Lavoisier Law ,Law of mass conservation. “The total amount of matter of the reaction compounds remains constant”

1. LAVOISIER LAW
CHEMICAL PRACTICE REPORT
By :
Adinda Melinda C.A
Amalia Prasiwi
Lazulfa Inda Lestari
Nurul Fatimah
Rifa Abriyanti
Tanza Lona Trista
X-8
SMA NEGERI 2 CIMAHI
2011

2. Experiment 1
I. Title : Lavoisier Law
II. Goal : To proof that the mass of the substance before and after reaction is
the same.
III. Theory : Based on Lavoisier Law ,Law of mass conservation. “The total
amount of matter of the reaction compounds remains constant”
IV. Materials and Equipment :
a. Y tube
b. O’hauss Balance
c. Pippet
d. Pb(NO3)2 liquid
e. KI liquid
V. Procedure :
a. Measure the Y tube into O’hauss Balance in empty condition.
b. Input ten drops of Pb(NO3)2 liquid to the one part of the Y tube.
c. Input ten drops of KI liquid on the other parts of the Y tube .
d. Measure again the Y tube wich is contains both of the liquid using O’hauss
Balance.
e. Mix both of the liquids in one side of Y tube.
f. Again, measure the Y tube wich is contains the result of reaction.
VI. Observation :
Mass before reaction Mass after reaction
32,05 gr 32,05 gr
VII. Question :
1. Write down the chemical equation from the experiment !
2. From the experiment ,explain the Lavoisier law !
Answer :
1. 2KI + Pb(NO3)2 PbI2 + 2KNO3
2. Lavoisier states that the total amount of matter of the reaction compounds
remains constant.

3. VIII. Conclution :
From this experiment we can proved the truth of the Lavoisier law, law of Mass
Conservation. “mass of the substance before and after reaction is the same”
Experiment 2
I. Title : Identification C,H,O
II. Goal : To identify C,H,O by evaporation
III. Theory : The existence of the elements carbon and hydrogen in organic
samples, more certain can be shown through chemical means, namely
by the combustion test.
IV. Materials and Equipment :
a. spirtus
b. matches
c. 2 reaction tubes
d. pippet
e. spatula
f. gas pipeline
g. Clamp
h. Sugar (C12H22O11)
i. Ca(OH)2 liquid
j. CuO powder
V. Procedure :
a. Pour a half spatula of CuO into the reaction tube.
b. Pour a spatula of sugar into the same reaction tube.
c. Mix both of the materials.
d. Input the Ca(OH)2 liquid into the second reaction tube.
e. Set all equipments based on the picture :

4. VI. Question :
1. Write down the chemical formula of burning sugar and functions of CuO !
2. Why does appears lime water become cloudly ?
3. Make chemical equation for number two !
4. What happened with cobalt paper after tested ?
5. Write the chemical equation for number four !
Answer :
1. C12H22 O11 (S) + CuO(S) CO2(g) + H2O(g) + C(s) . Function of CuO is as an
oxidant
2. Because the reaction produce CO2 ,then it reacts with limewater (Ca(OH)2)
.So,the colour of lime water become transparent and cloudly.
3. Ca(OH)2(aq) + CO2 (g) CaCO3(s) + H2O(g)
4. The colour of cobalt paper turns from blue to pink.
5. COCl2(s) + 6H2O(g) COCl.6H2O
VII. Conclution :
From this experiment we can prove that in the sugar compounds there are elements C
(carbon) ,H (Hydrogen),O (Oxygen). The existence of elements of C can be evidenced by
color changes that occur in the limewater solution becomes turbid. While the elements of H
can be proved by using cobalt blue paper. Compounds having these elements is an Organic
Compounds.

5. Experiment 3
I. Title : Redox reaction
II.Goal : To prove the redox reaction between Zn and CuSO4.
III.Theory : There are three concepts used to explain the definition of redox
reaction. Those three concepts are the release and uptake of oxygen
atom,the electrons transfer,and the change in oxidation number.
IV.Materials and Equipment :
a. Test tube
b. A piece of Zn (zinc)
c. CuSO4 Liquid
V. Procedure :
a. Pour the CuSO4 liquid into the test tube.
b. Put Zn into the same test tube.
c. Wait for some hours.
d. Observe that occur on Zn and CuSO4.
VI. Observation :
Subtance Before reaction After reaction
Zn  Had silver colour
 Solid
 Dissolved
 The colour changed to
black.
 Changed to ZnSO4
CuSO4  Had blue colour
 Liquid
 React with Zn
produce ZnSO4
 Cu be a red black
sediment.
 Had transparent
colour.
VII. Question :
1. Determine the oxidation number for each substance !
2. Determine the oxidizers and reducers !

6. Answer :
1. Zn + CuSO4 ZnSO4 + Cu
Reactant
 Zn = 0
 S in SO4
-2 = (1x S oxydation number) + (4x O oxydation number)
-2 = S + (-8)
-2+8 = S
S = 6
 Cu in CuSo4 = (1 x Cu) + (1 + S) + (4x O )
0 = Cu + 6 + (- 8)
Cu =2
Product
 Zn in ZnSO4 = (1xZn) + (1xS) + (4xO)
0 = Zn + 6 + (-8)
Zn = 2
 Cu = 0
2. Reducer : Zn
Oxidator : Cu
VIII. Conclution :
From this experiment we can prove that redox reaction happen between Zn and CuSO4 .With
Zn as Reducer and CuSO4 as oxidator.