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Chapter 7
Chemical Quantities
The Mole
Atomic Mass and
Formula Mass
Copyright © 2008 by Pearson Education, Inc.
Publishing as Benjamin Cummings
The Mole
Copyright © 2008 by Pearson Education, Inc.
Publishing as Benjamin Cummings
Collection Terms
A collection term states a specific number of items.
• 1 dozen donuts = 12 donuts
• 1 ream of paper = 500 sheets
• 1 case = 24 cans
Copyright © 2008 by Pearson Education, Inc.
Publishing as Benjamin Cummings
A Mole of Atoms
A mole (mol) is a collection that contains
• The same number of particles as there are
carbon atoms in 12.01 g of carbon.
• 6.022 x 1023 atoms of an element (Avogadro’s
number).
1 mol C = 6.022 x 1023 C atoms
1 mol Na = 6.022 x 1023 Na atoms
1 mol Au = 6.022 x 1023 Au atoms
A Mole of A Compound
A mole
• Of a covalent compound has Avogadro’s number
of _________________.
1 mol CO2 = 6.022 x 1023 CO2 molecules
1 mol H2O = 6.022 x 1023 H2O molecules
• Of an ionic compound contains Avogadro’s
number of ___________ units.
1 mol NaCl = 6.022 x 1023 NaCl formula units
1 mol K2SO4 = 6.022 x 1023 K2SO4 formula units
Samples of One Mole
Quantities
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Avogadro’s Number
Avogadro’s number 6.022 x 1023 can be written as
an equality and two conversion factors.
As an equality:
1 mol particles = 6.022 x 1023 particles
As conversion Factors:
6.022 x 1023 particles and 1 mol particles
1 mol particles
6.022 x 1023 particles
Using Avogadro’s Number
Avogadro’s number
• Converts moles of a substance to
the number of particles.
How many Cu atoms are in 0.50 mol
Cu?
0.50 mol Cu x 6.022 x 1023 Cu atoms
1 mol Cu
= 3.0 x 1023 Cu atoms
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Using Avogadro’s Number
Avogadro’s number
• Converts particles of a substance
to moles.
How many moles of CO2 are
2.50 x 1024 CO2 molecules?
2.50 x 1024 CO2 x 1 mol CO2
6.022 x 1023 CO2
= 4.15 mol CO2
Learning Check
The number of atoms in 2.0 mol Al is
2.0 mol Al x 6.022 x 1023 Al atoms =
1 mol Al
1.2 x 1024 Al atoms
Learning Check
The number of moles of S in 1.8 x 1024 atoms S is
1.8 x 1024 S atoms x
1 mol S
6.022 x 1023 S atoms
3.0 mol S atoms
=
Subscripts and Moles
The subscripts in a formula state
• The relationship of atoms in the formula.
• The moles of each element in 1 mol of
compound.
Glucose
C6H12O6
In 1 molecule: 6 atoms C 12 atoms H
In 1 mol:
6 mol C
12 mol H
6 atoms O
6 mol O
Subscripts State Atoms and
Moles
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1 mol C9H8O4
=
9 mol C 8 mol H
4 mol O
Factors from Subscripts
Subscripts used for conversion factors
• Relate moles of each element in 1 mol compound.
• For aspirin C9H8O4 can be written as:
9 mol C
1 mol C9H8O4
8 mol H
1 mol C9H8O4
4 mol O
1 mol C9H8O4
and
1 mol C9H8O4
9 mol C
1 mol C9H8O4
8 mol H
1 mol C9H8O4
4 mol O
Learning Check
How many moles O are in 0.150 mol aspirin
C9H8O4?
0.150 mol C9H8O4 x 4 mol O
= 0.600 mol O
1 mol C9H8O4
subscript factor
Learning Check
How many O atoms are in 0.150 mol aspirin C9H8O4?
0.150 mol C9H8O4 x 4 mol O x 6.022 x 1023 O atoms
1 mol C9H8O4 1 mol O
subscript
factor
= 3.61 x 1023 O atoms
Avogadro’s
number
Atomic Mass
Atomic mass is the
• Mass of a single atom in
atomic mass units (amu).
• Mass of an atom
compared to a 12C atom.
• Number below the
symbol of an element.
(the average atomic
mass)
Periodic Table and Atomic Mass
Ag has
atomic
mass =
107.9 amu
S has
atomic mass
= 32.07 amu
C has atomic mass
= 12.01 amu
Atomic Mass Factors
The atomic mass
• Can be written as an equality.
Example: 1 P atom = 30.97 amu
• Can be written as two ___________ ________.
Example: 1 P atom and 30.97 amu
30.97 amu
1 P atom
Uses of Atomic Mass Factors
The atomic mass factors are used to convert
• A specific number of atoms to mass (amu).
• An amount in amu to ___________ of atoms.
Using Atomic Mass Factors
1. What is the mass in amu of 75 P atoms?
75 P atoms x 30.97 amu = 2323 amu (2.323 x103 amu)
1 P atom
2. How many Cu atoms have a mass of 4.500 x 105
amu?
4.500 x 105 amu x
1 Cu atom = 7081 Cu atoms
63.55 amu
Learning Check
What is the mass in amu of 75 silver atoms?
75 Ag atoms x 107.9 amu = 8093 amu
1 Ag atom
Learning Check
How many gold atoms have a mass of
1.85 x 105 amu?
1.85 x 105 amu x 1 Au atom = 939 Au atoms
197.0 amu
Formula Mass
The formula mass is
• The mass in amu of a compound.
• The _____ of the atomic masses of the
elements in a formula.
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Calculating Formula Mass
To calculate the formula mass of Na2SO4
• Multiply the atomic mass of each element by its
subscript, then total the masses of the atoms.
2 Na x 22.99 amu = 45.98 amu
1 Na
1 S x 32.07 amu = 32.07 amu
1S
4 O x 16.00 amu = 64.00 amu
1O
142.05 amu
Learning Check
Using the periodic table, calculate the formula
mass of aluminum sulfide Al2S3.
2 Al x 26.98 amu = 53.96 amu
1 Al
3 S x 32.07 amu = 96.21 amu
1S
150.17 amu
Molar Mass
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Molar Mass
The molar mass
• Is the mass of one
mol of an element or
compound.
• Is the atomic mass
expressed in grams.
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Molar Mass from the Periodic
Table
Molar mass
• Is the
atomic
mass
expressed
in grams.
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Publishing as Benjamin Cummings
Learning Check
Give the molar mass for:
A. 1 mol K atoms
=
B. 1 mol Sn atoms =
Molar Mass of a Compound
The molar mass of a compound is the sum of the
molar masses of the elements in the formula.
Example: Calculate the molar mass of CaCl2.
Element
Number
of Moles
Atomic Mass Total Mass
Ca
1
40.08 g/mol
40.08 g
Cl
2
35.45 g/mol
70.90 g
CaCl2
110.98 g
Molar Mass of K3PO4
Calculate the molar mass of K3PO4.
Element
Number
of Moles
Atomic Mass Total Mass
in K3PO4
K
3
39.10 g/mol
117.3 g
P
1
30.97 g/mol
30.97 g
O
4
16.00 g/mol
64.00 g
K3PO4
212.3 g
Some One-Mol Quantities
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32.07 g
55.85 g
58.44 g
294.20 g
342.30 g
Learning Check
A. K2O 94.20 g/mol
2 mol K (39.10 g/mol)
78.20 g
= 94.20 g
+ 1 mol O (16.00 g/mol)
+ 16.00 g
B. Al(OH)3 78.00 g/mol
1 mol Al (26.98 g/mol) + 3 mol O (16.00 g/mol)
+ 3 mol H (1.008 g/mol)
26.98 g
+ 48.00 g + 3.024 g
= 78.00 g
Calculations Using
Molar Mass
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Molar Mass Factors
Molar mass conversion factors
• Are written from molar mass.
• Relate grams and moles of an element or
compound.
Example: Write molar mass factors for methane
CH4 used in gas cook tops and gas heaters.
Molar mass:
1 mol CH4 = 16.04 g
Conversion factors:
16.04 g CH4
and
1 mol CH4
1 mol CH4
16.04 g CH4
Learning Check
Acetic acid C2H4O2 gives the sour taste to vinegar.
Write two molar mass factors for acetic acid.
Calculate molar mass:
24.02 + 4.032 + 32.00 = 60.05 g/mol
1 mol of acetic acid =
Molar mass factors
1 mol acetic acid
60.05 g acetic acid
60.05 g acetic acid
and
60.05 g acetic acid
1 mol acetic acid
Calculations Using Molar Mass
Molar mass factors are used to convert between the
grams of a substance and the number of moles.
Grams
Molar mass factor
Moles
Moles to Grams
Aluminum is used to build lightweight bicycle
frames. How many grams of Al are 3.00 mol Al?
Molar mass equality: 1 mol Al = 26.98 g Al
Setup with molar mass as a factor:
3.00 mol Al x
26.98 g Al
1 mol Al
= 80.9 g Al
molar mass factor
for Al
Learning Check
Allyl sulfide C6H10S is a
compound that has the
odor of garlic. How many
moles of C6H10S are in
225 g C6H10S?
Action Plan: Calculate
the molar mass, and
convert 225 g to moles.
Grams
Molar mass factor
Moles
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Solution
Calculate the molar mass of C6H10S.
(6 x 12.01) + (10 x 1.008) + (1 x 32.07)
= 114.21 g/mol
Set up the calculation using a mole factor.
225 g C6H10S x 1 mol C6H10S
114.21 g C6H10S
molar mass factor(inverted)
= 1.97 mol C6H10S
Grams, Moles, and Particles
A molar mass factor and Avogadro’s number
convert
• Grams to particles
Avogadro’s
number
molar mass
(g
mol
• Particles to grams
particles)
Avogadro’s
number
(particles
molar mass
mol
g)
Learning Check
How many H2O molecules are in 24.0 g H2O?
A) 4.52 x 1023
B) 1.44 x 1025
C) 8.02 x 1023
Learning Check
How many H2O molecules are in 24.0 g H2O?
24.0 g H2O x 1 mol H2O x 6.022 x 1023 H2O molecules
18.02 g H2O
1 mol H2O
= 8.02 x 1023 H2O molecules
Learning Check
If the odor of C6H10S can be detected from
2 x 10-13 g in one liter of air, how many molecules
of C6H10S are present?
2 x 10-13 g x
1 mol
x 6.022 x 1023 molecules
114.21 g
1 mol
= 1 x 109 molecules C6H10S
Percent
Composition
and
Empirical
Formulas
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Percent Composition
Percent composition
• Is the percent by mass of each element in a
formula.
Example: Calculate the percent composition of CO2.
CO2 = 1 C(12.01g) + 2 O(16.00 g) = 44.01 g/mol)
12.01 g C
44.01 g CO2
x 100
=
27.29 % C
32.00 g O
x 100
44.01 g CO2
=
72.71 % O
Learning Check
What is the percent composition of lactic acid,
C3H6O3, a compound that appears in the blood
after vigorous activity?
→
Solution
STEP 1
C 3 H 6 O3
3C(12.01) + 6H(1.008) + 3O(16.00) = 90.08 g/mol
36.03 g C + 6.048 g H + 48.00 g O = 90.08 g/mol
STEP 2
%C = 36.03 g C x 100 = 40.00% C
90.08 g cpd
%H = 6.048 g H x 100 = 6.714% H
90.08 g cpd
%O = 48.00 g O x 100 = 53.29% O
90.08 g cpd
Learning Check
The chemical isoamyl
acetate C7H14O2
contributes to the odor of
pears. What is the
percent carbon in
isoamyl acetate?
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Solution
C7H14O2
Molar mass C7H14O2 = 7C(12.01) + 14H(1.008)
+ 2O(16.00) = 130.18 g/mol
Total C = 7C(12.01) = g
%C
= total g C
total g cpd
x 100
%C
= 84.07 g C x 100 = 64.58 % C
130.18 g cpd
Empirical Formulas
The empirical formula
• Is the simplest whole number ratio of the
atoms.
• Is calculated by dividing the subscripts in the
actual (molecular) formula by a whole number
to give the lowest ratio.
C5H10O5  5 = C1H2O1 = CH2O
actual (molecular)
empirical formula
formula
Some Molecular and Empirical
Formulas
• The molecular formula is the same or a multiple of
the empirical.
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Learning Check
1. What is the empirical formula for _________?
A) C2H4
B) CH2
C) CH
2. What is the empirical formula for _________?
A) C4H7
B) C6H12
C) C8H14
3. Which is a possible molecular formula for ______?
A) C4H4O4
B) C2H4O4 C) C3H6O3
Learning Check
A compound contains 7.31 g Ni and 20.0 g Br.
Calculate its empirical (simplest) formula.
Lets ask, what does the empirical formula tell us?
The ratio of atoms of each element,
ratio of moles of each element.
Action Plan: Convert mass (g) to moles.
Solution
Convert 7.31 g Ni and 20.0 g Br to moles.
7.31 g Ni x 1 mol Ni
= 0.125 mol Ni
58.69 g Ni
20.0 g Br x 1 mol Br
= 0.250 mol Br
79.90 g Br
Divide by smallest:
0.125 mol Ni = 1 Ni
0.250 mol Br = 2 Br
0.125
0.125
Write ratio as subscripts: NiBr2
Converting Decimals to Whole
Numbers
When the number of moles for an element is a
decimal, all the moles are multiplied by a small
integer to obtain whole number.
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Learning Check
Aspirin is 60.0% C, 4.5 % H and 35.5 % O.
Calculate its empirical (simplest) formula.
We have percentages, not grams.
However, the percentages are really an expression of
grams of an element in 100. (exact) grams of
compound.
Solution
STEP 1. Calculate the moles of each element in
100. g of the compound.
100. g aspirin contains 60.0% C or 60.0 g C,
4.5% H or 4.5 g H, and 35.5% O or 35.5 g O.
60.0 g C x
1 mol C =
12.01 g C
5.00 mol C
4.5 g H
1 mol H =
1.008 g H
4.5 mol H
x
35.5 g O x
1mol O
16.00 g
=
O
2.22 mol O
Solution (continued)
STEP 2. Divide by the smallest number of mol.
5.00 mol C
2.22
4.5 mol H
2.22
2.22 mol O
2.22
=
2.25 mol C (decimal)
=
2.0 mol H
=
1.00 mole O
Solution (continued)
3. Use the lowest whole number ratio as subscripts
When the moles are not whole numbers, multiply
by a factor to give whole numbers, in this case x 4.
C: 2.25 mol C
x 4 = 9 mol C
H: 2.0 mol H
x 4 = 8 mol H
O: 1.00 mol O
x 4 = 4 mol O
Using these whole numbers as subscripts the
simplest formula is
C9H8O4
Molecular Formulas
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Relating Molecular and
Empirical Formulas
A molecular formula
• Is a multiple (or equal) of its empirical formula.
• Has a molar mass that is the empirical formula
mass multiplied by a whole number.
molar mass
= a whole number
empirical mass
• Is obtained by multiplying the empirical formula by
a whole number.
Diagram of Molecular and
Empirical Formulas
A small integer links
• A molecular formula and its empirical
formula.
• A molar mass and its empirical formula
mass.
Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings
Finding the Molecular Formula
Determine the molecular formula of compound that
has a molar mass of 78.11 g and an empirical
formula of CH.
STEP 1. Empirical formula mass of CH = 13.02 g
STEP 2. Divide the molar mass by the empirical
mass.
78.11 g = 5.999 ~ 6
13.02 g
STEP 3. Multiply each subscript in C1H1 by 6.
molecular formula = C1x 6 H1 x 6 = C6H6
Some Compounds with
Empirical Formula CH2O
formaldehyde
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Learning Check
A compound has a molar mass of 176.1g and an
empirical formula of C3H4O3. What is the
molecular formula?
What do we need to do?
From the empirical formula, determine the
empirical mass.
Then determine the whole number integer.
Solution
A compound has a formula mass of 176.1 and an
empirical formula of C3H4O3. What is the
molecular formula?
C3H4O3 = 88.06 g/EF
176.1 g (molar mass)
88.06 g (empirical mass)
= 2.00
Molecular formula = 2 x empirical formula
C3 x 2H4 x 2O3 x 2 = C6H8O6
Molecular Formula
A compound contains C 24.27%, H 4.07%, and Cl
71.65%. The molar mass is about 99 g. What are
the empirical and molecular formulas?
STEP 1. Calculate the empirical formula.
Write the mass percents as the grams in a
100.00-g sample of the compound.
C 24.27 g
H 4.07 g
Cl 71.65 g
Finding the Molecular Formula
(Continued)
Calculate the number of moles of each element.
24.27 g C x 1 mol C
12.01 g C
= 2.021 mol C
4.07 g H
= 4.04 mol H
x 1 mol H
1.008 g H
71.65 g Cl x 1 mol Cl
35.45 g Cl
= 2.021 mol Cl
Finding the Molecular Formula
(Continued)
Divide by the smallest number of moles:
2.021 mol C
=
1 mol C
2.021
4.04 mol H
=
2 mol H
2.021
2.02 mol Cl
=
1 mol Cl
2.021
Empirical formula = C1H2Cl1 = CH2Cl
Calculate empirical mass (EM) CH2Cl = 49.48 g
Finding the Molecular Formula
(Continued)
STEP 2. Divide molar mass by empirical mass.
Molar mass
= 99 g = 2
Empirical mass
49.48 g
STEP 3. Multiply the empirical formula by the small
integer to determine the molecular formula.
2 x (CH2Cl)
C1 x2 H2 x 2 Cl1x 2
=
C2H4Cl2
Learning Check
A compound is 27.4% S, 12.0% N and 60.6
% Cl. If the compound has a molar mass of
351 g, what is the molecular formula?
Solution
In 100. g, there are 27.4 g S, 12.0 g N, and 60.6 g Cl.
27.4 g S x 1 mol S = 0.854 mol S
32.07 g S
12.0 g N x
1 mol N = 0.857 mol N
14.01 g N
60.6 g Cl x
1mol Cl = 1.71 mol Cl
35.45 g Cl
Solution (continued)
Divide by the smallest number of moles
0.854 mol S /0.854
= 1.00 mol S
0.857 mol N/0.854
= 1.00 mol N
1.71 mol Cl/0.854
= 2.00 mol Cl
empirical formula = SNCl2 = 116.98 g
Molar Mass/ Empirical mass
351 g
=3
116.98 g
molecular formula = (SNCl2)3 = S3N3Cl6