Read Ch. 3 pg. 55-74
Ch. 3 – Matter: Properties and Changes
Properties of Matter:
Pure substance –
Ex.
Physical properties –
Ex.
-
Extensive properties
o Ex.
-
Intensive Properties
o Ex.
Chemical Properties –
Ex.
States of matter –



Changes of Matter:
Physical changes –
Ex.
Chemical Changes –
Ex.
Conservation of Mass –
Ex.
Mercury (II) oxide yields mercury + oxygen (show formulas and masses)
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Properties of Matter:
Mixtures –
Ex.
Homogeneous mixtures – also called “
”
Heterogeneous mixtures –
An alloy is:
Ex. (find three that are not listed in the section)
Separating Mixtures
- Mixtures are ______________ combined, but the processes used to separate a
mixture are _______________ processes.
Filtration –
Distillation –
Crystallization –
Chromatography –
Elements and Compounds
Elements –
Ex.
Periodic table –
Compounds –
Ex.
- Separation of a compound into its elements…
- The properties of a compound are different…
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Fig. 3-17
The following to be completed in class
Identify the following as extensive or intensive properties:
Mass, boiling point, density, length, melting point, volume, amount of stored energy, ability to conduct
Extensive
Intensive
Identify each as a mixture (hetero/homo) or pure substance (element/cpd). Use columns below.
water
bronze
cookie dough
potassium oxide
kool-aid
copper
Homogenous Mixture
concrete
sulfur
pizza
pop
Heterogeneous Mixture
air
salt
sugar
oxygen
Element
Compound
Identify the following as chemical/physical changes:
Campfire
Baking a cake
Boiling water
Burning a chemistry book
rusting steel
ice melting
lighting a match
mixing kool-aid
rotting garbage
Evidence of a chemical change:
1)
2)
Dissolving sugar in water
separating sand and water w/filter
Forming a precipitate
Evaporating water
3)
4)
Read Ch. 3 pg. 75-77
Law of definite proportions –
Percent by mass “% composition” = ____________________
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Ex. Table 3-4
Element
Carbon
Analysis by mass (g)
8.44
Hydrogen
1.30
Oxygen
10.26
Percent by mass (%)
Total
Ex. Table 3-5
Element
Carbon
Analysis by mass (g)
211.0
Hydrogen
32.5
Oxygen
256.5
Percent by mass (%)
Total
Law of Multiple Proportions –
Exs. H2O vs. ______
Cu and Cl
(take time to understand conclusions drawn from these numbers)
To be completed in class
Why we believe in atoms
• Atomic theory was developed by indirect observation
…that is they did not see atoms directly
Atomic Theory explains the following:
1) Law of Conservation of Mass - mass is neither created nor destroyed
KEY:
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2) Law of definite (constant composition) proportions - pure compounds are found to have
a definite composition, no matter how they are made or where they are found.
KEY:
3) Law of Multiple Proportions - if there is more than one compound made of the same
elements, when the amount of one element is held constant the amount of the other
element will be in a small whole number ratio.
KEY:
Quantitative applications of the 3 Natural Laws….
Table 3-6: Analysis Data of Two Copper Compounds
Compound
% Cu
% Cl
I
II
64.20
47.27
35.80
52.73
Mass copper (g)
in 100.0 g of
compound
64.20
47.27
Mass Chlorine
(g) in 100.0 g of
compound
35.80
52.73
Mass Ratio
(Mass Cu
Mass Cl)
1.793 g Cu/1 g Cl
.8964 g Cu/1 g Cl
1. Using table 3-6 above, show how the data demonstrates the law of multiple proportions.
2. If the formula for compound I is CuCl, what is the formula for compound II? (Use table 3-6 and your
answer to number 5.)
3) A 75.00-g sample of compound made of iron and oxygen is analyzed to contain 52.46 grams of iron.
a. How many grams of oxygen must be present?
b. What is the percent composition by mass of each element?
c. A 56.00 g sample of another compound of iron and oxygen is found to contain 43.53 grams of iron and
12.47 grams of oxygen. Could this be the same compound? If the compounds are different, use the
law of multiple proportions to show the relationship between them.
4) Given: H2 + ½ O2  H2O
How many grams of water will be formed if 8.00 grams of hydrogen,
H2, react completely with 64.0 grams of oxygen, O2?
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5) Given: 2 A + B2  2 AB
12.0 grams of A completely react with B2 to form 18.0 grams of AB.
How many grams of B2 must have reacted?
6. Given the following data, a. what is the formula for compound II if Compound I is CO?
b. what is the formula for compound II if Compound I is C2O?
c. what is the formula for compound II if Compound I is CO2?
Cpd I
Mass Oxygen
w/ 1 g C
1.33 g
Cpd II
2.66 g
CO
C2O
CO2
7. Given the following data, what is the formula for compounds II and III if Compound I is A 2B?
Cpd I
1.0 g A: 2.0 g B
Cpd II
2.0 g A: 4.0 g B
Cpd III
1.0 g A: 4.0 g B
A2B
Which are the same compound? _______
What data exemplify the law of constant composition? __________
What data exemplify the law of multiple proportions? __________
8) Knowing what you do about the law of constant composition, determine which of the following
compounds could be the same.
Mass A: Mass B
Compound I
12 g A : 6 g B
Compound II
9gA
: 3gB
Compound III
10 g A : 5 g B
9) Given your answer to question 8, if the formula for compound I was A2B6, what is the formula for
Compound II? _______
Compound III? _______
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