PYQ ANSWER SCHEME CHAPTER 1

PYQ CHAPTER 1: MATTER MARKING SCHEME SESSION 2021/2022

2011 / 2012

1 (a) Mole of NaCl4 = 9.0 g = 0.1538 mol
58.5 g mol-1
1
Molarity of NaCl = 0.1538 mol = 0.3076 M 1
0.500 L 1

= 0.31 M 1
1
(b) 2 mol of NO react with 1 mol of O2 1
1
1x 0.53 1
So, 0.53 mol NO react with 2 = 0.265 mol of O2
1
Limiting reactant is NO
1
2 mol of NO produced 2 mol of NO2 1
1
2 x 0.53 1
So, 0.53 mol NO produced 2 = 0.53 mol of NO2
1
Volume of NO2 at STP = 0.53 mol x 22.4 L mol-1 1
= 11.87 L

(c) Mole of CCl4 = 130.0 g = 0.8442 mol (provided)
i 154.0 g mol-1

Mole of SbF3 = 95.0 g = 0.5313 mol
178.8 g mol-1

2 mol of SbF3 react with 3 mol of CCl4

0.5313 x 3
So, 0.5313 mol SbF3 react with 2 = 0.7970 mol of CCl4 (needed)

Since mole of CCl4 needed is less than that provided, therefore SbF3 is the
limiting reactant

2 mol of SbF3 produced 3 mol of CCl2F2

0.5313 x 3

So, 0.5313 mol SbF3 produced 2 = 0.7970 mol of CCl2F2

Mass of CCl2F2 = 0.797 mol x 121 g mol-1
= 96.4 g

(c) Mole of CCl4 in excess = 0.8442 mol – 0.7970 mol

ii = 0.0472 mol
Mass of CCl4 in excess = 0.0472 mol x 154 g mol-1

= 7.27 g

PYQ CHAPTER 1: MATTER MARKING SCHEME SESSION 2021/2022

2012 / 2013

1 (a) B is atom & C is ion of the same element 1
i because having same number of protons and neutrons but different number of 1
electrons.
1
(a) A & D are isotopes of same element 1
ii because having same number of protons and number of electrons but different
1
number of neutrons. 1
1
(b) Mass of C = 12.0 g x 21.5 g 1
44.0 g
1
= 5.86 g
1
Mass of H = 6.84 g – 5.86 g 1
= 0.980 g 1
1
Element C H 1
Mass (g) 5.86 0.980 1
Mole (mol) 5.86 g = 0.488 mol 0.98 g = 0.980 mol
12.0 g/mol 1.0 g/mol
Ratio
0.488 = 1 0.980 = 2
0.488 0.488

Empirical formula = CH2

(CH2) n = 128.2 g/mol
n = 128.2 g/mol
14.0 g/mol
= 9.16 ≈ 9

Molecular formula = (CH2)n
= (CH2)9
= C9H18

(c) Reduction : MnO4- + 4H+ + 3e MnO2 + 2H2O
+ 2e
i

Oxidation : C2O42- 2CO2

(c) Reduction : 2MnO4- + 8H+ + 6e 2MnO2 + 4H2O
+ 6e
ii

Oxidation : 3C2O42- 6CO2

Overall : 2MnO4- + 8H+ + 3C2O42- 2MnO2 + 4H2O + 6CO2

PYQ CHAPTER 1: MATTER MARKING SCHEME SESSION 2021/2022

2013 / 2014

1 (a)(i) C2H2 + HCl C2H3Cl 1

(a)(ii) mol of C2H2 = 70.0 1
26
1
= 2.692 mol C2H2
1
mol of HCl = 102.0 1
36.5 1
1
= 2.795 mol HCl
1
1 mol HCl react with 1 mol C2H2 1
2.795 mol HCl react with 2.795 mol C2H2 1
1
2.692 mol C2H2 (available) < 2.795 mol C2H2 (needed)
1
(a)(iii) Limiting reactant is C2H2 produce 1 mol C2H3Cl 1
produce 2.692 mol C2H3Cl
1 mol C2H2 1
2.692 mol C2H2 (labels)

Mass of C2H3Cl = 2.692 mol x 62.5 g mol-1 1
= 168.3 g (axis)

(a)(iv) Mol of excess reactant, HCl = 2.795 − 2.692
= 0.103 mol

Mass of excess HCl = 0.103 mol x 36.5 g mol-1
= 3.80 g

(b)(i) 24 Mg 25 Mg 26 Mg
(b)(ii) 12 12 12

Relative atomic mass of Mg
= (24 x 78.6) + (25 x 10.1) + (26 x 11.3)

78.6 + 10.1 + 11.3
= 24.3

(b)(iii) Abundance (%)
78.6

11.3
10.1

m/e

PYQ CHAPTER 1: MATTER MARKING SCHEME SESSION 2021/2022

2014 / 2015 2NO(g) + 2CO(g) N2(g) + 2CO2(g)
1 (a)

Mole of CO = 300 g
12 + 16 g mol-1

= 10.71 mol

2 mol CO produces 1 mol N2 1
10.71 mol CO produces 5.36 mol N2
1
Mass of N2 = 5.36 x (2 x 14)
= 150.08 g 1

2 mol CO produces 2 mol CO2 1
10.71 mol CO produces 10.71 mol CO2
1
Mass of CO2 = 10.71 x [ 12 + 2 (16) ] 1
= 471.24 g 1
1
% w/w of CO2 = 471.24 x 100 1
471.24 + 150.08 1
1
= 75.8 % 1

(b)(i) Reduction: 6e- + Cr2O72- + 14H+ 2Cr3+ + 7H2O 1

Oxidation : x 6 ( Fe2+ Fe3+ + e- ) 1
1
Redox equation: Cr2O72-+14H+ + 6 Fe2+ 2Cr3+ + 7H2O + 6 Fe3+

(b)(ii) Mole of Na2Cr2O7 = 0.025 x 50 x 10-3
= 1.25 x 10-3 mol

Mass of Na2Cr2O7 = 1.25 x 10-3 x [ 2(23) + 2(52) + 7(16)]
= 0.328 g

(b)(iii) Mole of Cr2O72- = 0.025 x 30 x 10-3
= 7.50 x 10-4 mol

1 mol Cr2O72- reacts with 6 mol Fe2+
7.5 x 10-4 mol Cr2O72- reacts with 6 x 7.5 x 10-4

1
= 4.5 x 10-3 mol Fe2+

Molarity = mole of solute

V of solution, L
= 4.5 x 10-3 mol

40 x 10-3 L

= 0.1125 M

PYQ CHAPTER 1: MATTER MARKING SCHEME SESSION 2021/2022

2015 / 2016 CxHyN2 + O2 CO2 + H2O + NO2 1
1 (a) 1
mass of C = 12 x 0.558 = 0.15218 g 1
(b) 44
(c) 1
mass of H = 2.0 x 0.344 = 0.0382 g
18 3
3
mass of N = 0.250 – (0.15218 – 0.0382) 1
= 0.0596 g
1
Mass C H N 1
0.15218 0.0382 0.0596 1
1
Mole(mol) 0.15218 0.0382 0.0596
12 1
14
= 0.0127 = 0.0382 = 4.26 x 10-3

Simplest ratio 3 9 1

 Empirical formula = C3H9N

If the empirical formula = molecular formula

Molar mass = 3 (12) + 9 (1) + 1 (14)
= 59 g mol-1

Mole of C3H9N = 0.250 = 4.237 x 10-3 mol
59

Number of H atoms in the sample = 4.237 x 10-3 x 9 x 6.02 x 1023
= 2.29 x 1022 atoms

PYQ CHAPTER 1: MATTER MARKING SCHEME SESSION 2021/2022

2016 / 2017

1. (a) Solute = Na2CO3 Solvent = ethanol

Molarity = 2.5 M Density = 1.430 g/ml

Assume V solution = 1 L 1

Molarity = no of mol of Na2CO3 / V solution 1
1
2.5 M = n / 1 L
1
n = 2.5 mol
1
Mass of Na2CO3 = 2.5 mol x 106 g / mol = 265 g
1
Density = mass / volume 1

mass = density x volume 1
1
= 1.430 g / ml x 1000 ml 1
1
= 1430 g
Mass of ethanol = 1430 g – 265 g 1
1
= 1165 g
1
Molality = no of mole of Na2CO3 (mol) / mass of ethanol (kg) 1

= 2.5 mol / 1.165 kg

(b) i. = 2.15 m
Oxidation : ( Zn → Zn2+ + 2e- ) x 4
Reduction : 10 H+ + NO3- + 8 e- → NH4+ + 3 H2O
Acidic solution : 4 Zn + 10 H+ + NO3- → 4 Zn2+ + NH4+ + 3 H2O

ii. Mol of HNO3 = 0.5 M x (25/1000) L = 0.0125 mol

From equation: 1 mol of NO3- produces 4 mol Zn

0.0125 mol of NO3- produces 0.05 mol Zn

Mass of pure Zinc = 0.05 mol x 65.4 g / mol = 3.27 g

% purity of Zn = (mass of pure zinc / mass of impure zinc) x 100
= (3.27 g / 4.0 g) x 100 %
= 81. 75 %

PYQ CHAPTER 1: MATTER MARKING SCHEME SESSION 2021/2022

2017 / 2018

1. (a) i. 2 AgNO3 + CaBr2 → 2 AgBr + Ca(NO3)2 1
ii. Mol of AgNO3 = 0.050 mol L-1 x (25/1000) L 1
= 1.25 x 10-3 mol 1
Mol of CaBr2 = 0.050 mol L-1 x (25/1000) L 1
= 1.25 x 10-3 mol

From equation: 2 mol AgNO3 produces 2 mol AgBr

1.25 x 10-3 mol AgNO3 produces 1.25 x 10-3 mol AgBr

1 mol CaBr2 produces 2 mol AgBr
1.25 x 10-3 mol CaBr2 produces 2.5 x 10-3 mol AgBr

AgNO3 limit the amount of AgBr, therefore AgNO3 is the limiting 1
reactant.
1
% yield of AgBr = (actual yield / theoretical yield) x 100% 1
Theoretical yield AgBr = 1.25 x 10-3 mol x (107.9 + 79.9) g/mol 1
1
= 0.2348 g
1
% yield = (0.015 g / 0.2348 g) x 100 %
1
= 6.39 %

(b)

C HO

Mass 41.40 g 3.47 g 55.13 g

Mol = 41.40 g = 3.47 g = 55.13 g

12 g mol-1 1 g mol-1 12 g mol-1

= 3.45 mol = 3.47 mol = 3.45 mol

Mol Ratio = 3.45 / 3.45 = 3.47 / 3.45 = 3.45 / 3.45

=1 = 1.006 =1

≈1

Empirical formula = CHO 1
1
0.05 mol = 5.80 g = 116 g mol-1
Mr

PYQ CHAPTER 1: MATTER MARKING SCHEME SESSION 2021/2022 1
1
Molecular formula [CHO]n = 116
(29) n = 116
n=4

Molecular formula = C4H4O4

2018 / 2019

1. (a) i. 79 Br 81Br 1
1
35 35 1
1
ii. Average mass of bromine
1
51(79)+49(81) 1
1
= 51+49 1
1
= 79.98 a.m.u
1
Relative atomic mass 1

79.98 1

= (112) 12.00 = 79.98 1

(b) i. Assume mass of solution = 100 g

Mass of solute (NaCl) = 0.90 g

0.90

No of mol of NaCl = (23+35.5) / = 0.01538 mol
Mass of H2O = 100 g – 0.90 g = 99.1 g

99.1

No of mol of H2O = 18 / = 5.50 mol

0.01538 = 2.7886 x 10-3

Mole fraction of NaCl, XNaCl = (0.01538+5.50)

ii. ( )

Molality of NaCl, m NaCl = ( )

0.01538

= (19090.10)

= 0.1552 m

iii. ( )
= ( )

0.01538

= (1100000)

= 0.1538 M

M1V1 = M2V2

(0.01 )(100 )

V1 = 0.1538 = 6.50 mL

PYQ CHAPTER 1: MATTER MARKING SCHEME SESSION 2021/2022

(c) i. 100%

% yield = ℎ
400
100% 1
42.5 = ℎ 1

= 941.176 g 1

941.176 1
1
No of mole of SiCl4 = [(4 35.5)+28.1] /
1
= 5.5331 mol 1
1
1 mol SiCl4 ≡ 1 mol Si

5.5331 mol SiCl4 ≡ 5.5331 mol Si

Mass of Si = 5.5331 mol x 28.1 g / mol

= 155.47 g

ii. 1 mol Si ≡ 2 mol Cl2
5.5331 mol Si ≡ 5.5531 mol Si x 2 2

1

= 11.0662 mol Cl2

No of moles of unreacted Cl2 = 15 mol – 11.0662 mol
= 3.9338 mol

2019 / 2020 1
1. (a) i. Isotopes are atoms of an element which have the same number of 1
protons but different number of neutrons 1
1
In this example, bromine has two isotope; 3759 and 8351 with the
same number of proton which is 35 1
1
Whilst, the number of neutrons for 79Br and 79Br are 44 and 46 1
respectively

ii. No of Br - electron : 36

b) CxHyOz + O2 → CO2 + H2O

12

Mass of Carbon : 44 x 2.52 g = 0.6873 g

2

Mass of Hydrogen : 18 x 0.443 g = 0.0492 g
Mass of Oxygen : 1.00 g - 0.6873 g - 0.0492 g = 0.2635 g

PYQ CHAPTER 1: MATTER MARKING SCHEME SESSION 2021/2022

Element C H O 1
Mass (g) 0.6873 0.0492 0.2635
0.6873 0.0492 0.2635 1
Mole 12 16
(mol) = 0.0573 1 = 0.0165 1
Mole ratio 0.0573 = 0.0492 0.0165 1
0.0165 0.0165 1
Simplest = 3.47 0.0492 = 1.00 1
ratio 0.0165 1
7 = 2.98 2 1
1
6 1
1
Empirical formula = C7H6O2
1
(c) mass of solution = 1.509 g mL-1 (250 mL)
= 377.25 g 1

mol of solute (Ca(C2H3O2)2) = 0.25 mol L-1 (0.25L) 1
= 0.0625 mol
1
Mass of solute = 0.0625 mol (158.1 g mol-1) 1
= 9.88 g 1
1
Mass of solvent (H2O) = 377.25 g – 9.88 g = 367.37 g

0.0625

Molality = 0.3674
= 0.17 mol kg-1

(d) MgCl2 + 2NaOH → Mg(OH)2 + 2NaCl

Mol of MgCl2 = 15.1 = 0.1584 mol
95.3 −1

Mol of NaOH= 9.35 = 0.2338 mol
40.0 −1

1 mol MgCl2 ≡ 1 mol Mg(OH)2

0.1584 mol MgCl2 ≡ 0.1584 mol Mg(OH)2

2 mol NaOH ≡ 1 mol Mg(OH)2

0.2338 mol NaOH ≡ 0.1169 mol Mg(OH)2

Since NaOH produces fewer mol of Mg(OH)2, therefore
NaOH is the limiting reactant
MgCl2 is the excess reactant

Mass of Mg(OH)2 = 0.1169 mol x 58.3 g mol-1
= 6.815 g

PYQ CHAPTER 1: MATTER MARKING SCHEME SESSION 2021/2022

2020/2021

1. (a) i. 2184 1
1
ii. Mol of Na2SiO3= 50 = 0.4095 mol
122.1 −1 1

Number of Si atom = 0.4095 mol x 1 Si atom x 6.02 x 1023 mol-1 1
= 2.465 x 1023 atoms 1
1
b) Oxidation : 5Fe2+ 5Fe3+ + 5e
Mn2+ + 4H2O
Reduction : MnO4- + 8H+ + 5e 5Fe3+ + Mn2+ + 4H2O

Net equation : 5Fe2+ + MnO4- + 8H+
(acidic medium)

c) KxCryOz → K + Cr + O2

Element K Cr O 1
Mass (g) 26.58 0.0492 0.2635 1
26.58 35.35 38.07
Mole 39.1 52 16 1
(mol) = 0.6798 = 0.6798 = 2.3794
0.6798 0.6798 0.0165 1
Mole ratio 0.6798 = 1 0.6798 = 1 0.6798 = 3.5 1
1
Simplest 2 2 7 1
ratio
1
Empirical formula = K2Cr2O7

[K2Cr2O7] n = 294.20
(294.20) n = 294.20
n=1

Molecular formula = K2Cr2O7
Name of X = Potassium dichromate (VI)

3NO2 + H2O → 2HNO3 + NO

(d) i)

Mol of NO2 = 46 100 = 2.1739 mol
−1

PYQ CHAPTER 1: MATTER MARKING SCHEME SESSION 2021/2022

Mol of H2O = 18 50 = 2.7778 mol
−1

3 mol NO2 ≡ 2 mol HNO3

2.1739 mol NO2 ≡ 1.4493 mol HNO3 1
1
1 mol H2O ≡ 2 mol HNO3
1
2.7778 mol H2O ≡ 5.5556 mol HNO3
1
Since NO2 produces fewer mol of HNO3 than H2O, therefore 1
NO2 is the limiting reactant 1
H2O is the excess reactant 1

Mass of HNO3 (theoretical) = 1.4493 mol x 63.0 g mol-1
= 91.31 g

100%

% yield = ℎ

80.0 100%

% yield = 91.31

= 87.61 %