Introductory Chemistry:
A Foundation
Chemical Composition
FOURTH EDITION
Chapter 8
by Steven S. Zumdahl
University of Illinois
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
1
2
Atomic Masses
Atomic Masses
• Dalton used the percentages of elements in
compounds and the chemical formulas to deduce
the relative masses of atoms
• Unit is the amu.
• Balanced equation tells us the relative numbers of
molecules of reactants and products
C + O2 → CO2
– atomic mass unit
– 1 amu = 1.66 x 10-24g
1 atom of C reacts with 1 molecule of O2
to make 1 molecule of CO2
• If I want to know how many O2 molecules I will
need or how many CO2 molecules I can make, I will
need to know how many C atoms are in the sample
of carbon I am starting with
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
• We define the masses of atoms in terms of atomic
mass units
– 1 Carbon atom = 12.01 amu,
– 1 Oxygen atom = 16.00 amu
– 1 O2 molecule = 2(16.00 amu) = 32.00 amu
3
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
4
Example #1
Atomic Masses
Calculate the Mass (in amu) of 75 atoms of Al
• Atomic masses allow us to convert weights
into numbers of atoms
•
If our sample of carbon weighs 3.00 x 1020 amu we
will have 2.50 x 1019 atoms of carbon
1 C atom
3.00 x 10 20 amu x
= 2.50 x 1019 C atoms
12.01 amu
•
Determine the mass of 1 Al atom
1 atom of Al = 26.98 amu
Use the relationship as a conversion factor
75 Al atoms x
26.98 amu
= 2024 amu
1 Al atom
Since our equation tells us that 1 C atom reacts
with 1 O2 molecule, if I have 2.50 x 1019 C
atoms, I will need 2.50 x 1019 molecules of O2
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
5
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
6
The MOLE
Chemical Packages - Moles
• We use a package for atoms and molecules
called a mole
• A mole is the number of particles equal to the
number of Carbon atoms in 12 g of C-12
• One mole = 6.022 x 1023 units
• The number of particles in 1 mole is called
Avogadro’s Number
• 1 mole of C atoms weighs 12.01 g and has 6.02 x 1023 atoms
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
7
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
The Mole
• When two chemicals react with each other, we want
to know how many atoms or molecules of each are
used do that a formula of the product can be
established.
• This means we need a way to count atoms, but
HOW?
• The solution to this problem is to define a convenient
unit of matter that contains a know number of
particles.
• 1 moles = 6.022136736×1023 particles
– This value is commonly known has Avogadro’s number.
• Remember that one mole always contains the same
number of particles, no matter what the substance.
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
•
2.
3.
16.032 g
6.022
4 × 6.022
=
– CCl4
» We have 1 carbon and 4 chlorines
» 12.001 g/mol + 4(35.4527 g/mol) = 153.81 g/mol
» Therefore, if we have 1 mole of CCl4 it would weigh 153.81 g.
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
10
Vol. of an
“ideal” gas @ STP
e
b
f
a
# of moles of X
a) Multiply by Avogadro’s
Number
d) Divide by the molecular
weight of X
How many hydrogen atoms do I have?
•
• Example:
e) Multiply by 22.4 L / mol
f) Divide by 22.4 L / mol
c
d
c) Multiply by the
molecular weight of X
×1023
×1023
– To determine the molecular mass of a molecule, we add up the
atomic masses of the elements that make up the molecule.
b) Divide by Avogadro’s
Number
How many carbon atoms do I have?
•
4.
I have 6.022 ×1023 molecules of CH4
How much does it weigh?
•
– 1 mole of sodium weighs 22.99 g
• This not only true for atoms, but also for molecules, but we
use the molecular mass instead of the atomic mass.
# of
particles
How many CH4 molecules do I have?
•
• The mass of one mole of atoms of any element is the molar
mass, which is numerically equal to the atomic mass, but
in grams.
• Example:
MOLES are the CENTER
Thinking about moles with respect to molecules
and compounds is not all that different than with
atoms.
If I have 1 mole of CH4:
1.
Molar Mass
9
Molecules, Compounds, and The
Mole
•
8
Mass
2.409×1024
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
11
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
12
Example #2
EXAMPLES
At first these concepts are difficult to understand, but
working problems is the best way to become comfortable
with these concepts.
• What is the mass, in grams, • Calculate the molar mass of
of 1.5 mols of silicon?
CaCO3 (limestone).
100.08 g/mol
42.1283 g
• If you have 454 g of CaCO3,
• How many moles are
represented by 454 g of
how many moles do you
sulfur? How many atoms?
have? How many
molecules? How many
14.2 mols
atoms of oxygen?
8.53×1024 atoms
4.54 mols
2.73×1023 molecules
Copyright©2000 by Houghton
8.20×1024 atom of O 13
Mifflin Company. All rights reserved.
Compute the number of moles
and number of atoms in 10.0 g of Al
‘ Use the Periodic Table to determine the
mass of 1 mole of Al
1 mole Al = 26.98 g
’ Use this as a conversion factor for
grams-to-moles
1 mol Al
10.0 g Al x
= 0.371 mol Al
26.98 g
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
14
Example #2
Example #3
Compute the number of moles
and number of atoms in 10.0 g of Al
“ Use Avogadro’s Number to determine the
number of atoms in 1 mole
1 mole Al = 6.02 x 1023 atoms
” Use this as a conversion factor for
moles-to-atoms
Compute the number of moles
and mass of 2.23 x 1023 atoms of Al
‘ Use Avogadro’s Number to determine the
number of atoms in 1 mole
1 mole Al = 6.02 x 1023 atoms
’ Use this as a conversion factor for
atoms-to-moles
0.371 mol Al x
6.02 x 1023 atoms
= 2.23 x 1023 Al atoms
1 mol Al
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
15
Example #3
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
16
• The molar mass is the mass in grams of one
mole of a compound
• The relative weights of molecules can be
calculated from atomic masses
water = H2O = 2(1.008 amu) + 16.00 amu
= 18.02 amu
• 1 mole of H2O will weigh 18.02 g, therefore
the molar mass of H2O is 18.02 g
• 1 mole of H2O will contain 16.00 g of oxygen
and 2.02 g of hydrogen
26.98 g
= 9.99 g Al
1 mol Al
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
1 mol Al
= 0.370mol Al
6.02 x 1023 atoms
Molar Mass
Compute the number of moles
and mass of 2.23 x 1023 atoms of Al
“ Use the Periodic Table to determine the
mass of 1 mole of Al
1 mole Al = 26.98 g
” Use this as a conversion factor for
moles-to-grams
0.370 mol Al x
2.23x 1023 Al atoms x
17
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
18
•
Percent Composition
Percent Composition
Percentage of each element in a compound
• Percent Composition
– By mass
– The percentage of the mass of a compound represented by each of its
constituent elements.
• Can be determined from
° the formula of the compound or
± the experimental mass analysis of the
compound
• The percentages may not always total to 100%
due to rounding
• According to the law of constant composition, any sample of a
pure compound always consists of the same elements combined
in the same proportion by mass.
• Example: What is the percent composition of N and H in
ammonia (NH3)?
part
Percentage =
× 100%
whole
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
Mass Percent N in NH3 =
Mass of N in 1 mol of NH3
14.01 g N
=
× 100 = 82.27%
Mass of 1 mol of NH3
17.03 g NH3
Mass Percent H in NH3 =
Mass of H in 1 mol of NH3 3(1.008) g H
=
× 100 = 17.76%
Mass of 1 mol of NH3
17.03 g NH3
19
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
Example
•
Empirical and Molecular Formulas
1.
x mol A
y mol B
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
Assume you have 100 g of the compound therefore your percentages are the number of grams of C, H,
and O.
–
–
–
3.
AxBy
21
Hydrated Compounds
• If ionic compounds are prepared water solution and then
isolated as solids, the crystals often have molecules of
water trapped in the lattice.
• Compounds in which molecules of water are associated
with the ions of the compound are called hydrated
compounds.
• Examples:
– NiCl2 • 6H2O
– CaSO4 • 2H2O
73.14 g of C
7.37 g of H
19.49 g of O
Change the grams into moles.
•
•
•
4.
73.14 g of C × 1 mole of C / 12.011g of C = 6.09 mols of C
7.37 g of H × 1 mole of H / 1.0079 g of H = 7.31 mols of H
19.49 g of O × 1 mole of O / 15.9994 g of O = 1.22 mols of O
Find the mole ratios (divide the large amount(s) of moles by the smallest amount of moles).
•
•
5.
6.09 mols of C / 1.22 mols of O = 5 mols of C to 1 mol of O
7.37 mols of H / 1.22 mols of O = 6 mols of H to 1 mol of O
So what does this mean?
•
6.
C5H6O is the empirical formula
How to get the molecular formula?
•
Ratio gives
formula
100% - (73.14% + 7.37%) = 19.49%
How many grams of C, H, and O?
•
• From percent compositions we can determine
empirical and molecular formulas by the
following procedure.
Find mole
ratio
What is the % of oxygen?
•
2.
– A molecular formula showing the simplest
possible ratio of atoms in a molecule.
% A → g A → x mols A
% B → g B → x mols B
Eugenol is the active component of oil of cloves. It has a molar mass of 164.2 g/mol and
is 73.14% C and 7.37% H; the remainder is oxygen. What are empirical and molecular
formulas for eugenol?
Where to start?!
•
• We have discussed molecular formulas
• Empirical formulas
20
Molecular weight of the empirical formula.
–
82 g/mol
•
Divide the molecular weight of eugenol by the molecular weight empirical formula.
•
So there are 2 units of the empirical formula.
–
(C5H6O)2 = C10H12O2
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
–
164.2 g/mol / 82 g/mol = 2
22
Determining the Units of Hydration
• Example: The following reaction takes place.
1.023 g of CuSO4 · x H2O + heat → 0.654 g CuSO4 + ? g H2O
• Find the units of hydration.
• Step 1: determine the mass of water
1.023 g of CuSO4 · x H2O – 0.654 g of CuSO4 = 0.369 g of water
• Step 2: determine the number of moles
0.369 g of water × 1 mol of water / 18.02 g of water = 0.0205 mols of water
0.654 g CuSO4 × 1 mols CuSO4 / 159.6 g CuSO4 = 0.00410 mols CuSO4
• Step 3: determine the molar ratio
• But what about the molar mass? (Do I add the water?)
– YES!
– NiCl2 • 6H2O → 237.7 g/mol
• So, we know that we started with 1.023 g of
CuSO4 · 5 H2O
• 129.6 g/mol from NiCl2
• 108.1 g/mol from 6H2O
– CaSO4 • 2H2O → 172.14 g/mol
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
0.0205 mol H2 O
5.00 mol H2 O
=
0.00410 mol CuSO 4 1.00 mol CuSO 4
23
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
24
Example #4
Example #4
Determine the Percent Composition
from the Formula C2H5OH
‘ Determine the mass of each element in 1
mole of the compound
Determine the Percent Composition
from the Formula C2H5OH
“ Divide the mass of each element by the molar
mass of the compound and multiply by 100%
24.02g
× 100% = 52.14%C
46.07g
6.048g
× 100% = 13.13%H
46.07g
16.00g
× 100% = 34.73%O
46.07g
2 moles C = 2(12.01 g) = 24.02 g
6 moles H = 6(1.008 g) = 6.048 g
1 mol O = 1(16.00 g) = 16.00 g
’ Determine the molar mass of the
compound by adding the masses of the
elements
1 mole C2H5OH = 46.07 g
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
25
• The simplest, whole-number ratio of atoms in a
molecule is called the Empirical Formula
• The Molecular Formula is a multiple of the
Empirical Formula
%B
100g
mass A (g)
mass B (g)
MMA
MMB
factors of 20 = (10 x 2), (5 x 4)
factors of 12 = (6 x 2), (4 x 3)
GCF = 4
moles A
moles A
moles B
moles B
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
27
Example #6
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
‘ Convert the percentages to grams by
assuming you have 100 g of the compound
– Step can be skipped if given masses
47gC
= 47gC
100g
47gO
6.0gH
100g ×
= 47gO
100g ×
= 6.0gH
100g
100g
100g ×
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
Determine the Empirical Formula of
Benzopyrene, C20H12
‘ Find the greatest common factor (GCF) of the
subscripts
– can be determined from percent composition or
combining masses
100g
26
Example #5
Empirical Formulas
%A
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
29
’ Divide each subscript by the GCF to get the
empirical formula
C20H12 = (C5H3)4
Empirical Formula = C5H3
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
28
Example #6
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
’ Convert the grams to moles
47g C ×
1 mol C
= 3.9 mol C
12.01g
1 mol H
= 6.0 mol H
1.008g
1 mol O
47 g O ×
= 2.9 mol O
16.00g
6.0 g H ×
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
30
Example #6
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
“ Divide each by the smallest number of moles
Example #6
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
” If any of the ratios is not a whole number, multiply
all the ratios by a factor to make it a whole number
– If ratio is ?.5 then multiply by 2; if ?.33 or ?.67 then
multiply by 3; if ?.25 or ?.75 then multiply by 4
3.9 mol C ÷ 2.9 = 1.3
6.0 mol H ÷ 2.9 = 2
Multiply all the
Ratios by 3
Because C is 1.3
2.9 mol O ÷ 2.9 = 1
3.9 mol C ÷ 2.9 = 1.3 x 3 = 4
6.0 mol H ÷ 2.9 = 2 x 3 = 6
2.9 mol O ÷ 2.9 = 1 x 3 = 3
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
31
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
32
Example #6
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
° Use the ratios as the subscripts in the
empirical formula
3.9 mol C ÷ 2.9 = 1.3 x 3 = 4
6.0 mol H ÷ 2.9 = 2 x 3 = 6
2.9 mol O ÷ 2.9 = 1 x 3 = 3
C4H6O3
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
33
Molecular Formulas
• The molecular formula is a multiple of the
empirical formula
• To determine the molecular formula you
need to know the empirical formula and the
molar mass of the compound
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
34
Example #7
Example #7
Determine the Molecular Formula of Benzopyrene
if it has a molar mass of 252 g and an
empirical formula of C5H3
‘ Determine the empirical formula
Determine the Molecular Formula of Benzopyrene
if it has a molar mass of 252 g and an
empirical formula of C5H3
“ Divide the given molar mass of the
compound by the molar mass of the
empirical formula
– Round to the nearest whole number
• May need to calculate it as previous
C5H3
’ Determine the molar mass of the empirical
formula
5 C = 60.05 g, 3 H = 3.024 g
C5H3 = 63.07 g
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
35
252 g
=4
63.07 g
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
36
Example #7
Determine the Molecular Formula of Benzopyrene
if it has a molar mass of 252 g and an
empirical formula of C5H3
” Multiply the empirical formula by the
calculated factor to give the molecular
formula
(C5H3)4 = C20H12
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
37
Homework
• Questions and Problems:
• 6, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32,
34, 36, 38, 40, 42, 44, 46, 48, 50, 54, 56, 58,
60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82,
84
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
38