# Chapter 3 Lectures

```Chapter 3
Stoichiometric
Atomic Masses, Mole concept, and Molar
Mass (Average atomic mass).
Number of atoms per amount of element.
Percent composition and Empirical formula
of molecules.
Chemical equations, Balancing equations,
and Stoichiometric calculations including
limiting reagents.
1
Micro World
atoms & molecules
Macro World
grams
Atomic mass is the mass of an atom in
atomic mass units (amu)
By definition:
1 atom 12C “weighs” 12 amu
On this scale
1H
= 1.008 amu
16O
= 16.00 amu
2
Atomic Masses
Elements occur in nature as mixtures of
isotopes
Carbon =
98.89% 12C
1.11% 13C
<0.01% 14C
Carbon atomic mass = 12.01 amu
3
Measuring Atomic Mass
Figure 3.1: (left) A scientist injecting a sample into a
mass spectrometer. (right) Schematic diagram of a mass
spectrometer.
4
Spectrum
Most Abundant Isotope
5
Atomic mass of Neon
Consider 100 atoms of neon
amu
( 90.92atoms x 20atom
)(
(
8.82 atom
0.257 atoms
amu
x 22atom
) 
amu
x 21atom
)
2018 amu
/ 100
 20.18
Mifflin Company. All rights
6
Natural lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
Average atomic mass of lithium:
7.42 x 6.015 + 92.58 x 7.016
= 6.941 amu
100
7
The Mole
The number equal to the number of carbon
atoms in exactly 12 grams of pure 12C.
1 mole of anything = 6.022  1023 units of
that thing
8
equals
23
6.022  10 units
9
Molar Mass
A substance’s molar mass (molecular weight) is
the mass in grams of one mole of the compound.
C=12
O=16
CO2 = 44.01 grams per mole
10
eggs
Molar mass is the mass of 1 mole of shoes in grams
marbles
atoms
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)
11
1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu
12
EX. 3.2: Calculate the mass of
6 atoms of americium in grams
• From periodic table, Am has a mass of 243
amu.
• 6.022x1023 atoms weigh 243 g.
243g
6.022 x10
23
x
atoms Am
6 atomsAm
Mifflin Company. All rights

2.42 x1021 g
13
Molar Mass
A chemical compound is a collection of atoms.
What is the mass of 1 mol of CH4?
Mass of 1 mol C = 12.01 g
Mass of 4 mol H = 4x1.008 g
Mass of 1 mol CH4 = 16.04 g
14
Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
1 mol C3H8O molecules = 8 mol H atoms
1 mol H = 6.022 x 1023 atoms H
1 mol C3H8O 8 mol H atoms 6.022 x 1023 H atoms
72.5 g C3H8O x
x
x
=
1 mol C3H8O
1 mol H atoms
60 g C3H8O
5.82 x 1024 atoms H
15
S.Ex. 3.7
Calculate Molar mass of CaCO3
4.86 moles of CaCO3 → g CaCO3 → g CO32-
16
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
2 x (12.01 g)
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
%C =
C2H6O
52.14% + 13.13% + 34.73% = 100.0%
17
Determining Elemental Composition
(Formula)
Figure 3.5:
A schematic diagram of the
combustion device used to analyze substances for
carbon and hydrogen.
18
The masses obtained (mostly CO2 and
H2O and sometimes N2)) will be used to
determine:
1. % composition in compound
2. Empirical formula
3. Chemical or molecular formula if the
Molar mass of the compound is known or
given.
19
Example of Combustion
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
20
Collect 22.0 g CO2 and 13.5 g H2O
16 x 2
12
1mol CO2
44gCO2
1mol C
22gCO2
mc
22gCO2 x 12gC
mc =
12g C
% C in CO2
collected
= 6g C
44gCO2
21
Convert g to mole:
1mol C
nmol C
12g C
6g
nc =
6g x 1molC
12 gC
= 0.5 mol
Repeat the same for H from H2O
1mol H2O
18g H2O
2 mol H
13.5g H2O
nmol H
nmol H =
2x13.5
2g H
mH
= 1.5 mol H
18 mol H
Faster H but
still need O 22
mH =
2x13.5
18
mO = 11.5g – mC – mH
= 11.5 – 6 – 1.5 = 4g
= 1.5 g H
nO =
m
MM
=
4
16
= 0.25 mol O
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
23
OR
for C
mC
1. Fraction of C in CO2 =
12.01
=
MMCO2
=
44.01
2. Mass of C in compound = mass of CO2 x Fraction of C
mC
3. % C in compound =
x 100
msample
4. If N then % N = 100 - % C - % H
24
Then
Empirical Formula
Using the previously calculated % in compound:
% in gram
a. Number of mole of C =
Atomic mass of C
% in gram
b. Number of mole of H =
Atomic mass of H
a
Then divide by the smallest number:
smallest
:
b
smallest
:
c
smallest
25
Note
If results are :
0.99 : 2.01 : 1.00
Then you have to convert to whole numbers:
1 :2
:1
CH2N
If results are :
1.49 : 3.01 : 0.99
Then you have to multiply by 2:
3
:6
:2
C3H6N2
Hence, empirical formula is the simplest formula of a compound
26
S. E.3.11
71.65% Cl;
24.27% C ;
4.07 % H
2.021 mol Cl
2.021 mol C
4.04 mol H
1
:
1
:
2
Empirical formula is ClCH2
Molar mass = 98.96 g/mol
Molar mass/ empirical formula mass =
98.96/49.48 = 3
Molecular formula – (ClCH2)2 = Cl2C2H4
27
Formulas
molecular formula = (empirical formula)n
[n = integer]
molecular formula = C6H6 = (CH)6
empirical formula = CH
Then
Molecular Mass
=
n
Empirical Mass
28
Figure 3.6:
Examples of substances whose empirical and molecular formulas differ. Notice
that molecular formula = (empirical formula)n, where n is a integer.
29
Exercise 3.12
A compound analyzed and found to contain
only P and O.
Mass % of P = 43.64%
Molar mass = 283.88 g/mol.
What are the empirical and molecular
formulas?
30
Figure 3.8 The Structure of P4O10.
31
Chemical Equations
Chemical change involves a
reorganization of the atoms in one or
more substances.
32
Chemical Equation
A representation of a chemical reaction:
C2H5OH + O2  CO2 + H2O
reactants
products
Unbalanced !
33
Chemical Equation
C2H5OH + 3O2  2CO2 + 3H2O
The equation is balanced.
1 mole of ethanol reacts with 3 moles of
oxygen
to produce
2 moles of carbon dioxide and 3 moles of
water
34
Methane
Reacts
with
Oxygen to
Produce
Flame
35
36
S.Ex. 3.14
Balance the equation
(NH4)2Cr2O7 (s) → Cr2O3(s) + N2(g) +
H2O(g)
N and Cr are balanced; 4 for H2O balances H
It also balances O.
(NH4)2Cr2O7 (s) → Cr2O3(s) +
37
Decomposition of Ammonium
Dichromate
38
Decomposition of Ammonium
Dichromate
39
S. Ex.3.15
Balance the equation
NH3(g) + O2(g)→ NO (g) + H2O (g)
All molecules are of equal complexity.
Balance H first; 2 for NH3 and 3 for H2O
N can be balanced with a coeff. of 2 for NO
O can be balanced by 5/2 for O2
2NH3(g) + 5/2 O2(g)→ 2NO (g) + 3H2O (g)
x2
4NH3(g) + 5O2(g)→ 4NO (g) + 6H2O (g)
40
Calculating Masses of Reactants
and Products
1.
2.
3.
4.
Balance the equation.
Convert mass to moles.
Set up mole ratios.
Use mole ratios to calculate moles of
desired substituent.
5. Convert moles to grams, if necessary.
41
Methanol burns in air according to the equation
2CH3OH + 3O2
2CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
grams CH3OH
moles CH3OH
molar mass
CH3OH
209 g CH3OH x
moles H2O
grams H2O
molar mass
coefficients
H2O
chemical equation
4 mol H2O
18.0 g H2O
1 mol CH3OH
=
x
x
32.0 g CH3OH
2 mol CH3OH
1 mol H2O
235 g H2O
42
OR
2CH3OH + 3O2
2 mol
2CO2 + 4H2O
4 mol
2x(12+4+16) g
209 g
4x(2+16) g
m
m=
209 x 4(2+16)
2(12+4+16)
43
S. Ex 3.16
What mass of gaseous carbon dioxide can be
absorbed by 1.00 kg of lithium hydroxide?
Unbalanced eq.
LiOH(s) + CO2 (g) → Li2CO3 (s) + H2O(l)
Balanced equation
2 LiOH(s) + CO2 (g) → Li2CO3 (s) + H2O(l)
23.95 g/mol 44.0 g/mol
44
Limiting Reactant
The limiting reactant is the reactant
that is consumed first, limiting the
amounts of products formed.
45
Limiting Reagents
6 red
green
leftused
overup
46
47
Figure 3.12 Hydrogen and
Nitrogen React to Form Ammonia
48
Solving a Stoichiometry Problem
1.
2.
3.
4.
Balance the equation.
Convert masses to moles.
Determine which reactant is limiting.
Use moles of limiting reactant and mole
ratios to find moles of desired product.
5. Convert from moles to grams.
49
Limiting Reactant
N2 (g) +
3 H2 (g)
25.0kg
5.00 kg
8.93x102 2.48x103
→
2 NH3 (g)
??
8.93x102 mol N2 x 3 mol H2 = 2.68x103 mol H2
1 mol N2
H2 is limiting; 2.48x103 mol H2 x 2 mol NH3 = 1.65x103
3 mol H2
50
Limiting Reactant Calculations
What mss of molten iron is produced by 1 kg each of the
reactants?
Fe2O3(s) + 2 Al(s)  Al2O3(s) + 2 Fe(l)
1 mol
2 mol
6.26mol
37.04
??
If all Fe2O3(s) reacts
If all Al reacts
2x6.26 = 12.5 mol Fe -Limiting
37.04 mol Fe -- Excess
The 6.26 mol Fe2O3 will Disappear first
51
S. Ex 3.18
2NH3(g) +3CuO(s) → N2(g) +3Cu(s) +3H2O(g)
1.06 mol 1.14 mol ??g
If CuO is limiting; 1.14 mol CuO x 1 mol N2
3 mol CuO
= 0.380 mol N2
If NH3 is limiting; 1.06 mol NH3 x 1 mol N2
2 mol NH3
= 0.530 mol N2
Least amount of product obtained with CuO; Cuo is the limiting reactant
52
Theoretical Yield is the amount of product that would
result if all the limiting reagent reacted. Its amount is
Calculated using the balanced equation.
Actual Yield is the amount of product actually obtained
from a reaction. It is usually given.
53
Percent Yield
Actual yield = quantity of product actually
obtained
Theoretical yield = quantity of product
predicted by stoichiometry
54
Percent Yield Example
14.4 g
0.104 mol
excess
Actual yield = 6.26 g
1 mol S.A. → 1 mol aspirin
0.104 mol S.A. → 0.104 mol aspirin = 18.7 g aspirin
Percent yield = 6.26 g x 100 = 33.5 %
18.7 g
55
Solving a
Stoichiomet
ry Problem
Invovling
Masses of
Reactants
and
Products
56
Sample exercise 3.19 calculation
% yield
Methanol can be manufactured by
combination of gaseous carbon monoxide
and hydrogen. Suppose 68.5Kg CO(g) is
reacted with 8.60Kg H 2(g). Calculate the
theoretical yield of methanol. If 3.57x104g
CH3OH is actually produced, what is the
percent yield of methanol?
57
2 H2 (g) + CO(g)
→ CH3OH (l)
8.60 kg
68.5 kg
35.7 kg(actual yield)
4.27x103mol
2.44x103 mol
If all H2 are converted to methanol,
4.27x103mol H2 x 1 mol methanol = 2.14x103 mol methanol
2 mol H2
If all CO are converted to methanol,
2.44x103 mol CO x 1 mol methanol = 2.44x103 mol methanol
1 mol CO
H2 is the limiting reactant; theoretical yield = 2.14x103 mol methanol
= 68.6 kg methanol
% yield = actual yield x 100
theoretical yield
=
35.7 kg x100 = 52.0%
68.6 kg
58
Sample Exercise
Titanium tetrachloride, TiCl4, can be made by
combining titanium-containing ore (which is often
impure TiO2) with carbon and chlorine TiO2(s) + 2 Cl2(g) + C(s)
TiCl4(l) + CO2(g)
If one begins with 125 g each of Cl2 and C, but
plenty of titanium-containing ore, which is the
limiting reagent in the reaction? What quantity of
TiCl4 can be produced?
59
Virtual Laboratory Project
60
Practice Example 1
A compound contains C, H, N. Combustion of
35.0mg of the compound produces 33.5mg CO2
and 41.1mg H2O. What is the empirical formula of
the compound?
Solution:
1. Determine C and H, the rest from 33.5mg is N.
2. Determine moles from masses.
3. Divide by smallest number of moles.
61
Practice Example 2
Caffeine contains 49.48% C, 5.15% H, 28.87%
N and 16.49% O by mass and has a molar mass
of 194.2 g/mol. Determine the molecular
formula.
Solution:
1. Convert mass to moles.
2. Determine empirical formula.
3. Determine actual formula.
C8H10N4O2
62
Practice Example 3
Nitrogen gas can be prepared by passing
gaseous ammonia over solid copper(II)
oxide at high temperatures. The other
products of the reaction are solid copper and
water vapor. If a sample containing 18.1g of
NH3 is reacted with 90.4g of CuO, which is
the limiting reactant? How many grams of
N2 will be formed.
63
Practice Example 5
SnO2(s) + 2 H2(g)  Sn(s) + 2 H2O(l)
a) the mass of tin produced from 0.211 moles of hydrogen
gas.
b) the number of moles of H2O produced from 339 grams of
SnO2.
c) the mass of SnO2 required to produce 39.4 grams of tin.
d) the number of atoms of tin produced in the reaction of 3.00
grams of H2.
e) the mass of SnO2 required to produce 1.20 x 1021
molecules of water.
64
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