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Similar to Mole Concept @Kidegalize Network_0772034474.pdf (20) More from BagalanaSteven (20) 2. © Kidegalize
Relative Atomic Mass, Ar
Relative atomic mass, RAM of an
element is the mass of one atom of
that element compared with 1/12th of
the mass of an atom of carbon-12.
e.g. one atom of oxygen weighs 16g
denoted O = 16
3. © Kidegalize
Relative Molecular Mass, Mr
The relative molecular mass, RMM (for
molecules) or relative formula mass,
RFM (for ionic/ giant covalent) is the
average mass of a molecule/ formula
unit compared to 1/12th of the mass
of an atom of carbon-12.
4. © Kidegalize
To find the RMM/RFM, add up the RAMs
of all the atoms present in the formula
of the molecule/ ion/ compound.
RMM/RFM has no units.
6. © Kidegalize
(i) RFM of NaOH = 23 + 16 + 1
= 40
(ii) RFM of Ca(NO3)2 = 40 + 2 (14 + 16 × 3)
= 164
7. © Kidegalize
Percentage composition by mass
Percentage composition is the mass of
each element present in a compound
expressed in terms of percent by mass.
Thus percentage of element B is
% of B =
Mass of B in formula
RFM of compound
100%
10. © Kidegalize
Example 2 (UNEB 1992/P1/3)
The percentage by mass of water of
crystallization in CuSO4.5H2O is
(Cu = 64, S = 32, O = 16, H = 1)
A.
90
250
100 B.
18
250
100
C.
90
160
100 D.
18
160
100
11. © Kidegalize
Possible Solution
RFM of CuSO4.5H2O = 64 + 32 + (16 × 4) + 5 (1× 2 + 16)
= 250
Mass of water = 5 (1× 2 + 16)
= 90
%age of water of crystallization =
90
250
100
Correct Option is A
12. © Kidegalize
Knowledge Check 1
Where necessary use:
Cu = 64, Fe = 56, Ca = 40, Cl = 35.5,
S = 32, P = 31, Na = 23, O = 16,
N = 14, C = 12, H = 1
13. © Kidegalize
Qn. 1 (UNEB 2015/P1/9)
Which one of the following sulphates
contains the highest percentage of
sulphur?
A. (NH4)2SO4 B. Na2SO4
C. CaSO4 D. CuSO4
14. © Kidegalize
Qn. 2 (UNEB 2014/P1/12)
Which one of the following is the
percentage of sodium carbonate in
28.2g of hydrated sodium carbonate,
Na2CO3.10H2O?
A. 9.86% B. 26.20%
C. 29.02% D. 37.06%
15. © Kidegalize
Qn. 3 (UNEB 2006/P1/13)
The percentage of water of crystallization
in iron(II) sulphate, FeSO4.7H2O is
A.
126 100
278
B.
278 100
126
C.
126 100
152
D.
152 100
126
16. © Kidegalize
Qn. 4 (UNEB 2005/P1/38)
Which one of the fertilizers has the
highest nitrogen content per mole of
the fertilizer?
A. CO(NH2)2 B. (NH4)2SO4
C. NH4Cl D. NH3NO3
17. © Kidegalize
Qn. 5 (UNEB 2004/P1/27)
The percentage by mass of phosphorus
in calcium phosphate, Ca3(PO4)2 is
A. 8 B. 10
C. 17 D. 20
18. © Kidegalize
Qn. 6 (UNEB 1990/P1/29)
The percentage of oxygen in baking
powder, NaHCO3 is
A.
48 100
84
B.
16 100
84
C.
16 100
102
D.
48 100
102
19. © Kidegalize
The Mole
A mole is the amount of any substance
that contains the same number of
particles (atoms/ions/molecules) as
there are atoms in exactly 12g of
carbon-12.
The number of particles in one mole is the
Avogadro constant (6.02 × 1023).
20. © Kidegalize
The unit of mole is mol.
The molar mass of a substance is its
mass per mole (gmol-1). It has the same
numerical value as the RFM.
Thus
Molar mass, Mr =
Mass of substance, m
Amount in moles, n
23. © Kidegalize
Paper 1 method
Mole Atoms
1 6.02 × 1023
0.06 y
1 × y = 0.06 × 6.02 × 1023
y = 3.612 × 1022
24. © Kidegalize
ALT: Paper 1 method
Number of moles =
Number of particles
Avogadro constant
Number of particles = 0.06× 6.02 × 1023
= 3.612 × 1022
25. © Kidegalize
Example 2 (UNEB 2005/P1/40)
The mass of 4 atoms of phosphorus is
[Avogadro constant = 6.02 x 1023, P = 31]
A.
6.02 1023
4 31
B.
4 31
6.02 1023
C.
31 6.02 1023
4
D.
31
4 6.02 1023
28. © Kidegalize
Possible Solution
Paper 2/3 method
RFM of Na2CO3 = (2 × 23) + 12 + (3 × 16)
= 106
106g of Na2CO3 contains 1 mole
10.6g of Na2CO3 contains
10.6
106
moles
= 0.1moles
29. © Kidegalize
Paper 1 method
Mass(g) Moles(mol)
106 1
10.6 y
106 × y = 10.6 × 1
y =
10.6
106
= 0.1moles
30. © Kidegalize
ALT: Paper 1 method
Molar mass =
Mass of substance
Amount in moles
106 =
10.6
Amount in moles
Amount in moles =
10.6
106
= 0.1moles
31. © Kidegalize
Example 4
0.2moles of a hydroxide, M(OH)2 weighed
11.6g. Which one of the following is the
relative formula mass of the hydroxide?
A. 24 B. 34
C. 58 D. 41
34. © Kidegalize
ALT: Paper 1 method
Molar mass =
Mass of substance
Amount in moles
=
11.6
0.2
= 58
Correct Option is C
35. © Kidegalize
Example 4 (UNEB 2004/P1/8)
Zinc reacts with hydrochloric acid
according to the following equation:
Zn(s) + 2HCI(aq) ZnC12(aq) + H2(g)
The number of moles of hydrochloric acid
required to react completely with 7.0 g
of zinc is (Zn = 65)
39. © Kidegalize
Qn.1 (UNEB 2015/P1/33)
0.2moles of a hydroxide, X(OH)2 weighed
11.6g. Which one of the following is the
relative atomic mass of X?
A. 24 B. 34
C. 58 D. 41
40. © Kidegalize
Qn.2 (UNEB 2015/P1/17)
Which one of the following has the same
mass as 0.05moles of sulphur?
A. 2moles of carbon
B. 1.5moles of sodium
C. 0.13moles of aluminium
D. 0.1moles of oxygen atoms
41. © Kidegalize
Qn.3 (UNEB 1997/P1/30)
Which one of the following contains the
same number of atoms as 8g of sulphur?
A. 20g of calcium B. 10g of calcium
C. 12g of carbon D. 4g of carbon
42. © Kidegalize
Qn.4 (UNEB 1989/P1/23)
The number of moles of nitrogen
molecules in 42g of nitrogen is
A. 0.33 B. 0.67
C. 1.50 D. 3.00
43. © Kidegalize
Qn.5 (UNEB 1998/P1/35)
Hydrogen peroxide decomposes according
to the equation:
2H2O2(l) H2O(l) + O2(g).
How many moles of oxygen are given out
when 17.0g of hydrogen peroxide is
45. © Kidegalize
Mass-to-mass calculations
Mass-to-mass calculations require
balanced chemical equation and data
about the mass of one reactant/ product
(X) . To find the mass of another reactant
/product(Y), follow the steps:
STEP 1: Use data about the mass of X to
find the moles of X.
46. © Kidegalize
STEP 2: Use the mole ratio from balanced
chemical equation to find the moles of Y.
STEP 3: Use the moles of Y in step 2 and
its RFM to find the mass of Y.
47. © Kidegalize
Alternatively
Step 1: Write down the amounts in
moles of the relevant reactants and
products in the equation.
Step 2: Convert these amounts in moles
of the relevant reactants and products
to masses.
Step 3: Scale the masses to the
quantities required.
48. © Kidegalize
Example 1
Lead(II) carbonate decomposes on heating
according to the following equation.
PbCO3(s) PbO(s) + CO2(g)
Calculate the mass of the residue when
5.34g of lead(II) carbonate was heated
(Pb = 207, O = 15, C = 12)
50. © Kidegalize
Moles of PbCO3 =
5.34
267
=0.02
1mole of PbCO3 produces 1mole of PbO
Moles of PbO = 0.02
Mass of PbO = Moles of PbO × RFM of PbO
= 0.02 × 223
= 4.46g
51. © Kidegalize
Paper 2 method (ALT)
RFM of PbCO3 = 207 + 12 + (3 × 16)
= 267
RFM of PbO = 207 + 16
= 223
52. © Kidegalize
1mole of PbCO3 produces 1mole of PbO
267g of PbCO3 produces 223g of PbO
5.34g of PbCO3 produces
5.34 × 223
267
g
= 4.46g
53. © Kidegalize
Paper 1 method
Mole ratio PbCO3 : PbO = 1 : 1
Mass of PbCO3 Mass of PbO
267 223
5.34 y
267y = 5.34 × 223
y =
5.34 × 223
267
= 4.46g
54. © Kidegalize
Example 2(UNEB 2011/P1/33)
Magnesium burns in air according to the
following equation:
2Mg(s) + O2(g) 2MgO(s)
The mass of oxygen required to burn 5g
of magnesium completely is
[O = 16 ; Mg = 24]
59. © Kidegalize
Qn.1 (UNEB 1987/P2/1(a))
5.0 g of calcium carbonate was heated
strongly until there was no further
change.
(i) Write equation for the reaction.
(ii) Calculate the mass of the solid left.
60. © Kidegalize
Qn.2 (UNEB 1992/P2/2)
2.5 g of zinc carbonate was heated
strongly until there was no further
change.
(a) State what was observed.
(b) Write an equation for the reaction.
(c) Calculate the mass of the residue.
61. © Kidegalize
Qn.3 (UNEB 2002/P2/8)
Excess sodium sulphate solution was added to a
solution containing 4.14 g of lead(II) nitrate.
(a) What was observed?
(b) Write equation for the reaction that took
place.
(c) Calculate the mass of the solid formed.
62. © Kidegalize
Qn.4 (UNEB 2010/P1/38)
Zinc carbonate decomposes according to
the following equation when heated:
ZnCO3 (s) ZnO(s) + CO2(g)
The mass of zinc oxide formed when 2.5 g of
zinc carbonate is heated is
A. 0.41g B. 0.81g
C. 1.62g D. 3.24g
63. © Kidegalize
Qn.5 (UNEB 2005/P1/32)
Copper (II) oxide reacts with hydrogen
according to the equation:
CuO(s) + H2(g) Cu(s) + H2O(l).
The mass of copper formed when 8.0g
of the oxide is reacted with excess
hydrogen is
65. © Kidegalize
Qn.6 (UNEB 2004/P1/4)
Lead nitrate decomposes according to the
equation:
2Pb (NO3)2 2PbO(s) + 4NO2(g) + O2(g)
The mass of lead monoxide that is produced
when 3.31 g of lead nitrate is completely
decomposed is
67. © Kidegalize
Qn.7 (UNEB 2002/P1/32)
Zinc displaces copper from an aqueous solution
of Copper(II) sulphate according to the
equation:
CuSO4(aq) + Zn(s) Cu(s) + ZnSO4(aq)
The mass of copper in g that is displaced
by 13.10g of zinc is [Zn= 65.4]
A. 6.35 B. 12.72
C. 19.07 D. 25.82
68. © Kidegalize
Qn.8 (UNEB 1988/P1/38)
Calcium hydroxide reacts with ammonium chloride
according to the equation:
Ca(OH)2(s) + 2NH4Cl(s) CaCl2(s) + 2H2O(l) + 2NH3(g)
If 14.8 g of calcium hydroxide was reacted completely with
ammonium chloride, what mass of ammonia gas would
be evolved?
A. 1.7 g B. 3.4 g
C. 6.8 g D. 9.0 g
70. © Kidegalize
and 24dm3/24l /24000cm3 at room
temperature and pressure.
Thus 1 mole (32g) of oxygen occupies
22.4dm3 at s.t.p.
Also 1 mole (44g) of carbon dioxide
occupies 22.4dm3 at s.t.p.
71. © Kidegalize
Mass-to-volume calculations follow the path:
STEP 1: Use data about the mass of the solid to
find the moles of solid.
STEP 2: Use the mole from balanced chemical
equation to find the moles of the gas.
STEP 3: Use the moles of the gas in step 2 and
its RFM/MGV to find the mass/volume of the gas.
72. © Kidegalize
Example 1
25g of zinc carbonate was heated strongly
until no further change.
(a)Write the equation for the reaction.
(b) Calculate the volume of carbon dioxide
evolved at stp (1mole of a gas occupies
22.4dm3 at stp, Zn = 65, O = 16, C = 12)
74. © Kidegalize
125g of ZnCO3 evolves 22.4dm3 of CO2
25g of ZnCO3 evolves
22.4 × 25
125
dm3 of CO2
= 4.48 dm3
75. © Kidegalize
(b) Paper 1 method
Mole ratio ZnCO3 : CO2 = 1 : 1
Mass of ZnCO3 Volume of CO2
125 22.4
25 y
125y = 22.4 × 25
y =
22.4 × 25
125
= 4.48dm3
76. © Kidegalize
Example 2 (UNEB 2007/P1/38)
When 2.5 g of a solid is heated, 250cm3
of a gas was produced at s.t.p and a
residue of 1.4 g was left. The molecular
mass of the gas is
(1 mole of gas occupies 22.4dm3 at
s.t.p)
79. © Kidegalize
Paper 1 method
Mass of gas = 2.5 – 1.4 = 1.1g
Volume of gas Mass of gas
250 1.1
22400 y
250y = 1.1 × 22400
y =
22400 × 1.1
250
Correct Option is C
80. © Kidegalize
Example 3 (UNEB 2009/P1/26)
On heating, sodium nitrate produces sodium
nitrite and oxygen according to the following
equation:
2NaN03(l) 2NaNO2(s) + O2(g)
The mass of sodium nitrite formed when
480cm3 of oxygen was evolved at room
temperature is
81. © Kidegalize
[N =14, O = l 6, Na = 23; 1 mole of gas at
room temperature occupies 24 litres]
A. 1.38g B. 2.76g
C. 5.52g D. 11.04g
83. © Kidegalize
Mass of NaNO2 Volume of O2
2 × 69 24000
y 480
24000y = 2 × 69 × 480
y =
2 × 69 × 480
24000
= 2.76
Correct Option is B
84. © Kidegalize
Knowledge Check 4
Where necessary use:
1 mole of gas occupies 24dm3 at room
temperature and 22.4dm3 at s.t.p.
Zn = 65, Ca = 40, Cl = 35.5, S = 32
Na = 23, O = 16, N = 14, C = 12, H = 1
85. © Kidegalize
Qn.1(UNEB 1999/P1/6)
Sulphuric acid reacts with ammonia according
to the following equation:
2NH3(g) + H2SO4(l) (NH4)2SO4(s)
The mass of ammonium sulphate formed when
6l of ammonia reacts with excess sulphuric
acid is
A. 8.25g B. 16.50g
C. 33.00g D. 66.00g
86. © Kidegalize
Qn.2(UNEB 2007/P1/6)
Hydrogen chloride reacts with ammonia
according to the following equation:
NH3(g) + HCl(g) NH4Cl(s)
The mass of ammonium chloride formed
when excess ammonia is reacted with
0.56dm3 of hydrogen chloride at s.t.p is
88. © Kidegalize
Qn.3(UNEB 2007/P1/22)
Hydrogen burns in oxygen to form steam
according to the following equation:
2H2(g) + O2(g) 2H2O(g)
The mass of steam formed when 100cm3 of
hydrogen is burnt in excess oxygen at s.t.p is
A. 0.04g B. 0.08g
C. 0.12g D. 0.16g
89. © Kidegalize
Qn.4(UNEB 2009/P1/38)
When 3.0g of X was heated, 210cm3 of a gas
were evolved at s.t.p and 2.4g of solid
remained. The relative molecular mass of the
gas is
A.
0.6 × 22.4
210
B.
3 × 22.4
210
C.
2.4 × 22.4
210
D.
5.4 × 22.4
210
90. © Kidegalize
Qn.5(UNEB 2014/P1/37)
Sodium hydrogencarbonate decomposes when
heated according to the equation:
2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(l)
The mass of sodium hydrogencarbonate which
must be heated to give off 200cm3 of carbon
dioxide at room temperature is
91. © Kidegalize
A.
2 × 84 × 200
24000
B.
24000 × 84
2 × 200
C.
84 × 200
24000
D.
200 × 84
2 × 24000
92. © Kidegalize
Qn.6(UNEB 2000/P1/23)
The mass of 560cm3 of a gas X at s.t.p is
1.10g. The relative formula mass of the gas
is
A.
1.4 × 22.4
560
B.
22.4 × 1.1
560
C.
1.1 × 560
22.4
D.
560 × 22.4
1.1
93. © Kidegalize
Qn.7(UNEB 1994/P1/9)
Propene burns in oxygen according to the equation:
2C3H6(g) + 9O2(g) 6CO2(g) + 6H2O(l)
When 2.1g of propene is completely burned in
oxygen, the volume of carbon dioxide produced at
room temperature is
A. 1.2dm3 B. 2.4dm3
C. 3.6dm3 D. 4.8 dm3
94. © Kidegalize
Qn.8(UNEB 1997/P1/3)
Zinc reacts with hydrochloric acid according to
the following equation:
Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)
The volume of hydrogen gas liberated at s.t.p
when 13g of zinc reacted with excess
hydrochloric acid is
A. 2.24dm3 B. 4.48dm3
C. 22.4dm3 D. 11.2dm3
95. © Kidegalize
Qn.9(UNEB 2009/P1/22)
Glucose burns in oxygen according to the following
equation:
C2H12O6(s) + 6O2(g) 6CO2(g) + 6H2O (l) + Energy
The volume of oxygen at s. t. p that is required to produce
150g of carbon dioxide is
A.
150 × 22.4
44
dm3 B.
50 × 22.4
44 × 6
dm3
C.
44
22.4 × 150
dm3 D.
44 × 6
150 × 22.4
dm3
96. © Kidegalize
Qn.9(UNEB 2002/P1/11)
Calcium carbide reacts with water to produce a
gas according to the equation:
CaC2(s) + 2H2O(l) Ca(OH)2(s) + C2H2(g)
The volume of the gas produced at s.t.p when 6.4g
of calcium carbide reacts completely is
A.
150 × 22.4
44
dm3 B.
50 × 22.4
44 × 6
dm3
C.
44
22.4 × 150
dm3 D.
44 × 6
150 × 22.4
dm3
97. © Kidegalize
Qn.10(UNEB 2003/P1/30)
Zinc nitrate decomposes on heating according
to the equation:
2Zn(NO3)2(s) 2ZnO(s) + 4NO2(g) + O2(g)
The maximum volume of oxygen evolved in this
reaction at s.t.p when 7.56 g of zinc nitrate is
heated is
99. © Kidegalize
Qn.11(UNEB 1991/P2/2)
2.5g of zinc carbonate was heated
strongly until there was no further
change.
(a) State what was observed.
(b) Write an equation for the reaction.
(c) Calculate the mass of the residue.
100. © Kidegalize
Qn.12(UNEB 1990/P2/9)
When 0.107 g of ammonium chloride
was heated with excess calcium
hydroxide, a gas was evolved.
(a) Write equation for the reaction.
(b) Calculate the volume of gas that
was evolved at room temperature.
102. © Kidegalize
Volume-to-volume calculations
Volume-to-volume calculations require
Avogadro’s principle which states that
equal volumes of different gases contain
the same number of molecules, under the
conditions of temperature and pressure
OR the volumes of gases that react (and
gaseous products) are proportional to
their mole ratio.
103. © Kidegalize
Example 1
Calculate the volume of nitrogen monoxide
required to completely react with 20cm3
of oxygen under the same conditions of
temperature and pressure.
104. © Kidegalize
Possible Solution
Paper 2 method
2NO(g) + O2 2NO2
1 mole of O2 reacts with 2moles of NO
1volume of O2 reacts with 2volumes of NO
20cm3 of O2 reacts with 2 × 20cm3 of NO
= 40cm3
106. © Kidegalize
Example 2 (UNEB 1996/P1/12)
Ammonia is obtained from hydrogen and nitrogen
according to the equation:
N2(g) + 3H2(g) NH3(g).
The volume of ammonia produced when 25l of
nitrogen reacts with excess hydrogen at
constant temperature and pressure is
A. 12.5l B. 25.0l C. 50.0l D. 75.0l
108. © Kidegalize
Example 3 (UNEB 1993/P1/11)
Steam reacts with methane according to the
equation:
CH4(g) + 2H2O(g) 4H2(g) + CO2(g)
What volume of gas will remain when 30cm3 of
methane is reacted with 20cm3 of steam?
A. 20 cm3 B. 50 cm3
C. 70 cm3 D. 80 cm3
111. © Kidegalize
Example 4 (UNEB 1992/P1/31)
Carbon monoxide burns in oxygen according to
the equation:
2CO(g) + O2(g) 2CO2(g).
20cm3 of carbon monoxide was mixed with
20cm3 of oxygen and exploded. If all volumes were
measured at the same temperature and
pressure, what was the final gaseous volume?
A. 20cm3 B. 30cm3
C. 40cm3 D. 50cm3
113. © Kidegalize
Vol of excess oxygen = 20 – 10
= 10cm3
Let volume of CO2 formed be y
CO2 CO
Mole 2 2
Volume y 20
2y = 2 × 20 giving y = 20
Final volume = 20 + 10 = 30
Correct Option is B
114. © Kidegalize
Knowledge Check 5
Qn.1(UNEB 2012/P1/20)
When 150cm3 of oxygen was mixed with
500cm3 of hydrogen and the mixture
exploded, water was formed according to the
following equation:
2H2(g) + O2(g) 2H2O(l)
The volume of hydrogen that was left
unreacted is
A. 75cm3 B. 200cm3
C. 350cm3 D. 425cm3
116. © Kidegalize
Qn.3(UNEB 2005/P1/37)
Methane burns in oxygen according to the
equation:
CH4(g) + 2O2(g) CO2(g) + 2H2O(g).
The volume of methane that remains unburnt
when 50cm3 of methane is reacted with
40cm3 of oxygen is
A. 10cm3 B. 20cm3
C. 30cm3 D. 45 cm3.
117. © Kidegalize
Qn.4(UNEB 2007/P1/36)
Sulphur dioxide reacts with oxygen to form sulphur
trioxide according to the following equation:
2SO2(g) + O2(g) 2SO3(g)
20cm3 of sulphur dioxide was mixed with 25cm3 of
oxygen at a certain temperature and pressure.
The volume of oxygen that reacted was
A. 10.0cm3 B. 12.5cm3
C. 20.0cm3 D. 25.0cm3
119. © Kidegalize
Qn.6(UNEB 1989/P1/9)
Propane burns in oxygen according to the
following equation:
C3H8(g) + 5O2(g) 4H2O(g) + 3CO2(s)
The volume of oxygen required for complete
combustion of 10dm3 of propane is
A. 75dm3 B. 50dm3
C. 25dm3 D. 15dm3
120. © Kidegalize
Qn.7(UNEB 2004/P1/10)
Under a certain temperature and pressure,
hydrogen reacted with nitrogen according to the
equation below:
3H2(g) + N2(g) 2NH3(g)
The volume of nitrogen required to react with
150cm3 of hydrogen under the same
temperature and pressure is
A. 15.0cm3 B. 50.0cm3
C. 300.0cm3 D. 450.0cm3
121. © Kidegalize
Empirical formula
Empirical formula is a chemical formula
that shows the simplest ratio of the
atoms of each element that combine to
form a molecule/compound.
To determine the empirical formula
(i) Divide the %age/mass of each element
by its RAM, (Moles of atoms)
122. © Kidegalize
(ii) Divide each of the figures obtained in
(i) by the smallest (mole ratio).
(iii) If the results of the calculations in (ii)
don’t approximate to whole numbers,
multiply them all by 2 (for 1.5, 2.5, …)
123. © Kidegalize
Molecular formula
Molecular formula is a chemical formula
that shows the actual numbers of atoms
of each type of element that combine to
form the molecule/compound.
To find the molecular formula:
(i) Solve for n in the equation:
(empirical formula)n = RFM
124. © Kidegalize
(ii) Multiply each atom in the empirical
formula by the value of n calculated in
(i) and write the molecular formula.
125. © Kidegalize
Example 1 (UNEB 1988/P1/21)
A compound contains 92.3% carbon
and 7.7% hydrogen by mass. What is
the empirical formula of the
compound? (C=12; H=1)
A. C2H B. CH2
C. C2H2 D. CH
128. © Kidegalize
Example 2(UNEB 2012/P1/9)
12.7g of metal R reacts completely with
11.3g of oxygen to form an oxide. Which
one of the following is the formula of R?
(O = 16, R = 27)
A. RO2 B. R2O
C. R2O3 D. R3O2
131. © Kidegalize
Example 3(UNEB 1996/P1/29)
6.5 g of an element X combine with oxygen
to give 8.1 g of oxide. The simplest
formula of the oxide is (O =16, X = 65)
A. X2O B. XO
C. XO2 D. X2O3
132. © Kidegalize
Possible Solution
Mass of oxygen = 8.1 – 6.5 = 1.6g
Element present X O
Composition 6.5 1.6
Moles of atoms 6.5
65
1.6
16
0.1 0.1
Mole ratio 0.1
0.1
0.1
0.1
Simplest ratio 1 : 1
134. © Kidegalize
Example 4(UNEB 1997/P1/14)
The empirical formula of a compound A is
C3H4. 25g of A occupies 14dm3 at s.t.p.
The molecular formula of A is (C=12; H=1;
1 mole of gas occupies 22.4dm3 at s.t.p)
A. C3H4 B. C3H8
C. C6H6 D. C6H8
136. © Kidegalize
(C3H4)n = 40
(3 ×12 + 4 ×1)n = 40
40n
40
=
40
40
n = 1
Molecular formula is C3H4
Correct Option is A
137. © Kidegalize
Example 5 (UNEB 2011/P1/10)
When 5.74g of a hydrated salt X was heated,
3.22g of the anhydrous salt, Y was formed. The
number of moles of water of crystallisation in X
is (Y = 161, O = 16, H = 1)
A. 2 B. 5
C. 7 D. 10
139. © Kidegalize
Compound present Y H2O
Composition 3.22 2.54
Moles of atoms 3.22
161
2.54
18
0.02 0.14
Mole ratio 0.02
0.02
0.14
0.02
Simplest ratio 1 : 7
141. © Kidegalize
Example 6(UNEB 2016/P2/6 (b),(c))
A gaseous organic compound J contains
82.76% carbon, the rest being hydrogen.
(b) Calculate the empirical formula of J.
(c)140cm3 of J weighed 0.363g at s.t.p.
Determine the molecular formula of J.
(1 mole of a gas occupies 22400cm3 at
s.t.p, C = 12, H = 1)
142. © Kidegalize
Possible Solution
Percentage of H2 = 100 – 82.76 = 17.24
Element present C H
Composition 82.76 17.24
Moles of atoms 82.76
12
17.24
1
6.896 17.240
Mole ratio 6.896
6.896
17.24
6.896
1 × 2 2.5 × 2
143. © Kidegalize
Simplest ratio 2 : 5
Empirical formula of J is C2H5
(c) 140cm3 of J weighs 0.363g
22400cm3 of J weighs
0.363 22400
140
g
= 58.08
144. © Kidegalize
(C2H5)n = 58.08
(2 ×12 + 5 × 1)n = 58.08
29n
29
=
58.08
29
n = 2
Molecular formula of J is C4H10
145. © Kidegalize
Knowledge Check 6
Qn.1(UNEB 1989/P1/40)
Compound R contains 15.8% of X and
84.2% of Y. The empirical formula of R is
(X =12, Y = 32)
A. XY3 B. X2Y
C. XY2 D. X3Y
148. © Kidegalize
Qn.4(UNEB 2003/P1/33)
A compound contains 53.3% oxygen,
6.7% hydrogen and 40% carbon. The
simplest formula of the compound is
[C=12, H=1, O=16]
A. CHO B. CH2O
C. C2H2O D. CH2O2
149. © Kidegalize
Qn.5(UNEB 2003/P1/34)
An oxide of P contains 50% by mass of
P. Its relative molecular mass is 64.
What is the formula of the oxide?
(P=32, O=16)
A. PO B. PO2
C. P2O D. PO3
150. © Kidegalize
Qn.5(UNEB 1996/P1/5)
An anhydrous salt R has a relative formula
mass of 158 and forms a hydrated salt
with the formula RnH2O. 79g of R combined
with 45g of water. The value of n is
(H=1, O=16)
A. 2 B. 3
C. 5 D. 10
152. © Kidegalize
Qn.7(UNEB 2011/P2/4)
A hydrated salt T, consist of 20.2% iron,
11.5%sulphur, 23% oxygen and 45.3%
water of crystallization.
(a) Calculate the empirical formula of T.
(Fe = 56, S = 32, O = 16, H = 1)
(b) Deduce the molecular formula of T.
(Relative formula mass of T = 278.)
153. © Kidegalize
Qn.8(UNEB 2009/P2/6)
A compound Z of molecular formula
AxBy .nH2O consists of 8.57% of A,
45.71% B and 45.72 of water.
(a) Determine the values of x, y and n.
(H = 1, O = 16, A = 27, B = 96)
(b) Write the molecular formula of Z
154. © Kidegalize
Qn.9(UNEB 1989/P2/3)
A gaseous hydrocarbon, X, contains 20%
hydrogen by mass.
7.5 g of X occupy 5.6 dm3 at STP.
(a) Calculate
(i) the empirical formula of X.
(ii) the molar mass of X.
(iii) the molecular formula of X.
155. © Kidegalize
Qn.10(UNEB 2018/P2/6)
Compound T contains 40.0% carbon, 6.7%
hydrogen and the rest being oxygen.
(a) (i) Calculate the empirical formula of T.
(ii) Determine the molecular formula of T.
(Relative formula mass of T = 60)
156. © Kidegalize
Moles in solutions
• Molar solution: is a solution that
contains 1mole of a substance
dissolved in 1litre/1000cm3.
• Molarity: is the number of moles of a
substance dissolved in 1litre/
1000cm3 of water. The unit of
molarity is moldm-3or moll-1 or M
157. © Kidegalize
The product of molarity and molar mass of a
substance is its concentration in grams per
litre (gl-1 or gdm-3).
Thus Molarity =
Concentration in gdm−3
Molar mass
• Standard solution: is a solution whose
concentration is known.
161. © Kidegalize
Paper 1 method
RFM of NaOH = 23 + 16 + 1
= 40
250cm3 of NaOH weighs 1.2g
1000cm3 of NaOH weighs
1.2 1000
250
g
= 4.8gdm-3
163. © Kidegalize
Example 2 (UNEB 1992/P1/11)
The molarity of a solution containing
40g of sodium hydroxide in 500cm3 of
the solution is
A. 0.2 M B. 0.5 M
C. 1.0 M D. 2.0 M
166. © Kidegalize
Example 3(UNEB 1990/P1/4)
The number of moles of sodium ions
contained in 100cm3 of 2M solution of
sodium carbonate is
A. 0.2 B. 0.4
C. 2.0 D. 4.0
168. © Kidegalize
1mole of Na2CO3 produces 2moles of Na+
0.2moles of Na2CO3 produces 2 × 0.2moles of Na+
= 0.4moles
171. © Kidegalize
Example 4 (UNEB 2000/P1/19)
What is the mass of sulphuric acid
(Mr=98) in 5cm3 of a 0.2 M solution of the
acid?
A.
98 5
0.2 1000
B.
98 0.2 5
1000
C.
98 0.2
5 1000
D.
98 5 1000
0.2
173. © Kidegalize
1000cm3 of H2SO4 contains 0.2moles
5cm3 of H2SO4 contains
0.2 5
1000
moles
1mole of H2SO4 weighs 98g
0.2 5
1000
moles of H2SO4 weighs
98 0.2 5
1000
g
Correct Option is B
174. © Kidegalize
Knowledge Check 7
Qn. 1(UNEB 2002/P1/19)
The mass of ammonium ion, NH4
+ in 0.5M
(NH4) 2SO4 solution is (H=1, N=14, O=16,
S=32)
A. 0.5 ×132 B.
132
0.5 × 2
C. 2 × 0.5 × 18 D.
0.5 × 2
36
175. © Kidegalize
Qn. 2(UNEB 2002/P1/25)
The mass of silver nitrate, AgNO3 in
0.2 M solution of the salt is
(Ag =108, O =16, N =14)
A. 17.0 B. 34.0
C. 85.0 D. 170.0
176. © Kidegalize
Qn. 3(UNEB 2009/P1/19)
The mass of sodium hydroxide present in
200cm3 of a 0.05M sodium hydroxide
solution is [H = 1, O = 16, Na 23]
A. 0.25g B. 0.40g
C 2.00g D. 10.00g
177. © Kidegalize
Qn. 4(UNEB 1987/P1/5)
The mass of potassium hydroxide, KOH,
contained in 250cm3 of 0.01M of
potassium hydroxide solution is
(K=39, H=1, O=16)
A. 0.056g B. 0.140g
C. 0.280g D. 0.560 g
178. © Kidegalize
Qn. 5(UNEB 1991/P1/8)
The mass of nitric acid (HNO3) required
to make 200cm3 of a 2M solution is
A. 31.5g B. 25.2g
C. 15.8g D. 12.6g
179. © Kidegalize
Qn. 6(UNEB 1993/P1/3)
A 0.2 molar solution of X contains
18.25 g of X per litre of the solution.
The relative molecular mass of X is
A. 18.25 B. 36.50
C. 45.63 D. 91.25
180. © Kidegalize
Qn. 7(UNEB 1996/P1/17)
What mass of sodium hydroxide is in
0.5 litre of 2M sodium hydroxide
solution:
A. 10g B. 20g
C. 40g D. 80g
181. © Kidegalize
Qn. 8(UNEB 2009/P1/11)
The molarity of a solution containing
49g of sulphuric acid in 250cm3 of
solution is [H = 1, O = 16, S = 32]
A. 0.125M B. 0.50M
C. 1.00M D. 2.00M
182. © Kidegalize
Qn. 9(UNEB 2010/P1/10)
5.72g of hydrated sodium carbonate,
Na2CO3.10H2O was dissolved in water to make
500cm3 of a solution. The molarity of the
soltuion is [Na = 23, O = 16, C = 12, H = 1].
A. 0.05M B. 0.02M
C. 0.04M D. 0.11M
183. © Kidegalize
Qn. 10(UNEB 1988/P1/9)
How many grams of pure sodium sulphate
crystals, Na2SO4.10H2O (relative
molecular mass=322) would be required to
make 250cm3 of 0.01M sodium sulphate
solution?
A. 0.40g B. 0.81g
C. 1.60g D. 3.22g
184. © Kidegalize
Qn. 11(UNEB 1997/P1/10)
0.02 moles of calcium chloride (CaCl2) is
dissolved to make 200cm3 of solution.
What is the concentration of chloride ions
in moles per litre, in this solution?
A. 0.05 M B. 0.1 M
C. 0.2 M D. 0.3 M
185. © Kidegalize
Qn. 12(UNEB 2005/P1/26)
The concentration, in grammes per litre,
of a 0.05M sodium carbonate solution
is [N=23, O=16, C=12]
A. 0.05 x 83 B. 0.05 x106
C.
106
0.05
D.
83
0.05
186. © Kidegalize
Qn. 13(UNEB 2018/P1/18)
Which one of the following is the number
of moles of hydrogen ions in 100cm3 of
a 0.05M sulphuric acid?
A. 0.0025 B. 0.01
C. 0.25 D. 1.00
187. © Kidegalize
Diluting solutions
When solutions are diluted, the amount
of solute in the diluted solution is the
same as was in the volume of the
original solution that was taken i.e.
amount of solute in original solution =
amount of solute in diluted solution
C1V1 = C2V2
190. © Kidegalize
Paper 1 method
Using CV = constant
C1V1 = C2V2
0.1 × 20 = 100 × C2
C2 =
0.1 20
100
= 0.02M
191. © Kidegalize
Example 2(UNEB 1999/P1/8)
25.0cm3 of a 0.4M NaOH solution was
diluted to 250cm3 with distilled water.
The molarity of the resultant solution is
A. 0.01 B. 0.04
C. 0.02 D. 0.4
193. © Kidegalize
Knowledge Check 8
Qn.1(UNEB 2006/P1/24)
200.0cm3 of a 0.1M sodium hydroxide
solution was diluted with water to make
two litres of solution.
The concentration of the dilute solution is
A. 0.002 M B. 0.050 M
C. 0.020 M D. 0.010 M
194. © Kidegalize
Qn.2(UNEB 2005/P2/7(b))
100.0 cm3 of a 0.1M sulphuric acid solution
was transferred to a 250cm3 volumetric
flask. The acid was then diluted with water
until the volume of the solution was exactly
250.0 cm3. Calculate:
(i) the number of moles of sulphuric acid in
the diluted solution.
(ii) the concentration of the diluted solution in
moles per dm3.
195. © Kidegalize
Qn. 3
Calculate the volume of water that
must be added to 150cm3 of 1.2M
sodium sulphate solution to produce a
solution whose concentration is 0.24M
197. © Kidegalize
Step 1: Calculate the number of moles of
the solution for which you know the molar
concentration and volume.
Step 2: Use the mole ratio from balanced
chemical equation to determine the
number of moles of the second solution.
198. © Kidegalize
Step 3: Calculate the molar concentration
of the second solution from its volume and
the number of moles from step 2.
199. © Kidegalize
Example 1
25cm3 of 0.08M sodium hydroxide was
found to require 20cm3 of hydrochloric
acid to be completely neutralised.
Calculate the molarity of hydrochloric
acid.
201. © Kidegalize
1mole of NaOH reacts with 1mole of HCl
0.002moles NaOH reacts with 1mole of HCl
20cm3 of HCl contains 0.002moles
1000cm3 of HCl contains
0.002 1000
20
moles
= 0.1M
202. © Kidegalize
Paper 1 method
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
Moles of acid
Moles of base
=
Conc of acid volume of acid
Conc of base volume of base
1
1
=
Ca 20
0.08 25
Ca = 0.1
203. © Kidegalize
Example 2(UNEB 1989/P1/13)
25.0cm3 of 0.1M sodium carbonate was
found to require 23.5cm3 of hydrochloric acid
to be completely neutralised. The molarity of
hydrochloric acid is
A.
23.5 0.1
2 25.0
B.
23.5 0.1 2
25.0
C.
2 25.0
23.5 0.1
D.
25 0.1 2
23.5
204. © Kidegalize
Possible Solution
Na2CO3(aq) + 2HCl(aq) NaCl(aq) + CO2(g) + H2O(l)
Moles of acid
Moles of base
=
Conc of acid volume of acid
Conc of base volume of base
2
1
=
Ca 23.5
0.1 25
Ca =
25 0.1 2
23.5
Correct Option is D
205. © Kidegalize
Knowledge Check 9
Qn.1(UNEB 1998/P1/4)
20cm3 of 0.2M HCl reacts with 25cm3
of sodium hydroxide solution. The
molarity of the hydroxide is
A.
25 0.2
20
B. 20 0.2
25
C. 25
20 0.2
D.
20
25 0.2
206. © Kidegalize
Qn.2(UNEB 1999/P1/30)
What is the molarity of sodium hydroxide
solution if 30cm3 of 0.2M hydrochloric acid
just neutralizes 20cm3 of the alkali?
A.
20
30 0.2
B.
20 0.2
30
C.
30
20 0.2
D.
30 0.2
20
207. © Kidegalize
Qn.3(UNEB 2003/P1/9)
Hydrochloric acid reacts with sodium
hydroxide according to the equation:
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
25.0cm3 of 0.10 M hydrochloric acid
reacted completely with 20cm3 of sodium
hydroxide. What is the molarity of sodium
hydroxide?
209. © Kidegalize
Qn.4(UNEB 2003/P1/26)
20cm3 of an acid HX neutralized 25cm3 of
0.05M sodium carbonate solution.
The molarity of the acid is
A.
25 0.05
20
B.
2 25 0.05
20
C.
2 20 0.05
25
D.
20 2
25 0.05
210. © Kidegalize
Qn.5(UNEB 2007/P1/26)
10cm3 of sulphuric acid reacted
completely with 25cm3 of 0.1M sodium
hydroxide solution. The molarity of the
sulphuric acid is
A. 0.125M B. 0.250M
C. 0.500M D. 1.000M
212. © Kidegalize
Qn.7(UNEB 2018/P1/22)
Which one of the following is the molarity
of a 25.0cm3 sodium carbonate solution
required to neutralize 20.0cm3 of a
0.15M dibasic acid?
A. 0.060M B. 0.120M
C. 0.188M D. 0.240M
213. © Kidegalize
Basicity of an acid
Basicity of an acid is the number of moles
of a base that reacts with one mole of the
given acid
OR the number of hydrogen ions produced
by one mole of the acid e.g. basicity
of hydrochloric acid is one [monobasic]
and that of sulphuric acid is two [dibasic].
214. © Kidegalize
To determine the basicity of a given acid,
calculate the mole ratio for the reaction between
the base and the acid by following the path:
Step 1: Calculate the number of moles of the acid
and the base that reacted.
Step 2: Determine the mole ratio by diving each
number of moles in step 1 by the smallest of the
two giving your answer in the form:
moles of acid : moles of base
215. © Kidegalize
Example 1
20cm3 of 0.2M sodium hydroxide
solution required 20cm3 of 0.1M
solution of an acid HnX. Determine the
basicity of the acid, HnX.
217. © Kidegalize
1000cm3 of HnX contains 0.1moles
20cm3 of HnX contains
0.1 20
1000
moles
= 0.002moles
HnX : NaOH = 0.002 : 0.004
Simplest ratio is HnX : NaOH = 1 : 2
Basicity of HnX is 2 or n = 2
218. © Kidegalize
Example 2(UNEB 2010/P1/15)
25cm3 of 0.12 M sodium hydroxide was
neutralized by 30.0cm3 of a solution of
a dibasic acid. The molarity of the acid is
A. 0.05M B. 0.06M
C. 0.01M D. 0.12M
219. © Kidegalize
Possible Solution
Moles of acid
Moles of base
=
Conc of acid volume of acid
Conc of base volume of base
1
2
=
Ca 30.0
0.12 25
Ca =
25 0.12
30.0 2
= 0.05
Correct Option is A
220. © Kidegalize
Knowledge Check 10
Qn.1(UNEB 2001/P1/26)
25cm3 of a 0.25M acid required 25cm3
of 0.5M sodium hydroxide solution for
neutralisation. The basicity of the acid is
A. 1 B. 2
C. 3 D. 4
222. © Kidegalize
Qn.3(UNEB 1993/P1/6)
10cm3 of a dibasic acid was neutralised by
20cm3 of a 0.2M sodium hydroxide. The
molarity of the acid is
A.
2 10
0.2 20
B.
0.2 20
2 10
C.
0.2 10
2 20
D.
2 0.2 20
10
223. © Kidegalize
Qn.4(UNEB 2006/P1/30)
15cm3 of a dibasic acid was neutralization by
30cm3 of a 0.4M potassium hydroxide
solution. The morality of the acid is?
A.
2 15
0.4 30
M B.
0.4 30
2 15
M
C.
0.4 15
2 30
M D.
2 0.4 30
15
M
224. © Kidegalize
Titration
Titration is a technique used to determine
the volume of one solution required to
react exactly with a known volume of
another solution.
Titrations frequently involve the reactions
of acids with bases.
225. © Kidegalize
Titrations involve three major components:
• A burette containing a solution of one of
the reagents.
• A flask containing an accurately known
volume of the other reagent solution
added using a pipette.
226. © Kidegalize
• An indicator (added to the contents of
the flask) that gives a visual indication
of when the reaction is complete.
Thus the change in colour of the
indicator marks the end point of the
titration.
Common acid-base indicators include:
227. © Kidegalize
Phenolphthalein is suitable for strong
acid–strong base and weak acid–strong
base whereas methyl orange is suitable
for strong acid–strong base and strong
acid–weak base.
Indicator Colour in acid Colour in base
Methyl orange Red Yellow
Phenolphthalein Colourless Pink
228. © Kidegalize
Practical examination 545/3-4
You will be required to determine:
• Concentration of a solution or
• RFM of a compound or
• Percentage purity or
• Mole ratio of a reaction or
• Basicity of an acid or
229. © Kidegalize
• RAM of an element or
• Formula of a compound or
• Number of moles of water of
crystallization.
For details of each see UCE LABORATORY
PACKAGE chapter 1
230. © Kidegalize
Table of results
• Volume of pipette: 1 dp
• Volume of burette: 2 dps with the last
digit being zero.
• Level of consistency: 0.1cm3
• Pick the most consistent values.
231. © Kidegalize
Sample table of results
Volume of pipette used ……………cm3
Values used for calculating average
volume ……………………………………...
Average volume of BA1 used ………………
Final burette reading(cm3) 26.00 27.20 25.20
Initial burette reading(cm3) 0.00 2.10 0.00
Volume of BA1 used(cm3) 26.00 25.10 25.20
25.0
25.10, 25.20
25.10 + 25.20
2
=25.15
232. © Kidegalize
Treatment of results
• Perform all calculations from first
principles (Paper 2/3 method).
• State mole ratio from balanced
chemical equation.
• Do not write units on titre values.
• Do not use Pencil to fill values.
233. © Kidegalize
Example (UNEB 2019/P3/1)
You are provided with the following:
BA1, which is a solution made by
dissolving 3.45g of a hydrated salt
X.nH2O in 250cm3 of water.
BA2, which is a 0.1M hydrochloric acid.
234. © Kidegalize
You are required to determine the value
of n in the salt.
Procedure:
Pipette 25cm3 (or 20cm3) of BA1 into
a conical flask. Add 2-3 drops of methyl
orange indicator and titrate with BA2
from the burette.
236. © Kidegalize
Volume of pipette used ……………cm3
Titre values of BA2 used for average
……………………………………...............
Average volume of BA2 used ………………
Final burette reading(cm3)
Initial burette reading(cm3)
Volume of BA2 used(cm3)
237. © Kidegalize
Questions
(a) Calculate the
(i) number of moles of hydrochloric
acid that reacted.
(ii) number of moles of X.nH2O that
reacted. (1mole of X.nH2O reacts
with 2moles of hydrochloric acid)
239. © Kidegalize
Possible Solution
Volume of pipette used ……………cm3
Titre values of BA2 used for average
……………………………………...
Average volume of BA2 used ………………
Final burette reading(cm3) 29.10 28.00 29.60
Initial burette reading(cm3) 1.50 0.50 2.10
Volume of BA2 used(cm3) 27.60 27.50 27.50
25.0
27.50 and 27.50
27.50 + 27.50
2
= 27.50
240. © Kidegalize
(a) (i) 1000cm3 of HCl contains 0.1mole
27.5cm3 of NaOH contains
0.1 27.5
1000
moles
= 0.00275moles
(ii) 2moles of HCl reacts with 1mole of X.nH2O
Moles of X.nH2O that reacted =
1
2
0.00275
= 0.001375
241. © Kidegalize
(iii) 25cm3 of BA1 contains 0.001375moles of X.nH2O
250cm3 of BA1 contains
1000 0.001375
25
moles
= 0.01375moles
(b) 0.01375moles of X.nH2O weighs 3.45g
1mole of X.nH2O weighs
1 3.45
0.01375
g
= 250.91g
242. © Kidegalize
Mass of nH2O = 250.91 – 106
= 144.91
RMM of H2O = (2 × 1) + 16 = 18
18n = 144.91
18n
18
=
144.91
18
n = 8
