Collections of first chapter of Chemistry for Malaysia Syllabus 1. MPM Specimen Papers 2. STPM 2013 Sem 1 3. STPM 2...
Al PM ex T C a he n m is try Atoms, Molecules and Stoichiometry
Contents
1. MPM Specimen Paper
2. STPM 2013 Semester 1 3. STPM 2014 Semester 1
ST
4. Malaysia Matriculation Paper 2013
All prepared by
[email protected]
Objective questions Q1 ( MPM Specimen Paper )
Al PM ex T C a he n m is try
\
Answer : D
Explanation
Particle
Number of protons
Number of electrons
Number of neutrons
7
7
9
8
10
10
9
9
10
35
34
44
2-
ST
+
All prepared by
[email protected]
Al PM ex T C a he n m is try
Question 2 ( MPM Specimen Paper )
Answer : D
Explanation
RAM =
(
) (
)
= 14.00
RMM of X2 is 28 .
Isotopes of X have same number of protons but different in number of
ST
neutrons .
All prepared by
[email protected]
Question 3 ( STPM 2013 Sem 1 ) Tetraethyllead(IV) , (CH3CH2)4Pb , is added to petroleum as an antiknock agent . (CH3CH2)4Pb is prepared from ethane , CH3CH3 , according to the following reactions : CH3CH3 + Cl2 → CH3CH2Cl + HCl
Al PM ex T C a he n m is try
4CH3CH2Cl + 4NaPb → (CH3CH2)4Pb + 3Pb + 4NaCl Which statement about the reactions is not true ?
A. One mole of Cl2 produces one mole of CH3CH2Cl .
B. Four moles of Cl2 produce one mole of (CH3CH2)4Pb .
C. One mole of HCl is formed for each mole of (CH3CH2)4Pb produced .
D. Four moles of CH3CH3 is required for each mole of (CH3CH2)4Pb produced . Answer : C
Explanation
Overall equation :
4CH3CH3 + 4Cl2 + 4NaPb → (CH3CH2)4Pb + 3Pb + 4NaCl + 4HCl
1 mole of Cl2 ≡ 1 mole of CH3CH2Cl
4 mole of Cl2 ≡ 1 mole of (CH3CH2)4Pb
1 mole of (CH3CH2)4Pb ≡ 4 mole of HCl
4 mole of CH3CH3 ≡ 1 mole of (CH3CH2)4Pb
ST
All prepared by
[email protected]
Question 4 ( STPM 2014 Sem 1 ) The proton number of ion Y is 16 , its nucleon number is 33 and it has the same number of electrons and neutrons . What is the charge of ion Y ? A. -2 B. -1
Al PM ex T C a he n m is try
C. +1 D. +2
Answer : B
Explanation Nucleon number
Number of protons
Number of electrons
Number of neutrons
33
16
17
17
Since Y has one electron more than proton , thus it carries a charge of -1 .
ST
All prepared by
[email protected]
Question 5 ( STPM 2014 Sem 1 ) An element X with a relative atomic mass of 28.1 consists of isotopes 30
X . If the percentage abundance of
29
X and
30
28
X , 29X and
X are the same , what is the
percentage abundance for 28X ? A. 0.32
Al PM ex T C a he n m is try
B. 3.33 C. 93.3 D. 96.7
Answer : C
Explanation
Let the percentage abundance of 29X and 30X be a .
Let the percentage abundance of 28X be ( 100 – a –a ) .
RAM = 28.1
( )
( )
(
)
= 28.1
a = 3.33 %
Percentage abundance of 29X and 30X = 3.33 %
Percentage abundance of 28X = (100 – 3.33 – 3.33 )% = 93.34 %
ST
= 93.3 %
All prepared by
[email protected]
Structured & Essays Question
Al PM ex T C a he n m is try
Question 1 ( MPM Specimen Paper )
(a) (i) Name the subatomic particle Y . [1m]
Answer : neutron
(ii) Draw the paths of the beams of electrons & hydrogen ions in the above diagram . [2m]
Answering skill : Beam of electrons deflect more than that of H+ ions .
Answer :
ST
All prepared by
[email protected]
(iii) If a beam of deuterium ions is passed through the electric field , explain the difference in deflection angle between the beam of hydrogen ions and that of deuterium ions .[2m] Answer : The deflection angle for deuterium ions is smaller than that of hydrogen ions .
This is because deuterium ion has a larger m/e value than hydrogen ions .
Al PM ex T C a he n m is try
* Alternative answer :
The deflection angle for deuterium ion is half of that for hydrogen ion .
This is because the deuterium ion is twice as heavy as hydrogen ions .
* Extra notes
The heavier the ions ( larger m/e value ) , the smaller the deflection angle for that ion .
Isotopes for hydrogen
Isotopes
Nucleon number
Number of protons
Number of electrons
Number of neutrons
1
1
1
0
2
1
1
1
3
1
1
2
( protium )
( deuterium )
ST
( tritium )
All prepared by
[email protected]
(b) P+ and Q- ions are isoelectronic with the (i) State the nucleon number of the
isotope .
isotope .
Answer : 40
(ii) Identify P+ & Q- ions . Answer : ~ P+ ion : K+ ion / potassium ions ~ Q- ion : Cl- ion / chloride ions
Al PM ex T C a he n m is try
*Extra Note
Isoelectronic species : species with same number of electrons
For P+ ions :
Number of Number of protons electrons 19 18 For Q ions : Number of electrons 18
ST
Number of protons 17
All prepared by
[email protected]
Question 2 ( STPM 2013 Sem 1 ) Chlorine gas exists as diatomic molecules , Cl2 . In an experiment , 326.6 g of chlorine was used to react with excess iron to produce iron (III) chloride , FeCl3 . (a) Write a balanced equation for this reaction . [1m]
Answer : 2 Fe (s) + 3 Cl2 (g) → 2 FeCl3 (s)
Al PM ex T C a he n m is try
(b) Calculate the theoretical yield of FeCl3 . [2m] Answer :
Number of moles of Cl2 =
mol
= 4.60 moles
Number of moles of FeCl3 produced =
× 4.60 mol
= 3.067 mol
Mass of FeCl3 produced = 3.067 × [ 55.8 + 3( 35.5) ] = 497.8 g ( 4 s.f )
*Extra !!! ( What if we don’t know how to calculate the number of moles of FeCl3
ST
produced ?? )
All prepared by
[email protected]
(c) If the percentage of the actual yield of FeCl3 obtained in the laboratory is 82.0% , calculate the mass of FeCl3 produced . [2m] Answer :
Mass of FeCl3 produced = 497.8 × 82.0% = 408.2 g
Al PM ex T C a he n m is try
(d) Explain why there is a difference between the calculated theoretical value yield of FeCl3 with that of the actual yield obtained in the laboratory . [2m] Answer :
Some of the chlorine gas escapes from the reacting vessel .
Some of FeCl3 will be decomposed and forms FeCl2 .
ST
2 FeCl3 (s)
All prepared by
[email protected]
2 FeCl2 (s) + Cl2 (g)
Question 3 ( Matriculation 2013 ) Silicon exists as three stable atoms with nucleon number of 28 , 29 and 30 . The proton number of silicon is 14 . (a) Write the symbol for any one of 3 silicon atoms . Indicating its proton and nucleon number in the symbol . [1m]
Al PM ex T C a he n m is try
Answer :
@
@
(b) If the amount of a sample of silicon with nucleon number of 28 , 29 and 30 are 92.2% , 4.68% and 3.12% , calculate the relative atomic mass of silicon . [2m] Answer :
RAM =
(
) (
) (
)
= 28.1
(c)Sketch and label the mass spectrum of silicon . [2m] Answer :
Answer :
ST
All prepared by
[email protected]
Question 4 ( Matriculation 2013 ) When 80.0g of ethene gas , C2H4 reacts with 110.0 g of hydrochloric acid , HCl , chloroethane , C2H5Cl is produced . (a) Write a balanced chemical equation for the reaction . [1m]
Answer : C2H4 (g) + HCl (aq) → C2H5Cl (g)
Al PM ex T C a he n m is try
(b) Determine the limiting reactant in the reaction . [3m] Answer :
Number of moles of ethene =
(
)
(
)
mol
= 2.86 mol
Number of moles of HCl =
mol
= 3.01 mol
Since the number of moles for ethene is less than that for hydrochloric acid , thus the limiting reactant is ethene gas .
(c) Calculate the mass of C2H5Cl produced . [3m] Answer :
Number of moles of C2H5Cl = number of moles of ethene gas = 2.86 mol
Mass of C2H5Cl produced = 2.86 × [ 2(12.0) + 5(1.0) + 35.5 ]
ST
= 184 g ( 3s.f )
All prepared by
[email protected]
(d) Calculate the mass of excess reactant remained at the end of the reaction . [3m] Answer :
Number of HCl used = number of moles of ethene gas = 2.86 mole
Number of moles of HCl remained = 3.01 - 2.86
Al PM ex T C a he n m is try
= 0.15 mol
Mass of excess HCl remained = 0.15 × ( 35.5 + 1.0 )
ST
= 5.48 g ( 3sf )d
All prepared by
[email protected]