LESSON PLAN
“FUNDAMENTAL LAWS OF
CHEMISTRY”
Written by :
ARNIDA DEWANTARI
(K3309016)
Chemistry Education Programm
Departement of Mathematics and Natural Science Education
Faculty of Teacher Training and Education
Sebelas Maret University
2012
LESSON PLAN FOR LEARNING
Education Unit : SMA Negeri 1 Kudus
Subject : Chemistry
Grade/Semester : X / I
Subject Matter : Stoichiometry
Sub Subject Matter : Fundamental Laws of Chemistry
Standard of competence : To understand the fundamental laws of chemistry
Base competence : To understand the Law of Conservation of Mass
(Lavoisier’s Law), the Law of Definite Proportions
(Proust’s Law), the Law of Multiple Proportions
(Dalton’s Law), the Law of Combining Volumes (Gay
Lussac’s Law) and the Avogadro’s Hypothesis.
Duration : 1 X 30 minutes (1 X meeting)
I. Indicator
A. Cognitive
1. Product
a. Explain principle of the law of conservation of mass (Lavoiser’s Law)
that mass of substance before reaction equals with mass after reaction
b. Explain principle of the law of definite proportions (Proust’s Law) that
mass of two substance which make compound
c. Explain the law of multiple proportions (Dalton’s Law) for elements
forming than one compound
d. Explain the law of combining volumes (Gay Lussac’s Law) for
chemical reactions that involved gases
2. Process
a. Calculate the mass of particles based on the law of conservation of
mass for chemical reactions
b. Determine mass of compound that produced in a chemical reaction
based on the law of definite proportion
c. Calculate the mass of particles based on the law of multiple
proportions for chemical reactions
d. Determine ratio volume of compound that produced in a chemical
reaction based on the law of combining volumes
B. Affective
1. Character
a. Honest
b. Curiosity
c. Toil
d. Creative
e. Discipline
f. Confidence
g. Responsibility
2. Social Skills
a. Skillful Cooperation
b. Communicative
II. Learning Objective
A. Cognitive
1. Product
a. Student can explain principle of the law of conservation of mass
(Lavoiser’s Law) that mass of substance before reaction equals with
mass after reaction
b. Student can explain principle of the law of definite proportions
(Proust’s Law) that mass of two substance which make compound
c. Student can explain the law of multiple proportions (Dalton’s Law)
for elements forming than one compound
d. Student can explain the law of combining volumes (Gay Lussac’s
Law) for chemical reactions that involved gases
2. Process
a. Calculate the mass of particles based on the law of conservation of
mass for chemical reactions
b. Determine mass of compound that produced in a chemical reaction
based on the law of definite proportion
c. Calculate the mass of particles based on the law of multiple
proportions for chemical reactions
d. Determine ratio volume of compound that produced in a chemical
reaction based on the law of combining volumes
B. Affective
1. Character
a. Honest
b. Curiosity
c. Toil
d. Creative
e. Discipline
f. Confidence
g. Responsibility
2. Social Skills
a. Skillful Cooperation / Teamwork
b. Communicative
III. Learning Matter
1. The Law Of Conservation Of Mass (Lavoisier’s Law)
Antoine Lavoisier (1743-1794),a French chemist, was one of the first to insist
on the use of the balance in chemical research. Lavoisier demonstrated that when
hydrogen gas (H2 ) burns and combines with oxygen gas (O2) in a closed
contAiner to yield water (H2O), the mass of water obtained is equal to the mass of
hydrogen and oxygen gases consumed.
2 H2 (g) + O2 (g) → 2H2O
The Lavoisier’s experiment straightened the Priestley’s observation
and brought down the phlosgiston theory. From this experiment and man others,
Lavoisier discovered that the total mass of the substances during a chemical
reaction experiences no change. Based on this observation, he then formulated the
law of conservation of mass (Lavoisier’s Law) which states :
“In a chemical reaction, the total mass of substances before and after the
reaction remains the same.”
Example :
254 g of copper and 128 g of sulfur react completely and form compound of
copper sulfide. According to the law of conservation of Mass, how much copper
sulfide can be obtained from the reaction?
Answer :
Mass of the substances before reaction = mass of the substances after reaction
Mass of copper + mass of sulfur = mass of copper sulfide
254 g + 128 g = mass of copper sulfide
Mass of copper sulfide = 382 g
2. The Law Of Definite Proportions (Proust’s Law)
The substances which were later called compounds, had elements with a fixed
ratio regardless of whether the compounds were natural or are synthesized. In
1779 a French chemist, Joseph Proust (1754-1826) made an attempt to prove the
general acceptance of this phenomenon. One of the experiments he conducted was
by reacting hydrogen and oxygen. Proust found out that hydrogen and oxygen
could combine and form water compound with a fixed ratio of 1 : 8.
Mass of hydrogen : mass of oxygen = 1 : 8
Proust soon discovered that the compounds always contain elements with a
certain fixed ratio. Based on this, he formulated the Law of Definite Proportions
(Proust’s Law) which states :
“A chemical compound always contains the same proportion of elements by
mass.”
Example :
In an electrolysis process, 18.0 g of water is decomposed into 2.0 g of hydrogen
and 16.0 g of oxygen.
a. Determine the masses of hydrogen and oxygen that can be obtained from
electrolysis of 50.0 g of water.
b. What is the mass of water needed to obtain 100.0 g of oxygen ?
Answer :
a. The proportion by mass of hydrogen and oxygen in water = 2 g : 16 g = 1 : 8
electrolysis of 50.0 g of water will produced :
- Mass of hydrogen =
19 x 50.0 g = 5.6 g
- Mass of oxygen =
89 x 50.0 g = 44.4 g
b. Mass of water needed for electrolysis
=
98 x 100.0 g = 112.5 g
3. The Law Of Multiple Proportions (Dalton’s Law)
John Dalton (1766-1844) discovered a new law; a development Proust’s
Law according to proust’s law, a compound is composed of elements could combine
and form more then one type of compound. Dalton observed a certain order related
to the ratio of elements in at he compounds.
To understand this, take a look at the experiment between nitrogen and
oxygen that produces two types of compounds : nitrogen oxide I and nitrogen oxide
II.
- in the first experiment, 0.875 g nitrogen reacts with 1.00 g of oxygen and
produces nitrogen oxide I
- in the second experiment, the mass of nitrogen is increased to 1.75 g while the
mass of oxygen remains the same. This reaction produces a different
compound is nitrogen oxide II.
In turned out that, with the same mass of oxygen, the ratio of the masses of
nitrogen in the two compounds take the form of simple whole numbers.
Mass of nitrogen in a nitrogen oxide I
Mass of nitrogen in nitrogen oxide II
= 0.875 g : 1.75 g
= 1: 2
Based on his observation, Dalton formulated the Law Of Multiple Proportions
(Dalton’s Law) :
“If two types of elements combine and form more than one compound, and it the
mass of one of the elements in the compounds is same, then the ratio of the masses
of the others element in a compounds will take the form of simple whole numbers.”
Example :
A chemist reacted carbon with oxygen and obtained two different compounds. The
composition of carbon and oxygen in the first compound was 42.9 % of carbon and
57.1 % of oxygen while the second compound contained 27.3 % of carbon and 72.7
% of oxygen. Verify that the proportion by mass of oxygen in both compounds
supports the law of Multiple Proportion.
Answer :
Assume there are 100 g of compound I and 100 g of compound II.
Mass of
compound
Mass of
carbon
Mass of
oxygen
Mass of carbon : mass of oxygen
Compound I 100 g 42.9 g 57.1 g 42.9 g : 57.1 g = 1 : 1.33
Compound II 100 g 27.3 g 72.7 g 27.3 g : 72.7 g = 1 : 2.66
The ratio by mass of oxygen in both compounds for every gram of carbon
= The ratio of oxygen in compound I : The ratio of oxygen in compound II
= 1.33 g : 2.66 g
= 1 : 2
The ratio by mass of the oxygen in both compounds takes the form of simple whole
numbers, as state in the Law of Multiple Proportion.
e. The Law Of Combining Volumes (Gay-Lussac’s Law)
Joseph Louis Gay-Lussac (1778-1850) is a French scientist who conducted a
study on gases with accurate quantitative measures. When studying the composition
of oxygen in air, he was interested in the chemical reaction between hydrogen and
oxygen gases formed water vapor. Gay-Lussac then observed if measured at
constant T and P, for every 2 volumes of hydrogen gas and 1 volume of oxygen gas,
2 volumes of water vapor is obtained.
Hydrogen gas + oxygen gas → water vapor
2 volumes : 1 volume : 2 volumes
From the observation, which was tested for its general validity, in 1808 Gay-Lussac
formulated the Law of Combining Volumes (Gay-Lussac’s Law) for reactions that
involve gases :
“ At the constant temperature and pressure, the ratio of the volumes of gases
consumed or produced in chemical reaction takes the form of the simple whole
numbers.”
Example :
100 volumes of gas X is decomposed into 50 volumes of gas Y and 75 volumes of
gas Z.
a. Calculated the proportion by volumes of the gases involved in the reaction.
b. For every 2.0 L of gas X, calculated the amount of gas Y and gas Z produced.
Answer :
a. Proportion by volume
Gas X Gas Y + Gas Z
100 volumes : 50 volumes : 75 volumes
4 volumes : 2 volumes : 3 volumes
b. Given that, gas X : gas Y : gas Z = 4 : 2: 3
For every 2.0 L gas X produced :
Volume of gas Y formed
=
24 x 2.0 L = 1.0 L
Volume of gas Z formed
=
34 x 2.0 L = 1.5 L
IV. Learning Method
Approach : Constructivism
Model / Learning Methode : Cooperative model with type of Make A Match
V. Learning Activities
Learning Step
No. Activity Time Character
A. Initial Activity
5 minutes
a. Apperception
Teacher burn a piece of paper
that already known the mass
and then asking the student,
“Are the mass of the dust is
same with the piece of paper?”.
We can finding the answer in
this material today.
Curiosity
b. Orientation
Inform learning objective today
Told to apply type of Make A
Match
Discipline
c. Motivation
Outlined the benefits of learning
that materials for the future
Creative, Toil
B. Main Activity
1. Exploration
Teacher explain material about
the Law of Conservation of
Mass (Lavoisier’s Law), the
Law of Definite Proportions
(Proust’s Law), the Law of
Multiple Proportions (Dalton’s
Law), and the Law of
Combining Volumes (Gay
10 minutes
Curiosity,
Toil
Lussac’s Law)
2. Elaboration
a. Teacher informing the way
of doing cooperative with
Make A Match type 10 minutes
Communicative
b. Teacher divide student into
some heterogeneous group
Confidence
c.Teacher prepare some of card
content concept, one part is
question card and the other is
answer card then give that cards
to the each group
Confidence,
Teamwork, Toil
d. Each of group thinking the
answer/question from the card
which is gotten
Toil, Teamwork,
Communicative
e.Each of student stick each of
card in the whiteboard
Discipline, Toil
f. Each of group which can
finish for matching that cards
firstly given the point
Toil, Discipline
3. Confirmation
a. Teacher check all the answer 2 minutes
Curiousity,
Responsibility
C. Final Activity
3 minutes
a. Students with guidance from
the teacher concludes lesson
material today
Creative,
Communicative
b.Teacher gives homework for
doing Evaluation page 145
book of Theory and Application
of Chemistry
Responsibility,
Toil, Discipline
c.Teacher suggest the student to
study about the material for the
Responsibility
next meeting
VI. Learning Sources, Tools and Materials
A. Learning SourcesProf. Effendy, Ph.D. A- Level Chemistry For Senior High School Students. published by Bayumedia Publishing.
Purba, Michael. 2004. Kimia untuk SMA utuk Kelas X. Jakarta: Erlangga
Sunardi. 2008. Kimia Bilingual SMA/MA Kelas X. C V Yrama Widya
Susilowati, Endang. 2009. Teory and application of Chemistry 1. Surakarta: PT. Tiga Serangkai
B. Tools and MaterialsPowerpoint of Fundamental Laws, Cards, Whiteboard, Markers, LCD
VII. Assessment
1. Kind of Assessment
- Essay question (Cognitive)
- Student activity in discussion and tourmanent (Affective)
2. Instrument
a. 1 gram of sodium exactly to react with 1.54 grams of chlorine gas
produce sodium chloride. Calculate mass of chlorine gas required to
produce 7.62 grams of sodium chloride. ( max score 5)
b. 4 grams of copper exactly to reacts with 2 grams of sulfur to form
copper sulfide. What is gram of copper sulfide can be formed if 10
grams of copper and 10 grams of sulphur are reacted? ( max score 7.5)
Answer
a. Because of sodium exactly to react with chlorine, then,
Sodium chloride produced = Sodium mass + Chlorine mass
= 1 g + 1.54 g = 2.54 g
To produced 7,62 grams of sodium chloride is required chlorine:
7 .622. 54 x 1.54 g = 4.62 g
Thus, chlorine requires as much as 4.62 grams.
If the answer is correct entirely, the point is 5
If the answer just correct partly, the point is 2.5 (just show the mass of
Cl)
b. The ratio of mass of copper : mass of sulfur = 4 g : 2 g = 2 : 1
So,
Mass copper: mass of sulfur
2 : 1
10 g : 5 g
Point 2.5
Mass of copper sulfide = mass of copper + mass of sulfur
= 10 g + 5 g
= 15 g
Point 2.5
Total point 5
Final value =
pq
x100, where p = number of score obtained
q = total score (10)
The assessment of cognitive also can be gotten from score in matching the
answer/question when doing games of Make A Match. Each of student in
group will be given score 100 if the match is correct et al.
Surakarta, April 3rd 2012
Lecture
Prof. Dr. Ashadi
NIP. 130 516 325
Student
Arnida Dewantari
K3309016
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