Transcript continued

Unit 2 – Section C
Conserving Matter
HW 1
Read & take notes on Section C.1
C.1 – Keeping Track of Atoms
The law of conservation of matter – in a
chemical reaction matter is neither created
nor destroyed.
+
C
1 Carbon
atom (C)

O2
1 oxygen
molecule (O2)
CO2
1 carbon dioxide
molecule (CO2)
Molecules can be converted and decomposed by chemical
processes: but atoms are forever.
C.1 – Keeping Track of Atoms
(continued)
Reactants are placed on the left of the arrow;
Products are placed on the right.
+
C
1 Carbon
atom (C)

O2
1 oxygen
molecule (O2)
CO2
1 carbon dioxide
molecule (CO2)
In a balanced chemical equation, the number of atoms for left
side equals the number for the right side.
C.1 – Keeping Track of Atoms
(continued)
Coefficients indicate the relative number of
each unit involved.
+
Cu (s)
2 Copper
atoms (Cu)

O2 (g)
oxygen
molecule (O2)
CuO (s)
2 copper oxide
molecules (CuO)
C.1 – Keeping Track of Atoms
(continued)
Chemists use the term formula unit when
referring to the smallest unit of an ionic
compound.
+
Cu (s)
2 Copper
atoms (Cu)

O2 (g)
oxygen
molecule (O2)
CuO (s)
2 copper oxide
molecules (CuO)
Classwork
Answer questions 1-5 in Section C.2 pg 155
C.2 – Accounting for Atoms
1) Methane burning with oxygen
CH4 + 2 O2  CO2 + 2 H2O
Reactants
Products
C
C
H
H
O
O
C.2 – Accounting for Atoms
(continued)
2) Hydrobromic acid reacting with magnesium
HBr + Mg  H2 + MgBr2
Reactants
Products
H
H
Br
Br
Mg
Mg
C.2 – Accounting for Atoms
(continued)
3. Hydrogen sulfide and metallic silver react
4 Ag + 4 H2S + O2  2 Ag2S + 4 H2O
Reactants
Products
Ag
Ag
H
H
S
S
O
O
C.2 – Accounting for Atoms
(continued)
4. Cellulose burns to form carbon dioxide and
water vapor.
C6H10O5 + 6 O2  6 CO2 + 5 H2O
Reactants
Products
C
C
H
H
O
O
C.2 – Accounting for Atoms
(continued)
5. Nitroglycerin decomposes explosively to form
nitrogen, oxygen, carbon dioxide and water vapor.
2 C3H5(NO3) 3  3 N2 + O2 + 6 CO2 + 5 H2O
Reactants
Products
C
C
H
H
N
N
O
O
HW 2
Read & take notes on Section C.3
Address & answer questions 1-6 in section C.4
C.3 – Nature’s Conservation:
Balancing Chemical Equations
If polyatomic ions (examples NO3-, CO32-)
appear as both reactants and product treat
them as units.
If water is involved, balance the hydrogen and
oxygen atoms last.
Recount all atoms after you think an equation
is balanced.
C.4 – Writing Chemical Equations
Writing to balance the chemical equations…
Methane
Chlorine
Chloroform
Hydrogen
chloride
__ CH4 + __ Cl2  __ CHCl3 + __ HCl
Reactants
Products
C
C
H
H
Cl
Cl
C.4 – Writing Chemical Equations
(continued)
1a.
__ C + __ O2  __ CO
Reactants
Products
C
C
O
O
C.4 – Writing Chemical Equations
(continued)
1b.
__ Fe2O3 + __ CO  __ Fe + __ CO2
Reactants
Products
C
C
Fe
Fe
O
O
C.4 – Writing Chemical Equations
(continued)
2.
__ CuO + __ C  __ Cu + __ CO2
Reactants
Products
C
C
Cu
Cu
O
O
C.4 – Writing Chemical Equations
(continued)
3.
__ O3  __ O2
Reactants
Products
O
O
C.4 – Writing Chemical Equations
(continued)
4.
__ NH3 + __ O2  __ NO2 + __ H2O
Reactants
Products
N
H
O
N
H
O
C.4 – Writing Chemical Equations
(continued)
5.
__ Cu + __ AgNO3  __ Cu(NO3)2 + __ Ag
Reactants
Products
Cu
Ag
N
O
Cu
Ag
N
O
C.4 – Writing Chemical Equations
(continued)
6.
__ C8H18 + __ O2  __ CO2 + __ H2O
Reactants
Products
C
C
H
H
O
O
HW 3
Read & take notes on Section C.5
C.5 – Introducing the Mole Concept
Chemist have created a counting unit for
elements called the mole (symbolized mol).
602 000 000 000 000 000 000 000 particles
6.02 x
23
10
One mole of ANY element or molecule contains
C.5 – Introducing the Mole Concept
(continued)
Furthermore, the atomic weight of elements
can be used to find the molar mass of a
substance.
One mole of boron
atoms (6.02 x 1023)
would have a molar
mass of 10.81 g
C.5 – Introducing the Mole Concept
More examples…
(continued)
One mole of carbon atoms
(6.02 x 1023) would have a
molar mass of _______ g
C.5 – Introducing the Mole Concept
More examples…
(continued)
One mole of copper atoms (6.02 x 1023)
would have a molar mass of ______ g
One mole of silver atoms (6.02 x 1023) would
have a molar mass of ______ g
One mole of gold atoms (6.02 x 1023) would
have a molar mass of ______ g
C.5 – Introducing the Mole Concept
(continued)
(Curve ball) How about the molar mass of
oxygen gas (O2)?
One mole of oxygen gas (O2)
molecules (6.02 x 1023) would
have a molar mass of _______ g
C.5 – Introducing the Mole Concept
(continued)
(Curve ball 2) How about the molar mass of
water (H2O)?
One mole of water (H2O) molecules (6.02 x 1023)
would have a molar mass of _______ g
C.6 – Molar Masses
HW Questions 1-4,6,8
pg 163
C.6 – Molar Masses
1. One mole of nitrogen (N) atoms (6.02 x 1023)
would have a molar mass of _______ g
2. One mole of nitrogen (N2) molecules
(6.02 x 1023) would have a molar mass of
_______ g
C.6 – Molar Masses
(continued)
3. One mole of table salt (NaCl)
molecules (6.02 x 1023) would have a
molar mass of _______ g
C.6 – Molar Masses
(continued)
4. One mole of table sugar (C H O )
12 22 11
molecules (6.02 x 1023) would have a
molar mass of _______ g
C.6 – Molar Masses
(continued)
6. One mole of magnesium phosphate
Mg3(PO4)2 molecules (6.02 x 1023)
would have a molar mass of ______ g
C.6 – Molar Masses
(continued)
8. One mole of calcium hydroxyapatite
Ca10(PO4)6(OH)2 molecules (6.02 x 1023) would
have a molar mass of ______ g
HW 5
Read & take notes on Section C.7
C.7 – Equations and Molar Relationships
Let’s revisit copper-refining…
2 CuO(s) + C(s)  2 Cu(s) + CO2(g)
Alternatively stated…
2 mol CuO + 1 mol C  2 mol Cu + 1 mol CO2
In this example, for every two
moles of CuO that react, one mole
of CO2 is produced.
C.7 – Equations and Molar Relationships
(continued)
Sample Problem: A refiner needs to convert
955.0 g CuO to pure Cu. What mass of C is
needed for this reaction?
Using . . .
2 mol CuO + 1 mol C  2 mol Cu + 1 mol CO2
We can reason that . . .
1 mol CuO
79.55 g CuO
X mol CuO
=
955.0 g CuO
C.7 – Equations and Molar Relationships
(continued)
Sample Problem: A refiner needs to convert
955.0 g CuO to pure Cu. What mass of C is
needed for this reaction? (continued)
Solving for X . . .
1 mol CuO
79.55 g CuO
X mol CuO
=
955.0 g CuO
955.0 g CuO X 1 mol CuO = 79.55 g CuO X X mol CuO
955.0 g CuO X 1 mol CuO
= X mol CuO
79.55 g CuO
NOTICE . . .
C.7 – Equations and Molar Relationships
(continued)
Sample Problem: A refiner needs to convert
955.0 g CuO to pure Cu. What mass of C is
needed for this reaction? (continued)
Solving for X . . .
955.0 g CuO X 1 mol CuO
= X mol CuO
79.55 g CuO
This all started with . . .
1 mol CuO
79.55 g CuO
12.01 mol CuO = X
A proportion we created called
a conversion factor, both
referring to the same number of
particles.
C.7 – Equations and Molar Relationships
(continued)
Sample Problem: A refiner needs to convert
955.0 g CuO to pure Cu. What mass of C is
needed for this reaction? (continued)
So . . .
The refiner knows there are 12.01 mol CuO in 955.0 g.
Remembering the equation we
started with . . .
2 mol CuO + 1 mol C  2 mol Cu + 1 mol CO2
12.01 mol CuO X
1 mol C
2 mol CuO
= 6.005 mol C
C.7 – Equations and Molar Relationships
(continued)
Sample Problem: A refiner needs to convert
955.0 g CuO to pure Cu. What mass of C is
needed for this reaction? (continued)
So . . . Once we know
The refiner knows we need 6.005 mol C to refine 955 g of
CuO we can calculate the actual mass of C needed . . .
Appropriate conversion factors . . .
12.01 g C
6.005 mol C X
1 mol C
=
72.12 g C
C.8 – Molar Relationships
HW 6 Questions 1-4 pg 166
C.8 – Molar Relationships
1.
CuO(s) + 2 HCl(aq)  CuCl2(aq) + H2O (l)
Molar mass of each…
C.8 – Molar Relationships
(continued)
2.
Mass (in grams) of:
a. 1.0 mol HCl
b. 5.0 mol HCl
c. 0.50 mol CuO
C.8 – Molar Relationships
(continued)
3.
# of moles represented by
a. 941.5 g CuCl2
b. 201.6 g CuCl2
c. 73.0 g HCl
C.8 – Molar Relationships
4.
CuO(s) + 2 HCl(aq)  CuCl2(aq) + H2O (l)
a. How many moles of CuO are needed to
react with 4 mol HCl?
a. How many moles of HCl are needed to react
with 4 mol CuO?
Mole Quiz (explained)
Start with
6 K + B2O3  3 K2O + 2 B
Please write out the conversion factor for the reactants and
products of the above chemical equation.
1)
3)
1 mol K
2)
1 mol B2O3
-------------------
-------------------
39.098 g K
69.619 g B2O3
1 mol K2O
4)
1 mol B
-------------------
-------------------
94.194 g K2O
10.811 g B
Mole Quiz (continued)
6 K + B2O3  3 K2O + 2 B
A processor needs to convert 955.0 g B2O3 to pure B. What mass
of K is needed for this reaction? (show all your work)
955.0 g B2O3 X
1 mol B2O3
------------------69.619 g B2O3
= 13.72 mol B2O3
6 mol K
39.098 g K
------------- X 13.72 mol B2O3 = 82.32 mol K X --------------- =
1 mol B2O3
1 mol K
3219 g K
Mole Quiz (continued)
16 Al + 3 S8  8 Al2S3
A processor plans to create 1000. g of Al2S3. What mass of
pure S is needed for this reaction? (show all your work)
1000. g Al2S3 X
1 mol Al2S3
------------------150.16 g Al2S3
= 6.660 mol Al2S3
3 mol S8
256.53 g S8
------------- X 6.660 mol Al2S3 = 2.498 mol S8 X ---------------- =
8 mol Al2S3
1 mol S8
640.8 g S8
HW 7
Read & take notes on Section C.9
C.9 – Compositions of Materials
The percent mass of each material found in
an item is called the percent composition.
Hint: remember solution concentration
Example: post-1982 penny has a mass of 2.500 g has 2.4375 g zinc &
0.0625 g copper. What is the percent composition?
2.4375 g zinc
2.500 g total
X 100% = 97.50 % zinc
C.9 – Compositions of Materials
(continued)
Example: post-1982 penny has a mass of 2.500 g has 2.4375 g zinc &
0.0625 g copper. What is the percent composition?
0.0625 g copper
2.500 g total
X 100% = 2.50 % copper
C.9 – Compositions of Materials
(continued)
Why are the ideas of molar mass & percentage
composition so important?
Some Copper-containing Minerals
Common Name
Formula
Chalcocite
Cu2S
Chalcopyrite
CuFeS2
Malachite
Cu2CO3(OH)2
Mass of copper
Mass of Cu2S
It helps us
determine which is
more profitable to
mine.
X 100% = % copper
C.10 – Percent Composition
HW 8 – Questions 1-2 on pg 168
C.10 – Percent Composition
1. An atom inventory for Cu3(CO3)2(OH)2
There are :
3 Cu atoms
2 C atoms
8 O atoms
2 H atoms
C.10 – Percent Composition
(continued)
2. Percent copper in ?
a. Chalcopyrite CuFeS2
Start with . . .
63.55 g + 55.85 g + (2 x 32.07g) = 183.54 g
63.55 g
183.54 g
x 100 % = 34.62 % Cu
C.10 – Percent Composition
(continued)
2. Percent copper in ?
b. malachite Cu2CO3 (OH)2
(2 x 63.55 g) + 12.01 g + (5 x 16.00 g) + (2 x 1.008)
(2 x 63.55 g)
221.13 g
= 221.13 g
x 100 %
= 57.48 % Cu
c. Chalcocite , at 79.85% copper, would be the
most profitable to mine.
C.11 – Retrieving Copper
HW 9 – please pre-read the lab.
HW 10
Read & take notes on Section C.12
C.12 – Conservation in the Community
Resources…
Renewable
Fresh water,
Air,
Fertile soil ,
Plants, and
Animals
Nonrenewable
Metals,
Natural gas,
coal and
petroleum
EVENTUALLY
replenished by
natural processes.
CANNOT be readily
replenished.
C.13 – Rethinking, Reusing,
Replacing & Recycling
HW 11 Questions 1 & 2 on pg. 176
C.14 – The Life Cycle of Material
C.15 – Copper Life-Cycle Analysis
Not covered