EXPERIMENT 1

Basic Laboratory Technique

OBJECTIVES

The objectives of this experiment are as follow:

(a) to use triple beam balance

(b) to measure the weigh using differentiating weighing method

(c) to measure the accurate and inaccurate measuring liquid volume apparatus

based on calculated volume from weight and density information

(d) to record data using the right apparatus

METHODOLOGY

APPARATUS:

1. Triple Beam Balance

2. Weighing bottle

3. Erlenmeyer flask 2 x 250 mL

4. Burette 2 x 50 mL

5. Pipette 1 x 25 mL

6. Beaker 1 x 50 mL or 25 mL

7. Measurement cylinder 1 x 50 mL or 25 mL

MATERIALS:

1. Water

2. Coins 1 cents x 20

PROCEDURES:

A. Weighing

1. The weight of one coin is measured for three times.

2. The weight of the coin is jotted down in the table provided.

3. Repeat step 1 until step 2 by replacing one coin with 20 coins.

4. The mean of the three reading obtained were calculated.

B. Volume Measurement

1. The apparatus were ensured clean before they were being used. Wash if necessary.

2. Two Erlenmeyer flasks were labeled as Flask A and Flask B. The weights of both

Erlenmeyer flasks were measured and the data were recorded in a table.

3. 25 mL of water was put into the Erlenmeyer flask A by using burette. The reading of

the burette was recorded to the nearest 0.05 mL.

4. Another 25 mL of water was transferred into Erlenmeyer flask B by using the pipette.

5. The weights of each flask were weighed and recorded.

6. 25 mL of water was transferred into Flask A by using the beaker and another 25 mL

of water was transferred into Flask B by using the measuring cylinder.

7. The accuracy of each of the apparatus (measuring cylinder, beaker, burette and

pipitte) were identified by using the volume and the weighs obtained.

8. The density of water was calculated by using the data obtained.

RESULTS

A. WEIGHING

Reading Mass of one coin (g)

First reading 1.8

Second 1.7

Third reading 1.7

Mean weight 1.7

Table 1: The weight of the mass of 1 coin

Reading Mass of 20 coins (g)

First reading 34.9

Second 34.9

Third reading 34.7

Mean weight 34.8

Table 2: The weights of the mass of 20 coins

B. VOLUME MEASUREMENT

Weight of the empty Erlenmeyer flask A: 113.3 g

Weight of the empty Erlenmeyer flask B: 106.1 g

When using a burette, the weight of the Erlenmeyer flask A with water is as follow:

Reading Mass (g)

First reading 138.3

Second 138.2

Third reading 138.3

Mean weight 138.3

Table 3: Weight of the Erlenmeyer flask A fill with water by using the burette.

When using a pipette, the weight of the Erlenmeyer flask B with water is as follow:

Reading Mass (g)

First reading 131.2

Second 131.3

Third reading 131.2

Mean weight 131.2

Table 4: Weight of the Erlenmeyer flask A fill with water by using the pipette

When using a beaker, the weight of the Erlenmeyer flask A with water is as follow:

Reading Mass (g)

First reading 136.9

Second 137.9

Third reading 137.5

Mean weight 137.4

Table 5: Weight of the Erlenmeyer flask A fill with water by using the beaker.

When using a measuring cylinder, the weight of the Erlenmeyer flask B with water is as follow:

Reading Mass (g)

First reading 130.5

Second 130.7

Third reading 130.6

Mean weight 130.6

Table 6: Weight of the Erlenmeyer flask B fill with water by using the measuring cylinder.

Calculation

Water density by using burette

ρ = m/v

= (138.3 – 113.3)g / 25.00 mL

= 1 g

Water density by using pipette

ρ = m/v

= (137.4 – .)g / 25 mL

= 1.004 g

Water density by using beaker

ρ = m/v

= (137.4 – 113.3)g / 25.0 mL

= 0.964 g

Water density by using measurement cylinder

ρ = m/v

= (130.6 – 106.1)g / 25 mL

= 0.98 g

DISCUSSION

In this experiment, students are exposed to the technique of measuring certain material

using the apparatus provided. Laboratory experimentation is a very important element in the

study of Chemistry. In most of the laboratory experiment, technique in measuring is one of the

most required and important laboratory techniques that must be mastered by the students.

In measuring, each of the given apparatus has its own accuracy. Not all of the apparatus

in the laboratory will have the same accuracy. In this experiment, student will learn how to

record the correct measuring number in order to show the accuracy of the apparatus. This

technique actually included the concept of significant number and round – up numbers. Not just

that, student will learn to report the calculated result using data gain from measurement with

correct significant number from this experiment as well.

The apparatus used in this experiment included the triple beam balance. It is an

apparatus that used to measure the mass of material. It was created to weigh with the accuracy

near to 0.0g (one decimal point). The reading error of the triple beam balance is up to 0.05

grams. It is named triple beam balance due to the presence of the three parallel beams which

each of the beams actually are supporting one weight. The beams are supported by a fulcrum

and at one end of the beam was a pan that used to put on the unknown weight material. In

order to measure the weight of the wanted materials, we must slide the three weights along the

three beams to increase their lever arm. Read the reading when the indicator was at the

balanced point.

The parts of the triple beam balance are as follow:

The other apparatus used in this experiment is the electronic balance which is able to

measure up to 0.0000 gram. It uses the electrical energy to measure the mass of the materials.

Thus, it is very accurate if the technique applied is correct. One of the most basic requirements

in using the electronic balance is that the surrounding must very calm. That is why the electronic

balance is usually attached with the glass cover on it. This glass is to prevent any movement of

air that will affect the measurement of the electronic balance.

In this experiment, we are going to measure the weight the solid material that is the coin

and to measure the volume. As for the volume measurement, a volume of liquid can be

measure using the measuring cylinder, burette, pipette and volumetric flask. Among all those

measurer for volume, the burette and the pipette will have the most accurate measurement.

In this experiment, we are experimenting about the accuracy of the apparatus. The

apparatus that are listed for comparison are burette, pipette, measuring cylinder and the beaker.

All these apparatus has their own accuracy. We can determine the accuracy of these apparatus

by calculating the density of the water based on the volume that we measure. From the result,

the pipette and the burette show the reading of the most accurate where the calculated density

of water only difference for 0.004 g / cm3 for pipette and 0.00 g / cm3 for burette. Theoretically,

Solution

Solution

pipette will have a more accurate measurement compare to the burette. However, this

experiment has give out a different result from the theory might be due to the parallax error

during reading the volume of water. It might as well is an effect due to the parallax error when

reading the mass measurement using the triple beam balance. Another one more reason of the

deviation from the theory might be because the calculation is based on the mean of the three

reading obtained.

CONCLUSION

From this experiment, we know that to get a more accurate measurement of volume, we

can use the pipette and burette. Other than that, we had learnt to use the triple beam balance to

measure the weight of certain material. Some tiny error might happen on the calculation and it

might deviate from the theory value due to the parallax error during reading of the measurement

of material using the apparatus.

REFERENCES

1. http://www.myweigh.com/3bb.gif

2. http://genchem.rutgers.edu/balance3b.html

3. http://www.physics.smu.edu/~scalise/apparatus/triplebeam/

4. http://www.wisc-online.com/objects/index_tj.asp?objID=GCH202

EXPERIMENT 2

DILUTION

OBJECTIVES

The objective of this experiment is to determine the concentration of the coloured

solution such as FeCl3 by using the dilution and colour differentiation (calorimetric) technique.

Other than that we might learn to calculate the concentration of the diluted solution by using the

correct formula.

METHODOLOGY

APPARATUS:

8. Volumetric flask 5 x 50ml

9. Burette 1 x 50 mL

10. Pipette 1 x 5 mL

11. Test tube 10 units

MATERIALS:

3. 0.10 M FeCl3 solution

4. 0.10 M KCNS solution

5. FeCl3 solution (B)

Procedures:

1. All of the material and apparatus were prepared.

2. The following solutions were prepared based on the formula given below:

Solution: (a) 10 mL FeCl3 solution (5.0 x 10-2M)

(b) 10 mL FeCl3 solution (1.0 x 10-2M)

(c) 10 mL FeCl3 solution (5.0 x 10-3M)

(d) 10 mL FeCl3 solution (1.0 x 10-3M)

(e) 10 mL FeCl3 solution (5.0 x 10-4M)

Formula:

M1V1 = M2V2

Where M1 is the Molarity of the concentrated solution

M2 is the molarity of the diluted solution

V1 is the volume of the concentrated solution

V2 is the volume of the diluted solution

3. Each of the solution is prepared by using the dilution method where the solution (b) is

prepared by using the solution (a) and solution (c) is prepared by using solution (b) and

so on. Each of the solution was prepared up to 50 mL in the 50mL volumetric flask.

4. After all of the solution has been prepared, 5 mL of each of the solution was transferred

into a test tube and 2 drops of KCNS (Potassium Thiocyanate) was added to each of the

test tube. The solution was swirled and shaken if necessary until it is homogeny and wait

for any colour changes.

5. 5 mL of B solution (FeCl3 of unknown concentration) is put into another solution and 2

drops of KCNS solution were added.

6. The test tube was swirl and sheken if necessary. The colour changes of the solution was

compared with the series of solution that have prepared previously(step 4). Suggest the

concentration of the B solution.

RESULT

A)The volume of the FeCl3 and distilled water needed to prepared each of the solution.

SOLUTION CONCENTRATION VOLUME OF FeCl3 VOLUME OF WATER, H2O

a 5.0 x 10−² M 25 ml of 0.1M FeCl3 25 mL of H2O

b 1.0 x 10−² M 10 mL of FeCl3 in (a) 40 mL of H2O

c 5.0 x 10−³ M 25 mL of FeCl3 in (b) 25 mL of H2O

d 1.0 x 10−³ M 10 mL of FeCl3 in (c) 40 mL of H2O

e 5.0x 10 ‾4 M 25 mL of FeCl3 in (d) 25 mL of H2O

B)The colour changes of the FeCl3 solutions after the 2 drops of KSCN were added.

TEST TUBE SOLUTION CONCENTRATION COLOUR CHANGES

A a 5.0 x 10−² M Colourless–Dark Red

B b 1.0 x 10−² M Colourless–Red

C c 5.0 x 10−³ M Colourless–Light Red

D d 1.0 x 10−³ M Colourless–Orange

E e 5.0x 10 ‾4 M Colourless–Light Orange

F B unknown Colourless–Light Red

DISCUSSION

Concentration of a solution is usually express in Molarity (M). There are still many other

ways to express the concentration of a solution including using the percentage volume and

weight expression, part per million and many others more. However, in this experiment we will

just concentrate on calculating the concentration of the solution using Molarity (M). The Molarity

of a solution can be calculated using the formula below:

Molarity (M) = mole of solute (n)

volume of solution (v)

When a solution is prepared, we can prepare it from a stock solution. Stock solution is

the solution with a known concentration. Although the concentration changes, it does not mean

that the number of mole of a solution has changed. It is still remain the same. The only

difference is that the volume of the distilled water of solvent increases if the solution is diluted.

Dilution is a process of preparing a less concentrated solution from a more concentrated

solution by adding distilled water. In this experiment, the less concentrated solution was

prepared by using the more concentrated solution prepared earlier. For example, solution b was

prepared by using the solution which is prepared from 0.1 M FeCl3. In other word, the other

solution from b to e was prepared from 0.1 M FeCl3.

This is done by using the formula provided that is M1V1 = M2V2. In this formula, M1V1

stands for the molarity and the volume of the concentrated solution while the M2V2 stands for the

diluted solution. For preparing the diluted solution, the volume of the concentrated solution can

be calculated by using this equation. Once the volume of the concentrated solution was

obtained, the distilled water is add in until it reaches the needed volume for the diluted solution.

After prepared the diluted solution, we are needed to identify the concentration of the B

solution. This can be done by observing the colour changes of the FeCl3 solution when it react

with the KSCN (Potassium Thiocyanate). The colourless solution of the FeCl3 will usually

change to reddish colour of orange colour depending in its concentration.

From the experiment, the colourless B solution of FeCl3 reacts with KSCN to produce

light – red colour. Since then, we can assumed that the concentration of the B solution can be

more or less the same like the solution prepared diluted solution that change into the same

colour as well. Thus, we can assumed that the B solution has the concentration that is almost

the same as the prepared c solution which has the concentration of 5.0 x 10−³ M.

CONCLUSION

From this experiment, a new concentration of a solution can be prepared by using the

stock solution that is with a known concentration. The dilution can be carried out by applying the

formula of M1V1 = M2V2 in calculation of the volume of the needed solution. We can as well has

an roughly estimation of the concentration of a solution by using the colour comparison method

based on the reaction that occur. In this experiment, the B solution is estimated to has the

concentration of the 5.0 x 10−³ M.

REFERENCES

1. http://en.wikipedia.org/wiki/Dilution

EXPERIMENT 3

ACID AND BASES TITRATION

OBJECTIVES

The objective of this experiment is to determine the concentration of the Sodium

Hydroxide solution through titration technique using the hydrochloric acid and sulphuric acid.

METHODOLOGY

APPARATUS:

12. Volumetric flask 100mL

13. Filter funnel

14. Beaker 1 x250mL

15. Burette 1 x 50mL

16. Pipette 1 x 25mL

MATERIALS:

6. Distilled water

7. 100mL of 1.000 x 10-2M HCl solution

8. 100mL of 1.000 x 10-2 H2SO4 solution

9. 10mL of C solution containing NaOH (with pipette)

10. Phenolphthalein

PROCEDURES:

1. 10mL of the C solution was pipette into a volumetric flask (100mL). Distilled water was

added into the volumetric flask until it reaches the mark to dilute the C solution.

2. The solution was mixed thoroughly by inverting the volumetric flask. Once it was done,

the solution was transferred into a clean beaker. The solution was labelled as C.

3. A burette was cleansed and it was rinsed with 5mL of HCl solution (1.000 x 10-2M) twice.

HCl solution (1.000 x 10-2M) was placed into the rinsed burette by using the filter funnel.

4. A pipette was cleansed and rinsed with C solution twice. 25mL of the C solution was

pipette into three different Erlenmeyer flasks. Two drops of phenolphthalein indicator

was added into the C solution in each of the Erlenmeyer flasks.

5. The initial reading of the burette containing the HCl solution was recorded to the nearest

two decimal points.

6. C solution was titrated using the HCl from the burette until the C solution changes to

colourless solution. This colourless is recognised as the end point of the titration.

7. The final reading of the burette is recorded and the volume of the HCl used was

calculated.

8. The titration was repeated trice and all of the result was recorded in the table.

9. The concentration of the C solution was calculated.

10. The whole experiment is repeated by using replacing the HCl solution with the H2SO4

solution.

RESULTS

Titration with HCl (1.000 x 10-2M)

READING VOLUME OF HCl (1.000 x 10-2M) USED (mL)

FIRST READING 48.30

SECOND READING 47.90

THIRD READING 47.80

MEANS OF SECOND

AND THIRD RREADING

47.90 + 47.80

2

Titration with H2SO4 (1.000 x 10-2M)

READING VOLUME OF HCl (1.000 x 10-2M) USED (mL)

FIRST READING 24.80

SECOND READING 25.30

THIRD READING 25.10

MEANS OF SECOND

AND THIRD RREADING

25.30 + 25.10

2

Calculation

Reaction Equation: HCl + Naoh → NaCl + H2O

1 mole of HCl will react with 1 mole of NaOH to produce 1 mole of NaCl and 1 mole of H2O.

47.85=

25.2=

11 =

1 mole of HCl1 mole of NaOH

1 mole of HCl = 1 mole of NaOH

1000MHCl

vHCl

1000MNaOH

vNaOh=

=1000MNaOH

25.001000(0.01)

47.85

Reaction Equation: H2SO4 + 2 NaOH → Na2SO4 + 2 H2O

I mole of H2SO4 will react with 2 mole of NaOH to produce 1 mole of Na2SO4 and 2 mole of H2O.

Concentration of C Solution

0.005224 + 0.0049602

0.005092 M

MNaOH

12 =

1 mole of H2SO4

2 mole of NaOH

1 mole of H2SO4 = 2 mole of NaOH

=2 x 1000MNaOH

25.001000(0.01)

25.20

=2 x 1000MNaOH

vNaOH

1000M H2SO4

vH2SO4han that,

another standard solution must be used as

well.omposition of

the unknown solution

was concentrat

ion solution.

MNaOH = 1000(0.01)(25.00) (25.20)(2)(1000)

= 0.004960 M

MNaOH = 1000(0.01)(25.00) (47.85)(1000)

= 0.005224 M

=

DISCUSSION

Titration is the process that a chemist used to determine the concentration of a particular

solution. In titration, it involves two types of reagent that is the solution with the unknown

concentration and the solution with known concentration or namely standard solution.

In this experiment, the unknown concentration solution is indicated as C solution which

is a solution of NaOH. Thus, we are able to carry out the process of titration to determine the

concentration of the C solution by using two types of acids that is the monoprotic acid

(Hydrochloric Acid, HCl) and diprotic acid the Sulphuric Acid (H2SO4).

Actually the reaction that occurs in between the C solution and the acid are the

neutralisation process. Thus, the reaction will produce a salt and water. In the monoprotic acid,

the Sodium Chloride (NaCl) salt was produced by one mole of each of the reactant that are the

HCL and the C solution (NaOH). In the other hand, the diprotic acid (H2SO4) produces Sodium

Sulphate (Na2SO4) when it reacts with 2 moles of C solution (NaOH).

In this experiment, the indicator used was phenolphthalein. This indicator will

show pink colour in the higher pH solution or in alkali or diluted base. When the pH of the

solution decreases, the density of the pink colour will also decrease. When it reaches he neutral

pH, it will become colourless. The titration is stop when the indicator turns into colourless

because this is the equivalence point. At this point, stoichiometrically equivalent quantities are

brought together. Thus, the quantity of the acid and base are the same at this point.

The volume of acid that is used in this point can be used to calculate the concentration

on the C solution. In the titration that uses the HCl, the concentration of the C solution

calculated is 0.005224 M while in the titration using the H2SO4, the concentration of the NaOH

used is 0.004960 M. Both of the value is very close. Thus, we can roughly estimate that the

concentration of the C solution is around the mean of the two values that is 0.005092 M.

There are a few precaution steps to take into account in this experiment. During the

titration process, make sure the C solution is being swirled all of the time until the equivalent

point is reached. Not just that, the Erlenmeyer flask used is supposed to have a large surface

area. Both of these both steps is to ensure that all of the molecule in the C solution react with all

=

of the acid molecules and minimize any error in the experiment. In the beginning of the

experiment, titration can be fast but when it come to nearly to the equivalent point, make sure

that the titration process is done drop by drop. This is to prevent the over titration.

CONCLUSION

As the conclusion, the titration process can be used to determine the concentration of

the unknown concentration solution. However, the concentration can be only determined if the

composition of the unknown concentration. Other than that, another standard solution must be

used as well.

EXPERIMENT 4

Hydrated Salt Formula

OBJECTIVES

1. To determine the volume of water content in hydrated salt.

2. To determine the formula of hydrated salt

METHODOLOGY

APPARATUS:

17. Crucible with lid

18. Tripod stand

19. Bunsen burner

20. Thermometer

21. Electrical balance

22. Stop watch

23. Dropper

24. Glass rode

MATERIALS:

11. Hydrated salt: CuSO4.5H2O

12. Distilled water

PROCEDURES:

1. The crucible was heated for a while.

2. The same crucible was weighed with its lid together

3. Approximately 2g of hydrated salt was added into the crucible.

4. The crucible with the hydrated salt was weighed and the data was recorded.

5. The salt was heated for around 3 – 5 minutes in the crucible with the lid opened.

6. Colour changes were observed and the data was recorded.

7. The crucible was stop from heating and it was covered with the lid until it cool down.

8. Observe the colour changes and weighed again the weight of the lid – covered crucible

with heated salt in it.

9. The salt was grinded using the glass rode and it was heated again along with the

opened crucible.

10. The salt was cooled to room temperature.

11. The whole apparatus of the crucible was weighed again. Heating was continued until a

constant weight was obtained.

12. The final reading of the salt temperature was recorded after the final weight.

13. A few drops of distilled water at room temperature were added into the crucible with the

heated salt in it.

14. The temperature changes were recorded and any changes were observed.

RESULTS

Weighing:

Weight of the crucible with lid: 61.74287g

Weight of the crucible with lid containing hydrated salt: 63.74386g

Weight of the crucible with lid containing hydrated salt after first heating: 63.02278g

Weight of the crucible with lid containing hydrated salt after second heating: 63.01018g

Weight of the crucible with lid containing hydrated salt after third heating: 63.00991g

Colour observation:

Colour changes from white to greenish – white colour throughout the heating process.

During the third time heating, the colour of the hydrated salt turn to brownish colour at the

circumference.

Temperature Changes:

The temperature of the hydrated salt after the final weight: 26⁰C.

The temperature of the hydrated salt when water droplets added: 44⁰C.

DISCUSSION

CALCULATION

1. Calculate the mole number of hydrated salt, results of unhydrated salt and weight of

leave water from this reaction.

Mass of the hydrated salt: 63.74386g - 61.74287g = 2.00099g

Molecular mass of CuSO4.5H2O = 250

Mole number of the hydrated salt:

Mass of the unhydrated salt: 63.00991g - 61.74287g = 1.26704g

Molecular mass of CuSO4 = 160

Mole number of the hydrated salt:

Mass of the leave water: 63.74386g - 63.00991= 0.73395.g

Molecular mass of H2O = 18

Mole number of the hydrated salt:

2. If the formula of the hydrated salt is CuSO4.xH2O, verify the x. Compare with the real

molecular formula.

Compound CuSO4 H2O

Mass (g) 1.26704 0.73395

Molecular Weight 160 18

Number Of Mole 7.919 x 10 -3 0.040775

Mass Of CuSO4.5H2OMolecular Mass Of CuSO4.5H2O =

2.00099250

8.00396 x 10-3mole=

Mass Of CuSO4

Molecular Mass Of CuSO4=

1.26704160

7.919 x 10-3mole=

Mass Of H2OMolecular Mass Of H2O =

0.7339518

0.040775 mole=

Simplest Ratio

1 5.149

The simplest ratio obtained for CuSO4 and H2O is 1:5. We assumed the little deviation is

error occur from the experiment. Since then, we change the simplest ratio into the

nearest round number that is 1:5. In other words, the molecular formula of the hydrated

salt is CuSO4.5H2O.

3. List any error sources. Comment the x value.

The error throughout the experiment:

1. The salt is not heated thoroughly since the circumference of the powder was

over – heated.

2. The crucible was not totally dried up of water before it was used. We just heat

it for a while.

3. The lid of the crucible was not properly close. Thus, during the cooling

process the unhydrated salt might react again with the water vapour in the

air.

4. Heating was not continued after the third heating although the weight is not

constant yet.

5. The sample is not cool in the decicator. Thus, water vapour in the

environment might fuse with the unhydrated salt again.

6. The powder was not properly grinded or spread before the next heating.

Thus, heating is just take account on the salt in the middle of the crucible.

7.919 x 10-3

7.919 x 10-3

=0.040775

7.919 x 10-3

=

All of the errors above actually lead to the little deviation of the value of x. This is

because the weight of the water calculated is not totally the weight of the water from the

salt. It might be originated from the crucible or from the surrounding.

4. What is the name of the process when water release from hydrated salt, endothermic or

exothermic process?

The name of the process when water released from the hydrated salt is endothermic.

This is because the heat from the environment was absorbed and used to break the

bonding between the water molecule and the CuSO4 molecule.

DISCUSSION

From this experiment, we know that the molecular formula of the hydrated salt can be

calculated through the heating and cooling down process. Heating will cause the water molecule

in the hydrated salt to evaporate. This evaporation enables us to calculate the weight of the

water molecule that left the hydrated salt. From there, we can find out the mass of the salt and

the mass of the water.

Once the mass is obtained, we can calculate the number of moles of the water molecule

and the number of moles of the salt itself. The formula used to calculate the mole number is as

follow:

Mass Of SubstanceMolecular Mass Of Substance=Mole Number

When the number of mole of each of the salt and the water has obtained, we can

calculate the ratio of the water and the salt itself. To get the ratio, we have to divide both the

mole numbers of the water and the salt with the smallest value on mole number obtained.

Convert the value of the ratio to the nearest round number.

After the heating process, the salt was unhydrated. When water droplets were add into

the salt, the temperature will increase. This is because the reaction of hydrating the salt was

exothermic reaction. Heat was released to when the water molecule create bonding with the salt

molecule. So, the heat that released had increased the temperature of the surrounding.

Before heating, the empty crucible was heated first. This step is done to evaporate the

water molecule on the crucible so that it will not affect the result of the experiment. When the

salt was heated, the lid was left opened. So, the water that evaporated can move into the

surrounding and not circulated in the crucible. When the heating was stop, the crucible was

close to prevent the water molecule in the environment to fused back with the salt molecule

while waiting for the crucible to cool down. Before the salt was reheated again, it was grinded

using a glass rode to spread the salt thoroughly in the crucible. This step can make sure that the

heating was not just concentrated on some part of the salt in the crucible. Heating of the

crucible with the salt in it was continued until a constant reading was obtained. Make sure that

the salt was not over – heated because the over - heating will cause another reaction to occur.

CONCLUSION

As the conclusion, the formula of the hydrated salt can be calculated through the heating

and cooling process. This technique is used because the heating process can eliminate the

water molecule in the salt and enable us to calculate the weight and mole number of the water

and the salt. The reaction of heating and evaporating the salt actually is the endothermic

reaction where the salt molecule absorbed the heat to breakdown the bonding between the salt

and the water. When, hydrating the salt, the reaction was exothermic because heat was release

during the formation of the bond between the water and the salt molecule.

EXPERIMENT 5

Charles Law

OBJECTIVES

The objectives of this experiment is to study the effect of the temperature to gas volume

METHODOLOGY

APPARATUS:

25. Stop watch

26. Erlenmeyer Flask equipped with rubber stopper, rubber tube with clip

27. Beaker 500mL x 1

28. Bunsen burner

29. Thermometer

30. Tripod stand

PROCEDURES:

1. The apparatus as shown as the figure above was prepared and set up.

2. The empty Erlenmeyer flask equipped with the rubber stopper and the clip was weighed.

3. The Erlenmeyer flask was placed in a beaker containing water.

4. The beaker containing water was boiled for ten minutes with the tube opened so that the

air in the flask will have similar temperature with the water boiling point.

5. The sink was filled with water while the air in the flask was still hot.

6. As the heating continue, the rubber tube was clipped and the position of the clip was

marked.

7. The flask was removed from the boiling water and the end of the tube together with the

clip was put into the sink that contained water.

8. The flask was left in the air.

9. The clip was opened and the water was flow into the flask. The air temperature in the

flask was ensured to decrease to the temperature of the water in the sink.

10. The tube was re – clipped again at the marked position and the flask was removed from

the sink with water.

11. The volume of the Erlenmeyer flask filled with water and equipped with the rubber

stopper and the tube along with the clip was weighed again.

12. After that, the whole apparatus was filled with water and weighed again. By using the

water density calculate the volume of the apparatus used.

13. A graph of volume versus the temperature was plotted and it was extrapolated to zero

volume.

RESULTS

Weighing:

Item weighted Weight (g)

a. Weight of equipped flask 151.0267

b. Weight of equipped flask + water 187.7072

c. Weight of equipped flask + full water 348.8200

Data:

i. Weight of water in full flask (c-a) 197.7933

ii. Weight of sucked water (b-a) 36.6805

iii. Volume of sucked water (ii ÷ water

density)

36.6805g ÷ 0.9978 g cm-3 = 36.6805 cm3

iv. Volume of air at 100 C⁰ Volume in the flask = (i) x water density

= 197.7933g x 1 g cm-3

= 197.7933 cm3

v. Volume of air in room temperature (Volume of air at 100 C) - (Volume of Sucked Water)⁰

= (iv) – (iii)

= 197.7933cm3 – 36.6805cm3

= 161.1128cm3

Volume of gas at the water temperature:

DISCUSSION

Charles’ law is also known as the law of the volumes. It is an experimental gas law in

which it describes how the gases tend to expand when it is heated. The Charles’ law was first

published by French natural philosopher name Joseph Louis Gay – Lussac in the year of 1802.

However, in these days we know that the Charles’ Law actually is talking about at a

constant pressure the volume of a given mass of an ideal gas will increase with the increase of

the temperature on the absolute temperature scale. In other word, the gas will expand when the

temperature increases.

This is because when a gas is heated, the particle in the gas received more kinetic

energy. Thus, they are able to move and vibrate faster and exert more pressure on the wall

containing them. Since the pressure is maintained at the same value, the pressure exerted had

caused the volume to increase since the volume is allowed to expand.

V1

27oC + 273 =197.7933

100oC + 273

V1 = 197.7933100oC + 273

300x

= 159.08 cm3

The Charles’ Law can be written as:

Where V is the volume and T is the absolute temperature. This law also can be expressed in

few other ways such as follow:

From the equation, we can clearly see that for each of the increase of the temperature in the

absolute scale will cause the volume to increase as well with a condition that is a constant

pressure. At the same time, these equations also shown us a constant value when it is being

expressed in below:

Charles’ Law is very useful in calculating the volume of gas in different temperature. For

example in this experiment, Charles’ Law was applied to obtain the volume of gas at room

temperature. However, the volume calculated and the volume obtained by filling up the flask

with water is different. The calculated value is 159.08 cm3 while the volume of gas obtained by

filling up the flask is 161.1128 cm3. There is a different of a few cm3 in volume. This error might

occur because the clip position on the tube was not really at the end. It is a few millimeter away

from the end of the tube. However, the volume that takes into account was the volume until the

end of the tube. Other than that, during filling up the flask with water, there are a few air bubbles

in the water. This might affect the weight of the water measured and thus affect the result of the

volume calculated. Other than that, another error in the experiment is that the flask was not

totally boiled in the water for 10 minutes. Heating was stop when the water was boiling. It is

supposedly to wait until the flask was in the boiling water for 10 minutes. Furthermore, the water

in the beaker that was heated did not reach the boiling point of water. It just reached 97 C⁰

because the water used is not distilled water bit the pipe water.

Volume Of GasAbsolute Temperature

Constant =

QUESTIONS

1. What is the temperature (in Celsius) obtained when the extrapolated graph cross at

temperature – axis? Compare with the theory result.

The temperature is –290⁰C when the graph is extrapolated to the temperature – axis.

The theory result stated that the temperature is supposedly to be at -273⁰C. This

situation occur because the experiment is not carried out in the standard pressure and

condition. Thus, the gas will cat like the real gas and not like the ideal gas.

2. Using Charles’ Law and the amount of the gas volume at 100⁰C from your experiment,

determine the volume theory at the water temperature in the sink.

V1:Volume of gas at 100⁰C T1: Temperature of 100⁰C

V2: Volume of gas at water temperature T2: temperature of water (26⁰C)

3. Illustrate a graph for volume theory of water in the sink in the same graph paper. Where

did graph cross the temperature axis? Explain your data.

197.7933100oC + 273

197.7933100oC + 273 =

V2

26oC + 273

V2 = (26⁰C + 300)x

= 158.5528 cm3

When extrapolated, the graph crosses at -265⁰C. This is because in this experiment real

gas is used and not the ideal gas. Real gas has attractive forces between its molecules.

Thus, it will make the volume in experiment less than volume theory. Thus, the

temperature obtained do not cross at -273⁰C.

CONCLUSION

As the conclusion, the volume of a gas will increase when the temperature increases.

This situation is just exactly the same as the theory stated in the Charles’s Law.

REFERENCES

1. BROWN, LEMAY, BURSTEN, MURPHY-CHEMISTRY:The Central Science

2. http://en.wikipedia.org/wiki/Charleslaw

EXPERIMENT 6

Dissimilarities Between Electrovalent and Covalent Bond

OBJECTIVES

The objectives of this experiment is to enable the student to differentiate the properties

between covalent compound (either pure covalent or ionic covalent) and electrovalent

compound.

METHODOLOGY

APPARATUS:

1. Test tubes

2. Beaker 150mL

3. Stopper

4. Tube

5. Battery

6. Wire

7. Bulb

MATERIALS:

13. NaCl solution

14. C2H5Br liquid

15. H2O2 liquid

16. AgNO3 Solution

17. NaOH solution

18. Phenolphthalein

19. HCl solution

20. Na2CO3 solid

21. Zn/Cu/C electrode

22. NH4OH solution

23. Ethanol

24. NaCl solid

PROCEDURES:

C. Precipitation

5. 3 clean test tube were prepared and were labeled as A, B and C

6. Test tube A was filled with 5mL of NaCl solution, test tube B was filled with 5mL of

C2H5Br liquid and test tube C with 5mL of H2O2 liquid.

7. 1mL of AgNO3 was added into eah of the test tube

8. The three test tubes were observed.

9. The precipitation time was recorded.

D. Gas Released

9. 5mL of NaOH solution was filled into a test tube and 2 drops of phenolphthalein was

added in too.

10. Another test tube was filled with 5mL of HCl. It was equipped with a rubber stopper

and a glass tube.

11. A small amount of Na2CO3 was added into the test tube containing HCl. Any

reactions that occur were observed and the gas released was determined by

inserting the glass tube into the test tube containing NaOH.

12. Any colour changes was observed and recorded in the result.

E. Electric Conductivity

1. A beaker was filled with HCl solution until half full.

2. Zn / Cu electrode was put into the solution.

3. The circuit was switched on when all of the connection was correct

4. The bulb was observed whether it does blinked or not.

5. The circuit was switched off and the HCl solution was replaced with the NaCl

solution, Naoh Solution, H2O2 liquid and NaCl solid in ethanol.

RESULTS

C. PRECIPITATION

SOLUTION OBSERVATION

5mL NaCl + 1mL AgNO3 White precipitation was formed immediately

5mL of C2H5Br + 1mL AgNO3

Two layers were formed

The first layer contained white precipitation

and the second layer was a little bit clear.

The precipitation takes a three minutes to

formed.

5mL H2O2 + 1mL AgNO3

There is no any presence of any precipitation

after 10 minutes.

D. GAS RELEASE

ACTION OBSERVATION

5mL NaOh + 2 drops of Phenolphtalein The solution changes from colourless to pink

colour

This indicates the solution is a basic solution

5mL of HCl + Na2CO3 Hisses sound was heard

Reaction occurred with the releases of air

bubles

Inserting the glass tube in the NaOH

Solution

The colour of the solution changes from pink

to colourless.

The gas releases neutralise the NaOH

solution.

E. ELECTRICAL CONDUCTIVITY

Solution Blink Clearity

HCl Blink Clear

NaCl solution Blink Not so clear

NaOH Blink Not clear

H2O2 No blink Not clear

NaCl solid in ethanol No blink Not clear

DISCUSSION

A. PRECIPITATION

Precipitation is the process of formation of the solid particles in a solution during the

chemical reaction. The chemical that formed is call as precipitates while the solution that

remains from the reaction is called as the supernatant. In a usual condition, the supernatant will

float on or stay above the precipitate. During the process of the precipitation heat was released

and the environment temperature was increased. This reaction is so called as exothermic

reaction. In each of the precipitation reaction, the more reactive reagent will replace the less

reactive reagent. The newly formed chemical will be in two state that is the solid state

(precipitate) and the liquid state (supernatant).

In the first reaction, the below process has occurred:

AgNO3 (aq) + NaCl (aq) NaNO3 (aq) + AgCl (s)

The white precipitate is the AgCl that formed from the reaction. In this reaction, both of the

chemical compounds seem to exchange their cations and anion. This situation is namely

metathesis.

In the second test tube, the C2H5Br react with the AgNo3 to formed two layers where one

of the layers were white precipitate while the other one is colourless precipitate. The white

precipitate was actually the Silver Bromide (AgBr) while the colourless precipitate was the ethyl

nitrate (C2H5NO3). These compounds are formed by the reaction below:

AgNO3 (aq) + C2H5Br (l) C2H5NO3 (aq) + AgBr (s)

As for the third test tube, no reaction actually occurs. This is because the H2O2 is not

reactive and it will only react with strong oxidant. In this case, the AgNo3 was not a strong

oxidant. Thus, no reaction occurred in the test tube when the solution was added together.

B. GAS RELEASED

In this experiment, the reaction that occurs in the test tube with the HCl was shown as

below:

Na2CO3 + 2HCl 2NaCl + CO2 + H2O

From the reaction, we can see that the carbon dioxide was released. Other than that, the NaCl

salt and water was released too. The gas released into the NaOH solution will neutralise the

basicity of the NaOH solution. Thus, solution will change from pink solution to pale pink to

colourless.

C. ELECTRIC CONDUCTIVITY

From the experiment, the result proved that the HCl, NaCl and NaOH solution will make

the light bulb blink. This is because all of those three solutions were good electrolytes. In other

words is that, all of those three solutions are able to produce the free moving cation and anion in

the liquid state either solution or the molten state. These free moving ions will enable the electric

to be conducted from the positive end to the negative end. Thus, the light bulb will blink because

current can pass through it. However, the H2O2 and the NaCl in ethanol are not able to make the

light bulb to blink because the solutions do not have the free moving ion. Both of the solutions

are covalent compound and they do not conduct electricity.

QUESTIONS

1. Describe whether this solution can be electrically conducting or have electrolyte

properties: H2S solution, melt AgCl, HCl, solution, and melt FeS

From the given solutions, all of them are able to conduct electricity because all of the

solutions are able to produce free moving ions in the liquid state either in solution or

molten state. However, the H2S solution is not able to conduct electricity because it is

not ionic compound. It is covalent compound. Thus, it is unable to produce free moving

ions.

2. Why NaCl solution can be electrically conducting and solid NaCl cannot?

When NaCl is in aqueous state, the solution will has 4 types of free moving ions that is

Na+, Cl-, H+ and OH-. Thus, the solution is able to transfer the electrons from the one

end to the other making the circuit being switch on. However, the solid NaCl does not

have the free moving ions. All of the particles are the NaCl molecule that is neutral. They

do not conduct electricity.

CONCLUSION

From the experiment, we can know that the electrovalent compound is the compound

formed when one or more electrons are being transferred from the metal element to non – metal

element particles. A covalent bond is a bond that formed when both particles shared one or

more electrons. Both of the compound form from the two different ways will show different

reactions.

REFERENCES

3. BROWN, LEMAY, BURSTEN, MURPHY-CHEMISTRY:The Central Science

4. http://en.wikipedia.org/wiki/electrovalentbond

5. http://en.wikipedia.org/wiki/covalentbond

EXPERIMENT 7

Molecular Geometry

OBJECTIVES

The objectives of this experiment are as follow:

(e) To determine the geometry of molecule

(f) to build 3 dimensional model for molecule

METHODOLOGY

APPARATUS:

1. Ball and Stick models

PROCEDURES:

F. Geometry Molecule Of HCl, BH3 and CO2

10. The HCl molecule was built by using ball and 1 stick.

11. The CO2 molecule was built by using ball with 2 sticks as the terminal atoms and the

balls with 2 stick as the central atom.

12. The BH3 molecule was built by using ball with 1 stick at the end atom and ball with 3

stick as the central atom.

13. The geometry and bond angle of each of the molecules were determined.

G. Geometry Molecule Of CH4, NH3, and H2O

1. CH4, NH3, and H2O geometry model were build using ball with one stick as the

terminal atom and ball with 4 stick as the central atom. One or two stick in the central

atom was left without balls to show the valence electrons where one stick is

corresponding to one pair of valence electrons.

2. The geometry of CH4. NH3 and H2O was determined

3. The real angle bonding for each molecule was determined using TTPE whether it is

less or equal with the basic angle.

H. Geometry Molecule Of PF5, AsCl5, ICl3, XeF2, SF4

1. AsCl5 ad PF5 molecule model was built using the ball with 5 sticks as the central

atom and ball with one stick as the terminal atom.

2. The actual angle of these molecules was determined

3. The molecules of ICl3, XeF2, and SF4 were build by using the same ball and the

stick was left without ball to show the lone pair electrons for central atom. You might

get variety of arrangement; identify the most stable geometry for each molecule and

determine their actual angle.

I. Molecular Geometry Of SF6, XeF6 and BrF6

1. The molecule above was built using ball with two stick as the terminal atom and ball

with 6 stick as the central atom.

2. The possible geometry for XeF4 and BrF5 molecules was restructure and the most

stable geometry was determined for that molecule and the angle of each of the

molecule was determined as well.

RESULTS

Bil Molecules Molecular geometry Predicted bond Angles

1. HCl linear 1800

2. BH3 Trigonal planar 1200

3. CO2 Linear 1800

4. CH4 Tetrahedral 109.50

5. NH3 Trigonal pyramidal <109.50

6. H2o V-shaped <109.50

7. PF5 Trigonal bipyrimidal 1200 and 900

8. ASCl5 Trigonal bipyramidal 1200 and 900

9. ICl3 Trigonal bipyramidal 1200 and 900

10. XeF2 Linear 1800

11. SF4 Trigonal bypiramidal 173.10 and 101.60

12. SF6 octahedral 1200 and 900

13. XeF4 Square planar 900

14. BrF5 Square pyramidal 84.80 and 900

DISCUSSION

Molecular geometry or molecular structure is the three-dimensional arrangement

of the atoms that constitute a molecule. It determines several properties of a substance

including its reactivity, polarity, phase of matter, color, magnetism, and biological activity. The

molecular geometry can be determined by various spectroscopic methods and diffraction

methods. IR, Microwave and Raman spectroscopy can give information about the molecule

geometry from the details of the vibrational and rotational absorbances detected by these

techniques. X-ray crystallography, neutron diffraction and electron diffraction can give molecular

structure for crystalline solids based on the distance between nuclei and concentration of

electron density. Gas electron diffraction can be used for small molecules in the gas phase.

NMR and FRET methods can be used to determine complentary information including relative

distances, dihedral angles, angles, and connectivity. Molecular geometries are best determined

at low temperature because at higher temperatures the molecular structure is averaged over

more accessible geometries (see next section). Larger molecules often exist in multiple stable

geometries (conformational isomerism) that are close in energy on the potential energy surface.

Geometries can also be computed by ab initio quantum chemistry methods to high accuracy.

The molecular geometry can be different as a solid, in solution, and as a gas.The position of

each atom is determined by the nature of the chemical bonds by which it is connected to its

neighboring atoms. The molecular geometry can be described by the positions of these atoms

in space, evoking bond lengths of two joined atoms, bond angles of three connected atoms, and

torsion angles (dihedral angles) of three consecutive bonds.

Valence shell electron pair repulsion (VSEPR) theory is a model in chemistry used to

predict the shape of individual molecules based upon the extent of electron-pair electrostatic

repulsion. The premise of VSEPR is that the valence electron pairs surrounding an atom

mutually repel each other, and will therefore adopt an arrangement that minimizes this

repulsion, thus determining the molecular geometry. The number of electron pairs surrounding

an atom, both bonding and nonbonding, is called its steric number.VSEPR theory is usually

compared and contrasted with valence bond theory, which addresses molecular shape through

orbitals that are energetically accessible for bonding. Valence bond theory concerns itself with

the formation of sigma and pi bonds. Molecular orbital theory is another model for

understanding how atoms and electrons are assembled into molecules and polyatomic ions.

VSEPR theory mainly involves predicting the layout of electron pairs surrounding one or

more central atoms in a molecule, which are bonded to two or more other atoms. The geometry

of these central atoms in turn determines the geometry of the larger whole.The number of

electron pairs in the valence shell of a central atom is determined by drawing the Lewis

structure of the molecule, expanded to show all lone pairs of electrons, alongside protruding and

projecting bonds. Where two or more resonance structures can depict a molecule, the VSEPR

model is applicable to any such structure. For the purposes of VSEPR theory, the multiple

electron pairs in a multiple bond are treated as though they were a single "pair".These electron

pairs are assumed to lie on the surface of a sphere centered on the central atom, and since they

are negatively charged, tend to occupy positions that minimizes their mutual electrostatic

repulsions by maximising the distance between them. The number of electron pairs therefore

determine the overall geometry that they will adopt.

For example, when there are two electron pairs surrounding the central atom, their

mutual repulsion is minimal when they lie at opposite poles of the sphere. Therefore, the central

atom is predicted to adopt a linear geometry. If there are 3 electron pairs surrounding the central

atom, their repulsion is minimized by placing them at the vertices of a triangle centered on the

atom. Therefore, the predicted geometry is trigonal. Similarly, for 4 electron pairs, the optimal

arrangement is tetrahedral

Question:

What are the most important factor in order to determine the ion or molecule geometry?

The most important factor in order to determine the ion or molecule geometry is type of bonds

and lone pair.

How to get the number of lone pair electron and the number of bonding pair electron so

that you can use TTPE?

To get the number of lone pair electron and the number of bonding pair electron we must know

the number of electron of the molecule. Then, we must use the Lewis structure to show the

number and types of bond between atoms. From that structure, we can indicate the shape

molecules depend on the lone pair and bonding.

Why the geometry molecule for H2O is non linear compared to CO2?

The H2O is non linear because it has an O that has negative charge in the centre and the H

atom which is positive charge in both sides. So, the molecule of water is polar which the bond

dipoles do not directly oppose each other and therefore do not cancel each other. As the result,

the structure of the molecule will bend. The structure of CO2 does not bend because it has

double bond. So it has the zero dipole moment so that the structure will be linear.

Give the correlations between the basic geometry and hybridization for central atom.

Both basic geometry and hybridization for central atom has the

CONCLUSION

From this experiment, we can determine the 5 molecular geometries. Each of the

molecular geometries can be determined by using the VSEPR theory.

REFERENCES

1. BROWN, LEMAY, BURSTEN, MURPHY-CHEMISTRY:The Central Science

2. http://www.csupomona.edu/~egoldstein/121/VSEPR6.HTM

3. http://en.wikipedia.org/wiki/Orbital_hybridisation

4. http://en.wikipedia.org/wiki/VSEPR