass
1. n  1.52  0.108 mol of N
N
14.01
3.47
nO 
 0.217 mol of O
16.00
2.
0.108
N:
1
0.108
↓
Empirical Formula
0.217
O
2
0.108
3. Thus the empirical formula is: NO2
↓
Molecular Formula
56
4. The molar mass of the empirical
formula NO2 = 14.01 + (2x16.00) =
46.01g
5. The ratio between the empirical
formula and the molecular formula:
90 1.956
molar mass of compound
Ratio 
2
Ratio 
empirical molar mass
46.01
6. The molecular formula is (NO2)2 = N2O4
57
H.W. Which of the following is an empirical formula:
a) C12H22O10.
‫ايا منهم اليمكن تبسيطه اكثر مما هو عليه؟‬
b) H2SO4.
c) Hg2Cl2
d) S8.
H.W. A sample of acid compound contains 40.1 percent of
C, 6.6 percent of H, and 53.3 percent of O. The molar
mass of this compound is 60 g/mol . What is the molecular
formula ?
C5H6O
(b) C2HO2
(c) C2H4O2
(d) CH2O4
(a)
58
3.7 Chemical Reactions and
Chemical Equations
 Chemical Reaction: is a
process in which one or more
substances is changed into one
or more new substances
 Chemical Equation: uses chemical symbols
to show what happens during a chemical
reaction
reactants
products
59
3
ways of representing the reaction of H2
with O2 to form H2O
60
How to “Read” Chemical Equations?
2 Mg + O2
2 MgO Read it!
√
2 moles Mg + 1 mole O2 makes 2 moles MgO √
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO √
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
Molar masses
2 grams Mg + 1 gram O2 makes 2 g MgO
X
61
Balancing Chemical Equations
1. Write the correct formula/s for the
reactant/s on the left side and the correct
formula/s for the product/s on the right side
of the equation.
‫نكتب الصيغه الصحيحه لكل متفاعل (على الطرف االيسر) ولكل‬
)‫ناتج (على الطرف االيمن‬
62
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.
2C2H6 NOT C4H12
‫وزن المعادله الكيميائيه يكون بتغير االرقام التي بجانب‬
‫الصيغه وليست التي تحتها بحيث يكون للعنصر نفس العدد‬
.‫على طرفي المعادله‬
63
3. Start by balancing those elements that appear
in only one reactant and one product.
‫توزن اوال العناصر االقل ظهورا‬
4. Balance those elements that appear in two or
more reactants or products.
‫ثم توزن العناصر االكثر ظهورا‬
5. Check to make sure that you have the same
number of each type of atom on both sides of
the equation.
‫الخطوه االخيره هي التأكد من ان لديك نفس العدد من‬
‫الذرات لكل عنصر على طرفي المعادله‬
64
Example
Nitrogen monoxide reacts with oxygen to form
nitrogen dioxide balance this reaction?
1. Write the correct formula/s for the
reactant/s on the left side and the correct
formula/s for the product/s on the right side
of the equation.
NO  O2  NO2
65
2. Start by balancing those elements that appear
in only one reactant and one product.
NO  O2  NO2
1 nitrogen
on left
1 nitrogen
on right
start with N not O
N is balanced
66
3. Balance those elements that appear in two or
more reactants or products.
NO  O2  NO2
3 oxygen
on left
2 oxygen
on right
multiply NO by 2 and
NO2 by 2
2NO  O2  2NO2
2 oxygen + 2 oxygen = 4
2 x 2 = 4 oxygen
on left
on right
O is balanced
67
4. Check to make sure that you have the same
number of each type of atom on both sides of
the equation.
2NO + O2
2NO2
2N
2N
2O+2O=4
4 O (2 x 2)
Reactants
2N
4O
Products
2N
4O
68
Example
Ethane reacts with oxygen to form carbon dioxide
and water balance this reaction?
1. Write the correct formula/s for the
reactant/s on the left side and the correct
formula/s for the product/s on the right side
of the equation.
C2H6 + O2
CO2 + H2O
69
2. Start by balancing those elements that appear
in only one reactant and one product.
C2H6 + O2
2 carbon
on left
C2H6 + O2
6 hydrogen
on left
C2H6 + O2
CO2 + H2O start with C or H but not O
1 carbon
on right
multiply CO2 by 2
2CO2 + H2O
2 hydrogen
on right
multiply H2O by 3
2CO2 + 3H2O
70
3. Balance those elements that appear in two or
more reactants or products.
C2H6 + O2
2 oxygen
on left
2CO2 + 3H2O
7
multiply O2 by 2
4 oxygen + 3 oxygen = 7 oxygen
(3x1)
on right
(2x2)
C2H6 + 7 O2
2
2CO2 + 3H2O
2C2H6 + 7O2
4CO2 + 6H2O
remove fraction
multiply both sides by 2
71
4. 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)
4C
12 H (2 x 6)
12 H (6 x 2)
14 O (7 x 2)
14 O (4 x 2 + 6)
Reactants
4C
12 H
14 O
Products
4C
12 H
14 O
72
Example 3.12 p98 : Balance the following reaction
Nitrogen monoxide reacts with oxygen to form nitrogen
dioxide and balance this reaction?
73
3
H.W. What is the coefficient of H2O when the equation
is balanced:
_ Al4C3 + _ H2O  _ Al(OH)3 + 3CH4
a. 13
b. 4
c. 6
d. 12
H.W. What are the coefficients of Al4C3 ,H2O and
Al(OH)3, respectively, when the equation is balanced:
_ Al4C3 + _ H2O  _ Al(OH)3 + 3CH4
a. 4,1,5
b. 1,12,4
c. 1,24, 4
d. 4,12,1
74
n = number of moles
m = mass (atom or molecule)
M = molar mass (atomic mass or molecular mass)
What is the relation between them?
m
g
n

 mol
M g / mol
75
3.8 Amounts of Reactants and
Products

Two important questions:

How much product will be formed from
specific amount of reactants?


e.g. 6.0 g reactant→ ? product
How much starting reactants must be used
to obtain a specific amount of product?

e.g. ? reactant → 6.0 g product
76
Mole
Method
m
n
M
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 units77
Example 3.13 p101:
C6 H12O6  6O2  6CO2  6H 2O
If 856g of C6H12O6 is consumed by a
person over a certain period, what is the
mass of CO2 produced?
78
Solution:
1.
Write balanced chemical equation
C6 H12O6  6O2  6CO2  6H 2O
Balanced!
2.
Convert quantities of known substances into
moles → convert grams of C6H12O6 to moles
of C6H12O6
m(C6 H12O6 )
n(C6 H12O6 ) 
M (C6 H12O6 )
856g

 4.750 mol
180.2g/mol
79
3. Use coefficients in balanced equation
to calculate the number of moles of
the sought quantity → mole ratio
(from the balanced equation):
1 mole C6H12O6 → 6 mole of CO2
4.754 mole → ? mole CO2
6 mol x 4.754 mol
n(CO2 ) 
 28.50 mol
1 mol
80
4. Convert moles of sought quantity into
desired units →
convert the moles of CO2 → grams of CO2
m(CO2 )
n(CO2 ) 
 m  nxM
M (CO2 )
m(CO2 )  28.50 mol x 44.01 g/mol  1.25 x10 g
3
Summary: grams of C6H12O6 → moles of
C6H12O6 → moles of CO2 → grams of CO2
81
Example 3.14 p101:
2Li(s)  2H2O(l )  2LiOH (aq)  H2 ( g )
How many grams of Li are needed to
produce 9.89g of H2?
Strategy:
grams of H2 → moles of H2 → moles of Li→ grams of Li
82
Solution:
1.
Write balanced chemical equation
2Li(s)  2H2O(l )  2LiOH (aq)  H2 ( g )
Balanced!
2.
convert grams of H2 to moles of H2
m( H 2 )
n( H 2 ) 
M (H 2 )
9.89g

 4.920 mol
2.016g/mol
83
3. Mole ratio (from the balanced
equation) → Moles of Li
2 mole of Li → 1 mole H2
?mole of Li → 4.927 mole H2
2 mol x 4.927 mol
n( Li ) 
 9.854 mol
1 mol
4. Convert the moles of Li → grams of Li
m
n
 m  nxM
M
m( Li )  9.854 mol x6.941 g/mol  68.1g
84
3.9 Limiting Reagent ‫الكاشف المحدد‬
Limiting Reagent: is the reactant used
up first in a reaction and thus
determine the amount of product
 Excess Reagent ‫الكاشف الفائض‬: is the
reactant present in quantities greater
than necessary to react with the
quantity of the limiting reagent (the one
that is left at the end of the reaction).
 → Limiting reagent is in a reaction of
more than one reactant!

85
Limiting Reagent:
Reactant used up first in
the reaction.
2NO + O2
2NO2
NO is the limiting reagent
O2 is the excess reagent
86
Questions in Limiting Reagent:
 First: we have to determine which
reactant is the limiting reagent and
which is the excess reagent!
 Second: after we know which one is the
limiting reagent, we could determine
the amount of the product!!
 Third: after we know the excess
reagent, we could determine how much
excess of it is left at the end of the
reaction!!!
87
Example 3.15 P102:
2 NH3 ( g )  CO2 ( g )  ( NH 2 )2 CO(aq)  H 2O(l )
637.2g of NH3 are treated with 1142g of
CO2.
(a) Which of the two reactants is the limiting
reagent?
(b) Calculate the mass of (NH2)2CO formed.
(c) How much excess reagent (in grams) is left
at the end of the reaction?
88
Solution:
(a) Which of the two reactants is the
limiting reagent? ‫المتفاعل اللذي يعطي موالت‬
‫ النه يحد من كميه الناتج‬,‫اقل من الناتج هو الكاشف المحدد‬
.‫التي يمكن ان تتكون‬
For the first reactant NH3: ‫تحول الى موالت‬
1. 637.2 g NH3 → convert to moles
m( NH 3 )
n( NH 3 ) 
M ( NH 3 )
637.2 g

 37.416 mol
17.03 g/mol
89
Solution:
2. Mole ratio:
2 moles NH3 → 1 mole of (NH2)2CO
37.416moles of NH3 → ?n (NH2)2CO
37.416 x1
n( NH 2 ) 2 CO 
 18.71 mol
2
90
For the second reactant CO2: ‫تحول الى موالت‬
1. 1142 g CO2→ convert to moles
n(CO2 ) 

m(CO2 )
M (CO2 )
1142 g
 25.949 mol
44.01 g/mol
2. Mole ratio:
1 mole CO2 → 1 mole of (NH2)2CO
25.949 mol CO2 → ?n (NH2)2CO
25.949 x1
n( NH 2 ) 2 CO 
 25.95 mol
1
Thus: the limiting reagent is NH3 because it
produces a smaller amount of (NH2)2CO
91
(b) Calculate the mass of (NH2)2CO formed.
We have 18.71 mol of (NH2)2CO using NH3 as
the limiting reagent → convert to grams
m
n
 m  nxM
M
m[( NH 2 ) 2 CO]  18.71mol x60.06 g/mol  1124 g
92
(c) How much excess reagent (in grams) is
left at the end of the reaction?
Excess reagent is CO2:
grams of CO2 left = initial grams – reacted grams
1. moles of CO2 left = initial moles – reacted moles
2. moles of CO2 left → grams of CO2 left
1.
Initial moles of CO2 = 25.95 mol
Reacted moles of CO2 (calculated as follows):
93
Reacted moles of CO2 (calculated as follows):
We know that 18.71 mol (NH2)2CO is produced
Thus: 1 mol CO2 →1 mole (NH2)2CO
?n CO2 → 18.71 mol (NH2)2CO
18.71x1
n(CO2 ) 
 18.71 mol
1
THUS: moles of CO2 reacted is 18.71 mol
94
moles of CO2 left = initial moles – reacted moles
= 25.95 – 18.71 = 7.24 mol
2. moles of CO2 left → grams of CO2 left
m
n
 m  nxM
M
m(CO2 )  7.24mol x 44.01 g/mol  319 g
Thus: the mass of CO2 remaining (left) = 319g
95
H.W. # 1: When 22.0 g NaCl and 21.0 g H2SO4 are mixed and
react according to the equation below, which is the limiting
reagent?
2NaCl + H2SO4  Na2SO4 + 2HCl
(a)
NaCl
(b) H2SO4
(c) Na2SO4
(d) HCl
(e)
No reagent is limiting.
H.W. # 2: 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.
The limiting Reagent is CO
(a) 7.20 mol
(b) 44.0 mol
(c) 3.60 mol
(d) 1.80 mol
96
3.10 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.
% Yield =
Actual Yield
x 100
Theoretical Yield
Actual Yield is always less.
97
3.10 Reaction Yield
Theoretical Yield is the amount of
product that would result if all the
limiting reagent reacted.
Limiting regent mass →
moles of limiting reagent→
moles of product →
grams of product (theoretical yield of
the product)
98
Example 3.16 p106:
TiCl4 ( g )  2Mg (l )  Ti(s)  2MgCl2 (l )
3.54 x107g of TiCl4 are reacted with 1.13
x 107g of Mg.
(a) Calculate the theoretical yield of Ti in
grams
(b) Calculate the percent yield if 7.91 x
106g of Ti are actually obtained.
99
Solution:
(a) Calculate the theoretical yield of Ti in
grams.
‫ احسبي وزنه التيتانيوم الناتجه؟‬:‫أي بمعنى اخر‬
Strategy:
1. Given: Masses of two reactants → limiting reagent
problem
2. Masses of reactants → moles of reactant
3. Moles of reactants → used to calculate the moles
of product
4. Moles of product is the less number of moles
(limiting reagent)
5. Moles of product → grams of product (theoretical
yield of the product)
100
grams of TiCl4 → moles of TiCl4 → moles of Ti
7
m
3.54 x10 g
5
n(TiCl4 ) 

 1.87 x10 mol
M 189.7 g / mol
Mole ratio:
1 mole TiCl4 → 1 mole Ti
1.97 x105 mole → ?n Ti
n(Ti) =1.87 x105 mol
101
grams of Mg → moles of Mg → moles of Ti
7
m
1.13x10 g
6
n( Mg ) 

 4.64 x10 mol
M 24.31g / mol
Mole ratio:
2 mole Mg → 1 mole Ti
4.64 x106 mole → ?n Ti
6
4.64 x10
5
n(Ti) 
 2.32 x10 mol
2
102
n(Ti) is the less number of moles (limiting
reagent) = 1.87 x105 mol
Moles of Ti → grams of Ti
Thus: mass of Ti = theoretical yield of
Ti = m =nM
m(Ti)  nM
 1.87 x10 molx 47.88 g / mol  8.95 x10 g
5
6
103
(b) Calculate the percent yield if 7.91 x
106g of Ti are actually obtained.
actual yield
% yield 
x100
theoretica l yield
6
7.91x10 g

x100  88.4%
6
8.95 x10 g
H.W. Solve the practice exercise p107
104
End of Chapter 3
105