Group 1 Experiment 14 and 15

EXPERIMENT 14
Chemical Properties of Water
OBJECTIVE REAGENTS
Calcium Oxide
To demonstrate Sulfur Powder
some of the Coper (II) Sulfate
chemical properties Iron (III) Chloride
Cobalt Chloride Hexahydrate
of water
Materials: 4 Test Tubes, Evaporating Dish, Bunsen Burner, Glass Rod, 10-mL Graduated
Cylinder, Test Tube Brush, Medicine Dropper, Spatula, Test Tube Rack, Test Tube Holder,
Red Litmus Paper
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GROUP 1
Anciro, Arafhel
Aquino Rosette
Aupe, Joanna
Bañez, Bea Coline
Casupang, Rhynell
Cervantes, Bryle
Chavez, Ronessa
Combalicer, Flordeliza
Davi, Shakirah
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PROCEDURE
4
PROCEDURE
#1
Place a lump of calcium oxide in an evaporating dish. Add
water dropwise until it is completely wet. Observe any
reaction that occurs. When the reaction is over, drop a piece
of red litmus paper and observe.
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Observations and Results
Calcium Oxide + Water
 The Calcium Oxide dissolves when a
drop of water was added.
With Red Litmus Paper
 When Red Litmus Paper was added in
Calcium Oxide with Water it turned
blue.
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PROCEDURE #2
Ignite a small amount of sulfur powder contained in one end of
the glass tubing. Insert the end of the glass tubing with the
burning sulfur in the mouth of a test tube. Quickly add 2-3 mL
of water and cover the test tube tightly. Shake vigorously for a
few minutes. Drop a piece of blue litmus paper and observe.
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Sulfur + Heat
 The Sulfur powder produce fumes and the residue of the
sulfur powder in the glass tube became slightly dark and
melt.
Sulfur Fumes + Water
 The Sulfur fumes and water did not intermixed.
With Blue Litmus Paper
 When the blue litmus paper was put in the test tube with
water and finished heating, the result became red.
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PROCEDURE #3
Place a few crystals of Copper (II) Sulfate into a test
tube. Observe their color. Heat the test tube gently until
the crystals are dehydrated. Allow the test tube to cool,
then add 2-3 drops of water. Observe what happens.
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Before Heating
 The texture is rough and is color blue. Have
crystalline structure.
After Heating
 The color became white
Heated CuSO4 + Water
 The white powder was dissolved completely into
water and the color turned to blue.
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PROCEDURE #4
Place a few crystals of FeCl3 into a test
tube. Add 5 mL of water and heat until the
solution boils.
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Ferric Chloride + Water
 When we put the water in the test tube with
Ferric Chloride, the color became light
yellow.
After Heating
 The Ferric Chloride became creamy orange.
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PROCEDURE #5
Heat a few crystals of Cobalt Chloride Hexahydrate in a
test tube. Let the residue cool, and then dissolve it in a
small amount of water. Using this solution as “ink”, write
something on a piece of bond paper. Dry the paper by
passing it over the Bunsen flame.
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Before Heating
 The color is red violet.
While Heating
 The color turns to dark violet.
After Heating
 The color became sky blue and the substance.
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EXPERIMENT 14
Questions
1. What class of compounds is produced when water reacts with a
metallic oxide? Which of the results of the experiment proves this?
✘ When compound react with water

with metallic oxide the result is
base. Then calcium is base.
✘ Calcium Hydroxide, the result of
calcium oxide in water. To form
hydroxide Ca(OH)2.
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2. What class of compounds is produced when water reacts with a
non-metallic oxide? Which of the results of the experiment proves
this?
✘ The class of compounds is

produced when water reacts with a
non-metallic oxide is Acids. It was
proved by the reaction of sulfur
oxide in water to form H2SO3 (aq)
acid.
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3. What change is observed when copper (ll) sulfate crystals
become dehydrated?
✘ It gives color changes into white

because of the lesser amount of
water of molecules in it.
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4. Write the balance chemical equation for each of the following:
a. Action of calcium oxide in water

Ans: CaO(s) + H2O(s) -> Ca(OH)2
(aq)
b. Action of sulfur dioxide in water

Ans: SO2(aq) + H2O(s) ->H2SO3(aq)
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4. Write the balance chemical equation for each of the following:
c: Action of heat in copper(ll) sulfate

crystals
Ans: CuSO4 + 5H2O -> CuSO4 + 5H2O
d: Action of heat on cobalt chloride

hexahydrate
Ans: CoCl + 6H2O -> CoCl + 6H2O
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EXPERIMENT 15
Solubility and Factors Affecting It
OBJECTIVES REAGENTS
Kerosene
1. To determine the effect of the
nature of solute and solvent on the Ammonium Chloride
solubility of a substance. Sodium Chloride
2. To analyze how pressure and Sugar
temperature affect the solubility of Napthalene
gases in liquid. Cooking Oil
3. To determine the solubility of salt Gasoline
at various temperatures and prepare Ice Cubes
the solubility curve.
Materials: 14 Test Tubes, Triple Beam Balance, Bunsen Burner, Spatula, Test Tube Rack,
Test Tube Holder, 10 mL Graduated Cylinder, Hot Plate, Stirring Rod, 250 mL Beaker,
Crubicle Tong, Pipette, 1 Graphing Paper, Test Tube Brush, Thermometer, Ice Cubes
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PROCEDURE
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PROCEDURE #1
Effect of Nature of Solute and Solvent on Solubility
Test the solubility of a pinch(if a solid) or about 10 drops(if a
liquid) of the following solutes in separate 1-mL portion of
water and kerosene: sodium chloride, sugar, naphthalene,
gasoline, and cooking oil.
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Data, Observations and Results
Water
Soluble
Kerosene
Insoluble
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PROCEDURE #2
Effect of Pressure and Tempreature on the Solubility of Gas in
Liquid
Open a bottle of carbonated soft drink. Observe what happens.
Transfer about 30 mL of it to a beaker. Allow the effervescence to
subside, and then heat gently on a hot plate.
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Upon opening the bottle of carbonated
soft drink
 It has a fizzing effect and the bubbles comes out from
the lid of the bottle because of the pressurized carbon
dioxide inside it.
Upon heating the carbonated soft
drink
 It loses it’s carbonation, it doesn’t fizz anymore so
the bubbles that came up when it was opened was
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replaced by the bubbles coming from the heat where
PROCEDURE #3
Effect of Temperature on the Solubility of Solid in Liquid
Measure exactly 5.0 mL of water into a test tube. Weigh exactly 1.0 g of NH4Cl on
a triple beam balance. Add this to the test tube. Heat the solution carefully without
boiling until all crystals dissolve. Insert a thermometer into the test tube and allow
the solution to cool. Use ice cubes to cool the solution.
✘ Repeat the experiment using
0.75 g, 0.5 g and 0.25 g of
NH4Cl, respectively.
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23 degree
Celsius
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21 degree
Celsius
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19 degree
Celsius
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17 degree
Celsius
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Weigh Weigh Solubility g Crystallizatio

Solute Solvent NH4Cl per n Begins
(g Nh4Cl) (g H2O) 100 g of
H2O
1.0 g 5.0 g 20 23°C
0.75 g 5.0 g 15 21°C
0.5 g 5.0 g 10 19°C
0.25 g 5.0 g 5 17°C
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GRAPH
The higher the temperature, the
higher the solubility of solid in liquid
(directly proportional)
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EXPERIMENT 15
Questions
1. Based on your observations, which solutes are
Polar ? Non-Polar ? Explain your answer.
✘ Polar - sodium chloride and sugar,

because it is dissolved in water
which is a polar molecule.
✘ Non - Polar - gasoline,
naphthalene, cooking oil, because
it is soluble in non-polar molecule
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which is the kerosene.
2. Define Polar molecule in Non-Polar
molecule.
✘ Polar molecules have a high

boiling point, low vapor pressure
and high surface tension, while
the non-polar molecules have a
low boiling point and surface
tension but high vapor pressure.
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3. Give the general statement correlating
solubility to the nature of solute and solvent.
✘ "like dissolves like" which means

polar substance dissolves at polar
molecules and non-polar to non-
polar. Additionally, ionic are also
expected to be dissolved at polar
substance.
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4. Why was there a sudden effervescence when
the softdrink water was opened?
✘ Because it contain carbon

dioxide which is dissolved under
pressure . The liquid cannot hold
as much carbon dioxide, so the
excess bubbles out of the
solution.
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5. Account for the observations seen when the
carbonated drink was heated.
✘ It started to lose its carbonation.

The fizzing effect vanishes , and
the solubility of gases in them is
decreased because of the high
temperature.
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6. What is the effect of pressure and temperature on
the solubility of gas in a liquid?
✘ The higher the temperature, the

lower the solubility of gas while
the high pressure. means an
increase in the solubility of gas.
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7. Describe the trend of the curve. Is the dissolving
process exothermic or endothermic ? Explain your
answer.
✘ The higher the temperature the higher the

solubility of solid in liquid. Exothermic,
because the NH + Cl solution gets colder as
if dissolves in H2O and absorbs energy from
water. However, the crystallization is
exothermic since the heat of solution is
released and gives and gives off energy to
the surroundings.
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THANK
S!
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EXPERIMENT 14

Chemical Properties of Water

✘ When compound react with water

✘ The class of compounds is

✘ It gives color changes into white

a. Action of calcium oxide in water

b. Action of sulfur dioxide in water

c: Action of heat in copper(ll) sulfate

d: Action of heat on cobalt chloride

Weigh Weigh Solubility g Crystallizatio

✘ Polar - sodium chloride and sugar,

✘ Polar molecules have a high

✘ "like dissolves like" which means

✘ Because it contain carbon

✘ It started to lose its carbonation.

✘ The higher the temperature, the

✘ The higher the temperature the higher the

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