Download Chapter7.1

Chapter 7
7.1
7.2
7.3
7.4
Chemical Quantities
The Mole
Molar Mass
Calculations Using Molar Mass
Percent Composition and Empirical
Formulas
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
1
7.1 A Mole



A mole contains 6.02 x 1023 particles (atoms,
ions, molecules, formula unit)
The number 6.02 x 1023 is known as Avogadro’s
number.
One mole of any element contains Avogadro’s
number of atoms.
1 mole Na = 6.02 x 1023 Na atoms
1 mole Au = 6.02 x 1023 Au atoms
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
2
A Mole of Molecules

One mole of a covalent compound contains
Avogadro’s number of molecules.
1 mole CO2 = 6.02 x 1023 CO2 molecules
1 mole H2O = 6.02 x 1023 H2O molecules

One mole of an ionic compound contains
Avogadro’s number of formula units.
1 mole NaCl = 6.02 x 1023 NaCl formula units
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
3
Samples of One Mole Quantities
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
4
Avogadro’s Number


Avogadro’s number is written as conversion
factors.
6.02 x 1023 particles and
1 mole
1 mole
6.02 x 1023 particles
The number of molecules in 0.50 mole of CO2
molecules is calculated as
0.50 mole CO2 molecules x 6.02 x 1023 CO2 molecules
1 mole CO2 molecules
= 3.0 x 1023 CO2 molecules
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
5
Learning Check
A. Calculate the number of atoms in 2.0 moles
of Al.
B. Calculate the number of moles of S in 1.8 x
1024 S.
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
6
Solution
A. Calculate the number of atoms in 2.0 moles of Al.
2.0 moles Al x 6.02 x 1023 Al atoms
1 mole Al
=1.2 x 1024 Al atoms
B. Calculate the number of moles of S in 1.8 x 1024 S.
1.8 x 1024 S atoms x 1 mole S
6.02 x 1023 S atoms
= 3.0 mole S atoms
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
7
7.2 Molar Mass


The mass of
one mole is
called molar
mass (g/mole).
The molar
mass of an
element is the
atomic mass
expressed in
grams.
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
8
Learning Check
Give the molar mass to the nearest 0.1 g.
A. 1 mole of K atoms
=
________
B. 1 mole of Sn atoms =
________
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
9
Solution
Give the molar mass to the nearest 0.1 g.
A. 1 mole of K atoms
=
39.1 g
B. 1 mole of Sn atoms
=
118.7 g
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
10
Molar Mass of CaCl2

For a compound, the molar mass is the sum of the
molar masses of the elements in the formula. We
calculate the molar mass of CaCl2 to the nearest 0.1 g
as follows.

Formula mass of CaCl2 = [40.1 + 2(35.45)] = 111.1amu
Formula mass = molar mass, so

111.1amu = 111.1g/mol
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
11
Molar Mass of K3PO4
Determine the molar mass of K3PO4 to 0.1 g.
Molar Mass = [3(39.1) + 31.0 + 4(16)]
= 212.3g/mol
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
12
One-Mole Quantities
32.1 g
55.9 g
58.5 g
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
294.2 g
342.3 g
13
Learning Check
A. 1 mole of K2O
= ______g
B. 1 mole of antacid Al(OH)3
= ______g
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
14
Solution
A. 1 mole of K2O
2 (39.1) + 1 (16.0) = 94.2 g/mol
1mole K2O x 94.2g/mol = 94.2g
B. 1 mole of antacid Al(OH)3
1 (27.0) + 3 (16.0) + 3 (1.0) = 78.0 g/mol
1mole Al(OH)3 x 78.0 g/mol = 78.0g
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
15
Learning Check
Prozac, C17H18F3NO, is an antidepressant that
inhibits the uptake of serotonin by the brain.
What is the molar mass of Prozac?
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
16
Solution
Prozac, C17H18F3NO, is a widely used antidepressant
that inhibits the uptake of serotonin by the brain.
What is the molar mass of Prozac?
17C (12.0) + 18H (1.0) + 3F (19.0) + 1N (14.0) + 1 O (16.0) =
204 + 18
+
57.0
+
14.0
+ 16.0
= 309 g/mole
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
17
Molar Mass Factors
Methane CH4 known as natural gas is used in
gas cook tops and gas heaters.
1 mole CH4 =
16.0 g
The molar mass of methane can be written as
conversion factors.
16.0 g CH4
and
1 mole CH4
1 mole CH4
16.0 g CH4
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
18
Learning Check
Acetic acid C2H4O2 gives the sour taste to vinegar.
Write two molar mass conversion factors for acetic
acid.
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
19
Solution
Acetic acid C2H4O2 gives the sour taste to vinegar.
Write two molar mass factors for acetic acid.
2(12.0) + 4(1.0) + 2(16) = 60.0g/mol
1 mole of acetic acid
1 mole acetic acid
60.0 g acetic acid
=
60.0 g acetic acid
and
60.0 g acetic acid
1 mole acetic acid
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
20
7.3 Calculations with Molar Mass
 Mole factors are used to convert between the
grams of a substance and the number of moles.
Grams
Mole factor
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
Moles
21
Calculating Grams from Moles
Aluminum is often used for the structure of
lightweight bicycle frames. How many grams
of Al are in 3.00 moles of Al?
3.00 moles Al x 27.0 g Al
1 mole Al
= 81.0 g Al
mole factor for Al
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
22
Learning Check
The artificial sweetener aspartame (Nutri-Sweet)
C14H18N2O5 is used to sweeten diet foods, coffee and
soft drinks. How many moles of aspartame are
present in 225 g of aspartame?
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
23
Solution
Calculate the molar mass of C14H18N2O5.
14 (12.0) + 18 (1.0) + 2 (14.0) + 5(16.0)
= 294 g/mole
Set up the calculation using a mole factor.
225 g aspartame x 1 mole aspartame
294 g aspartame
mole factor(inverted)
= 0.765 mole aspartame
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
24
7.4 Percent Composition


In a compound, the percent composition is the
percent by mass of each element in the formula.
In one mole of CO2 there are 12.0 g of C and
32.0 g of O (molar mass 44.0 g/mol),
12.0 g C
x 100
44.0 g CO2
32.0 g O
x 100
44.0 g CO2
=
27.3 % C
=
72.7 % O
100.0 %
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
25
Learning Check
What is the percent carbon in C5H8NNaO4
(MSG monosodium glutamate), a compound
used to flavor foods and tenderize meats?
1) 7.10 %C
2) 35.5 %C
3) 60.0 %C
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
26
Solution
2) 35.5 %C
Molar mass = 169.1 g
% = total g C
x 100
total g MSG
=
60.0 g C
x 100 = 35.5 % C
169.1 g MSG
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
27
Types of Formulas



The molecular formula is the true or actual
number of the atoms in a molecule.
The empirical formula is the simplest whole
number ratio of the atoms.
The empirical formula is calculated by dividing
the subscripts in the molecular formula by a
whole number to give the lowest ratio.
C5H10O5
5 = C1H2O1 = CH2O
molecular
formula
empirical formula
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
28
Some Molecular and Empirical
Formulas
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
29
Learning Check
A. What is the empirical formula for C4H8?
1) C2H4
2) CH2
3) CH
B. What is the empirical formula for C8H14?
1) C4H7
2) C6H12
3) C8H14
C. Which is a possible molecular formula for
CH2O?
1) C4H4O4 2) C2H4O2
3) C3H6O3
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
30
Solution
A. What is the empirical formula for C4H8?
2) CH2
C4H8
4
B. What is the empirical formula for C8H14?
1) C4H7
C8H14
2
C. Which is a possible molecular formula for
CH2O?
2) C2H4O2
3) C3H6O3
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
31
Learning Check
If the molecular formula has 4 atoms of N, what
is the molecular formula if SN is the empirical
formula? Explain.
1) SN
2) SN4
3) S4N4
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
32
Solution
If the molecular formula has 4 atoms of N, what
is the molecular formula if SN is the empirical
formula? Explain.
3) S4N4
If the molecular formula has 4 atoms of N, and
S and N are related 1:1, then there must also be
4 atoms of S.
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
33
Relating Empirical and Molecular
Formulas

A molecular formula is equal to or a multiple
of the empirical formula.

Thus, the molar mass is equal to or a multiple
of the empirical mass.
molar mass
= a whole number
empirical mass

Multiply the empirical formula by the whole
number to determine the molecular formula.
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
34
Finding the Molecular Formula
Determine the molecular formula of a compound
that has a molar mass of 78.0 and an empirical
formula of CH.
1. Empirical mass of CH = 13.0 g/mol
2. Divide the molar mass by the empirical mass.
3. 78.0 g/mol = 6.00
13.0 g/mol
4. Multiply the subscripts in CH by 6.
5. Molecular formula = (CH)6 = C6H6
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
35
Learning Check
A compound has a formula mass of 176.0 and an
empirical formula of C3H4O3. What is the
molecular formula?
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
36
Solution
A compound has a formula mass of 176.0 and an
empirical formula of C3H4O3. What is the molecular
formula?
C3H4O3 = 3(12.0) + 4(1.0) + 3(16.0) = 88.0 g/mol
176.0 g/mol (molar mass)
= 2.00
88.0 g/mol (empirical mass)
Molecular formula = 2 (empirical formula)
(C3H4O3 )2 = C6H8O6
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
37
Finding the Molecular Formula
A compound is C 24.27%, H 4.07%, and Cl
71.65%. The molar mass is known to be 99.0 g.
What are the empirical and molecular formulas?
1.
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
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
38
Finding the Molecular Formula
Continued
2. Calculate the number of moles of each element.
24.27 g C x 1 mole C
= 2.02 moles C
12.0 g C
4.07 g H
x 1 mole H
1.01 g H
= 4.03 moles H
71.65 g Cl x 1 mole Cl
35.5 g Cl
= 2.02 moles Cl
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
39
Finding the Molecular Formula
(continued)
3. Divide each by the smallest
2.02 moles C
=
1 mole C
2.02
4.03 moles H
=
2 moles H
2.02
2.02 moles Cl =
1 mole Cl
2.02
Empirical formula = C1H2Cl1 = CH2Cl
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
40
Finding the Molecular Formula
(continued)
4. Calculate empirical mass (EM)
empirical mass CH2Cl = 49.5 g/mol
5. Divide molar mass by empirical mass
Molar mass
= 99.0 g/mol = 2
Empirical mass
49.5 g/mol
6. Determine Molecular formula
(CH2Cl)2 = C2H4Cl2
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
41
Learning Check
Aspirin is 60.0% C, 4.5 % H and 35.5 % O.
Calculate its empirical (simplest) formula.
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
42
Solution (continued)
In 100 g of aspirin, there are 60.0 g C,
4.5 g H, and 35.5 g O.
60.0 g C x 1 mole C = 5.00 moles C
12.0 g C
4.5 g H
x
1 mole H =
1.01 g H
4.5 moles H
35.5 g O x
1mole O =
16.0 g O
2.22 moles O
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
43
Solution (continued)
Divide by the smallest number of moles.
5.00 moles C =
2.25 moles C
2.22
4.5 moles H
=
2.0 moles H
2.22
2.22 moles O =
1.00 mole O
2.22
Note that the results are not all whole
numbers. To obtain whole numbers,
multiply by a factor to give whole numbers.
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
44
Solution (continued)
Multiply each number of moles by 4
C: 2.25 moles C x 4 = 9 moles C
H: 2.0 moles H x 4 = 8 moles H
O: 1.00 mole O x 4 = 4 moles O
Use the whole numbers as subscripts to
obtain the simplest formula
C9H8O4
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
45
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?
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
46
Solution
In 100.0 g, there are 27.4 g S, 12.0 g N and
60.6 g Cl.
27.4 g S x 1 mole S = 0.854 mole S
32.1 g S
12.0 g N x
1 mole N = 0.857 moles N
14.0 g N
60.6 g Cl x
1mole Cl = 1.71 moles Cl
35.5 g Cl
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
47
Solution (continued)
Dividing by the smallest number of moles
0.854 mole S /0.854 = 1.00 mole S
0.857 mole N/0.854 = 1.00 mole N
1.71 moles Cl/0.854 = 2.00 moles Cl
Empirical formula = SNCl2 = 117.1 g/mol
Molar Mass/ Empirical mass
351
= 3
117.1
Molecular formula = (SNCl2)3 = S3N3Cl6
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
48