4. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…b¤gCh.E.D |@@ ٣
Safety
# Safety Equipment :
1. Safety Shower 4. Fire ExƟnguishers
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 .
5. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…b¤gCh.E.D |@@ ٤
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 AnalyƟcal 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
6. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…b¤gCh.E.D |@@ ٥
- Second equation : when the blame is yellow ( incomplete combustion ) the
equation is :
C4H10 + O2 ------------ C(S) + H2O
#@@@@@òÔib@paìä@òÜ÷c@òÔib@paìä@òÜ÷c@òÔib@paìä@òÜ÷c@òÔib@paìä@òÜ÷c#
Gن אFLLWE
QuesƟon 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
7. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…b¤gCh.E.D |@@ ٦
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 ) *
Gن אFLLWE
QuesƟon 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
* The correct Answer is ( A ) *
Gن אFLLWE
QuesƟon 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
* The correct Answer is ( A ) *
Gن אFLLWE
8. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…b¤gCh.E.D |@@ ٧
QuesƟon No.1 :
One of the following statements is correct :
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
* The correct Answer is ( A ) *
Gن אW
QuesƟon 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 ) *
10. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…b¤gCh.E.D |@@ ٨
# 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 .
א א א د ن א د א
(1) MelƟng 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. IdenƟficaƟon .
2. DeterminaƟon degree of purity .
- Note : Melting point is related to intermolecular bounds ( Ğא אØ א ) so Strong
intermolecular bounds ------- Higher Melting point .
- Note : When we measure the M.pt we well use capillary tube ( אª א ) and when
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
11. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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- The Procedure :
1. Introduce the finely powdered , dry solid sample into a capillary tube is sealed at
one end .
?وאú א אª א א Ğא א א د?
2. Place the capillary tube in the melting apparatus .
?אز אª א?
3. Adjust the rate of heating so that the temperature rises at moderate rate .
?אل אل Ĩ א ğא د ęĦ?
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 .
?א د Ĝ ğא א ğא دنאħęĦ אل א د
א א א Ħ دħ د א ğא د?
5. If the temperature is rising at a very fast rate during melting , repeat with a new
sample .
?ل Ĩ א ğא د ذאאم א אل?
6. Record the melting point ranges observed , and Identify the name of the
unknown .
?و אÙ د א אد و א אق?
@@
- Example on the outcomes of the experiment :
111 C° - 115 C° و د א
111 C° - 111.5 C° × و د א @@@@
א
א א
12. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…b¤gCh.E.D |@@ ١٠
- 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 .
?ª ً ن א س ز א?
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 .
?א א אª ً א אن ج אد úو אÙ אª א?
3. Heat the tube with its content over water bath gently until you notice the
evolution of air bubbles from the capillary tube .
?אאª א ج ĝ א ª م دא Ĩª?
13. General Chemistry Lab
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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
אª א א א?
- Note : Boiling point is related to intermolecular bounds .
So Strong intermolecular bounds
(3) Density :
Definition of Density : It is the mass of the
@òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa
Turn the burner off , and record the boiling point as the temperature at which
the unknown liquid enters the capillary tube .
א ğא د ن א و א אאª א א א
Boiling point is related to intermolecular bounds .
intermolecular bounds Higher Boiling point
ً ن א دس ز
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
?א ğא د ن א و א א
Boiling point
material per unit volume .
14. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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- Note : Usually the density is expressed in grams per mL or cc ( ) . 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 .
?א Ù و و Ù د א?
2. Determine the mass of the solid .
?א د א دFאאن אمTop loading?E
3. Into 100 mL graduated cylinder , pour about 50 ml of water . record the iniƟal
volume of water correctly .
?ę א אא Øو א ğא א Ø?
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 .
?א א אא د א دً و אª א د ª אم ص א
א ª د و ě?
5. Record the final volume of water .
?א ğא?
15. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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6. Calculate the density of the solid from the data collected .
?א د א אË אª א?
@@
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 .
?ę א ğא و אFØEא وאن وאFFume hoodEě?
2. Weigh the volumetric flask with stopper .
?אد א א ğא و אزن?
3. Fill the volumetric flask with the unknown liquid exactly to the mark .
?ğאو א ً ĩúو אÙ ğא و א א?
4. Weigh the volumetric flask with content and stopper .
?אد وא Ĩ ğא و אزن?
5. Calculate the density of the unknown liquid .
?úو אÙ א א?
(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 .
16. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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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 } . @
Non-Polar Compounds : such as { CH4 / CCL4 / C6H12 / C6H4 / CO2 } . @@@@
- Note : Polar Non-Polar compounds depends on the following factors :
1. Electronegativity difference ( و אق ) , the highest electronegativity atoms
are { F , O , N , CL } .
2. Electric dipole moment ( א م ) , and it depends on the direction of the
polar bonds .
3. ن Ù ن ً אª א א א א
و א Ğאوم א ن وذ ًאÙ א א אق אK
@@@@
18. General Chemistry Lab
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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 :
ً ĨEªא ذد ن ذא
- 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 observaƟon .
@òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa
) 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 :
אً ذא אن و ن ªא ذ ĜFً Ĩ
אªو אن ن אK
Introduce a few crystals of the solid material in a clean , dry test tube , then add
about 2.0 ml of the solvent . shake well and record your observaƟon .
Soluble In
Water
١٦
( C , H ) compounds including ( OH , NH ) are soluble in water on condition that :
ذ نFOHEאً ذא אن و ن ªא ذ Ĝ
אªو אن ن א
clean , dry test tube , then add
about 2.0 ml of the solvent . shake well and record your observaƟon .
Soluble In
Water
19. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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?א ú אª א د אªא د دوًא א
Ġא?
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 observaƟon .
?אאدħא ú אª אħ ªא א אא
Ġא وًא?
3. To approve your results , check the solubility of the different solvents used in this
part of experiment with each other .
?א א א אª א ذא Ĝ?
20. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…b¤gCh.E.D |@@ ١٨
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 ***
21. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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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 ( د ) [ 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 ***
22. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…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 :
אאħ
23. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…b¤gCh.E.D |@@ ٢١
#@@@@@òÔib@paìä@òÜ÷c@òÔib@paìä@òÜ÷c@òÔib@paìä@òÜ÷c@òÔib@paìä@òÜ÷c#
Gن אFLLWE
QuesƟon No.1 :
One of the following statements is correct :
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
* The correct Answer is ( D ) *
Gن אFLLWE
QuesƟon No.1 :
The physical property that increases with increasing atmospheric pressure is :
a. Melting point
b. Boiling point
c. Density
d. Solubility
e. Mass
* The correct Answer is ( B ) *
Question No.2 :
CCL4 dissolved in a certain solvent , this means that the solvent is :
a. Polar
b. Has a high boiling point
24. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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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
* The correct Answer is ( D ) *
Gن אFLLWE
QuesƟon 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
* The correct Answer is ( D ) *
Question No.2 :
One of the following is not a chemical change :
a. Burning of coal
b. Neutralization of HCL with NaOH
25. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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c. Dissolving salt in water
d. Production of MgO from Mg
e. Production of Mg3N2 from Mg
* The correct Answer is ( C ) *
Gن אFLLWE
QuesƟon 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
* The correct Answer is ( E ) *
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
* The correct Answer is ( A ) *
Question No.3 :
One of the following is insoluble in CH3OH :
a. NH3
b. H2O
c. H2SO4
26. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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d. C6H6
e. CH3NH3
* The correct Answer is ( D ) *
Gن אFLLEذج ĪW
QuesƟon 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
* The correct Answer is ( C ) *
Question No.2 :
One of the following is insoluble in hexane ( C6H14 ) :
a. NH3
b. C7H16
c. C6H6
d. CCL4
e. CH4
* The correct Answer is ( A ) *
Gن אFLLEW
QuesƟon No.1 :
An example of a chemical property is :
a. Density
b. Mass
c. Solubility
d. Acidity
27. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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e. Boiling point
* The correct Answer is ( D ) *
Gن אW
QuesƟon 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
* The correct Answer is ( B ) *
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
* The correct Answer is ( C ) *
29. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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# Introduction :
- Hydrated Salts [ אª א א Ĝ ] are chemical compounds ,
also called Hydrate . The general formula is :
where :
- MA : The Anhydrous salt [ אÙ א ]
- The Dot ( . ) : وא אن 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
JאאFEWא دא د א אªFא ĞאĦ אلE
ً و אÙ א و א وج אª ن א ل אĦ אً
אن Ù ú Ğאאل אW
אس אª ×س אª ×Hª
Water Of Crystallization
MA . XH2O
Blue Bluish white
ق زق
30. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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# Example of hydrates :
CaSO4.2H2O ĞאFאم אª ×E
1 mol of (CaSO4.2H2O) 2 mol of H2O
مً א א ً ن א
1 mol of CaSO4 2 mol of H2O
אÙ א ً ن א م
# Alums :
Alums are hydrated double sulfate salts [ ª × אª אªدو א א ] , that have
the general formula :
where :
- M+1
: mono-volent ions [ א د ª ] ,
like Na+1
, Ag+1
, NH4
+
……
- M+3
: tri-volent ions [ ªא ] , 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
التسخين قبل الوزن١غرام الوزنبعدالتسخين٠.٦غرام المتبقي٠.٤غرام
31. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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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 : ?Ğא?ًא א ª د ن ī و و
(2) Desiccator : א ٍل ٍ א × م ز
- The Procedure :
1. Obtain crucible and cover from your instructor and clean them thoroughly with
soap and water , then dry .
?وאن ًא و?
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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 .
?و Ğא ęم אĦ א وא Ğא ĞאFאدE?
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 Úא אאم Óא ن وذ وא د ×د ن و Úא
א א د ×د?
Wאل وא א وذ אق و א و א א Ğא
4. Weigh the empty crucible (without cover) accurately .
?Ğאزنאون?
5. Place about 1.0 g of the unknown hydrated salt in the crucible and weigh again .
?אªאمًאد ġوزن Ğאúو אÙ א א?
6. Place the crucible on the clay triangle as in step 2 . Adjust the cover such that
portion of the crucible is uncovered .
?Ğא ę א وא א Ġא ęم אĦ א Ğא?
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 .
? Ğא אزل و א ن و א א و Ğאد?
8. Allow the set to cool as described in step 3 , then weigh again , Take care to
handle the crucible with tongs only .
?Ġא ×د ن א و Úאאل Ğא ĩن ص وא زن?
33. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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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 unƟl you achieve constant mass .
?و Ğאًאد ġزن א Ġא ×د ن Úא ً ĦK و א א
א Ĩ ĝ ن وא א Ġא{ًאد ġ Ġא د ذ × ú אن ذא
Ĝ?
34. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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*** Pre – Laboratory Questions ***
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 .
35. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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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 ( א ) 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 ( א ) 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 .
*** Post – Laboratory Questions ***
37. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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#@@@@@òÔib@paìä@òÜ÷c@òÔib@paìä@òÜ÷c@òÔib@paìä@òÜ÷c@òÔib@paìä@òÜ÷c#
Gن אFLLWE
QuesƟon 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
* The correct Answer is ( C ) *
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
* The correct Answer is ( B ) *
Jאل אJW
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
38. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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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
Gن אFLLWE
QuesƟon 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
* The correct Answer is ( B ) *
Jאل אJW
- 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
39. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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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
* The correct Answer is ( E ) *
Jאل אJW
- 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
* The correct Answer is ( A ) *
40. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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Gن אFLLWE
QuesƟon 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
* The correct Answer is ( B ) *
Jאل אJW
- 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
* The correct Answer is ( A ) *
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
41. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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b. 4
c. 8
d. 6
e. 5
* The correct Answer is ( D ) *
J א א אل אאJ
Gن אFLLWE
QuesƟon 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
* The correct Answer is ( D ) *
Jħ א ً אאل وא א א אل א–
Gن אFLLEذج ĪW
QuesƟon 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 :
42. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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a. 5
b. 6
c. 7
d. 8
e. 11
* The correct Answer is ( A ) *
Jאאل אل אJ
Gن אFLLEW
QuesƟon 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
* The correct Answer is ( C ) *
J א א אل אħ אJ
Gن אFLLEW
QuesƟon No.1 :
The percentage composition of sulphate ions (SO4
2-
) in K2SO4.Al2(SO4)3.24H2O ( 948.2
g/mol ) equals :
a. 5.69%
b. 8.25%
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c. 45.56%
d. 54.43%
e. 40.50%
* The correct Answer is ( E ) *
Jאل אJW
- The M.wt of hydrous salt = 948.2 g/mol
- The M.wt of (SO4
2-
) = 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 .
Gن אW
QuesƟon 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
* The correct Answer is ( E ) *
Jאل אJW
w/w % = 46.8% = mass of water × 100% mass of water = 0.468
mass of hydrate mass of hydrate
46. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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Empirical Formula of an Ionic compoundEmpirical 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 ( د ) 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 ( Ħ א زن א )
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
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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
JWħ ª ً وط א Ġא אªאن ذא دF{
J{EلF{≈KEħ ª و× ªאن ذא دF{J{Eل
F{≈KEאن ذא وF{ħ{Eد אªאل א ī
F{≈{×Z{≈KE
- 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 .
?وאن ًא و?
2. Heat the clean dry crucible and cover over Bunsen burner to redness .
48. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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?א א ق و Ğא א Ğא?
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 .
?×د ن Úאدא دħ ×د Óא وא Ğא אل و
א א?
4. Weigh the empty crucible (without cover) accurately .
?Ğאزنאون?
5. Put about 0.25 g of magnesium turnings in the crucible and weigh again .
?אª{אمًאد ġوزن Ğאم 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 .
?א ÓאF{Eא א ً ĩ Ğא و אª אKص א
ً Ħ א ل د ً א?
7. Continue heating until all magnesium is converted to ash, then remove the cover
and heat the opened crucible to redness .
?א א Ğא و אزل د ħ م א Ĝ Ħ א?
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 .
?ğ א ªא د ĞאدĦ א אم אªאØ
א وز ج?
9. Reheat the opened crucible contents until the ash is completely dry ( notice any
color changes ) .
49. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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?ً ĩد א ًאد ġ و א ĞאĦFن אÙE?
10. Cool the crucible and contents as done in step3, then weigh again .
?ًאد ġزن א Ġא و Ğאد?
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 ) unƟl you achieve constant mass .
?و ĞאĦًאد ġزن א Ġא ×د ن ÚאK א א
Ħ Ġאوĝ ن {Ĝ Ġא ذ × ق אن ذא
?
50. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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*** Pre – Laboratory Questions ***
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
0.0235 0.0235
So The empirical formula is : W3O8
Que. No. 2 : Phenol is a compound composed of C,H and O . CombusƟon 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 :
Wא אא ª ğא ªאم א א ú 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
51. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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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%
52. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
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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 .
53. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…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 :
54. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…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 .
55. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…b¤gCh.E.D |@@ ٥١
#@@@@@òÔib@paìä@òÜ÷c@òÔib@paìä@òÜ÷c@òÔib@paìä@òÜ÷c@òÔib@paìä@òÜ÷c#
Gن אFLLWE
QuesƟon 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
* The correct Answer is ( A ) *
Jאل אJW
- 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 : S0.017813 O 0.026875 S1O1.5 ( mulƟply by 2 ) The empirical formula is S2O3
0.017813 0.017813
56. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…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
* The correct Answer is ( C ) *
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
* The correct Answer is ( A ) *
Jאل אJW
- 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
Gن אFLLWE
QuesƟon No.1 :
57. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…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
* The correct Answer is ( D ) *
Jħ א א א אل אJ
Gن אFLLWE
QuesƟon 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
* The correct Answer is ( A ) *
Jאאل אل אJ
Gن אFLLWE
Question No.1 :
During the experiment of determination of empirical formula of magnesium oxide , the
by-product formed is :
58. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…b¤gCh.E.D |@@ ٥٤
a. NH3
b. Mg3N2
c. MgN2
d. MgO
e. MgO2
* The correct Answer is ( B ) *
Gن אFLLWE
QuesƟon 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
* The correct Answer is ( E ) *
Gن אW
QuesƟon 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
* The correct Answer is ( A ) *
60. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…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 [ ] : 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
61. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…b¤gCh.E.D |@@ ٥٦
- The Procedure :
1. Clean small Erlenmeyer flask with soap and water thoroughly and then dry it
completely .
?ً ĩ ن وא Ù אÙ و?
2. Weigh the clean dry flask with the aluminum foil and the rubber band .
?אط وאم א و Ğא א و אزن?
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 .
?ØÓא אא אFאEא Ğא و א و?
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 .
?אم م א ول ًא Ù א אط אאم م א و א
د س د?
5. Assemble the apparatus , as shown in ( figure 6.1 ) , with the beaker half filled
with water .
?א ز Ğא { س א ª א?
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 .
?א אس אĦ א ج אFدًEא
א אنً ĩ و د?
7. Record the temperature of the boiling water and the atmospheric pressure in the
laboratory .
?× א Ğא אħ א א א د?
62. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…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 .
?א א دħ ×د ن Úא אم אس א و א ęج א
ً و א?
9. Weigh the flask , aluminum foil , rubber band and the condensed vapor .
?א وא אط وאم א و و אزن?
10. To determine the volume of the flask , fill it to the rim with water , then measure
the volume of water using graduated cylinder .
?א אאم א س א א ğאħ و א
א?@@@@@@@@
63. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…b¤gCh.E.D |@@ ٥٨
*** Pre – Laboratory Questions ***
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
64. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…b¤gCh.E.D |@@ ٥٩
*** Post – Laboratory Questions ***
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 wriƩen 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 .
65. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…b¤gCh.E.D |@@ ٦٠
#@@@@@òÔib@paìä@òÜ÷c@òÔib@paìä@òÜ÷c@òÔib@paìä@òÜ÷c@òÔib@paìä@òÜ÷c#
Gن אFLLWE
QuesƟon 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
* The correct Answer is ( E ) *
Jאل אJW
- 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
Gن אFLLWE
QuesƟon No.1 :
A sample of 0.25 g of a volaƟle 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
66. @òßbÈÛa@õbîàîØÛa@n«@OGeneral Chemistry Lab
@òî¹…b×þa@òävÜÛa@O@òîöbîàîØÛa@ò†äa@…b¤gCh.E.D |@@ ٦١
e. 158
* The correct Answer is ( C ) *
Jאل אJW
- 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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