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