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