9/9/2022
Chapter 3
Mass Relationships in
Chemical Reactions
C
Cu
S
Hg
Fe
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
Mass of one mole of these elements:
C
Hg
S
12.01 g
200.6 g
32.06 g
Cu
Fe
63.55 g
55.85 g
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
1
9/9/2022
Micro World
atoms & molecules
amu
Macro World
grams
Atomic mass is the mass of an atom in atomic mass units (amu)
By definition:
1 atom of 126C “weighs” 12.00 amu
1H
1
6
Li
3
= 1.008 amu
= 6.015 amu
16O
8
= 16.00 amu
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.1
Average amu
By definition:
1 atom of 12C “weighs” 12.00 amu
1H
1
6
Li
3
= 1.008 amu
= 6.015 amu
16O
8
= 16.00 amu
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.1
2
9/9/2022
Average amu = 6.941
Natural lithium is:
7.42% 66Li
Li (6.015
(6.015 amu)
amu)
92.58% 77Li
Li (7.016
(7.016 amu)
amu)
92.58%
7.42 x 6.015 + 92.58 x 7.016
= 6.941 amu
100
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
Amount of Substances
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
1 mol = NA = 6.0221367 x 1023
Avogadro’s number (NA)
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.2
3
9/9/2022
apples
eggs
atoms
Molar mass of a substance
is the mass of 1 mole of
23 eggs]
apple
atoms]
apple]
this
atom in grams [mass (g) of 6.022 x 1023
eggs
substance
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
1 mole 12C atoms = 12.00 g 12C
1 mole lithium atoms =
6.941 g of Li
For any element
atomic mass (amu) = molar mass (grams/mol)
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.2
One Mole of:
12.01 g
C
Hg
32.07 g
S
200.6 g
Fe
Cu
63.55 g
55.85 g
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
4
9/9/2022
m
m=n xM
n = m /M
Mass
(g)
n
m=n
M
# of substance
Molar mass
(mole)
(g/mole)
(mole)
(g)
x M (g/mole)
n=
(mole)
m (g)
M (g/mole)
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
N
N = n x NA
n
# of substance
NA
Avogadro’s #
(mole)
N = n (mole) x NA (#/mole)
(#)
n = N /NA
Number
of elements
(#)
(#/mole)
n=
(mole)
N (#)
NA (#/mole)
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
5
9/9/2022
Dealing with amount of Atoms and Molecules is similar to
dealing with the amount of any substance, such as eggs
Mole is just an expression of
number same like any other
expression (dozen, pair, kilo)
6.022 x 1023 mol-1
m
N
Number
of eggs
Mass
(g)
(#)
D
n
?
n
d
# of substance
d
# of substance
(dozen)
(dozen)
g/dozen
Number
in dozen
12 /dozen
#/dozen
Nd
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
1. How many dozen are in 1600 g of egg boxes
knowing that the mass of dozen is 800 g?
nd = m
D
=
1600 g
800 g/dozen
= 2 dozen
2. How many eggs are in 1600 g of egg boxes?
N
=n
d
x
Nd
The number of dozen is found from the mass, as answered
nd =2 dozen)
above (
N of eggs
=
2 dozen
x
12/dozen = 24 eggs
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
6
9/9/2022
m = 0.551g
N=?
How many atoms are in 0.551 g of potassium (K) ?
M of K= 39.10 g/mole
N = n x NA
N = n x 6.022 x1023 atoms/mole
&
n = m/M
?
N=
N=
Mass (g)
Molar mass (g/mole)
x 6.022 x1023 atoms/mole
0.551 g
x 6.022 x1023 atoms/mole
39.10 (g/mole)
= 8.49 x 1021 K atoms
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.2
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
For any molecule
molecular mass (amu) = molar mass (grams)
1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g SO2
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.3
7
9/9/2022
m = 72.5 g
N=?
How many H atoms are in 72.5 g of C3H8O ?
M = 60 g/mole
The molar mass of C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g/mole
1 mol C3H8O molecules contains 8 mol H atoms
N=
Mass (g)
Molar mass (g/mole)
N =
C3H8O
N =
CH3H8O
72.5 g
60 (g/mole)
72.5 g
60 (g/mole)
x 6.022 x1023 molecule/mole
x 6.022 x1023
molecule
mole
atom
x 6.022 x1023 molecule
x8
molecule
mole
= 5.8 x 1024 H atoms
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.3
Formula mass is the sum of the atomic masses (in amu) in a
formula unit of an ionic compound.
1Na
NaCl
22.99 amu
1Cl + 35.45 amu
NaCl
58.44 amu
For any ionic compound
formula mass (amu) = molar mass (grams)
1 formula unit NaCl = 58.44 amu
1 mole NaCl = 58.44 g NaCl
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.3
8
9/9/2022
What is the formula mass of Ca3(PO4)2 ?
1 formula unit of Ca3(PO4)2
3 Ca
3 x 40.08
2P
2 x 30.97
8O
+ 8 x 16.00
310.18 amu
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.3
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
C2H6O
%C =
2 x (12.01 g)
x 100% = 52.14%
46.07 g
%H =
6 x (1.008 g)
x 100% = 13.13%
46.07 g
%O =
1 x (16.00 g)
x 100% = 34.73%
46.07 g
52.14% + 13.13% + 34.73% = 100.0%
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.5
9
9/9/2022
Empirical Formula of an Oxide
Empirical Formula is the simplest formula, or the formula with
the smallest integer that shows the ratios of atoms in a
molecule.
Structural Formula: Shows
how atoms are attached in
a molecule
B
B
B
A
A
B
B
B
Molecular Formula: Shows
the actual number of each
type of atom in the molecule
A2B6
Empirical Formula
A1B3
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
Example: Determine the empirical formula of a compound
containing: 32.38% Na, 22.65% S, and 44.97% O.
First: Assume that the % composition represents the mass (in gram)

 # mole 

Second: Convert mass to number of moles.


( g / mole 
mass ( g )
molar mass
Third: Divide by the smallest number of moles
Na = 32.38 g / 22.99 g/mole = 1.408 mole / 0.7065 mole = 1.993 ( 2 Na)
S = 22.65 g / 32.06 g/mole = 0.7065 mole / 0.7065 mole = 1 S
O = 44.97 g / 16.00 g/mole = 2.811 mole / 0.7065 mole = 3.988 ( 4 O)
Therefore, the empirical formula of the compound is: Na 2SO4
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
10
9/9/2022
Example: Determine the empirical formula of a compound
containing: 24.75% K, 34.77% Mn, and 40.51% O.
First: Assume that the % composition represents the mass (in gram)

 # mole 

Second: Convert mass to number of moles.


( g / mole 
mass ( g )
molar mass
Third: Divide by the smallest number of moles
K = 24.75 g / 39.10 g/mole = 0.6330 mole / 0.6329 mole  1 K
Mn = 34.77 g / 54.94 g/mole = 0.6329 mole / 0.6329 mole = 1 Mn
O = 40.51 g / 16.00 g/mole = 2.532 mole / 0.6329 mole  4 O
Therefore, the empirical formula of the compound is: KMnO4
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.5
Percent Composition and Empirical Formulas
gram
%
mass
molar mass
mole
ratio
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.5
11
9/9/2022
XY2.5
Divide on 5
XY1.5
Divide on 5
X1.2Y1.8Z1
X12Y18Z10
Divide on 2
X6Y9Z5
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
Experimentally,
how will you find
the empirical
formula of organic
compound?
Combust
11.5 g
C?H?O?
Collect 13.5 g H2O
and 22.0 g CO2
C?H?O?
mass of
number of atom
x M of x
in compound
compound atom
mass of atom =
M of compound
mass of C =
22.0 g x 12
= 6.0 g
44
n=
m
M
6.0 g
= 0.5 mole C
12 g/mole
in H2O
mass of H =
13.5 g x 1.0 x 2
= 1.5 g
18
1.5 g
1 g/mole
= 1.5 mole H
mass of O = mass of sample – (mass of C + mass of H)
mass of O = 11.5 – ( 6.0 + 1.5) = 4.0 g
Empirical formula is C0.5H1.5O0.25
0.25 0.25 0.25
4.0 g
= 0.25 mole O
16 g/mole
C2H6O
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
12
9/9/2022
A chemical reaction: is a process in which one or more substances is
changed into one or more new substances
A chemical equation: uses chemical symbols to show what happens
during a chemical reaction
3 ways representing the reaction of H2 with O2 to form H2O
reactants
products
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.7
How to “Read” Chemical Equations
2 Mg + O2
2 MgO
2 Mg atoms + 1 O2 molecule makes 2 formula units of MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
IT IS NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.7
13
9/9/2022
Balancing Chemical Equations
1. 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.
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.
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + 7 O2
2 CO2 + 3H2O
2
2 C2H6 + 7O2
Multiply the
coefficients with 2
4 CO2 + 6 H2O
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.7
Strategies for Balancing Chemical Equations
1. 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
1 carbon
on right
start with C or H but not O
multiply CO2 by 2
2CO2 + H2O
2 hydrogen
on right
multiply H2O by 3
2CO2 + 3H2O
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.7
14
9/9/2022
Strategies for Balancing Chemical Equations
2. Balance those elements that appear in two or more
reactants or products.
C2H6 + O2
2 oxygen
on left
2CO2 + 3H2O
4 oxygen
(2x2)
multiply O2 by
+ 3 oxygen
(3x1)
C2H6 + 7 O2
2
2CO2 + 3H2O
2C2H6 + 7O2
4CO2 + 6H2O
7
2
= 7 oxygen
on right
remove fraction by
multiplying both sides
by 2
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.7
Strategies for Balancing Chemical Equations
3. 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
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.7
15
9/9/2022
Balance these chemical equations
C7H16
+
O2
→
CO2
+
H2O
C4H10
+
O2
→
CO2
+
H 2O
C4H10O +
O2
→
CO2
+
H 2O
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
Amounts of Reactants and Products
1. Write balanced chemical equation
2. Convert quantities (in gram) of limiting reactant into moles
3. Use coefficients in balanced equation to calculate the number of moles
of the product
4. Convert moles of the product into mass (gram)
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.8
16
9/9/2022
Amounts of Reactants and Products
Example: Synthesis of Alum from Aluminum (Al metal)
Al
AlK(SO4)212H2O
Mass of Al
Expected? Mass of Alum
x Mole Al
x Mole Alum
1. Write balanced chemical equation
2. Convert quantities (in gram) of limiting reactant into moles
3. Use coefficients to calculate the number of moles of the product
4. Convert moles of the product into mass (gram)
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.8
If 209 g of methanol are used up in the combustion, what mass of
water is produced?
2 CH3OH + 3 O2
2 CO2 + 4 H2O
235? g
209 g
molar mass
CH3OH=32.0 g/mol
x Moles CH3OH
209 g
x
32.0 g/mole
m of product =
coefficients
chemical equation
2 mol
1 mol
molar mass
H2O =18.0 g/mol
2 x Moles H2O
x 18.0 g/mole = 235 g H2O
m of react
2 mol of product
x
x M of product
M of react
1 mol of reactant
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.8
17
9/9/2022
Limiting Reagents
2NO + O2
2NO2
NO is the limiting reagent as long as it is less than twice the number
6
6
6NO2 + 3O2
6NO + 6O2
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.9
Example on Limiting Reagents: In one process, 124 g of Al reacted
with 601 g of Fe2O3. Calculate the mass of Al2O3 formed.
2Al
124 g
+
Fe2O3
Al2O3 + 2Fe
601 g Fe2O3 needed
x mol Al
124 g
x
27.0 g/mole
½ x mol
Fe2O3 needed
1 mol
2 mol
x 160.0 g/mole = 367 g Fe2O3
There is 601 gFe2O3 (it is more than the needed 367 g). Therefore,
Al is the limiting reagent
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.9
18
9/9/2022
Example on Limiting Reagents: In one process, 124 g of Al reacted
with 601 g of Fe2O3. Calculate the mass of Al2O3 formed.
2Al
124 g
+
Fe2O3
Al2O3 + 2Fe
?g
x mol Al
½ x Mol of Al2O3
124 g
x
27.0 g/mole
1 mol
2 mol
x 102 g/mole
= 234 g Al2O3
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.9
What is the theoretical yield of PI3 from the reaction of 27.0 g of
P and 68.0 g of I2?
2 P(s)
27.0 g
+
30.97 g/mol
3 I2(s)
68.0 g
→
2 PI3(s)
?g
253.81 g/mol
27.0
 30.97
 2
= 0.436
68.0
 253.81
 3
= 0.089
Therefore, the limiting reagent is I2
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
19
9/9/2022
What is the theoretical yield of PI3 from the reaction of 27.0 g of
P and 68.0 g of I2?
2 P(s)
27.0 g
+
30.97 g/mol
3 I2(s)
68.0 g
→
2 PI3(s)
73.5? g
253.81 g/mol
x Mole I2
68 g
x
253.81 g/mole
2 mol
3 mol
2/3x Mole PI3
x 411.69 g/mole = 73.5 g PI3
Therefore, the limiting reagent is I2
A)
B)
C)
D)
E)
70.5 g
71.5 g
73.5 g
75.5 g
None of the above
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
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
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.10
20
9/9/2022
Reaction Yield
Synthesis of Alum from Aluminum (Al metal)
Actual yield
Calculate the % yield
Mass produced
Experimentally
Al
AlK(SO4)212H2O
Mass of Al
Expected Mass of Alum
Theoretical yield
x Mole Al
x Mole Alum
From the Experiment
% Yield

Actual Yield
produced mass of Alum
 100 
 100
Theoratical Yield
Expected mass of Alum
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
3.10
End of Chapter 3
Modified By Dr. Khalid A. M. Al-Saad Al-Kuwari
21