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Chapter Two:
ATOMS, MOLECULES,
AND IONS
Performance goal
•
Students should be able to:
•
Show how the postulates of the atomic theory “ explain” the laws of
conservation of Mass, Constant Composition (definite proportions) and
multiple proportions.
•
Describe the atomic structure
•
Given the composition of a nucleus, give the atomic and mass
numbers. Perform the reverse calculation
•
Distinguish between atom, molecule and ion. Give example of each.
•
Understand and know how to use the periodic table
•
Name simple compounds
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2–2
2.2 Fundamental
Chemical Laws
Law of conservation of mass
Antoine Lavoisier (1743-1794) by carefully weighing
the reactants and products of various reactions made
the following suggestion
• Mass cannot be created nor destroyed.
Show Law of conservation video of mercury (II) oxide
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2–4
• You have a chemical in a sealed glass container filled
with air. The setup is sitting on a balance as shown
below. The chemical is ignited by means of a magnifying
glass focusing sunlight on the reactant. After the
chemical has completely burned, which of the following is
true? Explain your answer.
– A. the balance will read less than 250.0g.
– B. The balance will read 250.0g.
– C. The balance will read greater
than 250.g
– D. Cannot be determined without
knowing the identity of the chemical.
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2–5
Learning Check
• A 0.406 g sample of magnesium reacts
with oxygen, producing 0.674 g
magnesium oxide as the only product.
What mass of oxygen was consumed in
the reaction?
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2–6
Law of definite proportion
Joseph Proust (1754-1826), proposed the constant
composition of compounds that states
• A given compound always contains
exactly the same proportion of elements
by mass
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2–7
Example of the Law Definite Proportions
• Let’s consider water
Sample A weighed 10.000 g
1.119 g H
8.881 g O
Sample B weighed 27.000 g
3.021 g H
23.979 g O
Composition is always
% H = 11.19
% O = 88.81
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2–8
Learning Check
• 3.06 g of hydrogen react with oxygen where 27.35 g
of water are obtained. In a second experiment, water
is decomposed resulting in 1.45 g of hydrogen &
11.51 g of oxygen. Is this consistent with the law of
definite proportions?
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2–9
Law of multiple proportions
John Dalton was a strong proponent of the Law of Multiple
Proportions (which is the 3rd postulate of Dalton’s atomic theory.
• When two elements form a series of
compounds, the ratios of the masses of
the second element that combine with
a fixed mass (ex. 1 gram) of the first
element can always be reduced to
small whole numbers.
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2–10
Example of the Law of Multiple Proportions
• That of nitrogen and oxygen forming three different
compounds that contain different relative amounts of
nitrogen and oxygen.
Compound A
Compound B
Compound C
Mass of Nitrogen that
combines with 1 g of Oxygen
1.750 g
0.8750 g
0.4375 g
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2–11
Example of the Law of Multiple Proportions
continued
The ratios of the masses of nitrogen combining with 1
gram of oxygen in each pair of compound should be
small whole integer number
A = 1.750
= 2 for nitrogen
0.8750
B
0.8750 = 1 for nitrogen, a ratio 2:1
0.8750
B = 0.8750
= 2 for nitrogen
0.4375
C
0.4375 = 1 for nitrogen, a ratio 2:1
0.4375
A = 1.750
= 4 for nitrogen
0.4375
C
0.4375 = 1 for nitrogen, a ratio 4:1
0.4375
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2–12
Which of the following pairs of compounds can be used to
illustrate the law of multiple proportions?
•
•
•
•
•
•
CH4 and CO2
NH4 and NH4Cl
NO and NO2
H2O and HCl
ZnO2 and ZnCl2
CO and CO2
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2–13
Which of the following pairs of compounds can be used to
illustrate the law of multiple proportions?
•
•
•
•
•
•
CH4 and CO2
NH4 and NH4Cl
NO and NO2
H2O and HCl
ZnO2 and ZnCl2
CO and CO2
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2–14
QUESTION
Two common air pollutants when analyzed yield the following
information; Compound I shows that 1 gram of sulfur combines
with 1 gram of oxygen. Compound II shows that 1.5 grams of
oxygen combines with 1 gram of sulfur. This is consistent with
the law of multiple proportions because…
1. the analysis shows that sulfur and oxygen can combine in
more than one way.
2. the analysis shows that the combination of sulfur and
oxygen is very close to 1 in both cases.
3. the analysis shows that a small whole number (2) could
be used to make a ratio of 2:3 for the oxygen mass in
compound I compared to compound II.
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2–15
ANSWER
Choice 3 offers the same conclusion as the actual Law of
Multiple Proportions. If 1 and 1.5 are both multiplied by 2 the
ratio is still maintained (1.5 to 1 is the same as 3:2 so
multiplying by 2 does not change the ratio and does produce
whole numbers for both).
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2–16
Learning Check
• Two different compounds are formed by the
elements carbon & oxygen. The first
compound contains 42.9% by mass carbon
and 57.1% by mass oxygen. The second
compound contains 27.3% by mass carbon
and 72.7% by mass oxygen. Show that the
data are consistent with the Law of Multiple
Proportions. (assume 100 g is in each
compound)
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2–17
Show your work
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2–18