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Unit 1, Activity 2, Specific Assessment Rubric
Chemistry
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 1
Unit 1, Activity 2, Specific Assessment Rubric
3
Measurements
are to the correct
number of
significant
figures
Units included
Answers are
within the range
of acceptable
error
Measurements
finished within
the prescribed
time limit
Questions
Answered
2
0
All measurements
2 or 3 measurements
Less than 2
measurements
All measurements
2 or 3 measurements
Less than 2
measurements
All measurements
2 or 3 measurements
Less than 2
measurements
All measurements
All safety rules
followed
Answered
correctly
2 or 3 measurements
Less than 2
measurements
Answered incorrectly
but supported by
evidence
Answered incorrectly.
No supporting
evidence.
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 1
Unit 1, Activity 3, Accuracy and Precision Worksheet
Figure 1
Figure 2
Figure 3
1. Determine the accuracy and precision represented by each group of darts in the figures
above. Explain your choices using complete sentences.
Figure 1
Figure 2
Figure 3
Precision?
Accuracy?
2. A basketball player throws 100 free-throws; 95 of these balls go through the goal; 5 miss
the goal entirely. Describe the precision and accuracy of the free-throws.
3. The same player is having an off day; 5 balls go through the goal; the other 95 balls
bounce off of the rim. Describe the precision and accuracy of the throws.
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 2
Unit 1, Activity 3, Accuracy and Precision Worksheet Answers
Figure 1
Figure 2
Figure 3
1. Determine the accuracy and precision represented by each group of darts in the figures
above. Explain your choices using complete sentences.
Precision?
Accuracy?
Figure 1
Picture 2
Good
All of the darts are
grouped in the same
area.
Poor
None of the darts
are grouped in the
bull’s-eye.
Poor
None of the darts
are grouped in the
same area.
Poor
Few of the darts are
grouped in the
bull’s-eye.
Picture 3
Good
All of the darts are
grouped in the
same area.
Good
All of the darts are
grouped in the
bull’s-eye.
2. A basketball player throws 100 free-throws; 95 of these balls go through the goal;
5 miss the goal entirely. Describe the precision and accuracy of the free-throws.
The player has good precision and good accuracy because so many of the balls
go through the goal.
3. The same player is having an off day; 5 balls go through the goal; the other 95
balls bounce off of the rim. Describe the precision and accuracy of the throws.
The player has good precision because so many balls bounce off the rim but poor
accuracy because so few balls make it through the goal.
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 3
Unit 2, Activity 1, Card Sort Template 1
Matter
Homogeneous
Pure Substance
Heterogeneous
Element
Mixture
Compound
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 4
Unit 2, Activity 1, Card Sort Template 2
Muddy Water
Na
Solution
As
salt water
Cl
Metal
NaCl
nonmetal
Metalloid
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 5
Unit 2, Activity 1, Sample Concept Map
Elements, Compounds and Mixtures Concept Map
MATTER
Can be either
MIXTURE
PURE
SUBSTANCE
Is
Is
ELEMENT
Chemically
combine to form
Can be
COMPOUND
HOMOGENEOUS
HETEROGENEOUS
Is either
Example
METAL
Example
Na
METALLOID
Example
As
NONMETAL
Example
Cl
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
NaCl
Is called
SOLUTION
Example
MUDDY WATER
Example
SALT WATER
Page 6
Unit 2, Activity 2, Sample Word Grid
Sample:
Homogeneous
Can be separated
Heterogeneous into individual
components
The properties of
the individual
components are
the same as
properties of the
sample
Salt
Water
Copper
Salt and water
Copper and
water
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 7
Unit 2, Activity 2, Sample Word Grid Answers
Sample:
Salt
Water
Copper
Salt and water
Copper and
water
Homogeneous
Can be separated
Heterogeneous into individual
components
X
X
X
X
X
The properties of
the individual
components are
the same as
properties of the
sample
X
X
X
X
X
X
X
Conclusions:
1. Salt (NaCl) is a homogeneous material that can be decomposed into individual elements
(sodium and chlorine). The properties of the salt differ from the properties of the
elements. Salt is a compound.
2. Water (H2O) is a homogeneous material that can be decomposed into elements (hydrogen
and oxygen). Water is a compound.
3. Copper is a homogeneous material that cannot be separated into components. Copper is
an element.
4. Salt and water combine to form a homogeneous material that can be separated into
parts. When the salt and water are mixed, their properties do not change. Salt water is a
homogeneous mixture called a solution.
5. Copper shot and water not homogeneous because the copper and water are easily seen
as individual parts. These parts can be separated easily. When the copper and water are
mixed, their individual properties do not change. This is a heterogeneous mixture.
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 8
Unit 2, Activity 4, Three Worlds of Chemistry
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 9
Unit 2, Activity 5, Density
Each box has the same volume. If each
ball has the same mass, which box would
weigh more? Why?
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 10
Unit 2, Activity 6, Split-Page Notes
Physical and Chemical Changes
Effervescent tablet in water
Observations
1. numerous bubbles formed as soon as tablet
touched the water
2. bubbles rose to top of water and burst
3. tablet disappeared
4. bubbles stopped forming
5. looks like nothing else is happening
Conclusion
The bubbles contained a gas that escaped into the air. The
tablet was a solid that underwent a chemical change with
the water to produce the gas bubbles. Once the tablet
(reactant) was used up, no more gas bubbles (products)
were formed, and the reaction stopped. There has been a
change in the identity of the material. It is no longer an
effervescent tablet. The production of a gas is evidence of
a chemical change (reaction) taking place.
Cutting a piece of paper
Observations
1. smaller pieces of paper are formed
Conclusion
The smaller pieces of paper are exactly like the original
piece of paper (reactant). There has been no change in the
identity of the material. It is still paper (product).
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 11
Unit 3, Activity 3, Exploring the Periodic Table
1.01
H
4.00
He
6.94
Li
9.01
Be
10.81
B
12.01
C
14.01
N
16.00
O
19.00
F
20.18
Ne
22.99
Na
24.30
Mg
26.98
Al
28.08
Si
30.97
P
32.07
S
35.45
Cl
39.95
Ar
39.10
K
40.08
Ca
69.72
Ga
72.61
Ge
74.92
As
78.96
Se
79.90
Br
83.80
Kr
85.47
Rb
87.62
Sr
114.82
In
118.71
Sn
121.75
Sb
127.60
Te
126.90
I
131.29
Xe
132.90
Cs
137.33
Ba
204.38
Tl
207.2
Pb
208.98
Bi
(209)
Po
(210)
At
(222)
Rn
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 12
Unit 3, Activity 4, GISTing
GISTing
The individual Gists are limited to 15 words.
Sample paragraph from notes:
Atomic radii
The atomic radius is ½ the distance between the centers of neighboring atoms. It is the size
of the atom due to the size of the electron cloud.
Group trends
The atomic radii of the main group elements (s & p sublevels) generally increases down a
group. The outermost electrons occupy energy levels that are farther from the nucleus.
Period trends
Atomic radius generally decreases across a period. This is caused by the increasing nuclear
charge of the nucleus as you go across a period. More protons are in the nucleus and more
electrons are in the same energy level. The increasing nuclear charge attracts the electrons
and pulls them closer to the nucleus.
Class gist statements for each sentence of the paragraphs
1. Atomic radius means how big an atom is. _____ _____ _____ _____ _____ _____ _____
2. Atoms get bigger down a group because there are more energy levels. _____ _____ _____
3. Atoms get smaller across a period because more protons attract the electrons pulling them
closer.
Summary: Atomic radius (size of the atom) increases down a group because of more energy
levels and across a period because of a greater attraction between the larger number of protons
and the outer electrons.
After several gisting activities, you will be able to construct summaries. Gisting is a mental
process and not necessarily a written one.
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Unit 4, Activity 1, Vocabulary Self-Awareness
Term
+

-
Definition
Example
Chemical bond
Ionic bond
Covalent bond
Metallic bond
Electronegativity
Polar covalent
bond
Nonpolar
covalent bond
Formula unit
Molecule
Molecular
formula
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 14
Unit 4, Activity 2, Ion Cards
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 15
Unit 4, Activity 2, Ion Cards
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 16
Unit 4, Activity 2, Ion Cards
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 17
Unit 4, Activity 2, Ion Cards
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 18
Unit 4, Activity 3, Chemical Formulas and Nomenclature I
Write formulas for the following compounds:
1. copper (I) oxide
_______________
2. aluminum hydroxide
_______________
3. triphosphorus decasulfide
_______________
4. zinc nitrate
_______________
5. hydrobromic acid
_______________
6. mercury (I) bromide
_______________
7. boron tribromide
_______________
8. sodium hydride
_______________
9. barium perchlorate
_______________
10. tetraphosphorus hexasulfide
_______________
11. sulfuric acid
_______________
12. calcium hypochlorite
_______________
13. ammonium phosphite
_______________
14. chromium (III) acetate
_______________
15. hydrosulfic acid
_______________
16. carbonic acid
_______________
17. phosphorus pentafluoride
_______________
18. cobalt (II) nitrate
_______________
19. magnesium sulfate
_______________
20. strontium phosphate
_______________
21. dichlorine monoxide
_______________
22. phosphorous acid
_______________
23. disulfur dichloride
_______________
24. iron (III) carbonate
_______________
25. perchloric acid
_______________
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 19
Unit 4, Activity 3, Chemical Formulas and Nomenclature I Answers
Write formulas for the following compounds:
1. copper (I) oxide
__Cu2O_______
2. aluminum hydroxide
__Al(OH)3_____
3. triphosphorus decasulfide
__P3S10_______
4. zinc nitrate
__Zn(NO3)2____
5. hydrobromic acid
__HBr(aq)_____
6. mercury (II) bromide
__HgBr2_______
7. boron tribromide
__BBr3________
8. sodium hydride
__NaH________
9. barium perchlorate
__Ba(ClO4)2____
10. tetraphosphorus hexasulfide
__P4S6________
11. sulfuric acid
__H2SO4(aq)___
12. calcium hypochlorite
__Ca(ClO)2____
13. ammonium phosphite
__(NH4)3PO3___
14. chromium (III) acetate
__Cr(C2H3O2)3_
15. hydrosulfic acid
__H2S(aq)_____
16. carbonic acid
__H2CO3(aq)___
17. phosphorus pentafluoride
__PF5_________
18. cobalt (II) nitrate
__Co(NO3)2____
19. magnesium sulfate
__MgSO4______
20. strontium phosphate
__Sr3(PO4)2____
21. dichlorine monoxide
__Cl2O________
22. phosphorous acid
__H3PO4(aq)___
23. disulfur dichloride
__S2Cl2_______
24. iron (III) carbonate
__Fe2(CO3)3___
25. perchloric acid
__HClO4(aq)___
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 20
Unit 4, Activity 3, Chemical Formulas and Nomenclature II
Name the following compounds.
1. K2SO4
______________________________
2. N2O4
______________________________
3. BaClO4
______________________________
4. HNO2(aq)
______________________________
5. FE2(SO4)3
______________________________
6. NH4F
______________________________
7. BaI2
______________________________
8. CrO3
______________________________
9. Cu(C2H3O2)2
______________________________
10. Ag2CO3
______________________________
11. NaOH
______________________________
12. Ca3(PO4)2
______________________________
13. ClF3
______________________________
14. K2SO3
______________________________
15. AlBr3
______________________________
16. MgCl2
______________________________
17. HC2H3O2(aq)
______________________________
18. P2O5
______________________________
19. FePO4
______________________________
20. SrBr2
______________________________
21. Al2S3
______________________________
22. LiBr
______________________________
23. NH3
______________________________
24. PbO2
______________________________
25. MgO
______________________________
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 21
Unit 4, Activity 3, Chemical Formulas and Nomenclature II Answers
Name the following compounds.
1. K2SO4
__potassium sulfate______________
2. N2O4
__dinitrogen tetroxide____________
3. BaClO4
__barium perchlorate_____________
4. HNO2(aq)
__nitrous acid__________________
5. Fe2(SO4)3
__iron (III) sulfate_______________
6. NH4F
__ammonium fluoride____________
7. BaI2
__barium iodide_________________
8. CrO3
__chromium (IV) oxide___________
9. Cu(C2H3O2)2
__copper (II) acetate_____________
10. Ag2CO3
__silver carbonate_______________
11. NaOH
__sodium hydroxide______________
12. Ca3(PO4)2
__calcium phosphate_____________
13. ClF3
__chlorine trifluoride_____________
14. K2SO3
__potassium sulfite_______________
15. AlBr3
__aluminum bromide_____________
16. MgCl2
__magnesium chloride____________
17. HC2H3O2(aq)
__acetic acid___________________
18. P2O5
__diphosphorous pentoxide________
19. FePO4
__iron (III) phosphate____________
20. SrBr2
__strontium bromide_____________
21. Al2S3
__aluminum sulfide______________
22. LiBr
__lithium bromide_______________
23. NH3
__ammonia____________________
24. PbO2
__lead (IV) oxide_______________
25. MgO
__magnesium oxide_____________
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 22
Unit 4, Activity 3, Molecular Geometry of Simple Molecules Student Sheet 1
Note: A represents the central atom in the molecule. B represents atoms bonded to the central
atom. B can be identical atoms or different atoms.
Directions:
1. Find the other students who have the same color balloons as you. Have someone inflate a
balloon as much as possible without popping it. Inflate your balloon(s) to the same size.
2. Using the patterns below, tie the appropriate number and color balloons together. For
example, for the AB2E model, tie 2 blue balloons and a white balloon together. For
groups of 4 balloons, it is easier to tie 2 balloons together and then the other 2 balloons
together, then twist the two groups together. For five-balloon groups, make sets of 2 and
3 balloons and twist. For six balloons, use 3 sets of 2 balloons twisted together.
3. Attach a piece of string to hang the finished model from the ceiling.
Type of
Molecule
Balloons
Needed for
Model
One colored
balloon models
for electron
pair geometries
AB2
Number of
Atoms
Attached to
the Central
Atom
2
2 pink
*
AB3
3
3 blue
*
AB2E
3
2 blue, 1 white
AB4
4
4 red
AB3E
3
3 red, 1 white
AB2E2
3
2 red, 2 white
AB5
5
5 green
AB4E
4
4 green, 1 white
AB3E2
3
3 green, 2 white
AB2E3
2
2 green, 3 white
AB6
6
6 yellow
AB5E
5
5 yellow, 1
*
*
*
white
AB4E2
4
4 yellow, 2
white
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 23
Unit 4, Activity 3, Molecular Geometry of Simple Molecules Student Sheet 2
Number of
lone pairs
around the
Central
Atom
Number
of atoms
attached
to the
Central
Atom
Electron
Pair
Geometry
Bond
angle of
electron
pairs
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Type of
Molecule
Molecular
Geometry
(Shape of the
molecule)
Example
Page 24
Unit 4, Activity 3, Molecular Geometry of Simple Molecules Answer Sheet
Number
of lone
pairs on
the
Central
Atom
0
Number
of atoms
attached
to the
Central
Atom
2
0
3
0
4
Tetrahedral
109.5
AB4
1
3
Tetrahedral
<109.5
AB3E
2
2
Tetrahedral
<109.5
AB2E2
0
5
Trigonal
90,
Bipyramidal
120,180
1
4
Trigonal
90,
Bipyramidal
120,180
2
3
3
2
0
6
1
2
Electron Pair
Geometry
Bond
angle of
Electron
pairs
Type of
Molecule
Molecular
Shape
Example
of
Molecule
Linear
180
AB2
Linear
CO2
120
AB3
Trigonal
planar
AB5
Trigonal
planar
Tetrahedral
Trigonal
Pyramidal
Bent
Trigonal
Bipyramidal
BF3
CH4
NH3
H2O
PCl5
AB4E
*See-Saw
SF4
90, 180
AB3E2
*T- structure
IBr3
180
AB2E3
* Linear
XeF2
Octahedral
90, 180
AB6
Octahedral
SCl6
5
Octahedral
90, 180
AB5E
4
Octahedral
90, 180
AB4E2
Trigonal
Bipyramidal
Trigonal
Bipyramidal
*Pyramidal
Planar
*Square
Planar
IF5
XeF4
Note: Molecular Shapes marked * may be omitted if time is a factor.
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 25
Unit 4, Activity 4, Chemical Bond Type Lab
Purpose: To observe characteristics of ionic and covalent bonds and to classify compounds as
ionic or covalent based on those observations.
Modified from http://www.hse.k12.in.us/staff/ebutzin/Documents/ICP/Bonding/bond%20types%20lab.doc
Safety:
 Wear goggles.
 Do not taste or touch any chemicals.
 Follow guidelines pertaining to an open flame.
Materials
 Test tubes
 Thin stem pipettes
 Iron ring and stand
 Candle with foil
holder





Small foil pie pan
Calcium chloride
Citric acid
Phenyl salicylate
Potassium iodide




Sodium chloride
Sucrose
Conductivity probe
Safety goggles
Procedure:
1. Place a few crystals of sucrose, sodium chloride, phenyl salicylate, calcium chloride,
citric acid and potassium iodide in separate locations around the pie pan as shown in
Figure B. Make sure all of the samples are approximately the same size. Do not allow
the crystals to touch.

Write a brief description of each of the 6 substances in a data table.
2. Testing melting point

Place the pie pan on the iron ring. Position the ring so it is just above the tip of a
candle flame, as shown in Figure A. Light the candle to check that you have the
correct height.

Place the candle under the middle of the pan and heat. Record the order in which
the substances melt. If a compound doesn’t melt record N/A.
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 26
Unit 4, Activity 4, Chemical Bond Type Lab
3. Testing the solubility in water
 Place a few crystals of each substance in separate test tubes. Add about 1 mL of
distilled water and agitate each.
 Record the solubility in the data table (Yes – if it dissolves, No – if it does not
dissolve).
4. Testing the conductivity in water
 Use the conductivity probe for each of the substances that WERE SOLUBLE in water
to determine if they conduct electricity or not. If the compound didn’t dissolve, do
NOT try to measure the conductivity.
 Rinse and dry the probe after each test.
Cleanup
 Rinse all test tubes with water and scrub with a test tube brush.
 Rinse off the pie pan and scrub with a test tube brush. Dry with a clean cloth.
 Wash hands and put away goggles.
Data Table
Compound
Description
Melting
Point
(1, 2, 3, 4,
N/A)
Solubility in
Water (Y/N)
Conductivity
Calcium
chloride
Citric acid
Phenyl
salicylate
Potassium
iodide
Sodium
chloride
Sucrose
Write and defend a conclusion based on a logical analysis of your experimental data.
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 27
Unit 4, Activity 6, RAFTing
R – Role (role of the writer)
A – Audience (to whom or what the RAFT is being written)
F – Form (the form the writing will take, as in letter, song, etc.)
T – Topic (the subject focus of the writing)
R – H2O
A – Oil
F – Letter
T – Intermolecular Forces between molecules
Dear Oil,
I know you would really like for us to get together. Unfortunately, my intermolecular
forces are too strong and will always keep us apart.
I am a polar molecule. I am attracted to other polar molecules much more than I am
attracted to your nonpolar structure. I also have hydrogen bonding which really makes me
extremely attractive to other like molecules. I guess you could say that the only thing we really
have in common is a really weak dispersion force. Unfortunately, this will not be strong enough
for us to base any lasting relationship.
Please feel free to look for another molecule with whom to combine. Perhaps you should
look for a nonpolar molecule with no tendency to hydrogen bond.
Sincerely,
Water
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 28
Unit 5, Activity 1, How Large Is a Mole?
Materials:
1.
2.
3.
4.
Samples of 5 different types of beans
Container for measuring the mass of the beans
Balance
Calculator
Procedure:
1. Measure the mass of each type of bean.
2. Using a ratio, students are to calculate the relative masses of the other beans by dividing
the mass of the beans by the mass of the smallest bean of the five types used.
3. Count how many whole beans are needed to get the mass in grams equal to the relative
mass calculated in step 2 for each type of bean.
4. Using the data in the relative mass column, place the empty container on the balance and
zero (tare) the balance. Add beans one at a time to count how many whole beans are
needed to get a mass in grams equal to the relative mass for each type of bean. (Mass of
container in this example is 25.6g.)
Name of
bean
Mass of the container and the
beans (g)
Mass of
beans
(g)
Relative
mass
(g)
Number of
beans
from step 3
Average of
Last Column

Calculate the average number of whole beans in a container by adding the number of beans in 1
container for each type of bean and dividing by 5.
Note: Use the number from the average of the last column box for all calculations.
The following ratios can be derived from the data:
beans
relative mass of beans
container
container
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 29
Unit 5, Activity 1, How Large Is a Mole?
Use the data and the ratios to solve the following problems:
1. Calculate the number of containers given 350 beans of each type of bean.
2. Calculate the number of beans given 5.5 containers of each type of bean.
3. Calculate the mass of 350 containers of each type of bean.
4. Calculate the number of containers given 400 g. of each type of bean.
5. Calculate the number of beans given 400 g of each type of bean.
Write and defend a conclusion based on logical analysis of the data obtained from this activity.
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 30
Unit 5, Activity 1, How Large Is a Mole? Answer Sheet
This activity is designed to help students understand the concept of the mole as a definite number
of particles. Using five varieties of different type beans, students will determine the relative mass
of each type of bean and express the relative masses in grams.
 Have students work in groups and provide each group with five sets of 40 beans, a container,
and a balance.
 Have students determine the total mass of each type of bean. Enter the data into the table
provided.
 Using a ratio, students are to calculate the relative masses of the other beans by dividing the
mass of the beans by the mass of the smallest bean of the five types used.
 Using the data in the relative mass column, place an empty container on the balance and zero.
Add beans one at a time to count how many whole beans are needed to get a mass in grams
equal to the relative mass for each type of beans. (Mass of container in this example is
25.6g.)
Name of bean
Red beans
Large lima
beans
Chick peas
Lentils
Black eyes peas
Mass of the
container and the
beans (g)
Mass of beans
(g)
46.7
21.1
77.9
52.3
44.5
27.7
35.3
18.9
2.1
9.7
Relative mass
(g)
Number
of beans
from step
3
211
.
 10.0
2.1
52.3
 24.9
2.1
9.0
1.0
4.6
Average of last column
19
19
20
19
20
19

Calculate the average number of whole beans in a container.
(19  19  20  19  20)
 19.4  19 wholebeans
5
Note: Use the number from the average of the last column box for all calculations.
The following ratios can be derived from the data:
beans
relative mass of beans
container
container
Note: Use the average number of beans for the beans/container ratio and the relative mass of each type
of bean for the mass/container ratio.
Use the data to solve the problems as you would solve mole problems.
1. Calculate the number of containers given 350. beans of each type of bean.
2. Calculate the number of beans given 5.5 containers of each type of bean.
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 31
Unit 5, Activity 1, How Large Is a Mole? Answer Sheet
3. Calculate the mass of 350. containers of each type of bean.
4. Calculate the number of containers given 400.g. of each type of bean.
5. Calculate the number of beans given 400.g of each type of bean.
Sample Calculations:
For these calculations, the container was a cup.
1. 350. beans X
2. 5.5 cups X
1 cup
 18.4 cups
19 beans
19 beans
 104.5 beans
1 cup
3. 350 red beans x
1cup
21.1g
×
= 388.7 g
19 red beans
1cup
350 lima beans×
1cup
24.9 g
×
= 458.68g
19 lima beans 1cup
4. 400. g X
1 cup
= 19.0 cups
21.1 g
5. 400. g X
1 cup
19 red beans
X
= 360.2 red beans
21.1 g
1 cup
Write and defend a conclusion based on logical analysis of the data obtained from this activity.
Regardless of the type of bean used, the number of beans per cup is consistent (within an
acceptable margin of error). Although the mass of each bean is different, the average number of
beans per cup is also consistent. The data supports the idea that a cup of beans contains 19
whole beans regardless of the type of bean used or its relative mass.
When using this activity as an introduction to mole problems, the container will be used as an
analogy to a mole.
For example:
Calculate the number of containers when given 350 beans of each type of bean.
Calculate the number of moles when given 350 atoms of any element.
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 32
Unit 5, Activity 3, Observing Chemical Reactions
Lab Station 1: Copper wire and silver nitrate solution
Materials: small piece of copper wire, pipettes of 0.2 M silver nitrate
solution, test tube, test tube rack, waste container
Procedure:
1.
Record a description of all reactants and products.
2.
Make a hook on one end of the copper wire.
3.
Hang the wire by the hook in a test tube.
4.
Pour enough silver nitrate solution into the tube to cover most of the
wire.
5.
Place the test tube into the test tube rack and observe for one
minute.
6.
Record observations.
7.
Empty the test tube contents into the large beaker (waste
container).
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 33
Unit 5, Activity 3, Observing Chemical Reactions
Station 2: zinc + hydrochloric acid
Materials: mossy zinc, 0.1M hydrochloric acid, microplate, pipette,
waste container
1.
Record a description of all reactants and products.
2.
Place a piece of mossy zinc a well of a microplate.
3.
Add 10 drops of hydrochloric acid to the well.
4.
Record observations.
5.
Empty the contents of the microplate into the large beaker (waste
container).
Station 3: sodium chloride + silver nitrate
Materials: sodium chloride solution, 0.2 M silver nitrate solution, small
test tube, pipette, large beaker
1.
Record a description of all reactants and products.
2.
Fill a small test tube halfway with sodium chloride solution.
3.
Add 3 to 5 drops of the silver nitrate solution.
4.
Record observations.
5.
Empty the contents of the microplate into the large beaker (waste
container).
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 34
Unit 5, Activity 3, Observing Chemical Reactions
Station 4: Acetic acid + sodium hydrogen carbonate
Materials: Acetic acid (vinegar), sodium hydrogen carbonate
(baking soda), 250 ml beaker, waste container
1.
Record a description of all reactants and products.
2.
Place one teaspoon of sodium hydrogen carbonate into a
small beaker.
3.
Add three teaspoons of Acetic acid.
4.
Record observations.
5.
Empty the beaker contents into the large beaker (waste
container).
Station 5: Acetic acid + sodium hydrogen carbonate +
phenolphthalein
Materials: Acetic acid (vinegar), sodium hydrogen carbonate
(baking soda), phenolphthalein, 250 ml beaker, waste container
1.
Record a description of all reactants and products.
2.
Place one teaspoon of sodium hydrogen carbonate into a
small beaker.
3.
Add three teaspoons of Acetic acid.
4.
Record observations.
5.
Empty the beaker contents into the large beaker (waste
container).
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 35
Unit 5, Activity 4, Split-Page Notetaking
Patterns for the types of chemical reactions
1. Composition
(Synthesis)
A + X  AX
2. Decomposition
AX  A + X
Single Replacement
Reaction:
A + BX  B + AX
or
Y + BX  X + BY
2 reactants form one product:
element + element  one compound
2Na + Cl2  2NaCl
compound + compound  one compound
CO2 + H2O  H2CO3
one compound  two or more products
2NaCl 2Na + Cl2
H2CO3 CO2 + H2O
element + compound  different element + different
compound
Use the activity series of the elements to predict the products.
Generally elements will replace any element below it on the
chart.
metals replace less active metals or hydrogen from a
compound
Cu + AgNO3 Ag + Cu(NO3)2 (reverse reaction will
not occur)
nonmetals replace less active nonmetals from a compound
Double Replacement
Reaction:
AX + BY  AY + BX
I2 + NaCl  Cl2 + NaI (reverse reaction will not
occur)
compound + compound  different compound + different
compound
Use a solubility table to predict precipitates (solids)
NaCl + AgNO3  NaNO3 + AgCl(s) (reverse reaction will
not occur)
Neutralization:
acid + base → salt + water
HCl + NaOH → NaCl + H2O
Neutralization is a type of double replacement reaction. An
ionic salt is formed from the cation of the base and the anion
of the acid. Neutralization is the reaction of the hydronium
ions and hydroxide ions to form water molecules.
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 36
Unit 5, Activity 6, Can You Make Two Grams?
Possible combinations that form precipitates:
Reaction Number
1. MgSO4• 7H2O + Ca(C2H3O2)2 • H2O → CaSO4(s)
246.50 g/mol
3.62 g
176.19 g/mol
246.50 g/mol
5.85 g
5.85 g
4.22 g
4.59 g
4.59 g
2.51 g
2.00g
3.37 g
2.99 g
→ MgCO3(s) +
K2SO4
+ 7H2O
138.21 g/mol
142.02 g/mol 18.02g/mol
84.31 g/mol
3.28 g
174.27 g/mol
2.00g
176.19 g/mol
3.52g
3.52g
+ Zn(C2H3O2)2 + 8 H2O
183.48 g/mol
105.99 g/mol
125.38g/mol
1.69g
138.21 g/mol
+ K2SO4
125.38g/mol
2.20g
138.21 g/mol
18.02 g/mol
2.78g
→ CaCO3(s)
2.12g
2.01g
+ 7 H2 O
174.27 g/mol
2.00g
105.99 g/mol
+ 7 H2O
2.27g
ZnCO3(s)
2.12g
142.02 g/mol 18.02g/mol
2.00g
→
18.02 g/mol
2.70 g
ZnCO3(s) + Na2SO4
8. Ca(C2H3O2)2 • H2O + K2CO3
176.19 g/mol
2.99g
2.00 g
→
18.02 g/mol
4.13
136.15 g/mol
2.59g
7. Ca(C2H3O2)2 • H2O + Na2CO3
176.19 g/mol
2.12 g
84.31 g/mol
6. ZnSO4 • 7H2O + K2CO3
287.56 g/mol
2.09 g
105.99 g/mol
5. ZnSO4 • 7H2O + Na2CO3
287.56 g/mol
18.02g
+ Na2SO4 + 7H2O
4. ZnSO4 • 7H2O + Ca(C2H3O2)2 • H2O → CaSO4(s)
287.56 g/mol
142.38g/mol
2.00 g
→ MgCO3(s)
3. MgSO4• 7H2O + K2CO3
246.50 g/mol
136.15 g/mol
2.59 g
2. MgSO4• 7H2O + Na2CO3
+ Mg(C2H3O2)2 + 7H2O
2.01g
+ 2NaC2H3O2 + H2O
100.09 g/mol 82.03 g/mol 18.02 g/mol
2.00g
3.28g
→ CaCO3(s)
+ 2KC2H3O2
100.09 g/mol 98.15 g/mol
2.76g
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
2.00g
3.92g
036g
+ H2 O
18.02g/mol
0.36g
Page 37
Unit 5, Activity 7, Vocabulary Self-Awareness
Term
+

-
Definition
Example
Oxidation
Reduction
Redox reaction
Oxidizing agent
Reducing agent
Skeleton
equation
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 38
Unit 5, Activity 7, Introduction to Oxidation-Reduction Reactions
Half- fill each well in the column with the indicated solution.
Column 1: Zn(NO3)2
Column 2: Pb(NO3)2
Column 3: Cu(NO3)2
1. Place one piece of zinc shot in each filled well in row 1.
2. Place one piece of lead shot in each filled well in row 2.
3. Place one piece of copper shot in each filled well in row 3.
Column
1
2
3
Row
1
2
3
4. Watch for two minutes, then record observations.
5. Make a list of the elements in order of reactivity.
6. Write the redox equation for each chemical change.
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 39
Unit 5, Activity 7, Introduction to Oxidation-Reduction Reactions Answer Sheet
1. Half- fill each well in the column with the indicated solution.
Column 1: Zn(NO3)2
Column 2: Pb(NO3)2
Column 3: Cu(NO3)2
2. Place one piece of zinc shot in each filled well in row 1.
3. Place one piece of lead shot in each filled well in row 2.
4. Place one piece of copper shot in each filled well in row 3.
Column
1
2
3
Row
1
2
3
5. Watch for two minutes, then record observations.
6. Make a list of the elements in order of reactivity.
7. Write the redox equation for each chemical change.
8. Write a conclusion based on your observations.
Elements in order of reactivity:
Zn, Pb, Cu
LEO the lion says GER (loss of electrons is oxidation, gaining electrons is reduction)
Redox equations:
Zn + Pb(NO3)2 → Pb + Zn(NO3)2
Zn0 → Zn 2+ + 2e- (oxidation)
Pb 2+ + 2e- → Pb0 (reduction)
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 40
Unit 5, Activity 7, Introduction to Oxidation-Reduction Reactions Answer Sheet
Zn + Cu(NO3)2 → Cu + Zn(NO3)2
Zn0 → + Zn 2+ 2e- (oxidation)
Cu2++ 2e- → Cu0 (reduction)
Pb + Zn(NO3)2 → NR (no reaction)
Pb + Cu(NO3)2 → Cu + Pb(NO3)2
Pb0 → Pb2+ +2e- (oxidation)
Cu2++2e- → Cu0 (reduction)
Cu + Zn(NO3)2 → NR
Cu + Pb(NO3)2 → NR
Conclusion: Oxidation is the process by which electrons are removed from atoms or ions.
Zn is oxidized by the other two ions. Pb is only oxidized by the Cu2+ ion. Cu is not
oxidized by either ion. Zn gives up its electrons more easily than the other ions.The
element that is oxidized is the reducing agent therefore Zn is the strongest reducing
agent, followed by Pb and lastly, by Cu.
Reduction is the process by which electrons are added to atoms or ions. The element that
is reduced is the oxidizing agent. Cu is the strongest oxidizing agent, followed by Pb and
then Zn.
Oxidation and reduction must take place at the same time because the number of
electrons lost must equal the number of electrons gained.
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 41
Unit 5, Activity 7, Introduction to Oxidation-Reduction Reactions Answer Sheet
Answers to ionic equations:
Molecular equation: Zn + Pb(NO3)2 → Pb + Zn(NO3)2
Ionic Equation:
Zn0 + Pb2+ + 2NO3-1 → Pb0 + Zn 2+ + 2NO3-1
Net Ionic Equation: Zn0 + Pb 2+→ Pb0 + Zn 2+
Molecular equation: Zn + Cu(NO3)2 → Cu + Zn(NO3)2
Ionic equation:
Zn0 + Cu2+ + 2NO3-1 → Cu0+ Zn 2+ + 2NO3-1
Net Ionic equation: Zn0 + Cu2+ → Cu0+ Zn 2+
Molecular equation: Pb + Zn(NO3)2 → NR (no reaction)
Molecular equation: Pb + Cu(NO3)2 → Cu + Pb(NO3)2
Ionic equation:
Pb0 + Cu2+ +2NO3-1 → Cu0 + Pb2+ + 2NO3-1
Net Ionic equation:
Pb0 + Cu2+ → Cu0 + Pb2+
Molecular equation: Cu + Zn(NO3)2 → NR
Molecular equation: Cu + Pb(NO3)2 → NR
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 42
Unit 6, Activity 4, Heating Curve
Gas State
Cp (g)
Boiling
Point (Hv)
Liquid State Cp (l)
Melting
Point (Hf)
Solid State Cp(s)
Heat is added to a substance in the solid state. The energy added will increase the temperature of
the substance to its specific melting point. The amount of energy required to raise the
temperature depends on the specific heat (Cp) and the state of the substance. Specific heat is the
amount of energy needed to raise the temperature of one gram of a substance by one degree
Celsius.
At the melting point, the temperature stops rising and the substance starts to melt. The energy
supplied is used to weaken the intermolecular forces of attraction and the temperature remains
constant. The amount of energy needed to melt a substance depends on its heat of fusion (H f).
The molar heat of fusion is the amount of energy required to melt one mole of a substance at its
melting point.
After the phase change is complete, the temperature rise will follow a different rate than that of
the solid because the liquid state has a different heat capacity.
At the boiling point, the temperature stops rising and the substance starts to boil. The energy
supplied is used to break the intermolecular forces of attraction and the temperature remains
constant. The amount of energy needed to boil a substance depends on its heat of vaporization
(Hv). The molar heat of vaporization is the amount of energy required to boil one mole of a
substance at its boiling point. The Hv is higher than the Hf because of breaking the forces of
attraction.
After the phase change is complete, the temperature rise will follow a different rate than that of
the liquid because the gaseous state has a different heat capacity.
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 43
Unit 6, Activity 4, Phase Diagrams
A phase diagram is a graph of the conditions of temperature and pressure at which the solid,
liquid, and gaseous phases of a substance exist. The lines separating the phases are called phase
boundaries. Each point on the phase boundary show the conditions under which the two phases
exist in dynamic equilibrium.
Each point along the solid/liquid phase boundary represents the temperature and pressure
combinations where the rate of the solid melting is equal to the rate of the liquid freezing. Each
point represents a melting point.
Each point along the liquid/ vapor phase boundary represents the temperature and pressure
combinations where the rate of the liquid boiling is equal to the rate of the vapor condensing.
Each point represents a boiling point.
Each point along the solid/gas phase boundary represents the temperature and pressure
combinations where the rate of the solid subliming is equal to the rate of the vapor condensing to
a solid (called deposition). Each point represents a sublimation point.
The triple point indicates the only temperature and pressure conditions where the solid, liquid,
and vapor phases are all in equilibrium.
The critical point (Pc) is the point above which a substance will always be a gas regardless of the
pressure and temperature. The critical temperature is the highest temperature a substance can
exist as a liquid. The critical pressure is the lowest pressure required for the substance to be a
liquid at the critical temperature.
Phase Diagram for H2O
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Phase diagram for CO2
Page 44
Unit 6, Activity 6, Exothermic and Endothermic Energy Diagrams
Heat of reaction (H) is the amount of heat released or absorbed during a chemical reaction.
H = H products - H reactants
The heat content of the reactants is higher than
the heat content of the products. Energy in the
form of heat will be released when the products
form. The heat of reaction (H) is negative.
Example:
2 H2(g) + O2(g) → 2H2O(g) H = - 483.6 kJ
The heat content of the reactants is lower than
the heat content of the products. Energy in the
form of heat must be absorbed (added) to form
the products. The heat of reaction (H) is
positive.
Example:
2 H2O(g) → 2H2(g) + O2(g)
H = + 483.6 kJ
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 45
Unit 6, Activity 6, Energy Diagram (with activation energy)
For a reversible reaction, the activated complex is the same. The activated complex occurs at the
maximum-energy position along the reaction pathway. The activation energy of the forward
reaction is lower than the activation energy of the reverse reaction in this energy diagram. The
H is the same amount for both reactions but the sign of H is negative for the forward reaction
and is positive for the reverse reaction.
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 46
Unit 7, Activity 1, Vocabulary Self- Awareness
Term
+

-
Definition
Example
Solution
Solute
Solvent
Soluble
Electrolyte
Nonelectrolyte
Colloid
Solubility
Saturated
solution
Unsaturated
solution
Supersaturated
Solution
equilibrium
Miscible
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 47
Unit 7, Activity 1, Vocabulary Self- Awareness
Immiscible
chromatography
molarity
molality
Colligative
property
Vapor pressure
Nonvolatile
Volatile
Freezing-point
depression
Boiling-point
elevation
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 48
Unit 7, Activity 3, Solution Concentrations
Sample Problems using factor-label method.
Molarity (M) =
moles of solute
liter of solution
Example 1:
Calculate the molarity of a 1500 ml solution that contains 45.0 g of MgCl2.
M=
45.0 g MgCl 2
1 mol MgCl 2
1000 mL
X
X
 0.31 M
1500 mL of solution 95.3 g MgCl 2
1L
Example 2:
Calculate the mass of solute in 750.0 mL of a 0.500 M H2SO4 solution.
Mass of solute =
750 mL
X
1L
0.500mol H 2SO 4 98.1g H 2SO 4
X
X
 36.8g H 2SO 4
1000 mL
1L
1mol H 2SO 4
Example 3:
Calculate the volume of solution that can be made using a 6.00 M solution using 45.0 g C6H12O6.
Volume =
45.0g C6 H12 O 6
X
1mol C6 H12 O 6
1L
X
 0.400 L
180.0g C6 H 12 O 6 6.00 mol C6 H 12 O 6
Molality (m) =
moles of solute
kg of solution
Example 1:
Calculate the molality of a solution containing 50.0 g of HC2H3O2 dissolved in 500.0 g Hm=O.
m=
50.0g HC 2 H 3O 2 1mol HC 2 H 3O 2 1000g H 2O
X
X
=1.67 m
500.0g H 2O
60.0gHC 2 H 3O 2
1kg H 2O
Example 2:
Calculate the mass of solute needed to make a 0.450 m NaOH solution containing 750.0 g H2O.
mass solute =
750.0g H 2O
X
1kg
0.450 mol NaOH 40.0 g NaOH
X
X
=13.5g NaOH
1000g
1kg
1mol NaOH
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 49
Unit 7, Activity 3, Solution Concentrations
Example 3:
Calculate the mass of solvent needed to make a 2.50 m H2SO4 solution containing 150.0 g of the
acid.
mass H2O =
150.0g H 2SO 4
X
1mol H 2SO 4
1kg H 2O
1000g
X
X
= 612 g H 2O
98.1g H 2SO 4 2.50 mol H 2SO 4
1kg
Mole Fraction (X) =
moles of solute (or solvent)
moles of solute + solvent
Example:
Determine the mole fraction of glucose, C6H12O6, in a solution containing 425 g glucose
dissolved in 750.0 g H2O.
Moles of glucose:
425g C6 H12 O6
X
1mol C6 H12O6
= 2.36 mol C6 H12O6
180g C6 H12 O6
Moles of water:
750.0 g H 2O
X
1mol H 2O
= 41.7 mol H 2O
18.0 g H 2 O
Total moles of solute + solvent:
2.36 mol + 41.7 mol= 44.06 mol
X  molglu cos e
Mole fraction of C6H12O6:
XC6H12O6 =
molC6H12O6
mol C6H12O6 +H2O
=
2.36 mol
= 0.5
44.06 mol
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 50
Unit 7, Activity 5, Ice Cream Recipe
Recipe for 40 pint size bags of ice cream (1 bag per student):
1 gal whole milk
1 pint half & half
6 cups sugar
6 t vanilla
Additional materials needed: spoon, large pot, several boxes of ice cream salt; two rolls of duct
tape, several large bags of ice, enough newspaper for each student to have a section, plastic bags
from grocery or discount stores
Combine all ingredients in the pot to make the ice cream mixture and heat until the sugar is
dissolved. Stir the mixture often to prevent the mixture from scorching. Pour into the empty
milk container. There may some extra mixture, so have a smaller container or zip top bag handy
also.
Per student:
1 pint size freezer zip top bags
1 gallon size freezer bags
2 plastic bags
a section of newspaper
ice cream salt
ice
duct tape
Directions:
Fill the gallon bag half full of ice. Add 1/2 inch layer of ice cream salt. Put 1/2 cup of ice cream
mixture in small bag. Seal the small bag and place duct tape over the sealed end. Put the small
bag inside of large one. Add enough ice to fill the gallon bag. Seal and duct tape the sealed end
of the large bag. Wrap the large bag with several layers of newspaper. Place the wrapped bag in a
couple of plastic bags. Tie the ends of the plastic bags. Shake or rotate the bags gently for about
15 min. Have a large container such as a dish pan handy to empty the water/ salt mixture into.
The water can be evaporated and the salt reused, if desired.
Powdered drink mixes can be made according to the directions on the package and used in place
of the ice cream. A “slush” will be formed.
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 51
Unit 7, Activity7, pH Lab Carousels
Lab Carousel 1: At each station place: a microplate, red litmus paper, blue litmus paper,
universal indicator and phenolphthalein (and any other available indicators), and stirring rod.
Pipettes containing:
Station 1: vinegar
Station 4: household ammonia solution
Station 2: distilled water
Station 5: colorless soda
Station 3: KOH solution
Station 6: HCl solution
Instruct students to test each solution with the indicator papers and indicators at each station and
identify each of the solutions as acids or bases. Data should be recorded in a student –generated
data table.
Lab Carousel 2: At each station place a microplate, pH paper and/ or pH meter, and stirring rod.
Pipettes containing:
Station 1: vinegar
Station 4: household ammonia solution
Station 2: distilled water
Station 5: colorless soda
Station 3: KOH solution
Station 6: HCl solution
Instruct students to determine the pH of the solutions and rank them in order of increasing pH.
Data should be recorded in a student –generated data table.
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 52
Unit 8, Activity 2, Alkanes
Alkanes are saturated hydrocarbons (compounds containing only carbon and hydrogen) with the
formula CnH2n+2, where “n” represents the number of carbon atoms. “Saturated” means that all
C-C bonds are single bonds.
Names of organic compounds follow the rules of IUPAC (International Union of Pure and
Applied Chemistry). Notice that each compound differs from the previous one by a –CH2 group.
A homologous series in one in which the compounds differ from each other by a specific unit.
The pattern for the first 10 alkanes is shown below.
Stem name
Alkane name
Formula
meth-
methane
CH4
Number
of isomers
1
eth-
ethane
C2H6
1
prop-
propane
C3H8
1
but-
butane
C4H10
2
pent-
pentane
C5H12
3
hex-
hexane
C6H14
5
hept-
heptane
C7H16
9
oct-
octane
C8H18
18
non-
nonane
C9H20
35
dec-
decane
C10H22
75
Isomers are compounds with the same molecular formula but different structural formulas.
Draw the isomers for pentane and hexane.
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 53
Unit 8, Activity 2, Alkanes
Rules for naming alkanes:
1. Pick out the longest continuous chain of carbon atoms and name it.
2. Number the carbon atoms from the end that will give the lowest numbers possible to the
branches.
3. Name the branches by adding –yl to the stem name and adding a number to indicate the
carbon atom the branch is attached to. The number will be followed by a dash. All
branches must have a number with it. Numbers are separated by commas.
*If branches are different groups, they appear alphabetically in the name.
4. If more than one of an alkyl group appears, a number prefix is used to denote the total
number of groups.
5. Dashes between carbon atoms do not need to be shown.
Examples:
6
5
4
3
CH3  CH2 CH2 CH CH3

2 CH2

1 CH3
Name: 3-methylhexane
CH3
CH3


CH3CHCHCH2CHCH2 CH3

CH2

CH3
Name: 3-ethyl-2,5-dimethylheptane
1
2
3
4
5
CH3 CH CH2 CH CH3


CH3
CH3
Name: 2,4-dimethylpentane
CH3

CH2

CH3CHCHCH3

CHCH2CH3

CH3
Name: 3,4,5-trimethylheptane
Name the isomers for pentane and hexane that were drawn on the previous sheet.
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
Page 54
Unit 8, Activity 2, Alkanes Answer Sheet
Isomers of Pentane (formula C5H12):
1. CH3CH2CH2CH2CH3
n-pentane ( n means normal straight chain)
2. CH3CHCH2CH3
2- methylbutane

CH3
3.
CH3
2,2-dimethylpropane

CH3CCH3

CH3
Isomers of hexane (formula C6H14)
1. CH3CH2CH2CH2CH2CH3
n-hexane
2. CH3CHCH2CH2CH3
2-methylpentane

CH3
3. CH3CH2CHCH2CH3

3-methylpentane
CH3
4. CH3 CH CH CH3


2,3-dimethylbutane
CH3 CH3
CH3

5. CH3 C CH2 CH3

2.3-dimethylbutane
CH3
Blackline Masters, Chemistry
Louisiana Comprehensive Curriculum, Revised 2008
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