Download PP 17 The mole from book

Chapter 3
Stoichiometry
Chapter 3
Chemical Stoichiometry
• Stoichiometry – The study of quantities of materials
consumed and produced in chemical reactions.
Copyright © Cengage Learning. All rights reserved
2
Section 3.1
Counting by Weighing
•
•
Need average mass of the object.
Objects behave as though they were all
identical.
Return to TOC
Copyright © Cengage Learning. All rights reserved
3
Section 3.1
Counting by Weighing
Exercise
A pile of marbles weigh 394.80 g. 10 marbles
weigh 37.60 g. How many marbles are in the
pile?
Avg. Mass of 1 Marble =
37.60 g
= 3.76 g / marble
10 marbles
394.80 g = 105 marbles
3.76 g
Return to TOC
Copyright © Cengage Learning. All rights reserved
4
Section 3.2
Atomic Masses
Counting
by Weighing
•
•
Elements occur in nature as mixtures of
isotopes.
Carbon = 98.89% 12C
1.11% 13C
< 0.01% 14C
Return to TOC
Copyright © Cengage Learning. All rights reserved
5
Section 3.2
Atomic Masses
Counting
by Weighing
Average Atomic Mass for Carbon
98.89% of 12 amu + 1.11% of 13.0034 amu =
exact number
(0.9889)(12 amu) + (0.0111)(13.0034 amu) =
12.01 amu
Return to TOC
Copyright © Cengage Learning. All rights reserved
6
Section 3.2
Atomic Masses
Counting
by Weighing
Average Atomic Mass for Carbon
•
•
Even though natural carbon does not
contain a single atom with mass 12.01, for
stoichiometric purposes, we can consider
carbon to be composed of only one type of
atom with a mass of 12.01.
This enables us to count atoms of natural
carbon by weighing a sample of carbon.
Return to TOC
Copyright © Cengage Learning. All rights reserved
7
Section 3.2
Atomic Masses
Counting
by Weighing
Schematic Diagram of a Mass Spectrometer
Return to TOC
Copyright © Cengage Learning. All rights reserved
8
Section 3.2
Atomic Masses
Counting
by Weighing
Exercise
An element consists of 62.60% of an isotope
with mass 186.956 amu and 37.40% of an
isotope with mass 184.953 amu.
• Calculate the average atomic mass and
identify the element.
186.2 amu
Rhenium (Re)
Return to TOC
Copyright © Cengage Learning. All rights reserved
9
Section 3.3
The Mole by Weighing
Counting
•
•
•
The number equal to the number of carbon
atoms in exactly 12 grams of pure 12C.
1 mole of anything = 6.022 x 1023 units of that
thing (Avogadro’s number).
1 mole C = 6.022 x 1023 C atoms = 12.01 g C
Return to TOC
Copyright © Cengage Learning. All rights reserved
10
Section 3.3
The Mole by Weighing
Counting
Concept Check
Calculate the number of iron atoms in a 4.48
mole sample of iron.
2.70×1024 Fe atoms
Return to TOC
Copyright © Cengage Learning. All rights reserved
11
Section 3.4
Molar Mass
•
Mass in grams of one mole of the substance:
Molar Mass of N = 14.01 g/mol
Molar Mass of H2O = 18.02 g/mol
(2 × 1.008 g) + 16.00 g
Molar Mass of Ba(NO3)2 = 261.35 g/mol
137.33 g + (2 × 14.01 g) + (6 × 16.00 g)
Return to TOC
Copyright © Cengage Learning. All rights reserved
12
Section 3.4
Molar Mass
Concept Check
Which of the following is closest to the average
mass of one atom of copper?
a)
b)
c)
d)
e)
63.55 g
52.00 g
58.93 g
65.38 g
1.055 x 10-22 g
Return to TOC
Copyright © Cengage Learning. All rights reserved
13
Section 3.4
Molar Mass
Concept Check
Calculate the number of copper atoms in a
63.55 g sample of copper.
6.022×1023 Cu atoms
Return to TOC
Copyright © Cengage Learning. All rights reserved
14
Section 3.4
Molar Mass
Concept Check
Which of the following 100.0 g samples
contains the greatest number of atoms?
a) Magnesium
b) Zinc
c) Silver
Return to TOC
Copyright © Cengage Learning. All rights reserved
15
Section 3.4
Molar Mass
Exercise
Rank the following according to number of
atoms (greatest to least):
a) 107.9 g of silver
b) 70.0 g of zinc
c) 21.0 g of magnesium
b)
a)
c)
Return to TOC
Copyright © Cengage Learning. All rights reserved
16
Section 3.4
Molar Mass
Exercise
Consider separate 100.0 gram samples of
each of the following:
H2O, N2O, C3H6O2, CO2
 Rank them from greatest to least number
of oxygen atoms.
H2O, CO2, C3H6O2, N2O
Return to TOC
Copyright © Cengage Learning. All rights reserved
17
Section 3.5
Learning to Solve Problems
Conceptual Problem Solving
•
Where are we going?

•
How do we get there?

•
Read the problem and decide on the final
goal.
Work backwards from the final goal to decide
where to start.
Reality check.

Does my answer make sense? Is it
reasonable?
Return to TOC
Copyright © Cengage Learning. All rights reserved
18
Section 3.6
Percent Composition of Compounds
•
Mass percent of an element:
mass of element in compound
mass % =
× 100%
mass of compound
•
For iron in iron(III) oxide, (Fe2O3):
2( 55.85 g)
111.70 g
mass % Fe =
=
× 100% = 69.94%
2( 55.85 g)+ 3( 16.00 g) 159.70 g
Return to TOC
Copyright © Cengage Learning. All rights reserved
19
Section 3.6
Percent Composition of Compounds
Exercise
Consider separate 100.0 gram samples of
each of the following:
H2O, N2O, C3H6O2, CO2
 Rank them from highest to lowest percent
oxygen by mass.
H2O, CO2, C3H6O2, N2O
Return to TOC
Copyright © Cengage Learning. All rights reserved
20
Section 3.7
Determining the Formula of a Compound
Formulas
•
Empirical formula = CH
 Simplest whole-number ratio
• Molecular formula = (empirical formula)n
[n = integer]
• Molecular formula = C6H6 = (CH)6
 Actual formula of the compound
Return to TOC
Copyright © Cengage Learning. All rights reserved
21
Section 3.7
Determining the Formula of a Compound
Analyzing for Carbon and Hydrogen
•
Device used to determine the mass
percent of each element in a compound.
Return to TOC
Copyright © Cengage Learning. All rights reserved
22
Section 3.7
Determining the Formula of a Compound
Exercise
The composition of adipic acid is 49.3% C,
6.9% H, and 43.8% O (by mass). The molar
mass of the compound is about 146 g/mol.
 What is the empirical formula?
C3H5O2
 What is the molecular formula?
C6H10O4
Return to TOC
Copyright © Cengage Learning. All rights reserved
23