chapter 4
DESCRIPTION
Chapter 4. Recycling and Chemical Mathematics. Nature’s Recycling – The carbon cycle. Chemical Equations. Chemical reaction: the reorganization of the atoms in one or more substances to form a different substance(s). A chemical equation for respiration: - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/1.jpg)
1
Chapter 4Chapter 4
Recycling and Chemical MathematicsRecycling and Chemical Mathematics
![Page 2: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/2.jpg)
2
Nature’s Recycling –The carbon cycle
![Page 3: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/3.jpg)
3
Chemical Equations
Chemical reaction: the reorganization of the atoms in one or more substances to form a different substance(s).
A chemical equation for respiration:
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)
What does this equation mean?What does this equation mean?
![Page 4: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/4.jpg)
4
What does this equation mean?What does this equation mean?
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)
“One molecule of glucose solid reacts with six molecules of oxygen gas to give (or produce) six molecules of carbon dioxide gas and six molecules of water.”
![Page 5: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/5.jpg)
5
Accounting for atoms in the process of respiration
![Page 6: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/6.jpg)
6
Example:Example: Recognizing Balanced Equations
Is the following equation balanced?
CaCl2 + 2Na3PO4 → Ca3(PO4)2 + 3NaCl
![Page 7: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/7.jpg)
7
Solution:Solution: Recognizing Balanced Equations
This is the balanced chemical equation.
33CaCl2 + 22Na3PO4 → Ca3(PO4)2 + 66NaCl
![Page 8: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/8.jpg)
8
Producing Aluminum via the Hall-Heroult Process
2Al2O3(l) + 3C(s) → 4Al(l) + 3CO2(g)
![Page 9: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/9.jpg)
9
Example:Example: Atomic Masses
The mass of an atom is the weighted average of its naturally-occurring isotopes.Gallium has 2 isotopes,
60.10% 69Ga (= 68.93 amu) and
39.90% 71Ga (= 70.92 amu).
What is the average atomic massaverage atomic mass of Ga?
![Page 10: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/10.jpg)
10
Solution:Solution: Atomic Masses
60.10% 69Ga (= 68.93 amu) and
39.90% 71Ga (= 70.92 amu).
What is the average atomic massaverage atomic mass of Ga?
(0.6010 x 68.93) + (0.3990 x 70.92) = contribution of 69Ga isotope contribution of 71Ga isotope
69.72 amu69.72 amu
![Page 11: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/11.jpg)
11
Formula Masses
The sum of the atomic masses of all the atoms in the formula.
H2O = 2 x 1.008 + 1 x 16.00 = 18.02 amu hydrogen oxygen
Fe(OH)3 = 1 x 55.847 + 3 x (16.00 + 1.008) = 106.87 amu iron oxygen hydrogen
![Page 12: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/12.jpg)
12
Avogadro’s Number
1 mol of anything = 6.02 x 1023 things 1 mol of C atoms = 6.02 x 1023 C atoms 1 mol of Al atoms = 6.02 x 1023 Al atoms
1 mol of Al2O3 = 6.02 x 1023 Al2O3 formula units 1 mol of frosted flakes =
6.02 x 1023 frosted flakes (Yum!)(Yum!)
![Page 13: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/13.jpg)
13
Conversions: Moles to Grams
How many grams of Al2O3 are in 31.27 moles of Al2O3?
= 3188 g Al= 3188 g Al22OO33
32
3232 O Almol 1
O Alg 101.96 x O Almol 31.27
![Page 14: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/14.jpg)
14
Conversions: Grams to Moles
How many moles of (NH4)2Cr2O7 are in 586.2 grams of (NH4)2Cr2O7?
= 2.326 moles of NH= 2.326 moles of NH44CrCr22OO77
7224
72247224 OCr)(NH g 06252
OCr)(NH mol 1 x OCr)(NH g 586.2
.
![Page 15: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/15.jpg)
15
Using Equations for Stoichiometric Calculations
How many moles of iron can be formed from 2.8 moles of iron(III) oxide, Fe2O3?
Fe2O3 + 2Al → 2Fe + Al2O3
![Page 16: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/16.jpg)
16
Solution:Solution: Using Equations for Stoichiometric Calculations
How many moles of iron can be formed from 2.8 moles of iron(III) oxide, Fe2O3?
Fe2O3 + 2Al → 2Fe + Al2O3
= 5.6 moles Fe= 5.6 moles Fe
3232 OFe mol 1
Fe mol 2 x OFe mol 2.8
![Page 17: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/17.jpg)
17
Using Equations for Stoichiometric Calculations
How many grams of Al2O3 can be formed from 1162 grams of Fe2O3?
Fe2O3 + 2Al → 2Fe + Al2O3
![Page 18: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/18.jpg)
18
Using Equations for Stoichiometric Calculations
How many grams of Al2O3 can be formed from 1162 grams of Fe2O3?
Fe2O3 + 2Al → 2Fe + Al2O3
StrategyStrategy
g Fe2O3 → mol Fe2O3 → mol Al2O3 → g Al2O3
![Page 19: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/19.jpg)
19
Solution:Solution: Using Equations for Stoichiometric Calculations
How many grams of Al2O3 can be formed from 1162 grams of Fe2O3?
g Fe2O3 → mol Fe2O3 → mol Al2O3 → g Al2O3
= 741.8 g Al= 741.8 g Al22OO33
32
32
32
32
32
3232 O Almol 1
O Alg 101.96 x
OFe mol 1
O Almol 1 x
OFe g 159.7
OFe mol 1 x OFe g 1162
![Page 20: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/20.jpg)
20
Using Equations for Stoichiometric Calculations
How many grams of CO2 can be formed from 80.7 grams of C6H12O6 in respiration? Assume excess oxygen.
C6H12O6 + 6O2 → 6CO2 + 6H2O
![Page 21: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/21.jpg)
21
Using Equations for Stoichiometric Calculations
How many grams of CO2 can be formed from 80.7 grams of C6H12O6 in respiration? Assume excess oxygen.
C6H12O6 + 6O2 → 6CO2 + 6H2O
StrategyStrategy
g C6H12O6 → mol C6H12O6 → mol CO2 → g CO2
![Page 22: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/22.jpg)
22
Solution:Solution: Using Equations for Stoichiometric Calculations
How many grams of Al2O3 can be formed from 1162 grams of Fe2O3?
g C6H12O6 → mol C6H12O6 → mol CO2 → g CO2
= 118 g CO= 118 g CO22
2
2
6126
2
6126
61266126 CO mol 1
CO g 44.01 x
OHC mol 1
CO mol 6 x
OHC g 180.2
OHC mol 1 x OHC g 80.7
![Page 23: Chapter 4](https://reader036.vdocuments.us/reader036/viewer/2022062517/568138d7550346895da09034/html5/thumbnails/23.jpg)
23
Common Polymers and Their Monomers