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General Chemistry Lab

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Safety


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Safety

# Safety Equipment :

1. Safety Shower 4. Fire Ex nguishers

2. Eyewash Fountain 5. Fire Alarms

3. Fire Blanket 6. First Aid Cabinet

7. Fume Hood : Used for experiments that produced poisonous & irritating gases

# Some Of Safety Rules :

1. The following pictograms indicating chemical hazards . you should learn :

2. Flammable liquids such as ( Alcohols / Ethers / Acetone ) should be heated in

water bath , not over direct flame .

3. Always pour acids into water not vice versa .

4. Never return unused chemicals to stock bottles .

5. If an acid or corrosive chemical is spilled on your skin , wash immediately with

plenty of cold water and inform your instructor .


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6. If an acid is spilled on your clothes , wash with water then neutralize with dilute

ammonium hydroxide .

7. If base is spilled on your clothes , wash with water then neutralize with dilute

acetic acid followed by dilute ammonium hydroxide .

8. If an acid or base spilled on the desk or floor , wash with plenty amount of water

then add sodium bicarbonate followed by washing with water again .

Note : The top loading balance ( ± 0.01g ) and Analy cal balance ( ± 0.0001g ) used for

determine the mass of any material .

# Safety Equipment :

1. Beaker & Conical Flask ( Erlenmeyer Flask ) : This glassware used as a container ,

both give approximates volume , but do not use them for volume measurements

.

2. Graduated Cylinder : Used for volume measurements .

3. Burette : Gives accurate volume measurements .

4. Volumetric Flask : Used to prepare chemical solution of certain molarity .

5. Bunsen Burner : And it haves two equations :

- first equation : when the blame is blue ( complete combustion ) the equation is :

C4H10 + O2 ------------> CO2 + H2


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- Second equation : when the blame is yellow ( incomplete combustion ) the

equation is :

C4H10 + O2 ------------> C(S) + H2O

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Ques on No.1 :

One of the following statements is correct :

a. Never smell directly any chemical

b. Always pour water into acids

c. Return unused chemicals to stock bottles

d. Taste any chemical

e. Heat volatile liquids over direct flame , don't use water bath

* The correct Answer is ( A ) *

Question No.2 :

One of the following is correct :

a. Graduated cylinders are used to measure the masses of liquids

b. Desiccators are used to cool substances


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c. Burettes flasks are used to measure pressure of solutions

d. Test tubes are used to measure the density of solutions

e. Bunsen burner is used to cool solutions

* The correct Answer is ( B ) *

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Ques on No.1 :

Which one of the following equipments is used to measure the volume of a liquid ? :

a. Graduated cylinder

b. Beaker

c. Crucible

d. Thermometer

e. Desiccator


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Ques on No.1 :

Which one of the following is wrong :

a. Weighing hot objects over the balance won't affect the mass reading

b. Fire extinguishers are usually filled with CO2

c. Never taste any chemical

d. Flammable liquids are heated in water bath

e. Always pour acids into water


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Ques on No.1 :


a. When dealing with poisonous chemicals , it is safe to do the experiments in the

fume hoods

b. When volatile liquids catch fire , use water as fire extinguisher

c. To protect your eyes , wear eye lenses

d. Always directly smell any chemicals

e. Taste any material found in the laboratory


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Ques on No.1 :

Circle the most correct statements :

a. Smoking , eating and drinking are allowed in the laboratory

b. Always pour water into acids

c. Don't heat flammable liquids over direct flame

d. Taste any chemical to identify it

e. Smell directly the chemicals in order to identify them

* The correct Answer is ( C ) *

Question No.2 :

Which equipment is used to measure the temperature of a liquid :

a. Graduated cylinder

b. Desiccator

c. Balance

d. Thermometer

e. Bunsen burner

* The correct Answer is ( D ) *

� �

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F٣E� �

Identification of

Chemical Compounds

Using Their Physical

Properties� �


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# Physical & Chemical Properties :

- Chemical Properties : " Is a characteristic of material involving its chemical change " .

- Physical Properties : " Is a characteristic that can be observed for a material without

changing its chemical identity " .

Such as : Solubility , Melting Point , Boiling Point , Density .

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(1) Mel ng Point :

Definition of Melting Point (M.pt) : " It is the temperature at which a particularly solid

changes to liquid or melt " .

- Why do we find the melting point ?

1. Iden fica on .

2. Determina on degree of purity .

- Note : Melting point is related to intermolecular bounds ( ��1א�mא�Tא�&%! ) so Strong

intermolecular bounds -------> Higher Melting point .

- Note : When we measure the M.pt we well use capillary tube ( E��Qא�G%� ) and when א(.

we use it , we must commitment to the following notes :

1. The height of the solid sample should be approximately 1 cm .

2. The solid sample should be well-stacked or well packing .

3. The solid sample should be heated slowly .

4. The solid sample & the thermometer should be at the same level .

Identification of chemical compounds using

their physical prosperities


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- The Procedure :

1. Introduce the finely powdered , dry solid sample into a capillary tube is sealed at

one end .

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2. Place the capillary tube in the melting apparatus .

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3. Adjust the rate of heating so that the temperature rises at moderate rate .

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4. When the temperature is 15° - 20° below the expected melting point , decrease

the rate of heating so that the temperature rises only one to two degrees per

minute .

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5. If the temperature is rising at a very fast rate during melting , repeat with a new

sample .

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6. Record the melting point ranges observed , and Identify the name of the

unknown .

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- Example on the outcomes of the experiment :

111 C° - 115 C° +!�gو�\.� gد�Iא"}F

111 C° - 111.5 C° 6B�gو�\.� gد�Iא"}F@@@@

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- Note : If the melting range increase then the impurity of the sample increase , and vice

versa .

- What is the effect of the impurities on melting point ?

If the impurities :

- Soluble �� then the melting point decrease ( M.pt ) .

- Insoluble �� such as ( sand / glass … etc ) have no effect .

(2) Boiling Point :

Definition of Boiling Point : " It is the temperature at which the vapor pressure of a

liquid equals the external atmospheric pressure " .

- Note : The boiling point is affected greatly by the external pressure exerted on the

liquids surface but slightly by the presence of impurities .

Boiling point α external pressure

- The Procedure :

1. Construct up the boiling point apparatus as directed by your instructor .

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2. In a test tube filled with the unknown liquid , insert a thermometer to which a

capillary tube sealed from the upper end is attached .

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3. Heat the tube with its content over water bath gently until you notice the

evolution of air bubbles from the capillary tube .

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4. Turn the burner off , and record the boiling point as the temperature at which

the unknown liquid enters the capillary tube

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- Note : Boiling point is related to intermolecular bounds .

So Strong intermolecular bounds

(3) Density :

Definition of Density : " It is the mass of the



Turn the burner off , and record the boiling point as the temperature at which

the unknown liquid enters the capillary tube .

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Boiling point is related to intermolecular bounds .

intermolecular bounds �� Higher Boiling point

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Density : " It is the mass of the material per unit volume " .

١١

Turn the burner off , and record the boiling point as the temperature at which

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Boiling point

material per unit volume " .


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- Note : Usually the density is expressed in grams per mL or cc ( <٣ ) . cc is a cubic

centimeter and is equal to a ( mL ) therefore :

The Density = mass / volume = g/mL or g/cc or kg/L

- Note (Just for knowing) : The definition of density on theory is : " a physical property

of a matter , as each element and compound has a unique density associated with it .

density defined in a qualitative manner as the measure of the relative " heaviness " of

objects with a constant volume " .

- The Procedure :

First : finding the density of irregular solid :

1. Obtain from your instructor an unknown solid of irregular shape .

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2. Determine the mass of the solid .

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3. Into 100 mL graduated cylinder , pour about 50 ml of water . record the ini al

volume of water correctly .

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4. Hold the cylinder at 45° angle and slide the solid into the water gently . take care

not to let the solid hit the cylinder bottom too hard . also avoid the presence of

any air bubbles .

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5. Record the final volume of water .

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6. Calculate the density of the solid from the data collected .

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Second : finding the density of unknown liquid :

1. Clean up a 10 mL volumetric flask . rinse with acetone and leave it in the fume

hood to dry .

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2. Weigh the volumetric flask with stopper .

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3. Fill the volumetric flask with the unknown liquid exactly to the mark .

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4. Weigh the volumetric flask with content and stopper .

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5. Calculate the density of the unknown liquid .

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(4) Solubility :

Definition of Solubility : " It's the amount of material that dissolves in a given amount

of solvent at a given temperature to give a saturated solution " .

- Note : The solubility of any compounds in a certain solvent depends on many factors

like :

1. Temperature .


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2. The nature of bonds in the chemical substance ( intermolecular & intermolecular

forces ) .

- Note : The general accepted rule in solubility is : " Like Dissolves Like " .

- Note : When we study the solubility ( solute & solvent ) . we have to study the polar &

non-polar compounds , so that we can determine that the solute will dissolve in the

solvent or not , according to the general rule " like dissolve like " , where :

1. Polar solute dissolved in polar solvent .

2. Non-polar solute dissolved in non-polar solvent .

3. Polar solute insoluble in non-polar solvent .

4. Non-polar solute insoluble in polar solvent .

- Note :

Polar Compounds : such as { H2O / NH3 / HCL / KCL / H2SO4 / KOH / NaCL } . ��<D: @

Non-Polar Compounds : such as { CH4 / CCL4 / C6H12 / C6H4 / CO2 } . @@@@��<D:

- Note : Polar & Non-Polar compounds depends on the following factors :

1. Electronegativity difference (�� the highest electronegativity atoms , ( ��قא��Dو

are { F , O , N , CL } .

2. Electric dipole moment ( JH\א�8��12م ) , and it depends on the direction of the

polar bonds .

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- Note :

1. Acetone ( CH3COCH3 ) is soluble in water .

2. Chloroform ( CHCL3 ) is slightly

- Note :

1. All ( C , H ) compounds are insoluble in water .

2. ( C , H ) compounds including ( OH , NH ) are soluble in water on condition that :

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- The Procedure :

1. Introduce a few crystals of the solid material in a

about 2.0 ml of the solvent . shake well and record your observa on .



) is soluble in water .

slightly soluble in water .

All ( C , H ) compounds are insoluble in water .

( C , H ) compounds including ( OH , NH ) are soluble in water on condition that :

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0�Iא@Gو}�]JB�Iאن��^B/��KKא��%ن

Introduce a few crystals of the solid material in a clean , dry test tube , then add


Soluble In

Water

١٦

( C , H ) compounds including ( OH , NH ) are soluble in water on condition that :

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0�Iא@Gو}�]JB�Iאن��^B/��Kא��%ن

clean , dry test tube , then add


Soluble In

Water


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�h�fאZ��X�3:?

2. For liquid material place about 2-3 drops of the liquid solute in a clean dry test

tube . add about 3mL of the solvent . Shake well and record your observa on .

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3. To approve your results , check the solubility of the different solvents used in this

part of experiment with each other .

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Que. No. 1 : Which of the following compounds is water – soluble ?

Solution :

a. K2SO4 �� [ Soluble ]

b. C2H5OH �� [ Soluble ]

c. CH3(CH2)10CH2OH �� [ Insoluble ]

d. C10H8 �� [ Insoluble ]

e. CH2(OH)CH(OH)CH2OH ��[ Soluble ]

Que. No. 2 : In which of the following solvents [ KCL ] is expected to dissolve ? explain

your answer .

Solution :

a. H2SO4 ( Diluted ) �� H2SO4 is polar compound and KCL also polar compound , and

the general rule says " like dissolves like " , So KCL [ Dissolve ] in H2SO4 .

b. HCL ( Concentrated ) �� Like the branch (a) before . HCL & KCL are polar

compounds , So it supposed that KCL dissolve in HCL , But because HCL is

Concentrated . then KCL is [ Insoluble ] in HCL concentrated .

c. C6H6 Benzene �� C6H6 is non-polar compound and KCL is polar , and according to

the rule " like dissolves like " , KCL [ Insoluble ] in Benzene .

*** Pre – Laboratory Questions ***


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ١٩

Que. No. 1 : Explain how the pressure changes affect the boiling point of a pure liquid ?

Solution :

The relationship between boiling point & the pressure is direct ( �4د�� ) [ B.pt α pressure

] . so when the pressure increase , the B.pt will increase .

Que. No. 2 : Define each of the following terms :

Solution :

a. Solubility �� " It's the amount of material that dissolves in a given amount of

solvent at a given temperature to give a saturated solution " .

b. Melting point �� " It is the temperature at which a particularly solid changes to

liquid or melt " .

c. Boiling point �� " It is the temperature at which the vapor pressure of a liquid

equals the external atmospheric pressure " .

d. Density ��" It is the mass of the material per unit volume " .

Que. No. 3 : What is the effect of each of the following cases on the measured ( melting

point ) of a solid . ( increase , decrease or no effect ) giving good explanation for each

answer ?

Solution :

a. Rapid heating �� [ Increase ] , the temperature will increase very fast so we

won't be able to read the accurate melting temperature & so it will be above the

real melting temperature .

b. A large quantity of the compound is introduced into the capillary tube �� [

Increase ] , because the large quantity need a lot of time to melt so it will

resulting too wide and possibly high melting point range .

*** Post – Laboratory Questions ***


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٢٠

c. If the capillary tube and the thermometer bulb are not at the same level �� The

answer depends on where to place the thermometer . Because if you put it on

lower level from the capillary tube , the temperature of the thermometer will

increase more rapidly than the tube , so the melting point will be higher than the

correct reading . and vice versa .

Que. No. 4 : Explain how the pressure of impurities is expected to affect the measured

melting point of a solid substance ?

Solution :

The Impurities will [ Decrease ] the melting point . because the impurities weaken the

intermolecular bounds that holding the solid together , so it takes less energy to pull

the molecules apart .

Que. No. 5 : Will measured density of a solid material be affected if part of the solid

dissolved in the liquid used ? why ?

Solution :

Yes , because the volume will differ , and density depends on volume .

Que. No. 6 : Draw the apparatus used to measure the boiling point of a volatile liquid ?

Solution :

�X.א<!S�_�`א��9١١


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٢١



Ques on No.1 :


a. Melting point is the temperature at which the solid is converted to liquid

b. Density is mass divided by volume

c. Molarity is the number of moles of solute divided by the volume of solution

d. Solubility is the amount of solute dissolved in a certain gas at a certain

temperature to give a unsaturated solution

e. Boiling point is the temperature at which the vapor pressure of the liquid equals

the external atmospheric pressure



Ques on No.1 :

The physical property that increases with increasing atmospheric pressure is :

a. Melting point

b. Boiling point

c. Density

d. Solubility

e. Mass


Question No.2 :

CCL4 dissolved in a certain solvent , this means that the solvent is :

a. Polar

b. Has a high boiling point


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٢٢

c. Has high density

d. Has a low density

e. Non-polar

* The correct Answer is ( E ) *

Question No.3 :

Which one of the following substances has a density below the density of water ?

a. CO

b. Cu

c. Ni

d. C2H5OH

e. Zn


Gw�S�K�ن�_K:אF١٥L٤L٢٠٠٩WE

Ques on No.1 :

To determine the density of unknown solid using water displacement method , the

solid was introduced into an equipment called :

a. Crucible

b. Flask

c. Desiccator

d. Cylinder

e. Balance


Question No.2 :

One of the following is not a chemical change :

a. Burning of coal

b. Neutralization of HCL with NaOH


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٢٣

c. Dissolving salt in water

d. Production of MgO from Mg

e. Production of Mg3N2 from Mg


Gw�S�K�ن�_K:אF٩L١٢L٢٠٠٩WE

Ques on No.1 :

Which one of the following changes is not a physical change :

a. Freezing of a liquid

b. Condensation of vapor

c. Melting of NaCL

d. Evaporation of water

e. Converting of Mg to MgO


Question No.2 :

The boiling point of a liquid decreases as :

a. The intermolecular forces between liquid molecules decrease

b. The external pressure increase

c. The liquid's mass decrease

d. The liquid's volume decrease

e. The liquid's density increase


Question No.3 :

One of the following is insoluble in CH3OH :

a. NH3

b. H2O

c. H2SO4


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٢٤

d. C6H6

e. CH3NH3


Gw�S�K�ن�_K:אF٩L١٢L٢٠٠٩E�'¡ذج%£W

Ques on No.1 :

Which one of the following changes is not a chemical change :

a. Combustion of butane gas

b. Neutralization of vinegar with NaOH

c. Melting of NaCL

d. Converting of Mg to Mg3N2

e. Reaction of Mg with O2


Question No.2 :

One of the following is insoluble in hexane ( C6H14 ) :

a. NH3

b. C7H16

c. C6H6

d. CCL4

e. CH4


Gw�S�K�ن�_K:אF١٣L٤L٢٠١١EW

Ques on No.1 :

An example of a chemical property is :

a. Density

b. Mass

c. Solubility

d. Acidity


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٢٥

e. Boiling point


Gن�_K:א��D.٢٠١١W

Ques on No.1 :

Circle the correct statement :

a. Solubility is the amount of solute dissolved in a certain amount of solvent at a

given temperature to give unsaturated solution

b. Boiling point is the point at which the vapor pressure of the substance equals the

external pressure

c. Density describes the volume per unit mass

d. Boiling point determines the purity of a substance

e. Melting point is independent on the external pressure


Question No.2 :

In an experiment , NaCL dissolves in a certain solvent , this means that the solvent is :

a. Non-polar

b. Has high density

c. Polar

d. Has low B.pt

e. Has low density


� �

�א��א��א��

F٤E� �

Water of

Crystallization


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٢٦

# Introduction :

- Hydrated Salts [ ��D،א):�3א��4�B��@0�IאV:��.>�2E%K�E� ] are chemical compounds ,

also called Hydrate . The general formula is :

where :

- MA : The Anhydrous salt [ J4א��r��Iא ]

- The Dot ( . ) : ��Mmא��وא�ن��2<ل��

- X : moles of water ( in one mole of hydrates )

- Note : The water molecules are held loosely within the crystal of the salt ( through

intermolecular forces ) , such that moderate heating of the hydrated salt will drive off

water molecules , producing the anhydrous salt . A color change sometimes

accompanies the dehydration process as in the following example :

CuSO4.5H2O (s) --------------> CuSO4 (s) + 5H2O

JIא�X�،EV:'3لא�\%&�*א��1א�,��F5��1$א0�IאI%$%دgدא'+��%gSא��I�\�¥��%DWJ��EFא��\�3

��20��8��.��^3א�7�K*אK�I�ل��א��J4V:�/.��ن��L5��1$א0�I��7وج:VאJB�Iو��%�Vא��I�rא��J4،و�

@��B،��Iאن%�r~K�"}Fא��Kא�ل�ÎאW

J4א��א��_�س5�K�6B��:א��_�س5�K�6BHg�7�K:0�:5��1$

Water Of Crystallization

MA . XH2O

Blue Bluish white

�%.��:S1ق �%.��زSق


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٢٧

# Example of hydrates :

CaSO4.2H2O §��:E5�K�6Bא��%مא��J4Fא�

1 mol of (CaSO4.2H2O) �� 2 mol of H2O

"}Fم�7K�.3:�BJ4א��Iא��KB��:%��:%ن��:��2�א�`�~�

1 mol of CaSO4 �� 2 mol of H2O

m\�J4א��r��Iא��KB��:%��:%ن��:��2�א�`�~�"}Fم�7K�.

# Alums :

Alums are hydrated double sulfate salts [ 5�K�6א��3:��F،��Qא�1دو$��:�3Iא��Iא ] , that have

the general formula :

where :

- M+1

: mono-volent ions [ ¨��Kא�د��5�.%�� ] ,

like Na+1

, Ag+1

, NH4+ ……

- M+3

: tri-volent ions [ 5�.%��838¨��Kא� ] , like

Al+3

, Fe+3

, Cr+3

……

Example of Alums : KAL(SO4)2.9H2O س��%��א��

# How to find (X) for a hydrated salt ?

Let's take the following hydrates salts (CuSO4.XH2O) like example :

CuSO4.XH2O CuSO4 + H2O

�� Moles of CuSO4 ( M.wt = 159.5 g/mol ) = 0.6 = 3.761 × 10 -3

mole

159.5

M+1

M+3

(SO4)2.XH2O

غرام ١الوزن قبل التسخين غرام ٠.٦التسخين بعدالوزن غرام ٠.٤المتبقي


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٢٨

�� Moles of H2O ( M.wt = 18 g/mol ) = 0.4 = 22.222 × 10 -3

mole

18

�� Moles of Hydrate �� couldn't be determined so :

From formula :

1 mole of CuSO4 �� X mole of H2O

3.761 × 10 -3

�� 22.222 × 10 -3

So : X = 5.9 ≈ 6 { X must always be integer }

%H2O = mass of H2O × 100% = 0.4 × 100% = 40%

mass of Hydrate 1.0

- Note : Some of the tools & devices that you will use in this experiment :

(1) Crucible : ?��א��א?�$��2gSא��5�$Sد+�_K�ن��D��Aو��Fو،

(2) Desiccator : ��%4א��V:'�ل$%@א��6K�م�7K��ز�D$%F

- The Procedure :

1. Obtain crucible and cover from your instructor and clean them thoroughly with

soap and water , then dry .

?�D��$<8،0�Iوא`��%ن��א��$�D�X.<8،�D��Hو��$Z��:V:}'?


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٢٩

2. Place the clean dry crucible and cover on a clay triangle supported by a ring .

Heat the crucible and cover to redness ( for about 5 minutes ) over the Bunsen

burner , using the non-luminous flame .

?�D��Hو�א��V7�%א��FLMא��א��وא��X�2<:^��א�H*אI�2%م��\�،8>٥Y��!دEV��!%:>�2?

3. Allow the crucible and cover to cool for one minute in their position before

placing them into a Desiccator using the tongs provided , allow them to cool at

room temperature .

��ن?�ªא8>،m\�Iאאم�7K�+�ن�,��D@א»��!Zوذ�،�D.��:@g��א�ª��C��و��D��H�/ن�6دI�gد!�\�وא

��אgSא�~��$Sد�2<6د�?

�X�3:W�H��V��Iאل��K�]وא¬%�Kא�L�IZوذ�،אقSא)و�2<و�JK�Iא�2<א��'��א��L,�]

4. Weigh the empty crucible (without cover) accurately .

?��ون��D��H��!�א��Sزنא��?

5. Place about 1.0 g of the unknown hydrated salt in the crucible and weigh again .

?:Vא��Iא��J4�rא��Iو�@א��وزن��دא��אمLM١�:�GS�\א��?

6. Place the crucible on the clay triangle as in step 2 . Adjust the cover such that

portion of the crucible is uncovered .

�mא�~0�H��<K�QB01$V:א��?Mوא،א�^�.��g%Hfא@��B�\��2%م�Iא*Hא�:^��2<א��LM?

7. Carefully heat the crucible and it's contents gently at the beginning after the salts

has expanded and frothed , remove the cover and heat the crucible strongly for

another 5 minutes .

?g�Ig%\��א��V7<80�H~א�زل�،�و��1��IאL%K�ن��و��،א��٥V7��Hא��و�D��%Kk@א�Y��!د&�'�?

8. Allow the set to cool as described in step 3 , then weigh again , Take care to

handle the crucible with tongs only .

?<!Sg%Hfא@��B6د�ن��D�'א��و:��C��ª٣אm\�m\�Iא3ل'V:��א��Z��ن��2<ص��?،8>زن:�g�'�&،وא


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٣٠

9. Heat the crucible and contents for another two minutes then allow to cool as

done in the previous step , then weigh again . The values obtained in step 8 and

step 9 should not differ by more than 0.05 g . If the difference is grater than 0.05 ,

repeat step 9 un l you achieve constant mass .

?Kkوא��V7 ��/ن�6د��B@אg%Hfא��\�،8>زن��دא�ªא8>،�,��*K\�!��D��%K@�D��$وא��א�\��

Sא�\O]9��K�]ن�Je��Kوא�א�^�:��g%Hf٠א{٠٥دא��g%Hfא"}Fg�2د��<\�،Zذ�V:6B��3K']אن�B9ذא�:�،<�

�K��8��KB>�2+`�>K�?


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٣١


Que. No. 1 : What are hydrated salts ? Give some examples .

Solution :

Hydrated salts is a crystalline salt molecular that is loosely attached to a certain

number of water molecules .

Ex. Na2SO4.10H2O / CuCL3.2H2O / CuSO4.5H2O

Que. No. 2 : The formula of magnesium sulfate hepto-hydrate is MgSO4.7H2O , calculate

the mass of water in a 6.5 g sample of this compound .

Solution :

The M.wt for MgSO4.7H2O = 246.4755 g/mol & The M.wt for water = 18 g/mol

• Then : Moles of MgSO4.7H2O = 6.5 = 0.02637 mole

246.4775

• From the formula , we find that :

1 mol of MgSO4.7H2O �� 7 mol of H2O

0.02637 mol of MgSO4.7H2O �� Z moles of water

Then : Z = 7 × 0.02637 �� Z = 0.18459 mole of water

Then : Mass of water = No. of moles × M.wt of water

Mass of water = 0.18459 × 18

Mass of water = 3.32262 g

Que. No. 3 : List at least three common uses of hydrated salts .

Solution :

1. Industries : Such as salt factories & chemical industries .

2. Treatment of diseases & Skin problems : Such as psoriasis and eczema .

3. The plaster ( Gypsum ) industry .


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٣٢

Que. No. 1 : What is the effect of each of the following , on the calculated value of X

( increase / decrease / no effect ) ? Justify your answer .

Solution :

a. Incomplete dehydration of the hydrated salts �� [ Decrease ] , because not all of

hydration ( ��%4א�� ) is removed , and since not all of mass of H2O is removed ,

then the calculated moles of that mass is too small , so X is calculated to be

number that is too small .

b. If mass of ( Crucible + Anhydrous salt ) was +0.1 more than the actual value ��

[ Decrease ] , because that increase in mass of anhydrous salt will reduce the

mass of water and thus will reduce the No. of water moles . And while the

relation between No. of water moles & X is direct , then X will decrease .

c. If the yellow flame was used in heating process instead of the blue non-luminous

flame �� [ Decrease ] , because the yellow flame is caused by incomplete

combustion and will leave soot on the crucible , so it will increase the mass of

anhydrous salt . And also the yellow flames are cooler than the blue flames so

that will not have full hydration removed .

Que. No. 2 : Is there any effect on the results of this experiment , if you weighted the

crucible while it's still hot ?

Solution :

Yes it will effect , because the hot object will create convection ( ESא��+� ) current on

the air around the balance . This in-fluctuating force reduce the air pressure on the

balance & can make it difficult to obtain stable reading .



@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٣٣

Que. No. 3 : Was it necessary to cover the crucible all the time while heating ? Why ?

Solution :

No it's not necessary , but it should open slightly so the hydration and moisture in the

sample come out of it .


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٣٤



Ques on No.1 :

Which one of the following is an alum ?

a. Na2SO4.10H2O

b. Na2S2O3.5H2O

c. AgAL(SO4)2.12H2O

d. CuSO4.5H2O

e. MgCL2.6H2O


Question No.2 :

The % w/w of H2O in the hydrated salt AB.XH2O is 36% . If the molar mass ( molecular

weight - M.wt - ) of the hydrated salt is 249.5 g/mol , then the number of moles of H2O

is :

a. 3

b. 5

c. 6

d. 12

e. 10


Jא��¨אل+��\��4 JW

w/w % = 36% = mass of water × 100% �� mass of water = 0.36

mass of hydrate mass of hydrate

And : - M.wt of AB.XH2O = 249.5 g/mol

- M.wt of Water = 18 g/mol

Remember : n ( No. of moles ) = Mass / M.wt


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٣٥

1 mole of AB.XH2O �� X mole of water

mass of AB.XH2O �� mass of water

249.5 18

Then : X × mass of AB.XH2O = mass of water

249.5 18

Then : X = mass of water × 249.5 �� X = 0.36 × 249.5

mass of AB.XH2O × 18 18

So : X = 4.99 ≈ 5


Ques on No.1 :

The mass ( in grams ) of water in 47.4 g of KAl(SO4)2.12H2O equals :

a. 2.16

b. 21.6

c. 0.216

d. 216

e. 4.16


Jא��¨אل+��\��4 JW

- The M.wt of [ K = 39 / Al = 27 / S = 32 / O = 16 / H = 2 ]

- The M.wt of KAl(SO4)2.12H2O = 39 + 27 + ( 32 + 16 × 4 ) × 2 + 12 × 18 = 474 g / mol

- Then :

1 mole of KAl(SO4)2.12H2O �� 12 mole of Water

47.4 �� mass of water

474 18

So : 12 × 0.1 = mass of water / 18 �� Then mass of water = 21.6 g


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٣٦

Question No.2 :

A 1.84 g sample of a certain hydrated salt was dried completely to give 1.3 g of the

anhydrous salt ( molar mass of the hydrous salt = 368 g / mol ) . The value of X equals :

a. 4

b. 3

c. 12

d. 2

e. 6


Jא��¨אل+��\��4 JW

- The M.wt of hydrous salt = 368 g/mol & it's mass = 1.84 g , then n = 0.005 mole

- The mass of water = 1.84 – 1.3 = 0.54 g & it's M.wt = 18 g/mol , then n = 0.03 mole

- Then :

1 mole of hydrous salt �� X mole of water

0.005 �� 0.03

So : 0.005X = 0.03 �� Then X = 6

Question No.3 :

One of the following chemicals is an alum :

a. NH4Al(SO4)2.12H2O

b. CuSO4.5H2O

c. CuSO4

d. NH4Al(SO4)2

e. NaHCO3.3H2O



@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٣٧


Ques on No.1 :

The % by mass of H2O in the salt Cu(NO3)2.6H2O equals :

a. 38.92

b. 36.55

c. 36.21

d. 37.34

e. 45.34


Jא��¨אل+��\��4 JW

- The M.wt of hydrous salt = 295.5 g/mol & it's moles = 1 mole , then m = 295.5 g

- The M.wt of water = 18 g/mol & it's moles = 6 mole , then m = 108 g

- Then :

w/w % = mass of water × 100% �� 108 × 100% = 36.55%

mass of hydrate 295.5

Question No.2 :

One of the following chemicals is an alum :

a. KCr(SO4)2.12H2O

b. CuSO4.5H2O

c. CrSO4.5H2O

d. ZnSO4.7H2O

e. (NH4)2Fe(SO4)2.6H2O


Question No.2 :

A sample weighing 8.480 g of a hydrated salt was dried to 6.360 g if the molar mass of

the dehydrated salt is 324 g/mol , then the value of X equals :

a. 7


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٣٨

b. 4

c. 8

d. 6

e. 5


J�F��א)��א��(�D��Q:א��¨אل+��\��4�_�h�D��9L$Sא+�!V:٣٦ J


Ques on No.1 :

A 0.990 g sample of the hydrated salt : MA.XH2O was dried to 0.5211 g . If the molar

mass of the anhydrous salt is 160 g/mol , the value of X ( moles of water in the

hydrated salt ) equals :

a. 5

b. 6

c. 7

d. 8

e. 11


J�_�h�9L$Sא،�,��Yא��وא��¨אل+�!V:�F��א)��א��(�D��Q:א��¨אل+��\��4٣٦–

Gw�S�K�ن�_K:אF٩L١١L٢٠٠٩E�'¡ذج%£W

Ques on No.1 :

A 0.648 g sample of the hydrated salt : MA.XH2O was dried to 0.4143 g . If the molar

mass of the anhydrous salt is 160 g/mol , the value of X ( moles of water in the

hydrated salt ) equals :


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٣٩

a. 5

b. 6

c. 7

d. 8

e. 11


JYא��אل¨��D��Q:א��¨אل+��\��4 J

Gw�S�K�ن�_K:אF٢١L٤L٢٠١٠EW

Ques on No.1 :

The mass ( in grams ) of H2O , in 6.66 g of BaCl2.2H2O ( 244.3 g/mol ) equals :

a. 0.491

b. 0.818

c. 0.981

d. 0.654

e. 1.14


J،+�!V:�F��א)��א��(�D��Q:א��¨אل+��\��4�_�h�9L$S٣٥א J

Gw�S�K�ن�_K:אF١٣L٤L٢٠١١EW

Ques on No.1 :

The percentage composition of sulphate ions (SO42-

) in K2SO4.Al2(SO4)3.24H2O ( 948.2

g/mol ) equals :

a. 5.69%

b. 8.25%


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٤٠

c. 45.56%

d. 54.43%

e. 40.50%


J+�א��¨אل��4\� JW

- The M.wt of hydrous salt = 948.2 g/mol

- The M.wt of (SO42-

) = 96 g/mol

- No. of Sulphate ions are 4 , So : 4×96 = 384 g/mol

- Then : 384 × 100% = 40.50%

948.2

- Note : X – the no. of water moles – are also known as water of crystallization value .

G��D.ن�_K:٢٠١١אW

Ques on No.1 :

The % by mass of H2O in a hydrated salt is 46.8% and X in this salt equals 6 . The

molecular weight ( in g/mol ) of this hydrated salt equals :

a. 474

b. 249.6

c. 278.1

d. 225

e. 230.8


Jא��¨אل+��\��4 JW

w/w % = 46.8% = mass of water × 100% �� mass of water = 0.468

mass of hydrate mass of hydrate


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٤١

- mass of water = No. of moles × M.wt of water

= 6 × 18 = 108 g

Then : mass of hydrate salt = 108 = 230.8 g

0.468

- Now : M.wt for hydrous salts = mass of hydrate = 230.8 = 230.8 g/mol

No. of moles 1

� �

�א��א��

F٥E� �

Empirical

Formula of an

Ionic Compound� �


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٤٢

Empirical Formula of an Ionic compound

Empirical Formula of an Ionic compound

# Introduction :

There are many ways in chemistry to represent chemical compounds . And the most

famous ( common ) ways are :

(1) Molecular formula : Shows the actual number of atoms in a compound . Ex. : C6H6

(2) Empirical formula : Shows the simplest whole no. ratio of atoms in a compound .

Ex. : CH

So the molecular formula being a multiplication of the empirical formula :

Where :

n : integer ( ��_hد�2 ) �� n = M.wt of real compound

M.wt of empirical compound

- Note : For ionic compounds always ( n = 1 ) .

# How to find empirical formula ?

To learn how to find empirical formula by ( synthesis method ) , we will take the

Magnesium Oxide ( MgxOy ) like an example :

Mg + O2 MgxOy

0.1 exist 0.7 ( *7�Kא��א�%زن )

So :

1. First we find the no. of moles of Mg :

Moles of Mg = 0.1 = 4.11 × 10-3

24.3

Molecular formula = [ empirical formula ] × n


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٤٣

2. Then we find the mass of O :

Mass of O = mass of MgxOy – mass of Mg = 0.7 – 0.1 = 0.6 g

3. Now we find the no. of moles of O :

Moles of O = 0.6 = 0.03 g/mol

18

4. Now after we found the no. of moles for both Mg & O . Devide by smallest

number of moles :

Mg 4.11 × 10-3

O 0.03 �� Mg 1 O 9.4

4.11 × 10-3

4.11 × 10-3

5. Rounding ≈ Mg 2O19

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J٢}٠Eل�^:،m\�F١}٩≈٩KE�9G�\.6وB�G�!�¯��א�ن�B9ذא��_hد�2F٨}٠ J٩}٠Eل�^:،m\�

F٨}٩≈١٠KE¯��א�ن�B9ذאVو��FV:٣}٠�9٧}٠E��.��_hد��א��¯G�,.Yא��ل�Îא��،��\�V�A>K�،

F٤}٩≈٤}٩×٢Z٨}١٨≈١٩KE

- Note : During the main rxn ( reaction ) , there is a side rxn occurs and its :

Mg (S) + N2 (g) Mg3N2

So (Mg3N2) is a by-product , and to get rid of it , add few drops of water :

Mg3N2 + H2O MgXOy + NH3

- The Procedure :

1. Obtain crucible and cover from your instructor and clean them thoroughly with

soap and water then dry them .

?��D��$<8،0�Iوא`��%ن��א��$�D�X.<8،�D��Hو��$Z��:V:}'?

2. Heat the clean dry crucible and cover over Bunsen burner to redness .


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٤٤

?Sא��Rא>K�V��!%:ق%��D��Hو�א��X�א�א��V7?

3. Allow the set to cool in its position for one minute, then using tongs, place the

crucible and cover in a desiccators and allow them to cool to room temperature .

��אgSد!�\�.�DI�gא�ª��C%��2/ن�6د@:��?�$Sد�96د�>K�،8>و��Kلאm\�I،LMא��وא�~0�H@א»��

��Hא�א�~��?

4. Weigh the empty crucible (without cover) accurately .

?��ون��D��H��!�א��Sزنא��?

5. Put about 0.25 g of magnesium turnings in the crucible and weigh again .

?:VאI~��%م@א��وزن��دא��אم٢٥}LM٠�:�GS�\א��?

6. Construct the set up , covering the crucible completely this time. Heat gently at

the beginning. Take care to lift the cover partially in order to introduce oxygen

occasionally to the reaction mixture .

?+�Qא�@��B�2%�«אLMF١}٥E�_�h٢٨א��א��ص>�2K@א��G�Kא)��h،و�mא��:�،8>V7��H@א�

+2��K��.��*C�B(א��9د'�ل+$�V:��1$0�H~א�L�S?

7. Continue heating until all magnesium is converted to ash, then remove the cover

and heat the opened crucible to redness .

?Sא��Rא>K��>K�<K�+�%��LאI~��%م�9S:�د،8>�زلא�~0�HوV7א��*7�Kא�+�B�?

8. Cool the crucible in position for few minutes, add few drops of water to

decompose any magnesium nitride ( Mg3N2 ) formed during heating. Notice the

production of any gas and smell .

?�0�IאV:5א�H!g�2�M�<8،Y��!دg��D.��:@��א��±د��،7*�Kא�8��0�;��Q�א��م%��~Iאא5T.V:E�+��]

?�K.9ج�E��زو�ESא�_�

9. Reheat the opened crucible contents until the ash is completely dry ( notice any

color changes ) .


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٤٥

��و�D��%Kk��دא�>K�eא��:�د��:�?%K�Iאא��7*��2�Fא��%ن@r�~�E��]E?

10. Cool the crucible and contents as done in step3, then weigh again .

?��±دא��و�D��%Kk��B��;@אg%Hfא�^��^�،8>زن��دא?

11. Reheat the crucible and contents for another tow minutes then allow cooling as

done in the previous step, then weighting again. The values obtained in steps 10

and 11 should note differ more than 0.05g. If the difference is greater than 0.05g,

repeat step ( 11 ) un l you achieve constant mass .

?�D��%Kkوא��V7*K\�!��<8،دא��زن8>،א��\�g%Hfא@��B6د�ن�� ªאK@�D��2;�`�א�\�>א��

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�K��8?


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٤٦


Que. No. 1 : An oxide of tungsten ( symbol W ) is a bright yellow solid . If 5.34 g of the

compound contains 4.32 g of tungsten , what is its empirical formula ?

Solution :

W + O2 WxOy

4.32 g 1.02 g 5.34 g

- The M.wt of tungsten = 183.85 g/mol

- Moles of W ( tungsten ) = 4.32 = 0.0235 mole

183.85

- Moles of O ( oxygen ) = 1.02 = 0.0637 mole

16

So : W 0.0235 O 0.0637 �� W1O2.71 ��D��,.٣

0.0235 0.0235

So The empirical formula is : W3O8

Que. No. 2 : Phenol is a compound composed of C,H and O . Combus on of 5.23 mg of

phenol yields 14.6 mg of CO2 and 3.01 mg of H2O . Find its empirical formula and the

percentage of each element in the substance .

Solution :

�X�3:Wא��¨אא}F@5א��+�D��5��~א�~�אم�g��%�+Kא��}'/.�%ل،

- Mass of Phenol ( CXHYOZ ) = 5.23 g

- Mass of CO2 = 14.6 g

- Mass of H2O = 3.01 g

- M.wt of CO2 = 44 g/mol

- M.wt of H2O = 18 g/mol


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٤٧

Now :

- Moles of CO2 = 14.6 = 0.332 mole

44

- Moles of H2O = 3.01 = 0.167 mole

18

So , we have to calculate :

- Moles of C in CO2 = 1 × 0.332 = 0.332 mole

- Moles of H in H2O = 2 × 0.167 = 0.334 mole

Now , we calculate the masses of H & C :

- Mass = M.wt × n ( No. of moles )

- Mass of H = 1 × 0.334 � mass of H = 0.334 g

- Mass of C = 12 × 0.332 � mass of C = 3.984 g

After that , we calculate :

Mass of Oxygen ( O ) = Mass of Phenol – Masses of ( C & H )

= 5.23 – ( 0.334 + 3.984 )

= 0.912 g

Moles of O = 0.912 = 0.057 mole

16

So :

C 0.332 H 0.334 O 0.057 �� C 5.82 H 5.86 O 1

0.057

0.057

0.057

The empirical formula of Phenol is : C6H6O

% C = mass of C × 100% = 3.984 × 100% = 76.17%

mass of Phenol 5.23

% H = mass of H × 100% = 0.334 × 100% = 6.39%

mass of Phenol 5.23

%O = 100% - ( 76.17 + 6.39 ) % = 17.44%


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٤٨

Que. No. 3 : Why should crucible tongs , not fingers , be always used for handling the

crucible after being heated .

Solution :

(1) Because sometimes the crucible is too hot , and this will lead to burn our fingers .

(2) Because if we touch it by our hands , the mass will be different because the

balance is too sensitive . And it will differ because the moisture that on our

fingers .


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٤٩

*** Post – Laboratory Questions *** *** Post – Laboratory Questions *** *** Post – Laboratory Questions ***

Que. No. 1 : How would the calculated empirical formula of magnesium oxide be

affected if :

Solution :

a. If incomplete dehydration was done after water was added �� The mass of

oxygen in magnesium oxide will Increase , so the number of oxygen moles will

also increase and that will make the empirical formula more than it should be

[ increase ] .

b. You forget to perform step (a) and hence didn't get rid of magnesium nitride

formed Mg3N2 �� The mass of oxygen will increase and in the same time the

mass of magnesium will decrease and that because the reaction with nitrogen .

So the ratio between oxygen & magnesium will be bigger than it should be , so

there will be [ increase ] in empirical formula .

c. If magnesium was not allowed to react completely with oxygen �� The mass of

oxygen will be less than the actual , so the ratio between oxygen & magnesium

will be also less than the actual , then the empirical formula will be smaller than

is should be [ decrease ] .

Que. No. 2 : Write the balanced chemical equation that represents the formation of

magnesium nitride .

Solution :

3Mg (S) + N2 (g) Mg3N2 (S)

Que. No. 3 : How can you get rid of the by-product formed in this experiment ? write

balanced chemical equations .

Solution :


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٥٠

By adding a few drops of water , to decompose and magnesium nitride formed :

Mg3N2 + 3H2O 3MgO + 2NH3

Que. No. 4 : list the errors that might have occurred during this experiment .

Solution :

(1) Weighing errors ( for Ex. The burning of magnesium oxide wasn't complete ) .

(2) The color of Bunsen flame is yellow .

(3) The dehydration wasn't complete .


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٥١



Ques on No.1 :

A compound SXOY contains 57% S while the rest is O . The empirical formula of this

compound is :

a. S2O3

b. SO3

c. SO2

d. SO

e. S2O


Jא��¨אل+��\��4 JW

- We suppose that the mass of the whole compound = 1 g

Then :

- Mass of ( S ) in the compound = 0.57 g

- Mass of ( O ) in the compound = 0.43 g

- The M.wt of S = 32 g/mol

- The M.wt of O = 16 g/mol

Now , we calculate :

- Moles of S = 0.57 = 0.017813 mole

32

- Moles of O = 0.43 = 0.026875 mole

16

So : S 0.017813 O 0.026875 �� S1O1.5 ( mul ply by 2 ) �� The empirical formula is S2O3

0.017813 0.017813


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٥٢

Question No.2 :

The empirical formula gives :

a. The actual whole number mole ratio of atoms in the compound

b. The number of moles in a compound

c. The simplest whole number mole ratio of atoms in a compound

d. The number of atoms in a molecule

e. The number of molecules in a molecule


Question No.3 :

The empirical formula of compound is C6H13 . If the molecular weight of this compound

is 170 g/mol , the actual formula of this compound is :

a. C12H26

b. C10H22

c. C4H10

d. C8H18

e. C6H14


Jא��¨אل+��\��4 JW

- From page No. 42 , n = M.wt of real compound = 170 g/mol

M.wt of empirical compound ( 6 × 12 + 13 × 1 ) g/mol

n = 170 g/mol = 2

85 g/mol

So : The actual formula = 2 × C6H13 = C12H26


Ques on No.1 :


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٥٣

0.46 g sample of nitrogen oxide contains 0.14 g of nitrogen , the empirical formula of

this oxide is :

a. N2O

b. N2O4

c. NO3

d. NO2

e. NO


J�_�h�9L$Sא،+�!V:�F��א)��א��(�D��Q:א��¨אل+��\��4٤٦ J


Ques on No.1 :

A sample of Manganese weighing 1.234 g was allowed to react completely with excess

oxygen to form 1.592 g of manganese oxide , the empirical formula of this oxide is :

a. MnO

b. Mn2O7

c. Mn2O3

d. MnO2

e. Mn3O4


JYא��אل¨��D��Q:א��¨אل+��\��4 J


Question No.1 :

During the experiment of determination of empirical formula of magnesium oxide , the

by-product formed is :


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٥٤

a. NH3

b. Mg3N2

c. MgN2

d. MgO

e. MgO2


Gw�S�K�ن�_K:אF٢١L٤L٢٠١٠WE

Ques on No.1 :

One of the following equations represents the elimination of the side product formed

during empirical formula experiment :

a. Mg + O2 �� MgO

b. 3MgO + 2NH3 �� Mg3N2 + 6H2O

c. 3Mg(OH)2 + 2NH3 �� Mg3N2 + 6H2O

d. 3Mg + N2 �� Mg3N2

e. Mg3N2 + 6H2O �� 3Mg(OH)2 + 2NH3


Gن�_K:א��D.٢٠١١W

Ques on No.1 :

One of the following is not an empirical formula :

a. HOOCCOOH �� C2H2O4 this can divide by 2

b. CH3COOCH3 �� C3H6O2 you can't divide it

c. CH3CH2COOH �� C3H6O2 you can't divide it

d. CH3COCH3 �� C3H6O you can't divide it

e. CH3OH �� CH4O you can't divide it


� �

�א��د�א��

F٦E� �

Molecular

Weight of a

Volatile Liquid


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٥٥

Molecular Weight Of a Volatile Liquid

P V = n Ru T

M.wt = m × Ru × T

P × V

# Introduction :

- Volatile Liquid [ ��HK:+�� ] : According to Dumas method that its boiling point is less

than 100 C° { B.pt < 100° C } .

- The Molecular weight of a volatile liquids can be found using the Ideal gas law :

Where :

- P : Pressure of gas ( in atm )

- V : Volume of gas ( in L )

- n : Number of moles of gas

- Ru : Universal gas constant = 0.0821 atm.L/mol.K

- T : Temperature of gas ( in Kelvin )

- Note : We know that [ n = mass / M.wt ] , so we can rewrite the law of ideal gas in the

form :

Where :

- M.wt : Molecular weight of volatile liquid

- m : Mass of the gas

- Remarks :

(1) The temp in K ( Kelvin ) = Temp in C ( Celsius ) + 273.15

(2) 1 atm = 760 mm Hg

(3) 1 L = 1000 mL


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٥٦

- The Procedure :

1. Clean small Erlenmeyer flask with soap and water thoroughly and then dry it

completely .

?�:��D��$<8،وא�`��%ن0�I��gr~`א�r��S9gSوS�!�X.?

2. Weigh the clean dry flask with the aluminum foil and the rubber band .

?�4�HIאوא��طم%��I(א�!SوL:א��X�א�gSوS�\א�ز³ن²?

3. Obtain from your instructor 5.0 mL of an unknown sample of the volatile liquid

( record its number ) and pour into the clean dry flask .

?Z��:V:}'٥T��:��%D«אא��V:��HKIאא��+V:F��א��<!S+±CE��X�א�א��gSوS�\א�@�D�hو?

4. Cover the top of the flask with the aluminum foil and secure it with the rubber

band . Then make a tiny hole through the aluminum foil with a sharp pin .

7K�אمא��طא�4�HI،8>אL�h��_Kgr~h$�א'3لوS!�א)I��%م��7K�אم?��DK��m!��א�\�SوgS�S%!�א)I��%م<88

?د�%س��د

5. Assemble the apparatus , as shown in ( figure 6.1 ) , with the beaker half filled

with water .

?+�Qא�@��Bز�Dא�Lµ١}�٦0�I��D�`.א��/س��L:،G�Kא��@?

6. Add boiling chips and start heating the beaker slowly until water starts to boil

( 10 minutes approximately ) . Allow gentle boiling to make sure that the liquid in

the flask has completely evaporated .

?��~��0�Iא��>K��H��د!��YI١٠�gF��MY��!Sא�1$�ج8>א��7*א��/سא��$�$1��\�E�B/K��H��Dא��،

%Iאא��+/ن�:��7��!gSوS�\��د%$?

7. Record the temperature of the boiling water and the atmospheric pressure in the

laboratory .

?6K7Iא@E%א�m~,א��9��MR��~Iא0�IאgSא��$Sد+±C?


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٥٧

8. Turn of the burner , remove carefully the flask from the baker using a towel and

allow to cool to room temperature . Then dry the flask completely .

��אgSא�~��،8>?�$Sد�96د�ن/�� ªא8>،��Q�:אم�7K��4א��Q،8>�'�ج�{Sא�\�SوgSV:א��/سא��$�$1��

��BgSوS�\א��$?

9. Weigh the flask , aluminum foil , rubber band and the condensed vapor .

?�8��KIאS�7�?ز³نא�\�SوgSL:وS!�א)I��%موא��طא�4�HIوא�

10. To determine the volume of the flask , fill it to the rim with water , then measure

the volume of water using graduated cylinder .

7K�אمא[H%א.�?��0�Iא<C��C>א�\�SوgS�F�:9�9א��0�I��D~V:א��+א��HKI،8>!>�\��س��_K�

�$±S�Iא?@@@@@@@@


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٥٨


Que. No. 1 : For which of the following liquids , the molecular weight can be

determined by the method described in this experiment :

Solution :

a. Glycerin (180° C) �� [ No ] we can't , because its B.pt > 100° C .

b. Benzene (80° C) �� [ Yes ] we can , because its B.pt < 100° C .

c. Dichloromethane (40° C) �� [ Yes ] we can , because its B.pt < 100° C .

Que. No. 2 : A gas cylinder containing 45g of CO2 . What is the mass of O2 that occupies

the same volume of CO2 at the same pressure and temperature ?

Solution :

Mass of CO2 = 45 g

M.wt of CO2 = 44 g/mol

M.wt of O2 = 32 g/mol

According to ideal gas low , and because the two gases have the same volume ,

pressure & temperature . Then :

Moles of CO2 = Moles of O2

Mass of CO2 = Mass of O2

M.wt of CO2 M.wt of O2

Mass of O2 = mass of CO2 × M.wt of O2

M.wt of CO2

Mass of O2 = 45 g × 32 g/mol

44 g/mol

Mass of O2 = 32.73 g


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٥٩


Que. No. 1 : List the sources of errors in this experiment :

Solution :

(1) Measurements errors ( Ex. measurement of mass , volume , temp & pressure )

(2) Errors in Calculations

(3) The gas didn't evaporate completely

(4) The flask wasn't dry or clean completely

Que. No. 2 : Was the measured volume of the flask equal to the one written on the

label ? Account for any differences .

Solution :

No , Because the flask is graduated to a certain point , for an example the flask used in

this experiment , it was wrièn on the label that the volume is 100 ml but in the real its

150 ml .

Que. No. 3 : If you mistakenly used 1.0 atm instead of the actual atmospheric pressure

in your calculations , will the value of the calculated molecular weight of the liquid

change ? Explain .

Solution :

Yes of course , The molecular weight will decrease according to the Ideal gas law .


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٦٠



Ques on No.1 :

The volume ( in ml ) of 0.18 g of C2H5OH vapor at 77° C and 0.6 atm is :

a. 161

b. 225

c. 125

d. 141

e. 187


Jא��¨אل+��\��4 JW

- The M.wt of C2H5OH = 46 g/mol

- The no. of moles of C2H5OH = 0.18 / 46 = 0.00391 mole

- T = 273.15 + 77 = 350.15 K

Then , According to Ideal gas law :

P V = n Ru T �� V = n Ru T �� V = 0.00391 × 0.0821 × 350.15 �� V = 187 mL

P 0.6


Ques on No.1 :

A sample of 0.25 g of a vola le liquid occupies 60.0 cm3 at 0.80 atm and 116° C . The

molecular weight of the volatile liquid ( in g/mol ) equals :

a. 170

b. 148

c. 166

d. 119


@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò�†ä�a@…b¤gCh.E.D |@@ ٦١

e. 158


Jא��¨אل+��\��4 JW

- T = 273.15 + 116 = 389.15 K

- V = 60 / 1000 = 0.06 L

According to Ideal gas law :

M.wt = m Ru T �� M.wt = 0.25 × 0.0821 × 389.15 �� V = 166 g/mol

P V 0.8 × 0.06

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