atoms , molecules & stoichiometry ( stpm + matriculation )

8/12/2019 Atoms , Molecules & Stoichiometry ( STPM + Matriculation )

1/14

All prepared [email protected]

Atoms,

Molecules

and

Stoichiometry

Contents

1. MPM Specimen Paper2. STPM 2013 Semester 1

3. STPM 2014 Semester 1

4. Malaysia Matriculation Paper 2013
mailto:[email protected]:[email protected]:[email protected]:[email protected]


2/14


Objective questions

Q1 ( MPM Specimen Paper )

\

Answer : D

Explanation

Particle Number of protons Number of

electrons

Number of

neutrons

7 7 9

2- 8 10 10

9 9 10

+ 35 34 44


3/14


Question 2 ( MPM Specimen Paper )

Answer : D

Explanation

RAM = ( )( )

= 14.00

RMM of X2is 28 . Isotopes of X have same number of protons but different in number of

neutrons .


4/14


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

4CH3CH2Cl + 4NaPb (CH3CH2)4Pb + 3Pb + 4NaCl

Which statement about the reactions is not true ?

A. One mole of Cl2produces one mole of CH3CH2Cl .

B. Four moles of Cl2produce one mole of (CH3CH2)4Pb .

C. One mole of HCl is formed for each mole of (CH 3CH2)4Pb produced .

D. Four moles of CH3CH3is required for each mole of (CH3CH2)4Pb produced .

Answer : C

Explanation

Overall equation :4CH3CH3+ 4Cl2+ 4NaPb (CH3CH2)4Pb + 3Pb + 4NaCl + 4HCl

1 mole of Cl21 mole of CH3CH2Cl 4 mole of Cl2 1 mole of (CH3CH2)4Pb 1 mole of (CH3CH2)4Pb 4 mole of HCl 4 mole of CH3CH31 mole of (CH3CH2)4Pb


5/14



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

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 .


6/14



An element X with a relative atomic mass of 28.1 consists of isotopes28

X ,29

X and

30X . If the percentage abundance of

29X and

30X are the same , what is the

percentage abundance for28

X ?

A. 0.32

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 ( 100aa ) . RAM = 28.1 () () ()

= 28.1

a = 3.33 %

Percentage abundance of 29X and 30X = 3.33 % Percentage abundance of 28X= (1003.333.33 )%

= 93.34 %

= 93.3 %


7/14


Structured & Essays Question

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 :


8/14


(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 .* 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 forthat ion .

Isotopes for hydrogenIsotopes Nucleon

number

Number of

protons

Number of

electrons

Number of

neutrons

( protium )

1 1 1 0

( deuterium )

2 1 1 1

( tritium )

3 1 1 2


9/14


(b) P+and Q

-ions are isoelectronic with the

isotope .

(i) State the nucleon number of the isotope .

Answer : 40(ii) Identify P

+& Q

-ions .

Answer :~ P

+ion : K

+ion / potassium ions

~ Q-ion : Cl

-ion / chloride ions

*Extra Note

Isoelectronic species : species with same number of electrons For P+ions :

Number of

protons

Number of

electrons

19 18

For Q-ions :Number of

protons

Number of

electrons

17 18


10/14


11/14


(c) If the percentage of the actual yield of FeCl3obtained in the laboratory is 82.0% ,

calculate the mass of FeCl3produced . [2m]

Answer :

Mass of FeCl3produced = 497.8 82.0%= 408.2 g

(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 FeCl3will be decomposed and forms FeCl2.

2 FeCl3(s) 2 FeCl2(s) + Cl2(g)


12/14


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]

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 :


13/14


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)(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 ]

= 184 g ( 3s.f )


14/14


(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= 0.15 mol

Mass of excess HCl remained = 0.15 ( 35.5 + 1.0 )= 5.48 g ( 3sf )d

atoms , molecules & stoichiometry ( stpm + matriculation )

Documents