CHAPTER 1: MATTER
SUBJECTIVE QUESTIONS
1.1 Atoms and Molecules
Q1
Isotopes are atoms of the same element with the same proton number but different nucleon
numbers. These three isotopes of silicon have different number of neutrons and therefore
different nucleon number. They have 14 protons and 14 electrons but the number of neutrons
of 28Si, 29Si and 30Si are 14, 15 and 16 respectively.
Q2
A
41
19 K
Q3
Species Number of protons Number of neutrons Number of electrons
235
92U
92 143 92
238
92 U 92 146 92
24
Mg2+ 12 12 10
79
Br 35 44 36
Q4
A sample enters the main body of the mass spectrometer as a gas. Liquid or solid samples are
vaporised in the vaporisation chamber. The gaseous sample is admitted into the ionisation
chamber. In the ionisation chamber, the sample is bombarded with high energy electrons to
produce positive ions. The positive ions are accelerated to high speeds by an electric field.
The accelerated ions are passed through a magnetic field. This field will deflect the beam of
positive ions according to their m/e ratio. Lastly, these ions reach the collector. The collector
measures the current due to the different ions and the data is recorded as a mass spectrum.
Q5
The unknown element is chromium because the most abundant isotope (83.79%) has the
atomic mass of 52.
By referring to the Periodic Table, the element which has an atomic mass 52 is chromium, Cr.
Q6
Average atomic mass of thallium
= (0.295)(203 amu) + (0.705)(205 amu)
= 204.41 amu
Q7
( 7.5 x 63 )+(2.5 x 65)
Ar of Cu = 7.5+2.5
= 63.5
Q8
( 69 x 3 ) + ( 71 x 2 )
Ar of X = ( 3+ 2 )
= 69.8
Q9
( 106.904 x 48.17 )+ (108.905 x 51.83 )
Ar of Ag = 48.17+51.83
= 107.94
Q10
The percentage abundance of 121Sb = x
The percentage abundance of 123Sb = 100 x
Average atomic mass of Sb
120.904 x +122.904 ( 100x )
= 100 = 121.80
x = 55.2
Therefore, percentage abundance of 121Sb = 55.2%,
percentage abundance of 123Sb = 44.8%
Q11
The number of neutrons in the nucleus = 19 9 = 10.
The mass of the nucleus = 19.0 amu.
1.2 Mole Concept
Q12
A
1 mol of NH3 consists of 6.02 x 1023 molecules.
0.5 mol of NH consists of (0.5 mol x 6.02 x 1023 molecules mol1) = 3.01 x 1023
3
Q13
58.7 g
Number of moles of NH4NO2 = 64 g /mol = 0.917
1 mol of NH4NO2 = 8 x 6.02 x 1023 atoms
0.917 mol NH NO
4 2
= 0.917 mol x (8 x 6.02 x 1023 atoms mol1)
= 4.42 x 1024 atoms
Q14
7.83 x 10 molecules
Number of moles of K3PO4 = 6.02 x 10 molecules /mol
1 mol of K3PO4 contains 3 mol of K atoms.
1.30 mol of K PO contains (3 x 1.30) mol = 3.90 mol K atoms
3 4
Q15
25.0 g
Number of moles of Na2CO3 = 106 g /mol
= 0.236
Number of moles of Na+ ion = 0.236 mol x 2
= 0.472
Q16
25.0 g
Number of moles of aspirin = 180 g/mol
= 0.139
1 mol of aspirin, C9H8O4 contains 9 x 6.02 x 1023 carbon atoms.
0.139 mol of aspirin contains 0.139 mol x (9 x 6.02 x 1023 atoms/mol)
= 7.53 x 1023 carbon atoms
Q17
5.3 x 10 molecules
Number of moles of ethanol = 6.02 x 10 molecules /mol
Number of hydrogen atoms
= 8.8 x 103 mol x (6 x 6.02 x 1023 atoms mol1)
= 3.2 x 1022
Q18
Mass of H O = (1.0 g cm3)(0.05 cm3)
2
= 0.05 g
0.05 g
Number of moles of H2O = 18 g/mol
= 2.78 x 103
Number of H2O molecules
= (2.78 x 103 mol)(6.02 x 1023 molecules mol1)
= 1.67 x 1021
Q19
0.53 g
Number of moles of H2O = 18 g/mol = 0.0294
BF3 and H2O have the same number of molecules.
Number of moles of H O = number of moles of BF
2 3
= 0.0294
Mass of BF = 0.0294 mol x 67.8 g mol1
3
= 1.99 g
Q20
Element N O
Mass (g) 4.2 12.0
Number of moles 4.2 12.0
14.0 = 0.30 16.0 = 0.75
Mole ratio 0.30 0.75
0.30 = 1.0 0.30 = 2.5
1.0 x 2 = 2.5 x 2 =
2.0 5.0
Empirical formula N2O5
Q21
Element Fe Cl
Mass (g) 3.528 6.709
Number of moles 3.528 6.709
55.9 = 35.5 =
0.0631 0.1890
Mole ratio 0.0631 0.1890
0.0631 =1 0.0631 =3
Empirical formula FeCl3
Q22
Element C H O
Mass (g) 3.528 6.709 31.4
Number of moles 58.8 9.8 31.4
12.0 = 4.9 1.0 = 9.8 16.0 =
1.9625
Mole ratio 2.5 x 2 = 5 5 x 2 = 10 1x2=2
Empirical formula C5H10O2
Molecular formula:
(C5H10O2)n = 102
[5(12.0) + 10(1.0) + 2(16.0)]n = 102
102.0n = 102
n=1
Therefore, the molecular formula of the organic compound is C5H10O2.
Q23
A
Mass of carbon = 85.6 g
Mass of hydrogen = (100 85.6) g = 14.4 g
Element C H
Mass (g) 85.6 14.4
Number of moles 85.6 14.4
12.0 = 1.0 = 14.4
7.133
Mole ratio 7.133 14.4
7.133 = 1.0 7.133 = 2.0
Empirical formula CH2
(CH2)n = 84
[12.0 + 2(1.0)]n = 84
14.0n = 84
n=6
The molecular formula of the hydrocarbon is C H .
6 12
Q24
44.0 g of CO2 contains 12.0 g of carbon.
0.75 g CO contains
2 (x12.0
44 )
0.75
g = 0.2045 g C
18.0 g of H2O contains 2.0 g of H.
2.0
0.205 g of H O contains
2 (x18.0 ) g = 0.0228 g H
0.205
Mass of oxygen
= mass of sample (mass of C + mass of H)
= [0.50 (0.2045 + 0.0228)] g
= 0.2727 g
Element C H O
Mass (g) 0.2045 0.0228 0.2727
Number of moles 0.2045 0.0228 0.2727
12.0 = 1.0 = 16.0 =
0.0170 0.0228 0.0170
Mole ratio 1x3=3 1.3 x 3 = 4 1x3=3
Empirical formula C3H4O3
Molecular formula:
(C3H4O3)n = 176.12
[3(12.0) + 4(1.0) + 3(16.0)n = 176.12
88.0n = 176.12
n=2
The molecular formula of X is C H O .
6 8 6
Q25
Mass of H2O = (2.545 1.625) g = 0.92 g
Mass of CuSO4 = 1.625 g
Element CuSO4 H2O
Mass (g) 1.625 0.92
Number of moles 1.625 0.92
1 59.7 = 18.0 = 0.0511
0.0102
Mole ratio 1 5
x is 5.
Q26
number of moles of NaCO2 (mol)
3
[Na2CO3] = 3
volume of solution (dm )
2 3
number of moles of NaCO
0.5 = 0.25
Number of moles of Na2CO3 = (0.5 mol dm3)(0.25 dm3)
= 0.125
Mass of Na CO = (0.125 mol)(106 g mol1)
2 3
= 13.25 g
Q27
Moles of solute = (1.55 mol dm3)(0.25 dm3) = 0.39
0.39 mol
Molarity = 0.35 dm = 1.11 mol dm3
Q28
5.5 g
Number of moles of NH3 = 17 g /mol = 0.323
moles of solute (mol)
Molality = mass of solvent (kg)
0.323 mol
Mass of solvent = = 0.202 kg
1.6 mol kg
= 202 g H2O
Q29
moles of solute(mol)
Molarity = volume of solution(dm ) = 1.15 mol dm3
Assume that: Moles of solute = 1.15
Volume of solution = 1000 cm3
Mass of solute = 1.15 mol x 34 g mol1 = 39.1 g
mass of solution(g)
Density = volume of solution( cm) = 1.11 g cm3
Mass of solution = (1.11 g cm3)(1000 cm3)
= 1110.0 g
Mass of solution = mass of solute + mass of solvent
Mass of solvent = (1110.0 39.1) g = 1070.9 g
1070.9 g
Moles of solvent, H2O = 18.0 g/mol = 59.5
H2O2
H2O2 n
x = Hn+n
O
2 2 H2O
1.15
= 1.15+59.5 = 0.019
Q30
moles of solute (mol)
Molality = mass of solvent ( kg ) = 1.60 mol kg1
Assume that:
Moles of solute = 1.60 mol
Mass of solvent = 1 kg = 1000 g
1000 g
Moles of solvent = = 55.56
18.0 g/mol
NaCl
n 1.60
XNaCl = nT = 1.60+55.56 = 0.0280
Q31
mass of solution (g)
Density = volume of solution ( cm3 ) = 1.25 g cm3
Assume:
Mass of solution = 1.25 g
Volume of solution = 1.0 cm3
= 1.00 x 103 dm3
moles of solute ( mol )
Molarity = volume of solution ( dm 3 ) = 2.69 mol dm3
Moles of solute = (2.69 mol dm3)(1.00 x 103 dm3)
= 2.69 x 103
Mass of solute = (2.69 x 103 mol)(342 g mol1)
= 0.920 g
Mass of solute = mass of solute + mass of solvent
Mass of solvent = (1.25 0.920) g = 0.33 g
moles of solute ( mol )
Molality = mass of solvent ( kg )
2.69 x 10
= 3.3 x 104 = 8.15 mol kg1
Q32
w mass of solute ( g)
% w = mass of solution (g) x 100% = 19.0%
Assume that:
Mass of solute = 19.0 g
Mass of solution = 100 g
19.0 g
Moles of solute = = 0.3016
63.0 g /mol
mass of solution(g)
Density = volume of solution( cm) = 1.11 g cm3
100 g
Volume of solution = = 90.1 cm3 = 0.0901 dm3
1.11 g cm
moles of solute ( mol ) 0.302
Molarity = volume of solution ( dm ) 3 = 0.0901 = 3.35 mol dm3
M1 V 1 = M2 V 2
(3.35 mol dm3)V1 = (3.00 mol dm3)(2.00 dm3)
V1 = 1.79 dm3
1.3 Stoichiometry
Q33
Zn(s) + H2SO4(aq) ZnSO4(aq) + H2(g)
mass of solution( g)
Density = volume of solution( cm3 ) = 0.182 g cm3
Assume that in 1.0 dm3 of solution:
Mass of solution = 182 g
Volume of solution = 1000 cm3
w mass of solute ( g )
% w = mass of solution (g) = 100 = 96.0
96.0
Mass of solute = x 182 g = 174.7 g
100
174.7 g
Moles of solute = 98.0 g mol = 1.783
1.783 mol
Molarity of H2SO4 = 1.0 dm = 1.783 mol dm3
Zn(s) + H2SO4(aq) ZnSO4(aq) + H2(g)
0.2015 dm
Number of moles of H2 gas = 24.0 dm = 8.396 x 103
From the equation: 1 mol of H2 gas is produced from 1 mol of H2SO4.
Number of moles of H SO = number of moles of H
2 4 2
= 8.396 x 103
8.396 x 10 mol
Volume of H2SO4 needed = 1.783 mol/dm
= 4.709 x 103 dm3
= 4.71 cm3
Q34
Pb(NO3)2(aq) + 2 KI (aq) PbI2(s) + 2KNO3(aq)
15.0 g
Number of moles of KI = 166.0 g/mol = 0.0904
25.0 g
Number of moles of Pb(NO3)2 = 331.2 g/ mol = 0.0755
From the equation: 2 mol KI reacts with 1 mol Pb(NO3)2.
0.0904 mol
0.0904 mol KI reacts with = 0.0452 mol Pb(NO3)2.
2
Moles of Pb(NO3)2 provided > Moles of Pb(NO3)2 needed
(0.0755 mol) (0.0452 mol)
KI is the limiting reactant and Pb(NO3)2 is in excess.
From the equation: 2 mol of KI produce 1 mol of PbI2.
0.0904 mol
0.0904 mol KI produces = 0.0452 mol PbI2.
2
Mass of PbI2 = 0.0452 mol x 461.0 g mol1 = 20.8 g
Q35
1 mol of gas occupies 22.4 dm3 at s.t.p.
0.2dm
Number of moles of O gas = = 8.93 x 103
2 22.4 dm /mol
From the equation: 5 mol of O2 is produced from 4 mol of KNO3.
4
Number of moles of KNO needed = x 8.93 x 103 mol = 7.14 x 103 mol
3 5
Mass of KNO = (7.14 x 103 mol)(101.1 g mol1) = 0.722 g
3
Q36
Na2CO3(aq) + 2HCI(aq) 2NaCl(aq) + CO2(g) + H2O(l)
Number of moles of Na2CO3 = (0.102 mol dm3)(0.025 dm3) = 2.55 x 103
From the equation: 1 mol Na2CO3 reacts with 2 mol HCl.
Number of moles of HCl needed = 2 x 2.55 x 103 mol = 5.10 x 103 mol
5.10 x 10 mol
[HCl] = 0.0382 dm = 0.134 mol dm3
Q37
1.756 g
Number of moles of BaSO4 = 233.3 g/ mol = 7.53 x 103
From the chemical equation: 1 mol Na2SO4 = 7.53 x 103 mol
7.53 x 10 mol
[Na2SO4] = 0.04 dm
= 0.188 mol dm3
Q38
H2A(s) + 2 NaOH(aq) Na2A(aq) + 2H2O(l)
Number of moles of NaOH = (2.05 mol dm3)(0.0232 dm3)
= 0.0476
From the equation: 2 mol NaOH react with 1 mol H2A.
0.0476 mol
Number of moles of H A =
2 2
= 0.0238
4.17 g
Molar mass of H2A = 0.0238mol
= 175.2 g mol1
Q39
Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)
Number of moles of HCl = (0.75 mol dm3)(0.15 dm3)
= 0.1125
Number of moles of HCl reacted with Mg
= 0.1125 0.0514
= 0.0611
From the equation: 2 mol HCl react with 1 mol Mg.
0.0611 mol
Number of moles of Mg =
2
= 0.0306
Mass of Mg = 0.0306 mol x 24.3 g mol1
= 0.744 g
0.744 g
% mass of Mg = 2.95 g x 100%
= 25.2%
Q40
Pb(NO3)2(aq) + 2KI(aq) PbI2(s) + 2KNO3(aq)
Number of moles of Pb(NO3)2 = (0.5 mol dm3)(0.1 dm3)
= 0.05
Number of moles of KI = (0.15 mol dm3)(0.2 dm3) = 0.03
From the equation:
1 mol of Pb(NO3)2 reacts with 2 mol of KI.
0.05 mol of Pb(NO ) reacts with (2 x 0.05 mol) = 0.1 mol of KI
3 2
Moles of KI provided < Moles of KI needed
(0.03 mol) (0.1 mol)
Therefore, limiting reactant is KI.
From the equation:
2 mol KI react with 1 mol Pb(NO3)2.
0.03mol
0.03 mol KI react with = 0.015 mol Pb(NO3)2
2
Moles of Pb(NO ) remaining = 0.05 0.015
3 2
= 0.035
Mass of Pb(NO3)2 remaining = (0.035 mol)(331.2 g mol1)
= 11.59 g
From the equation:
2 mol KI produced 1 mol PbI2.
0.03 mol KI produced 0.015 mol PbI2.
Mass of PbI formed = (0.015 mol)(461.0 g mol1)
2
= 6.92 g
From the equation:
2 mol KI produced 2 mol KNO3.
Moles of KI = moles of KNO = 0.03 mol
3
Mass of KNO3 formed = (0.03 mol)(101.1 g mol1)
= 3.03 g
CHALLENGE QUESTIONS
Structured Question
1. (a) Number of moles of AgCl
0.125 g
= 143.4 g mol = 8.72 x 104
(b) Ag+(aq) + Cl(aq) AgCl(s)
From the equation:
nAgCl = nCl = 8.72 x 104 mol
8.72 x 10 mol
[Cl] = [NaCl] = 0.025 dm
= 0.0349 mol dm3
Number of moles of NaCl in the stock solution
= (0.0349 mol dm3)(0.25 dm3)
= 8.72 x 103
(c) Mass of NaCl in the sample
= (8.75 x 103 mol)(58.5 g mol1)
= 0.512 g
(d) % NaCl in the sample
0.512 g
= 12.89 g x 100% = 3.97%
2. (a) Zn(s) + 2AgNO3(aq)
2Ag(s) + Zn(NO3)2(aq)
(b) nAgNO3 = (0.25 mol dm3)(0.05 dm3)
= 0.0125 mol
5.3
nZn = 65.4 = 0.081 mol
From the chemical reaction:
1 mol Zn reacts with 2 mol AgNO3
0.081 mol Zn reacts with 2 x 0.081 mol
= 0.162 mol AgNO3
nAgNO3 needed > nAgNO3 provided
(0.162 mol) (0.0125 mol)
Therefore, AgNO3 is the limiting reactant.
(c) From the equation:
2 mol AgNO3 produce 2 mol Ag.
0.0125 mol AgNO produces
3
0.0125 mol Ag.
Mass of Ag
= (0.0125 mol)(107.9 g mol1)
= 1.35 g
(d)
The excess reactant is Zn.
From the reaction:
2 mol AgNO3 react with 1 mol Zn.
0.0125
Number of moles of Zn reacted
2
= 6.25 x 103 mol Zn
Moles of excess reactant remaining
= Moles of Zn provided moles of Zn reacted
= 0.081 (6.25 x 103)
= 0.075
Mass of Zn = (0.075 mol)(65.4 g mol1)
= 4.91 g
3. (a) CH3COOH + NaOH CH3COONa + H2O
(b) Number of moles of NaOH
= (0.915 mol dm3)(0.0173 dm3)
= 3.37 x 103
From the equation:
1 mol NaOH reacts with 1 mol CH3COOH.
Moles of NaOH
= moles of CH3COOH
= 3.37 x 103 mol
Concentration of the diluted solution
3.37 x 10 mol
= 0.025 dm = 0.1348 mol dm3
(c) M1V1 = M2V2
M1 (10.0 cm3) = (0.1348 mol dm3)(250 cm3)
M1 = 3.37 mol dm3
mass of solution( g)
(d) Density = 3
volume of solution( cm ) = 1.05
Mass of solution
= (1.05 g/cm3)(10.0 cm3)
= 10.5 g
moles of solute (mol)
Molarity = volume of solution ( dm 3 )
= 3.37 mol dm3
Number of moles of solute
= (3.37 mol dm3)(0.01 dm3)
= 0.0337
Mass of solute
= (0.0337 mol)(60.0 g mol1)
= 2.02 g
w mass of solute ( g )
% w = mass of solution ( g ) x 100%
2.02
= 10.5 x 100%
= 19.3%
4. (a) 5H2O2 + 2MnO4 + 6H+
5O2 + 2Mn2+ + 8H2O
(b) Number of moles of MnO4
= (0.215 mol dm3)(0.0235 dm3)
= 5.05 x 103
(c) From the equation:
2 mol MnO4 react with 5 mol H2O2
Number of moles of H O
2 2
5
2 x 5.05 x 103
= 0.0126
Mass of H2O2
= (0.0126 mol)(34 g mol1)
= 0.429 g
0.429 g
% H O in the sample = x 100%
2 2 19.8 g
= 2.17%
5. (a) Element Y is chlorine.
35
Isotope notation of Y: 17Cl
Species Number of Number of Number of
protons neutrons electrons
(b) 37
Y 17 20 17
17
(c)
Y 17 20 18
Essay Question
1. (a) The relative atomic mass of an element is the average mass of one atom of the
1
element relative to 12 the mass of the one atom of carbon-12.
(b) To produce positive ions when a sample is bombarded by high-energy electrons
from a heated filament.
(c) The sample need to be ionised because ions can be deflected whereas neutral
atoms are not affected by a magnetic field.
(d) The deflection of ions depends on the m/e ratio of the ions. The ions with a
smaller m/e ratio will be deflected more.
