Chapter 3

 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
 3.7 Chemical Reactions and Chemical Equations
 3.9 limiting reagents
 3.10 reaction yield
Dr. Laila Al-Harbi

 Each atom have more than one isotope with different
abundance
 Average atomic Mass: the average mass of all of the
isotopes of an element, each one weighted by its
proportionate abundance
 Science each atom have more than one isotope with
different abundance
% abundance of isotope 1 % abundance of isotope 2
verage Atomic Mass = (mass of isotope 1) + (mass of isotope 2) + ...
100 10
A
0
   
   
   
Dr. Laila Al-Harbi

Average atomic mass of
Lithium
Average atomic mass of
carbon
 Natural lithium is:
 7.42% 6Li (6.015 amu)
 92.58% 7Li (7.016 amu
 Natural Carbon is:
 1.1% 13C (6.015 amu)
 98.9% 12C (7.016 amu)
Dr. Laila Al-Harbi
(7.42% x 6.015) + (92.58% x 7.016)
100
= 6.941 amu
(98.9 % x 12) + (1.18% x 13)
100
= 12.01 amu
The average atomic mass is between the atomic masses of the isotopes
And near the value of the highest abundance

EXAMPLE 3.1
PRACTIES EXERICISE 3.1
 65Cu (30.91percent)
Atomic mass 64.9278
 63Cu (69.091percent)
Atomic mass 62.93
 10B (19.78 percent)
Atomic mass 10.0129
 11B (80.78percent)
Atomic mass 11.0093
Dr. Laila Al-Harbi
(30.91% x 64.9278) + 69.091% x 62.93)
100
= 63.55 amu
(19.78 % x 10.0129) + 80.78% 11.0093)
100
=10.81amu

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
equal abundance MEAN each atom
has abundance 50% .

 Atomic mass is the mass of an atom in atomic mass units (amu)
 On this scale1H = 1.008 amu 16O = 16.00 amu
 Avogadro's Number , Is the number of atoms in exactly 12
grams of carbon-12 (NA = 6.022 x 1023)
 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
 One mole of a substance contains an Avogadro's Number of units
Dr. Laila Al-Harbi
By definition:
1 atom 12C “weighs” 12 amu

THUS:
one mole of H atoms has
6.022 x 1023 atoms
&
One mole of H2 molecules has
6.022 x 1023 molecules

C S
Cu Fe
Hg
C
One mole of these substances contain = 6.022 x 1023 atoms
but is not equal because they have different molar masses
Dr. Laila Al-Harbi

10
 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

11
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

12
 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

 What is the molar mass of the following compound ?
 NH3 , CH3COOH , Na2SO4 , C6H12O6
 NH3 = (1×14)+(3×1) = 17 g/mol
 C2H4O2 = (2×12)+(4×1) )+(2×16) = 60 g/mol
 Na2SO4 = (2×23)+(1×32) )+(4×16) = 142 g/mol
 C6H12O6 = (6×12)+(12×1) )+(6×16) = 180 g/mol
Dr. Laila Al-Harbi

 Calculate the molecular masses ( in amu) of the following
compounds ?
 Sulfur dioxide SO2 = 32.07+2 (16) = 64.07 amu
 Caffeine C8H10 N4O2
= 8(12.01)+ 10 (1.008) + 4(14.01)+ 2(16) = 194.20amu
Practice exercise3.5
Calculate the molecular masses of methanol?
 methanol C H4O
= 1(12.01)+ 4 (1.008) + 1(16) = 32.4 amu
Dr. Laila Al-Harbi

15
n = number of moles
m = mass (atom or molecule)
M = molar mass (atomic mass or molecular
mass)
What is the relation between them?
mol
molg
g
M
m
n 
/
n = number of moles
N = number of atoms or molecules
NA = Avogadro's number (atoms (or
molecules)/mol)
What is the relation between them?
mol
mol/molecules)or(atoms
molecules)(oratoms

AN
N
n

EXAMPLE 3.2
How many moles of He atoms are in 6.46 g
of He ?
Dr. Laila Al-Harbi
mol
molg
g
M
m
Hen 61.1
/003.4
46.6
)( 
How many grams of Zn are in 0.356 mole of Zn?
g3.23g/mol39.65mol356.0
)(


xm
nMm
M
m
Znn

 Methane is the principle component of
natural gas . How many CH4 are in 6.07 g of
CH4?
Dr. Laila Al-Harbi
mol
molg
g
M
m
Hen 378.0
/04.16
06.6
)( 

18
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

19
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
We know: 1 mol of S → 6.022 x1023 S atoms
0.508 mole → ? S atoms
There is 3.06 x1023 atoms of S in 16.3 g of S
H.W. Solve the Practice Exercise p85
mol508.0
g07.32
g3.16xmol1
n
atomsS1006.3
mol1
mol508.0xatoms106.022
atomsSofnumber
23
23
x
x



20
How many S atoms are in 16.3 g of S?
mol508.0
g/mol07.32
g3.16
)( 
M
m
Sn
atoms1006.3
atoms/mol10022.6mol508.0
)(
23
23
x
xx
nxNN
N
N
Sn A
A



‫مختصر‬ ‫حل‬

Dr. Laila Al-Harbi
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
21
mol011.0
g30.068
g0.334xmole1
HCofmolesofnumber 62 
molecules10624.6
mole1
molecules6.022x10xmol011.0
HCofmoleculesofnumber
21
23
62
x

1 mole of C2H6 → 6.022 x 1023 molecules of C2H6
0.011 mole of C2H6 → ? molecules of C2H6

22
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.
atoms1003.1
molecule1
moleculex10atomx2.5674
atomsHofnumber
atoms?Hmolecules])[(NH2.567x10
atomsH4])[(NHmolecule1
molecules10567.2
molmolecules/10022.6mol426.0
])[(
mol0.426
/60.06
25.6
])[(
24
23
22
23
22
23
23
22
22
x
CO
CO
xN
xxnxNN
N
N
CONHn
molg
g
M
m
CONHn
A
A








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 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
gx 22
23
10055.1
atom6.022x10
63.55gxatom1
Cuofgrams 


Percent composition of an
element in 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
C2H6O
%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%
3.5 Percent composition of compounds
Dr. Laila Al-Harbi
52.14% + 13.13% + 34.73% = 100.0%Check the
answer!

Example 3.8 PRACTIES EXERICISE 3.8
 Calculate the percent
composition by mass of H , P,
and O in H3PO4 acid ?
 Molar mass of H3PO4
 = 3(1.008)+ 1 (30.97) + 4(16)
= 97.99 amu
 Calculate the percent
composition by mass of H ,
P, and O in H2SO4 acid ?
 Molar mass of H2SO4
 = 2(1.008)+ 1 (32.7) + 4(16)
= 98.72 amu
Dr. Laila Al-Harbi
%H =
3(1.008)
97.99
x 100% = 3.0864%
%P =
1(30.97)
97.99
x 100% = 31.61 %
%O=
4(16)
97.99
x 100% = 65.31%
%H =
2(1.008)
98.72
x 100% = 2.026 %
%S =
1(32.07)
98.72
x 100% = 32.486 %
%O=
4(16)
98.72
x 100% = 64.83%

26
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
(c) Guanidine, HNC(NH2)2
(d) Ammonia, NH3
‫اساس‬ ‫على‬ ‫للنيتروجين‬ ‫مصدر‬ ‫هواغنى‬ ‫المواد‬ ‫هذه‬ ‫من‬ ‫ايا‬
‫النيتروجين؟‬ ‫من‬ ‫وزنيه‬ ‫نسبه‬ ‫اكبر‬ ‫على‬ ‫احتوائه‬
(a) %N = 46.6%
(b) %N = 58%
(c) %N = 71.1%
(d) %N = 82.2%

27
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.
K Mn O
% 100g 24.75g 34.77g 40.51g
n=m/M 24.75/39.10
=0.633mol
34.77/54.94
=0.6329mol
40.51/16.00
= 2.532mol
 on smallest no. of mole 0.633/0.632
=1
0.6329/0.632
= 1
2.532/0.632
=4
The empirical formula is K1 Mn1 O4
KMnO4
‫الحل‬ ‫خطوات‬
.1‫السؤال‬ ‫في‬ ‫المذكورة‬ ‫العناصر‬ ‫فيه‬ ‫نضع‬ ‫جدول‬ ‫ننشأ‬
.2‫عندنا‬ ‫كان‬ ‫فلو‬ ‫بالجرام‬ ‫عنها‬ ‫معبر‬ ‫المئوية‬ ‫النسبة‬ ‫أن‬ ‫نعتبر‬100‫ال‬ ‫فهذه‬ ‫المركب‬ ‫من‬ ‫جرام‬100‫على‬ ‫موزعة‬ ‫جرام‬
‫نسبتها‬ ‫حسب‬ ‫العناصر‬.
.3‫الموالت‬ ‫عدد‬ ‫نوجد‬n‫القانون‬ ‫باستخدام‬ ‫عنصر‬ ‫لكل‬n=m/M.
.4‫العناصر‬ ‫من‬ ‫مول‬ ‫أصغر‬ ‫على‬ ‫الموالت‬ ‫عدد‬ ‫نقسم‬.
.5‫تمثل‬ ‫عليها‬ ‫نحصل‬ ‫التي‬ ‫األرقام‬empirical formula‫السابق‬ ‫المثال‬ ‫في‬ ‫كما‬ ‫صحيحة‬ ‫أعداد‬ ‫تكون‬ ‫أن‬ ‫بشرط‬.
.6‫من‬ ‫بدأ‬ ‫بأعداد‬ ‫الصيغة‬ ‫في‬ ‫الموجودة‬ ‫األسفل‬ ‫في‬ ‫التي‬ ‫األرقام‬ ‫بضرب‬ ‫نقوم‬ ‫عشرية‬ ‫أعداد‬ ‫ظهور‬ ‫حالة‬ ‫في‬2،3.......
‫صحيحة‬ ‫أعداد‬ ‫على‬ ‫نحصل‬ ‫حتى‬.Courtesy of Dr. Fawzia Albelwi

28
C H O
% 100g 40.92g 4.58g 54.50g
n=m/M 40.92/12.01
= 3.407mol
4.58/1.008
=4.54.mol
54.50/16.00
= 3.406 mol
 on smallest no. of
mole
3.407/3.406
= 1
0.4.54/3.406
= 1.33
3.406/3.406
= 1
Convert into integer x
3
3 3.99 = 4 3
The empirical formula
is
C3 H4 O3
C3H4O3
‫الحل‬ ‫خطوات‬
.1‫السؤال‬ ‫في‬ ‫المذكورة‬ ‫العناصر‬ ‫فيه‬ ‫نضع‬ ‫جدول‬ ‫ننشأ‬
.2‫عندنا‬ ‫كان‬ ‫فلو‬ ‫بالجرام‬ ‫عنها‬ ‫معبر‬ ‫المئوية‬ ‫النسبة‬ ‫أن‬ ‫نعتبر‬100‫ال‬ ‫فهذه‬ ‫المركب‬ ‫من‬ ‫جرام‬100‫موزعة‬ ‫جرام‬
‫نسبتها‬ ‫حسب‬ ‫العناصر‬ ‫على‬.
.3‫الموالت‬ ‫عدد‬ ‫نوجد‬n‫القانون‬ ‫باستخدام‬ ‫عنصر‬ ‫لكل‬n=m/M.
.4‫العناصر‬ ‫من‬ ‫مول‬ ‫أصغر‬ ‫على‬ ‫الموالت‬ ‫عدد‬ ‫نقسم‬.
.5‫تمثل‬ ‫عليها‬ ‫نحصل‬ ‫التي‬ ‫األرقام‬empirical formula‫صحيحة‬ ‫أعداد‬ ‫تكون‬ ‫أن‬ ‫بشرط‬
.6‫عشرية‬ ‫أعداد‬ ‫ظهور‬ ‫حالة‬ ‫في‬‫السابق‬ ‫المثال‬ ‫في‬ ‫كما‬‫في‬ ‫الموجودة‬ ‫األسفل‬ ‫في‬ ‫التي‬ ‫األرقام‬ ‫بضرب‬ ‫نقوم‬‫الصيغة‬
‫من‬ ‫بدأ‬ ‫بأعداد‬2،3.......‫صحيحة‬ ‫أعداد‬ ‫على‬ ‫نحصل‬ ‫حتى‬.
Example 3.9 p90:
Ascorbic acid composed of 40.92% C, 4.58% H, and 54.50%
O by mass. Determine its empirical formula.

29
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 92g
. Determine the molecular formula.
Solution:
1.
2.
3. Thus the empirical formula is: NO2
Oofmol217.0
00.16
47.3
Nofmol108.0
01.14
52.1


O
N
n
n
2
108.0
217.0
1
108.0
108.0
:  ON
Present Composition
by Mass
↓
Empirical Formula
↓
Molecular Formula

30
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:
6. The molecular formula is (NO2)2 = N2O4massmolarempirical
compoundofmassmolar
Ratio  2
46.01
90
Ratio 
1.956

 A sample of a compound containing born (B) and hydrogen
(H) contains 6.444g of B and 1.803 g of (H). The molar mass
of the compound is about 30g. What is its molecular formula?
Dr. Laila Al-Harbi
nB = 6.444 g B x =0.5961 mol B
1 mol B
10.81 g B
nH = 1.803g H x =1.7888 mol H
1 mol H
1.008 g H
B :
=1.0
0.5961
0.5961
H :
1.7888
0.5961 = 3
BH3
Molar mass of empirical formula = 10.81 + 3 x 1.008 = 13.834g
The ratio between molar mass and the molar mass of empirical formula
= molar mass / empirical formula = 30 g / 13.834 g ≈ 2
B2H6

 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
Dr. Laila Al-Harbi
3 ways of representing the reaction of H2 with O2 to form H2O
reactants products

2 Mg + O2 2 MgO
Dr. Laila Al-Harbi
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 grams Mg + 1 gram O2 makes 2 g MgO
IS NOT

 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
Dr. Laila Al-Harbi
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

‫نكتب‬‫الصحيحة‬ ‫الصيغة‬‫متفاعل‬ ‫لكل‬(‫ا‬ ‫الطرف‬ ‫على‬‫اليسر‬)
‫ناتج‬ ‫ولكل‬(‫االيمن‬ ‫الطرف‬ ‫على‬)
‫وزن‬‫الكيميائية‬ ‫المعادلة‬‫بتغير‬ ‫يكون‬‫األرقام‬‫التي‬‫بجانب‬
‫الصيغة‬‫تحتها‬ ‫التي‬ ‫وليست‬‫للعنصر‬ ‫يكون‬ ‫بحيث‬‫العدد‬ ‫نفس‬
‫المعادله‬ ‫طرفي‬ ‫على‬.
‫العناصر‬ ‫اوال‬ ‫توزن‬‫ظهورا‬ ‫األقل‬,‫العناصر‬ ‫توزن‬ ‫ثم‬
‫ظهورا‬ ‫االكثر‬
‫األخيره‬ ‫الخطوة‬‫من‬ ‫التأكد‬ ‫هي‬‫أن‬‫من‬ ‫العدد‬ ‫نفس‬ ‫لديك‬
‫طرفي‬ ‫على‬ ‫عنصر‬ ‫لكل‬ ‫الذرات‬‫المعادلة‬
Dr. Laila Al-Harbi

 Start by balancing those elements that appear in
only one reactant and one product.
Dr. Laila Al-Harbi

 Balance those elements that appear in two or more
reactants or products.
Dr. Laila Al-Harbi

 Check to make sure that you have the same
number of each type of atom on both sides of the
equation.
Dr. Laila Al-Harbi

Example 3.12 PRACTIES EXERICISE 3.12
Al + O2 → Al2O3
2Al + O2 → Al2O3
2Al + 3/2O2 → Al2O3
2(2Al + 3/2O2 → Al2O3)
4Al + 3O2 → 2Al2O3
 Fe2O3 + CO → Fe +CO2
 Fe2O3 + CO → 2Fe +CO2
 Fe2O3 +1/3CO →
2Fe+1/3CO2
 3(Fe2O3 +1/3CO →
2Fe+1/3CO2)
 Fe2O3 + 3CO → 2Fe +3CO2
Dr. Laila Al-Harbi

40
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

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
Dr. Laila Al-Harbi

‫ال‬ ‫او‬ ‫المتفاعالت‬ ‫أحد‬ ‫بمعلومية‬ ‫الناتجة‬ ‫أو‬ ‫المتفاعلة‬ ‫المواد‬ ‫لمعرفة‬‫ناتجة‬
‫باالتي‬ ‫نقوم‬:
‫موزونه‬ ‫المعادلة‬ ‫تكون‬ ‫أن‬ ‫البد‬
‫المعطاه‬ ‫المادة‬ ‫حددي‬given‫المطلوبة‬ ‫الماده‬ ‫ثم‬Required‫اعملي‬ ‫و‬
‫تماما‬ ‫الباقي‬ ‫تجاهلي‬ ‫و‬ ‫بينهم‬ ‫عالقة‬
‫موالت‬ ‫عالقة‬ ‫الموزونه‬ ‫المعادله‬ ‫في‬ ‫العالقة‬
‫اذاكانت‬ ‫و‬ ‫موالت‬ ‫الى‬ ‫نحولها‬ ‫بالجرامات‬ ‫المعطاه‬ ‫الماده‬ ‫فلوكانت‬
‫الخطوة‬ ‫هذه‬ ‫الى‬ ‫نحتاج‬ ‫ال‬ ‫بالموالت‬
‫الحد‬ ‫بهذا‬ ‫نكتفى‬ ‫بالموالت‬ ‫المطلوبة‬ ‫الماده‬ ‫كانت‬ ‫اذا‬
‫المطلوبة‬ ‫الماده‬ ‫كانت‬ ‫اذا‬‫جرامات‬ ‫الى‬ ‫الموالت‬ ‫نحول‬ ‫بالجرامات‬.
Dr. Laila Al-Harbi

If 2 mol of C6H12O6 is burned , what is the number
of moles of CO2 produced?
OHCOOOHC 2226126 666 
Given Required
From the equation mole of C6H12O6 → produce 6 mol CO2
From the equation 2 mol C6H12O6 → x mol CO2
the number of moles of CO2 produced = 2 × 6/ 1=12 mol

If 2 mol of C6H12O6 is burned , what is the mass of
CO2 produced?
 If 2 mol of C6H12O6 is burned , how many grams of
CO2 produced?
OHCOOOHC 2226126 666 
Given Required
From the equation mole of C6H12O6 → produce 6 mol CO2
From the equation 2 mol C6H12O6 → x mol CO2
the number of moles of CO2 produced = 2 × 6/ 1=12 mol
the mass of CO2 produced = n × molecular mass of CO2
the mass of CO2 produced = 12 ×44.01 = 528.12 g

 A general over all equation for this very complex process
represents the degradation of glucose (C6H12O6) to CO2
and water. If 856 g of C6H12O6 is consumed by person over
a certain period, what is the mass of CO2 produced?
 n = m/M = 856/180.2 = 4.75 mol
 From the equation mole of C6H12O6 → produce 6CO2
 From the equation 4.75 mol C6H12O6 → x CO2
 From the equation = 4.75 × 6 / 1= 28.5
 the mass of CO2 produced = n × M
 the mass of CO2 produced = 28.5 × 44.01 = 1254.35 g
Dr. Laila Al-Harbi
C6H12O6 + 6O2 → 6 H2O +6CO2

Methanol burns in air according to the equation
If 209 g of methanol are used up in the combustion , what mass of
water is produced?
2CH3OH + 3O2 2CO2 + 4H2O
Dr. Laila Al-Harbi
Given Required
From the equation 2 moles of CH3OH → produce 4 mol H2O
From the equation 6.53 mol CH3OH → x mol H2O
the number of moles of H2Oproduced = 6.53 × 4/ 2=13.06 mol
the mass of H2O produced = n × molecular mass of CO2
the mass of H2O produced = 13.06 ×18 = 235g
n = m/M = 209/32= 6.53 mol

 All alkali metals react with water to produce hydrogen gas
and the corresponding alkali metal hydroxide. A typical
reaction is that between lithium and water
How many grams of Li are needed to produce 9.89g of H2 ?
 From the equation 2 mole of Li → produce mole of H2
 From the equation 2× 6.941 g Li → 2.016g H2
 From the equation x g Li → 9.89 g CO2
 the mass of CO2 produced = 2× 6.941× 9.89 g / 2.016g
= 68.1g Li
Dr. Laila Al-Harbi
Li (s) + 2 H2O (l) → 2 Li OH (aq) + H2 (g)

48
 Limiting Reagent: is the reactant used up first
in a reaction and thus determine the amount of
product
 Excess Reagent ‫الفائض‬ ‫:الكاشف‬ 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!

49
Limiting Reagent:
2NO + O2 2NO2
NO is the limiting reagent
O2 is the excess reagent
Reactant used up first in
the reaction.

‫الناتجه‬ ‫الماده‬ ‫كمية‬ ‫يحدد‬ ‫الذي‬ ‫الكاشف‬ ‫هو‬ ‫المحدد‬ ‫الكاشف‬
‫مره‬ ‫كل‬ ‫نفسه‬ ‫هو‬ ‫المحدد‬ ‫الكاشف‬ ‫يكون‬ ‫أن‬ ‫يشترط‬ ‫ال‬
‫موجوده‬ ‫االخرى‬ ‫الماده‬ ‫و‬ ‫أقل‬ ‫موالت‬ ‫بعدد‬ ‫موجود‬ ‫المحدد‬ ‫الكاشف‬ ‫دائما‬
‫بزياده‬
‫ك‬ ‫يعطيك‬ ‫أنه‬ ‫السابقة‬ ‫المسائل‬ ‫عن‬ ‫تختلف‬ ‫المحدد‬ ‫الكاشف‬ ‫مسألة‬‫ال‬
‫الناتج‬ ‫يطلب‬ ‫و‬ ‫المتفاعلين‬
‫التالية‬ ‫بالخطوات‬ ‫نقوم‬ ‫المحدد‬ ‫الكاشف‬ ‫نحدد‬ ‫لكل‬
1-‫موالت‬ ‫الى‬ ‫المتفاعالت‬ ‫المواد‬ ‫جرامات‬ ‫نحول‬
2-‫الموزونه‬ ‫المعادلة‬ ‫في‬ ‫الماده‬ ‫معامل‬ ‫على‬ ‫الناتجة‬ ‫الجرامات‬ ‫نقسم‬
3-‫المحدد‬ ‫الكاشف‬ ‫هي‬ ‫موالت‬ ‫عدد‬ ‫أقل‬ ‫الماده‬
4-‫السابق‬ ‫الجزء‬ ‫في‬ ‫تعلمنا‬ ‫ما‬ ‫حسب‬ ‫الناتجه‬ ‫الماده‬ ‫نوجد‬
Dr. Laila Al-Harbi

 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 + 1H2SO4  Na2SO4 + 2HCl
 n NaCl = 22/58.5 = 0.376/2=0.188 mol
 n H2SO4 = 21/98 = 0.214/1 = 0.241 mol
 n NaCl (0.188 mol)<n H2SO4 (0.241 mol)
 So NaCl is the limiting reagent
Example:
Dr. Laila Al-Harbi

Consider the combustion of carbon monoxide (CO) in oxygen
gas: 2CO(g) + O2(g) → 2CO2(g)
Starting with 3.60 moles of CO and 4 moles of O2, calculate the
number of moles of CO2 produced ?
 n CO = 3.6/2=1.88 mol
 n O2 = 4/1 = 4 mol
 n CO (1.88 mol)<n O2 (4 mol)
 So CO is the limiting reagent
From chemical eq. 2 mole CO = 2 mol CO2
3.6 mol = x
number of moles of CO2 produced = 3.6 x 2/2 = 3.6 mol
Dr. Laila Al-Harbi

 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce
aluminum chloride. Which reactant is limiting, which is in excess,
and how much product is produced?
2 Al + 3 Cl2  2 AlCl3
 n Al = 10/27=0.37 /2mol= 0.185 mol
 n Cl2 = 35/ 71= 0.493 /3 = 0.164 mol
 n Cl2 (0.185 mol)<n Al (0.164 mol)
 So Cl2 is the limiting reagent
From chemical eq. 3 mole Cl2 = 2 mol AlCl3
0.493 mol = x
number of moles of AlCl3 produced = 0.493 x 2/3 = 0.329 mol
mass of AlCl3 = 0.329 x 133.5 = 43.877 g
Dr. Laila Al-Harbi

Science Cl2 is the LR so Al is the excess the amount remain
From chemical eq. 3 mole Cl2 = 2 mol Al
0.493 mol = x
number of moles of Al react = 0.493 x 2/3 = 0.329 mol
mass of Al = 0.329 x 27 = 8.8 83 g
The excess mass of Al = total mass Al – reacted Al
= 10 -8.883 = 1.117 g
Dr. Laila Al-Harbi

 Urea (NH2)2CO is prepared by reacting ammonia with carbon
dioxide
2NH3 (g) + CO2 (g) → (NH2)2CO (aq) + H2O (ι)
 In on process 637.2 g of NH3 are treated with 1142 g of CO2 a)
which of the two limiting reagents? b) calculate the mass of
(NH2)2CO formed ? C) how much excess reagent ( in gram) is
left at the end of the reaction
 n NH3 = 637.2 /17=37.482 /2mol= 18.74 mol
 n CO2 = 1142/ 44= 25.95 /1 = 25.95 mol
 n NH3 (18.74 mol)<n CO2 (25.95 mol)
 So NH3 is the limiting reagent
From chemical eq. 2 mole NH3 = 1 mol (NH2)2CO
37.482 mol = x
mass of AlCl3 =18.94 x 60.06 = 1125.6 g
Dr. Laila Al-Harbi

 C) how much excess reagent ( in gram) is left at the end of
the reaction
Science NH3 is the LR so CO2 is the excess the amount
remain
From chemical eq. 2 mole NH3 = 1 mol CO2
37.482 mol = x
number of moles of CO2 react = 0.493 x 1/2 = 18.74 mol
mass of CO2 = 18.74 x 44 = 824.56 g
The excess mass of CO2 = total mass CO2– reacted CO2
= 1142 -823.4 = 318.6 g
Dr. Laila Al-Harbi

 n Al = 124/27=4.59 /2mol= 2.296 mol
 n Fe2O3 = 601/ 159= 3.78 /1 = 3.78 mol
 n Al <n Fe2O3 >>>> So Al is the limiting reagent
From chemical eq. 2 mole Al = 2 mol Al2O3
4.59 mol = x
mass of AlCl3 = 2.296 x 102 = 234 g
Dr. Laila Al-Harbi

In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O3 Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
Dr. Laila Al-Harbi
 n Al = 124/27=4.59 /2mol= 2.296 mol
 n Fe2O3 = 601/ 159= 3.78 /1 = 3.78 mol
 n Al <n Fe2O3 >>>> So Al is the limiting reagent
From chemical eq. 2 mole Al = 2 mol Al2O3
4.59 mol = x
mass of AlCl3 = 2.296 x 102 = 234 g

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
Actual Yield is always lees .
Dr. Laila Al-Harbi

 When 22.0 g NaCl mixed with excess H2SO4 and 8.95 g
HCl is formed .what is the %yield of HCl?
2NaCl + H2SO4  Na2SO4 + 2HCl
n NaCl = 22/58.5 = 0.376/2=0.188 mol
From chemical eq. 2 mole NaCl = 2 mol HCl
0.376 mol = x
number of moles of HCl produced = 0.376 x 2/2 =0.376 mol
mass of HCl produced = n x MM(HCl) = 0.376 x 36.5 =13.724 g
 %yield of HCl = practical/theoriotical x 100
 %yield of HCl = 8.95 /13.724 x 100 = 65.21%
Dr. Laila Al-Harbi

 When 22.0 g NaCl and 21.0 g H2SO4 are mixed and react according to
the equation below 8.95 g HCl is formed .what is the %yield of HCl?
2NaCl + 1H2SO4  Na2SO4 + 2HCl
 n NaCl = 22/58.5 = 0.376/2=0.188 mol
 n H2SO4 = 21/98 = 0.214/1 = 0.241 mol
 n NaCl (0.188 mol)<n H2SO4 (0.241 mol)
 So NaCl is the limiting reagent
From chemical eq. 2 mole NaCl = 2 mol HCl
0.376 mol = x
number of moles of HCl produced = 0.376 x 2/2 =0.376 mol
mass of HCl produced = n x MM(HCl) = 0.376 x 36.5 =13.724 g
 %yield of HCl = practical/theoriotical x 100
 %yield of HCl = 8.95 /13.724 x 100 = 65.21%
Dr. Laila Al-Harbi

 In one process, 3.54×107 g of TiCl4 are reacted with 1.13×107 g
of Mg a) Calculate the theoretical yield of the Ti? b) calculate
the percent yield if 7.91×106 g of Ti are obtained ?
TiCl4 (g) + 2Mg (ι)→ Ti (s)+ 2MgCl2 (ι)
n TiCl4 = 3.54×107 g /189.68 = 1.87x105 /1= 1.87x105 mol
 n Mg = 1.13×107 g / 24.3 = 1.87x105 /2 = 2.32 x105mol
 n TiCl4 <n Mg ,So TiCl4 is the limiting reagent
From chemical eq. 1 mole TiCl4 = 1 mol Ti
1.87x105 mol = x
number of moles of Ti produced = 1.87x105 x 1/1 = 1.87x105 mol
mass of Ti produced = n x MM(HCl) = 1.87x105 x 47.88 =
8.93x109 g
 %yield of Ti = practical/theoriotical x 100
 %yield of HCl = 7.91×106 / 8.93x109 x 100 = 88.52%
Dr. Laila Al-Harbi

Chapter 3

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Chapter 3

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