Chemical Equations and Stoichiometry
description
Transcript of Chemical Equations and Stoichiometry
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New Way Chemistry for Hong Kong A-Level Book 11
Chemical Equations Chemical Equations
and Stoichiometryand Stoichiometry3.13.1 Formulae of CompoundsFormulae of Compounds
3.23.2 Derivation of Empirical FormulaeDerivation of Empirical Formulae
3.33.3 Derivation of Molecular FormulaeDerivation of Molecular Formulae
3.43.4 Chemical EquationsChemical Equations
3.53.5 Calculations Based on Chemical EquationsCalculations Based on Chemical Equations
3.63.6 Simple TitrationsSimple Titrations
33
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New Way Chemistry for Hong Kong A-Level Book 12
3.3.11 Formulae of Formulae of
CompoundsCompounds
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New Way Chemistry for Hong Kong A-Level Book 13
Formulae of compoundsFormulae of compounds
How can you describe the composition of compound X?
1st way = by chemical formula1st way = by chemical formula
C?H?
ratio of no. of atoms
3.1 Formulae of compounds (SB p.43)
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New Way Chemistry for Hong Kong A-Level Book 14
How can you describe the composition of compound X?
Compound X
2nd way = by percentage by mass
2nd way = by percentage by mass
Mass of carbon atoms inside = …. g
Mass of hydrogen atoms inside = …. g
carbon atomshydrogen atoms
3.1 Formulae of compounds (SB p.43)
Check Point 3-1Check Point 3-1
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New Way Chemistry for Hong Kong A-Level Book 15
3.1 Formulae of compounds (SB p.44)
Compound Empirical formula
Molecular formula
Structural formula
Carbon dioxide
CO2 CO2 O = C =O
Water H2O H2O
Methane CH4 CH4
Glucose CH2O C6H12O6
Sodium fluoride
NaF Not applicable
Na+F-
Different types of formulae of some Different types of formulae of some compoundscompounds
O
H H
C
H
H
HH
O
OH
HH
H
OH
OH
H OH
H
OH
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New Way Chemistry for Hong Kong A-Level Book 16
3.3.22 Derivation of Derivation of
Empirical Empirical FormulaeFormulae
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New Way Chemistry for Hong Kong A-Level Book 17
From combustion From combustion datadata• During complete combustion, elements in a co
mpound are oxidized.
• e.g. carbon to carbon dioxide, hydrogen to water, sulphur to sulphur dioxide
• From the masses of the products formed, the number of moles of these atoms originally present can be found
3.2 Derivation of empirical formulae (SB p.45)
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New Way Chemistry for Hong Kong A-Level Book 18
3.2 Derivation of empirical formulae (SB p.45)
The laboratory set-up used for determining the empirical formula of a gaseous hydrocarbon
Example 3-2AExample 3-2A Example 3-2BExample 3-2B Check Point 3-2ACheck Point 3-2A
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New Way Chemistry for Hong Kong A-Level Book 19
Composition by mass
Empirical formula
3.2 Derivation of Empirical Formulae (SB p.48)
From combustion by massFrom combustion by mass
Example 3-2CExample 3-2C Check Point 3-2BCheck Point 3-2BExample 3-2DExample 3-2D
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New Way Chemistry for Hong Kong A-Level Book 110
3.3.33 Derivation of Derivation of
Molecular Molecular FormulaeFormulae
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New Way Chemistry for Hong Kong A-Level Book 111
What is molecular What is molecular formulae?formulae?
Molecular formula
= (Empirical formula)n
?
3.3 Derivation of molecular formulae (SB p.49)
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New Way Chemistry for Hong Kong A-Level Book 112
Empirical formula Molecular mass
Molecular formula
3.3 Derivation of Molecular Formulae (SB p.49)
From empirical formula and known From empirical formula and known relative molecular massrelative molecular mass
Example 3-3AExample 3-3A
Example 3-3BExample 3-3B
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New Way Chemistry for Hong Kong A-Level Book 113
3.3 Derivation of Molecular Formulae (SB p.51)
Water of Crystallization Derived from Water of Crystallization Derived from Composition by MassComposition by Mass
Hydrated salt Anhydrous salt
CuSO45H2O
(Blue crystals)
Anhydrous CuSO4
(White powder)
Na2CO310H2O
(Colourless crystals)
Anhydrous Na2CO3
(White powder)
CoCl2 2H2O
(Pink crystals)
Anhydrous CoCl2(Blue crystals)
Example 3-3CExample 3-3C
Check Point 3-3ACheck Point 3-3A
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New Way Chemistry for Hong Kong A-Level Book 114
Formula of a compound
Composition by mass
3.3 Derivation of Molecular Formulae (SB p.52)
Find composition by mass from Find composition by mass from formulaformula
Example 3-3DExample 3-3D
Example 3-3EExample 3-3E
Check Point 3-3BCheck Point 3-3B
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New Way Chemistry for Hong Kong A-Level Book 115
3.3.44 Chemical Chemical
EquationsEquations
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New Way Chemistry for Hong Kong A-Level Book 116
Chemical equationsChemical equations
a A + b B c C + d D
mole ratios
(can also be volume ratios for gases)
Stoichiometry
= relative no. of moles of substances involved
in a chemical reaction
3.4 Chemical equations (SB p.53)
Check Point 3-4Check Point 3-4
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New Way Chemistry for Hong Kong A-Level Book 117
3.3.55 Calculations Calculations
Based on Based on Chemical Chemical EquationsEquations
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New Way Chemistry for Hong Kong A-Level Book 118
Calculations based on equationsCalculations based on equations
3.5 Calculations Based on Equations (SB p.65)
Calculations involving reacting masses
Example 3-5AExample 3-5A Example 3-5BExample 3-5B
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New Way Chemistry for Hong Kong A-Level Book 119
Calculations based on equationsCalculations based on equations
3.5 Calculations Based on Equations (SB p.66)
Calculations involving volumes of gases
Example 3-5CExample 3-5C
Example 3-5DExample 3-5D
Check Point 3-5Check Point 3-5
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New Way Chemistry for Hong Kong A-Level Book 120
3.3.66 Simple TitratiSimple Titrati
onsons
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Simple titrationsSimple titrations
Acid-base titrationsAcid-base titrations
Acid-base titrationswith indicators
Acid-base titrationswith indicators
Acid-base titrationswithout indicators
Acid-base titrationswithout indicators
(to be discussed in later chapters)
3.6 Simple titrations (SB p.58)
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New Way Chemistry for Hong Kong A-Level Book 122
Finding the concentration of a solutionFinding the concentration of a solution
+solute
solvent
solution
Copper(II) sulphate
Water
Copper(II) sulphate solution
3.6 Simple titrations (SB p.59)
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New Way Chemistry for Hong Kong A-Level Book 123
Finding the concentration of a solutionFinding the concentration of a solution
~50 cm3
3.6 Simple titrations (SB p.59)
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New Way Chemistry for Hong Kong A-Level Book 124
~50 cm3
3.6 Simple titrations (SB p.59)
Finding the concentration of a solutionFinding the concentration of a solution
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~50 cm3
3.6 Simple titrations (SB p.59)
Finding the concentration of a solutionFinding the concentration of a solution
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~50 cm3
3.6 Simple titrations (SB p.59)
Finding the concentration of a solutionFinding the concentration of a solution
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~50 cm3
3.6 Simple titrations (SB p.59)
Finding the concentration of a solutionFinding the concentration of a solution
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New Way Chemistry for Hong Kong A-Level Book 128
~50 cm3
3.6 Simple titrations (SB p.59)
Finding the concentration of a solutionFinding the concentration of a solution
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New Way Chemistry for Hong Kong A-Level Book 129
50 cm3
Solution A
3.6 Simple titrations (SB p.59)
Finding the concentration of a solutionFinding the concentration of a solution
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New Way Chemistry for Hong Kong A-Level Book 130
~50 cm3
3.6 Simple titrations (SB p.59)
Finding the concentration of a solutionFinding the concentration of a solution
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New Way Chemistry for Hong Kong A-Level Book 131
~50 cm3
3.6 Simple titrations (SB p.59)
Finding the concentration of a solutionFinding the concentration of a solution
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New Way Chemistry for Hong Kong A-Level Book 132
~50 cm3
3.6 Simple titrations (SB p.59)
Finding the concentration of a solutionFinding the concentration of a solution
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New Way Chemistry for Hong Kong A-Level Book 133
~50 cm3
3.6 Simple titrations (SB p.59)
Finding the concentration of a solutionFinding the concentration of a solution
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New Way Chemistry for Hong Kong A-Level Book 134
~50 cm3
3.6 Simple titrations (SB p.59)
Finding the concentration of a Finding the concentration of a solutionsolution
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New Way Chemistry for Hong Kong A-Level Book 135
~50 cm3
3.6 Simple titrations (SB p.59)
Finding the concentration of a solutionFinding the concentration of a solution
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New Way Chemistry for Hong Kong A-Level Book 136
50 cm3
Solution B
3.6 Simple titrations (SB p.59)
Finding the concentration of a solutionFinding the concentration of a solution
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New Way Chemistry for Hong Kong A-Level Book 137
~100 cm3
3.6 Simple titrations (SB p.59)
Finding the concentration of a solutionFinding the concentration of a solution
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New Way Chemistry for Hong Kong A-Level Book 138
~100 cm3
3.6 Simple titrations (SB p.59)
Finding the concentration of a solutionFinding the concentration of a solution
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New Way Chemistry for Hong Kong A-Level Book 139
~100 cm3
3.6 Simple titrations (SB p.59)
Finding the concentration of a solutionFinding the concentration of a solution
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New Way Chemistry for Hong Kong A-Level Book 140
~100 cm3
3.6 Simple titrations (SB p.69)
Finding the concentration of a solutionFinding the concentration of a solution
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New Way Chemistry for Hong Kong A-Level Book 141
Finding the concentration of a solutionFinding the concentration of a solution
~100 cm3
3.6 Simple titrations (SB p.59)
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New Way Chemistry for Hong Kong A-Level Book 142
Finding the concentration of a solutionFinding the concentration of a solution
~100 cm3
3.6 Simple titrations (SB p.59)
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New Way Chemistry for Hong Kong A-Level Book 143
Finding the concentration of a solutionFinding the concentration of a solution
Solution C
100 cm3
3.6 Simple titrations (SB p.59)
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New Way Chemistry for Hong Kong A-Level Book 144
Comment on the concentrations of solutions Comment on the concentrations of solutions A, B and C !A, B and C !
contain the same amount of solute (same concentration)
2 x the amount of solute
Concentration of solution B is 2 times that of the concentrations of solutions A & B.
3.6 Simple titrations (SB p.59)
Concentration is the amount of solute in a unit volume of solution.
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New Way Chemistry for Hong Kong A-Level Book 145
Comment on the concentrations of Comment on the concentrations of solutions solutions A, B and C !A, B and C !
Concentration is the amount of solute in a unit volume of solution.
no. of spoons mass
no. of moles
3.6 Simple titrations (SB p.59)
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New Way Chemistry for Hong Kong A-Level Book 146
MolarityMolarity
Molarity is the number of moles of solute dissolved in 1 dm3 (1000 cm3) of solution.
A way of expressing concentrations
)3dm (in solution of volumesolute of moles of number Molarity
Unit: mol dm-3 (M)
3.6 Simple titrations (SB p.59)
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New Way Chemistry for Hong Kong A-Level Book 147
What does this mean?
1 dm3
contains 2 moles of HCl
“In every 1 dm3 of the solution, 2 moles of HCl is dissolved.”
3.6 Simple titrations (SB p.59)
Example 3-6AExample 3-6A Example 3-6BExample 3-6B
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New Way Chemistry for Hong Kong A-Level Book 148
Titration without an indicatorTitration without an indicator
3.6 Simple titrations (SB p.62)
By change in pH value
Example 3-6CExample 3-6C
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New Way Chemistry for Hong Kong A-Level Book 149
Titration without an indicatorTitration without an indicator
3.6 Simple titrations (SB p.62)
By change in temperature
Example 3-6DExample 3-6D
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New Way Chemistry for Hong Kong A-Level Book 150
1. Iodometric titration
I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6
2-(aq)brown colourless
in conical flask
in burette
3.6 Simple Titrations (SB p.65)
Redox titrationsRedox titrations
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New Way Chemistry for Hong Kong A-Level Book 151
1. Iodometric titration
I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6
2-(aq)
brown colourless
Add starch
During titration : brown yellow
in conical flask in burette
3.6 Simple Titrations (SB p.65)
Redox titrationsRedox titrations
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New Way Chemistry for Hong Kong A-Level Book 152
1. Iodometric titration
I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6
2-(aq)
brown colourless
During titration : brown yellow
3.6 Simple Titrations (SB p.65)
Redox titrationsRedox titrations
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New Way Chemistry for Hong Kong A-Level Book 153
1. Iodometric titration
I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6
2-(aq)
brown colourless
End point : blue black colourless(after addition of starch indicator)
During titration : brown yellow
3.6 Simple Titrations (SB p.65)
Redox titrationsRedox titrationsExample 3-6EExample 3-6E
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New Way Chemistry for Hong Kong A-Level Book 154
2. Titrations involving potassium permanganate
MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + H2O(aq) + 5Fe3+(aq)
purple pale green
In burette
In conical flask
3.6 Simple Titrations (SB p.66)
Redox titrationsRedox titrations
Example 3-6FExample 3-6F Check Point 3-6Check Point 3-6
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The END
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New Way Chemistry for Hong Kong A-Level Book 156
Give the empirical, molecular and structural formulae for the following compounds:
(a) Propene
(b) Nitric acid
(c) Ethanol
(d) Glucose
Back
Answer
3.1 Formulae of compounds (SB p.45)
C6H12O6C6H12O6(d) Glucose
C2H5OHC2H6O(c) Ethanol
HNO3HNO3(b) Nitric
acid
C3H6CH2(a) Propene
Structural formula
Molecular formula
Empirical
formula
Compound
O N
O
H
O
C
H
H
CH
H
H
OH
O
OH
HH
H
OH
OH
H OH
H
OH
C
H
H
CH C
H H
H
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New Way Chemistry for Hong Kong A-Level Book 157
A hydrocarbon was burnt completely in excess oxygen. It was found that 1.00 g of the hydrocarbon gives 2.93 g of carbon dioxide and 1.80 g of water. Find the empirical formula of the hydrocarbon.
3.2 Derivation of empirical formulae (SB p.46)
Answer
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New Way Chemistry for Hong Kong A-Level Book 158
3.2 Derivation of empirical formulae (SB p.46)
The relative molecular mass of CO2 = 12.0 + 16.0 2 = 44.0
Mass of carbon in 2.93 g of CO2 = 2.93 g = 0.80 g
The relative molecular mass of H2O = 1.0 2 + 16.0 = 18.0
Mass of hydrogen in 1.80 g of H2O = 1.80 g = 0.20 g
Let the empirical formula of the hydrocarbon be CxHy.
Mass of carbon in CxHy = Mass of carbon in CO2
Mass of hydrogen in CxHy = Mass of hydrogen in H2O
0.440.12
0.180.2
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New Way Chemistry for Hong Kong A-Level Book 159
Therefore, the empirical formula of the hydrocarbon is CH3.
3.2 Derivation of empirical formulae (SB p.46)
Back
Carbon Hydrogen
Mass (g) 0.80 0.20
No. of moles (mol)
Relative no. of moles
Simplest mole ratio
1 3
0667.00.12
80.0 20.00.120.0
10667.00667.0 3
0667.020.0
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New Way Chemistry for Hong Kong A-Level Book 160
Compound X is known to contain carbon, hdyrogen and oxygen only. When it is burnt completely in excess oxygen, carbon dioxide and water are given out as the only products. It is found that 0.46 g of compound X gives 0.88 g of carbon dioxide and 0.54 g of water. Find the empirical formula of compound X.
3.2 Derivation of empirical formulae (SB p.46)
Answer
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New Way Chemistry for Hong Kong A-Level Book 161
3.2 Derivation of empirical formulae (SB p.47)
Mass of compound X = 0.46 g
Mass of carbon in compound X = 0.88 g = 0.24 g
Mass of hydrogen in compound X = 0.54 g = 0.06 g
Mass of oxygen in compound X = 0.46 g – 0.24 g – 0.06 g = 0.16 g
Let the empirical formula of compound X be CxHyOz.
0.440.12
0.180.2
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New Way Chemistry for Hong Kong A-Level Book 162
Therefore, the empirical formula of compound X is C2H6O.
3.2 Derivation of empirical formulae (SB p.47)
Back
0667.00.12
80.0 06.00.106.0
601.006.0 1
01.001.0
Carbon Hydrogen Oxygen
Mass (g) 0.24 0.06 0.16
No. of moles (mol)
Relative no. of moles
Simplest mole ratio
2 6 1
01.00.16
16.0
201.002.0
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New Way Chemistry for Hong Kong A-Level Book 163
(a)5 g of sulphur forms 10 g of an oxide on complete combustion. What is the empirical formula of the oxide?
3.2 Derivation of empirical formulae (SB p.47)
Answer(a) Mass of sulphur = 5 g
Mass of oxygen = (10 – 5) g = 5 g
21Simplest mole ratio
Relative no. of moles
No. of moles (mol)
55Mass (g)
OxygenSulphur
156.01.32
5 313.00.16
5
1156.0156.0 2
156.0313.0
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New Way Chemistry for Hong Kong A-Level Book 164
(b) 19.85 g of element M combines with 25.61 g of oxygen to form an oxide. If the relative atomic mass of M is 31.0, find the empirical formula of the oxide.
3.2 Derivation of empirical formulae (SB p.47)
Answer(b)
The empirical formula of the oxide is M2O5.
52Simplest mole ratio
Relative no. of moles
No. of moles (mol)
25.6119.85Mass (g)
OM
64.00.31
85.19 6.10.1661.25
164.064.0 5.2
64.06.1
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New Way Chemistry for Hong Kong A-Level Book 165
(c) Determine the empirical formula of copper(II) oxide using the following results.
Experimental results:
Mass of test tube = 21.430 g
Mass of test tube + Mass of copper(II) oxide = 23.321 g
Mass of test tube + Mass of copper = 22.940 g
3.2 Derivation of empirical formulae (SB p.47)
Answer
![Page 66: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/66.jpg)
New Way Chemistry for Hong Kong A-Level Book 166
3.2 Derivation of empirical formulae (SB p.47)
(c) Mass of Cu = (22.940 – 21.430) g = 1.51 g
Mass of O = (23.321 – 22.940) g = 0.381 g
Therefore, the empirical formula of copper(II) oxide is CuO.
0238.05.63
51.1
11Simplest mole ratio
Relative no. of moles
No. of moles (mol)
0.3811.51Mass (g)
OxygenCopper
0238.00.16
381.0
10238.00238.0 1
0238.00238.0
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New Way Chemistry for Hong Kong A-Level Book 167
Compound A contains carbon and hydrogen atoms only. It is found that the compound contains 75 % carbon by mass. Determine its empirical formula.
3.2 Derivation of empirical formulae (SB p.48)
Answer
![Page 68: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/68.jpg)
New Way Chemistry for Hong Kong A-Level Book 168
3.2 Derivation of empirical formulae (SB p.48)
Back
Let the empirical formula of compound A be CxHy, and the mass of the compound be 100 g. Then, mass of carbon in the compound = 75 g
Mass of hydrogen in the compound = (100 – 75) g = 25 g
Therefore, the empirical formula of compound A is CH4.
41Simplest mole ratio
Relative no. of moles
No. of moles (mol)
2575Mass (g)
HydrogenCarbon
25.60.12
75 250.1
25
125.625.6 4
25.625
![Page 69: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/69.jpg)
New Way Chemistry for Hong Kong A-Level Book 169
The percentages by mass of phosphorus and chlorine in a sample of phosphorus chloride are 22.55 % and 77.45 % respectively. Find the empirical formula of the phosphorus chloride.
3.2 Derivation of empirical formulae (SB p.48)
Answer
![Page 70: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/70.jpg)
New Way Chemistry for Hong Kong A-Level Book 170
3.2 Derivation of empirical formulae (SB p.48)
Back
Let the mass of phosphorus chloride be 100 g. Then,
Mass of phosphorus in the compound = 22.55 g
Mass of chlorine in the compound = 77.45 g
Therefore, the empirical formula of the phosphorus chloride is PCl3.
31Simplest mole ratio
Relative no. of moles
No. of moles (mol)
77.4522.55Mass (g)
ChlorinePhosphorus
727.00.3155.22 182.2
5.3545.77
1727.0727.0 3
727.0182.2
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New Way Chemistry for Hong Kong A-Level Book 171
(a)Find the empirical formula of vitamin C if it consists of 40.9 % carbon, 54.5 % oxygen and 4.6 % hydrogen by mass.
3.2 Derivation of empirical formulae (SB p.49)
Answer(a) Let the mass of vitamin C analyzed be 100 g.
The empirical formula of vitamin C is C3H4O3.
343Simplest mole ratio
Relative no. of moles
No. of moles (mol)
54.54.640.9Mass (g)
OxygenHydrogenCarbon
41.30.129.40 60.4
0.16.4 41.3
0.165.54
141.341.3 35.1
41.36.4 1
41.341.3
![Page 72: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/72.jpg)
New Way Chemistry for Hong Kong A-Level Book 172
(b) Each 325 mg tablet of aspirin consists of 195.0 mg carbon, 14.6 mg hydrogen and 115.4 mg oxygen. Determine the empirical formula of aspirin.
3.2 Derivation of empirical formulae (SB p.49)
Answer(b) The masses of the elements are multiplied by 1000 first.
The empirical formula of aspirin is C9H8O4.
489Simplest mole ratio
Relative no. of moles
No. of moles (mol)
115.414.6195.0Mass (g)
OxygenHydrogenCarbon
25.160.120.195 6.14
0.16.14 21.7
0.164.115
25.221.725.16 02.2
21.76.14 1
21.721.7
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New Way Chemistry for Hong Kong A-Level Book 173
A hydrocarbon was burnt completely in excess oxygen. It was found that 5.0 g of the hydrocarbon gave 14.6 g of carbon dioxide and 9.0 g of water. Given that the relative molecular mass of the hydrocarbon is 30.0, determine its molecular formula.
3.3 Derivation of molecular formulae (SB p.50)
Answer
![Page 74: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/74.jpg)
New Way Chemistry for Hong Kong A-Level Book 174
3.3 Derivation of molecular formulae (SB p.50)
Let the empirical formula of the hydrocarbon be CxHy.
Mass of carbon in the hydrocarbon = 14.6 g = 4.0 g
Mass of hydrogen in the hydrocarbon = 9.0 g = 1.0 g
31Simplest mole ratio
Relative no. of moles
No. of moles (mol)
1.04.0Mass (g)
HydrogenCarbon
333.00.120.4 1
0.10.1
1333.0333.0 3
333.01
0.440.12
0.180.2
![Page 75: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/75.jpg)
New Way Chemistry for Hong Kong A-Level Book 175
3.3 Derivation of molecular formulae (SB p.50)
Therefore, the empirical formula of the hydrocarbon is CH3.
Let the molecular formula of the hydrocarbon be (CH3)n.
Relative molecular mass of (CH3)n = 30.0
n (12.0 + 1.0 3) = 30.0
n = 2
Therefore, the molecular formula of the hydrocarbon is C2H6.
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![Page 76: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/76.jpg)
New Way Chemistry for Hong Kong A-Level Book 176
Compound X is known to contain 44.44 % carbon, 6.18 % hydrogen and 49.38 % oxygen by mass. A typical analysis shows that it has a relative molecular mass of 162.0. Find its molecular formula.
3.3 Derivation of molecular formulae (SB p.50)
Answer
![Page 77: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/77.jpg)
New Way Chemistry for Hong Kong A-Level Book 177
3.3 Derivation of molecular formulae (SB p.50)
Let the empirical formula of compound X is CxHyOz, and the mass of the compound be 100 g. Then,
Mass of carbon in the compound = 44.44 g
Mass of hydrogen in the compound = 6.18 g
Mass of oxygen in the compound = 49.38 g
The empirical formula of compound X is C6H10O5.
5106Simplest mole ratio
Relative no. of moles
No. of moles (mol)
49.386.1844.44Mass (g)
OxygenHydrogenCarbon
70.30.1244.44 18.6
0.118.6 09.3
0.1638.49
2.109.370.3 2
09.318.6 1
09.309.3
![Page 78: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/78.jpg)
New Way Chemistry for Hong Kong A-Level Book 178
3.3 Derivation of molecular formulae (SB p.50)
Let the molecular formula of compound X be (C6H10O5)n.
Relative molecular mass of (C6H10O5)n = 162.0
n (12.0 6 + 1.0 10 + 16.0 5) = 162.0
n = 1
Therefore, the molecular formula of compound X is C6H10O5.
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![Page 79: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/79.jpg)
New Way Chemistry for Hong Kong A-Level Book 179
The chemical formula of hydrated copper(II) sulphate is known to be CuSO4 · xH2O. It is found that the percentage of water of crystallization by mass in the compound is 36 %. Find the value of x.
3.3 Derivation of molecular formulae (SB p.51)
Answer
![Page 80: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/80.jpg)
New Way Chemistry for Hong Kong A-Level Book 180
3.3 Derivation of molecular formulae (SB p.51)
Relative formula mass of CuSO4 · xH2O
= 63.5 + 32.1 + 16.0 4 + (1.0 2 + 16.0)x
= 159.6 + 18x
Relative molecular mass of water of crystallization = 18x
1800x = 5745.6 + 648x
1152x = 5745.6
x = 4.99
5
Therefore, the chemical formula of the hydrated copper(II) sulphate is CuSO4 · 5H2O.
10036
x186.159x18
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![Page 81: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/81.jpg)
New Way Chemistry for Hong Kong A-Level Book 181
(a)Compound Z is the major ingredient of a healthy drink. It contains 40.00 % carbon, 6.67 % hydrogen and 53.33 % oxygen.
(i) Find the empirical formula of compound Z.
(ii) If the relative molecular mass of compound Z is 180, find its molecular formula.
Answer
3.3 Derivation of molecular formulae (SB p.52)
![Page 82: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/82.jpg)
New Way Chemistry for Hong Kong A-Level Book 182
(a) (i) Let the mass of compound Z be 100 g.
Therefore, the empirical formula of compound Z is CH2O.
3.3 Derivation of molecular formulae (SB p.52)
Carbon Hydrogen Oxygen
Mass (g) 40.00 6.67 53.33
No. of moles (mol)
Relative no. of moles
Simplest mole ratio
1 2 1
33.30.1200.40 67.6
0.167.6 33.3
0.1633.53
133.333.3 2
33.367.6 1
33.333.3
![Page 83: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/83.jpg)
New Way Chemistry for Hong Kong A-Level Book 183
(ii)Let the empirical formula of compound Z be (CH2O)n.
n (12.0 + 1.0 2 + 16.0) = 180
30n = 180
n = 6
Therefore, the molecular formula of compound Z is C6H12O6.
3.3 Derivation of molecular formulae (SB p.52)
![Page 84: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/84.jpg)
New Way Chemistry for Hong Kong A-Level Book 184
(b)(NH4)2Sx contains 72.72 % sulphur by mass. Find the value of x.
Answer
3.3 Derivation of molecular formulae (SB p.52)
(b)
Since the chemical formula of (NH4)Sx is (NH4)2S3, the value of x is 3.
32Simplest mole ratio
Relative no. of moles
No. of moles (mol)
72.7227.28Mass (g)
S(NH4) unit
52.10.1828.27 27.2
1.3272.72
152.152.1 49.1
52.127.2
![Page 85: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/85.jpg)
New Way Chemistry for Hong Kong A-Level Book 185
(c) In the compound MgSO4 · nH2O, 51.22 % by mass is water. Find the value of n.
Answer
3.3 Derivation of molecular formulae (SB p.52)
(c)
Since the chemical formula of MgSO4 · nH2O is MgSO4 · 7H2O, the value of n is 7.
71Simplest mole ratio
Relative no. of moles
No. of moles (mol)
51.2248.78Mass (g)
H2OMgSO4
405.04.120
78.48 846.20.1822.51
1405.0405.0 7
405.0846.2
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![Page 86: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/86.jpg)
New Way Chemistry for Hong Kong A-Level Book 186
The chemical formula of ethanoic acid is CH3COOH. Calculate the percentage of mass of carbon, hydrogen and oxygen respectively.
3.3 Derivation of molecular formulae (SB p.52)
Answer
![Page 87: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/87.jpg)
New Way Chemistry for Hong Kong A-Level Book 187
3.3 Derivation of molecular formulae (SB p.52)
Relative molecular mass of CH3COOH
= 12.0 2 + 1.0 4 + 16.0 2
= 60.0
% by mass of C =
= 40.00 %
% by mass of H =
= 6.67 %
% by mass of O =
= 53.33 %
The percentage by mass of carbon, hydrogen and oxygen are 40.00 %, 6.67 % and 53.33 % respectively.
%1000.60
20.12
%1000.6040.1
%1000.60
20.16
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New Way Chemistry for Hong Kong A-Level Book 188
Calculate the mass of iron in a sample of 20 g of hydrated iron(II) sulphate, FeSO4 · 7H2O.
3.3 Derivation of molecular formulae (SB p.53)
AnswerRelative formula mass of FeSO4 · 7H2O
= 55.8 + 32.1 + 16.0 4 + (1.0 2 + 16.0) 7 = 277.9
% by mass of Fe =
= 20.08 %
Mass of Fe = 20 g 20.08 % = 4.02 g
%1009.277
8.55
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![Page 89: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/89.jpg)
New Way Chemistry for Hong Kong A-Level Book 189
(a)Calculate the percentages by mass of potassium , chromium and oxygen in potassium chromate(VI), K2Cr2O7.
Answer
3.3 Derivation of molecular formulae (SB p.53)
(a) Molar mass of K2Cr2O7 = (39.1 2 + 52.0 2 + 16.0 7) g mol-1
= 294.2 g mol-1
% by mass of K =
= 26.58 %
% by mass of Cr =
= 35.35 %
% by mass of O =
= 38.07 %
%100gmol2.294
gmol)21.39(1
1
%100gmol2.294
gmol)20.52(1
1
%100gmol2.294
gmol)70.16(1
1
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New Way Chemistry for Hong Kong A-Level Book 190
(b)Find the mass of metal and water of crystallization in
(i) 100 g of Na2SO4 · 10H2O
(ii) 70 g of Fe2O3 · 8H2OAnswer
3.3 Derivation of molecular formulae (SB p.53)
![Page 91: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/91.jpg)
New Way Chemistry for Hong Kong A-Level Book 191
3.3 Derivation of molecular formulae (SB p.53)
(b) (i) Molar mass of Na2SO4 · 10H2O = 322.1 g mol-1
Mass of Na =
= 14.28 g
Mass of H2O =
= 55.88 g
(ii) Molar mass of Fe2O3 · 8H2O = 303.6 g mol-1
Mass of Fe =
= 25.73 g
Mass of H2O =
= 33.20 g
g100gmol1.322
gmol)20.23(1
1
g100gmol1.322
gmol)100.18(1
1
g70gmol6.303
gmol)28.55(1
1
g70gmol6.303
gmol)80.18(1
1
Back
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New Way Chemistry for Hong Kong A-Level Book 192
Give the chemical equations for the following reactions:
• Zinc + steam zinc oxide + hydrogen
(b) Magnesium + silver nitrate silver + magnesium nitrate
(c) Butane + oxygen carbon dioxide + water
Answer
3.4 Chemical equations (SB p.54)
(a) Zn(s) + H2O(g) ZnO(s) + H2(g)
(b) Mg(s) + 2AgNO3(aq) 2Ag(s) + Mg(NO3)2(aq)
(c) 2C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(l)
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New Way Chemistry for Hong Kong A-Level Book 193
Calculate the mass of copper formed when 12.45 g of copper(II) oxide is completely reduced by hydrogen.
3.5 Calculations based on chemical equations (SB p.55)
Answer
![Page 94: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/94.jpg)
New Way Chemistry for Hong Kong A-Level Book 194
3.5 Calculations based on chemical equations (SB p.55)
CuO(s) + H2(g) Cu(s) + H2O(l)
As the mole ratio of CuO : Cu is 1 : 1, the number of moles of Cu formed is the same as the number of moles of CuO reduced.
Number of moles of CuO reduced =
= 0.157 mol
Number of moles of Cu formed = 0.157 mol
= 0.157 mol
Mass of Cu = 0.157 mol 63.5 g mol-1 = 9.97 g
Therefore, the mass of copper formed in the reaction is 9.97 g.
1gmol)0.165.63(g45.12
1mol g 63.5Cu of Mass
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New Way Chemistry for Hong Kong A-Level Book 195
Sodium hydrogencarbonate decomposes according to the following chemial equation:
2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(l)
In order to obtain 240 cm3 of CO2 at room temperature and pressure, what is the minimum amount of sodium hydrogencarbonate required?
(Molar volume of gas at R.T.P. = 24.0 dm3 mol-1)
3.5 Calculations based on chemical equations (SB p.55)
Answer
![Page 96: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/96.jpg)
New Way Chemistry for Hong Kong A-Level Book 196
3.5 Calculations based on chemical equations (SB p.55)
Number of moles of CO2 required
= = 0.01 mol
From the chemical equation, 2 moles of NaHCO3(s) give 1 mole of CO2(g).
Number of moles of NaHCO3 required = 0.01 2 = 0.02 mol
Mass of NaHCO3 required
= 0.02 mol (23.0 + 1.0 + 12.0 + 16.0 3) g mol-1
= 0.02 mol 84.0 g mol-1
= 1.68 g
Therefore, the minimum amount of sodium hydrogencarbonate required is 1.68 g.
13
3
mol cm 24000cm 240
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New Way Chemistry for Hong Kong A-Level Book 197
Calculate the volume of carbon dioxide formed when 20 cm3 of ethane and 70 cm3 of oxygen are exploded, assuming all volumes of gases are measured at room temperature and pressure.
3.5 Calculations based on chemical equations (SB p.56)
Answer
![Page 98: Chemical Equations and Stoichiometry](https://reader034.fdocuments.us/reader034/viewer/2022051402/568157bc550346895dc54087/html5/thumbnails/98.jpg)
New Way Chemistry for Hong Kong A-Level Book 198
3.5 Calculations based on chemical equations (SB p.56)
2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l)
2 mol : 7 mol : 4 mol : 6 mol (from equation)
2 volumes : 7 volumes : 4 volumes : - (by Avogadro’s law)
It can be judged from the chemical equation that the mole ratio of CO2 : C
2H6 is 4 : 2, and the volume ratio of CO2 : C2H6 should also be 4 : 2 according to the Avogadro’s law.
Let x be the volume of CO2(g) formed.
Number of moles of CO2(g) formed : number of moles of C2H6(g) used = 4 : 2
Volume of CO2(g) : volume of C2H6(g) = 4 : 2
x : 20 cm3 = 4 : 2
x = 40 cm3
Therefore, the volume of CO2(g) formed is 40 cm3.
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New Way Chemistry for Hong Kong A-Level Book 199
10 cm3 of a gaseous hydrocarbon was mixed with 80 cm3 of oxygen which was in excess. The mixture was exploded and then cooled. The volume left was 70 cm3. Upon passing the resulting gaseous mixture through concentrated sodium hydroxide solution (to absorb carbon dioxide), the volume of the residual gas became 50 cm3. Find the molecular formula of the hydrocarbon.
3.5 Calculations based on chemical equations (SB p.57)
Answer
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New Way Chemistry for Hong Kong A-Level Book 1100
3.5 Calculations based on chemical equations (SB p.57)
Let the molecular formula of the hydrocarbon be CxHy.
Volume of hydrocarbon reacted = 10 cm3
Volume of O2(g) unreacted = 50 cm3 (the residual gas after reaction)
Volume of O2(g) reacted = (80 – 50) cm3 = 30 cm3
Volume of CO2(g) formed = (70 – 50) cm3 = 20 cm3
CxHy + O2 xCO2 + H2O
1 mol : mol : x mol
1 volume : volumes : x volumes
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New Way Chemistry for Hong Kong A-Level Book 1101
3.5 Calculations based on chemical equations (SB p.57)
=
=
x = 2
=
=
= 3
As x = 2, = 3
y = 4
Therefore, the molecular formula of the hydrocarbon is C2H4.
(g)HC of Volume
(g)CO of Volume
yx
2
1x
3
3
cm 10cm 20
1x
(g)HC of Volume
(g)O of Volume
yx
2
14y
x
3
3
cm 10cm 30
14y
x
4y
x
4y
2
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New Way Chemistry for Hong Kong A-Level Book 1102
(a)Find the volume of hydrogen produced at R.T.P. when 2.43 g of magnesium reacts with excess hydrochloric acid.
(Molar volume of gas at R.T.P. = 24.0 dm3 mol-1)Answer
3.5 Calculations based on chemical equations (SB p.58)
(a) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)
No. of moles of H2= No. of moles of Mg
=
Volume of H2 = 2.4 dm3
1-3
2
mol dm 24.0
H of Volume1-mol g 24.3
g 2.43
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New Way Chemistry for Hong Kong A-Level Book 1103
(b)Find the minimum mass of chlorine required to produce 100 g of phosphorus trichloride (PCl3).
Answer
3.5 Calculations based on chemical equations (SB p.58)
31
21
(b) 2P(s) + 3Cl2(g) 2PCl3(l)
No. of moles of Cl2= No. of moles of PCl3
=
Mass of Cl2 = 77.45 g
The minimum mass of chlorine required is 77.45 g.
1-
2
mol g 2) (35.5
Cl of Mass
31
1-mol g 3)35.5 1.03(
g 10021
31
21
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New Way Chemistry for Hong Kong A-Level Book 1104
(c) 20 cm3 of a gaseous hydrocarbon and 150 cm3 of oxygen (which was in excess) were exploded in a closed vessel. After cooling, 110 cm3 of gases remained. After passing the resulting gaseous mixture through concentrated sodium hydroxide solution, the volume of the residual gas became 50 cm3. Determine the molecular formula of the hydrocarbon.
Answer
3.5 Calculations based on chemical equations (SB p.58)
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New Way Chemistry for Hong Kong A-Level Book 1105
3.5 Calculations based on chemical equations (SB p.58)
(c) CxHy(g) + O2(g) xCO2(g) + H2O(l)
Volume of CxHy used = 20 cm3
Volume of CO2 formed = (110 – 50) cm3 = 60 cm3
Volume of O2 used = (150 – 50) cm3 = 100 cm3
Volume of CxHy : Volume of CO2 = 1 : x
= 20 : 60
x = 3
Volume of CxHy : Volume of O2 = 1 :
= 20 : 100
= 5
)4y
x( 2y
4y
x
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New Way Chemistry for Hong Kong A-Level Book 1106
3.5 Calculations based on chemical equations (SB p.58)
(c) As x = 3,
= 5
= 2
y = 8
The molecular formula of the hydrocarbon is C3H8.
4y
3
4y
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New Way Chemistry for Hong Kong A-Level Book 1107
(d)Calculate the volume of carbon dioxide formed when 5 cm3 of methane was burnt completely in excess oxygen, assuming all volumes of gases are measured at room temperature and pressure.
Answer
3.5 Calculations based on chemical equations (SB p.58)
(d) CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
1 mol : 2 mol : 1 mol : 2 mol (from equation)
1 volume : 2 volumes : 1 volume: - (from Avogadro’s law)
5 cm3 x cm3
It can be judged from the equation that the mole ratio of CO2 : CH4 is 1 : 1, the volume ratio of CO2 : CH4 should also be 1 : 1.
=
x = 5 cm3
The volume of CO2(g) formed is 5 cm3.
3cm 5x
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New Way Chemistry for Hong Kong A-Level Book 1108
25.0 cm3 of sodium hydroxide solution was titrated against 0.067 M sulphuric(VI) acid using methyl orange as an indicator. The indicator changed colour from yellow to red when 22.5 cm3 of sulphuric(VI) acid had been added. Calculate the molarity of the sodium hydroxide solution.
3.6 Simple titrations (SB p.61)
Answer
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New Way Chemistry for Hong Kong A-Level Book 1109
3.6 Simple titrations (SB p.61)
2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l)
=
Number of moles of NaOH(aq) = Number of moles of H2SO4(aq)
Number of moles of H2SO4(aq) = 0.067 mol dm-3 22.5 10-3 dm3
= 1.508 10-3 mol
Number of moles of NaOH(aq) = 2 1.508 10-3 mol
= 3.016 10-3 mol
Molarity of NaOH(aq) =
= 0.121 mol dm-3
Therefore, the molarity of the sodium hydroxide solution was 0.121 M.
(aq)SOH of moles of NumberNaOH(aq) of moles of Number
42 12
21
33-
-3
dm 10 25.0mol 103.016
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New Way Chemistry for Hong Kong A-Level Book 1110
2.52 g of a pure dibasic acid with formula mass of 126.0 was dissolved in water and made up to 250.0 cm3 in a volumetric flask. 25.0 cm3 of this solution was found to neutralize 28.5 cm3 of sodium hydroxide solution.
(a)Calculate the molarity of the acid solution.
3.6 Simple titrations (SB p.61)
Answer
(a) Number of moles of acid = = 0.02 mol
Molarity of acid solution = = 0.08 M
1-mol g 26.01g 2.52
33 dm 10250mol 0.02
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New Way Chemistry for Hong Kong A-Level Book 1111
(b) If the dibasic acid is represented by H2X, write an equation for the reaction between the acid and sodium hydroxide.
3.6 Simple titrations (SB p.61)
Answer
(b) H2X(aq) + 2NaOH(aq) Na2X(aq) + 2H2O(l)
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New Way Chemistry for Hong Kong A-Level Book 1112
(c) Calculate the molarity of the sodium hydroxide solution.
3.6 Simple titrations (SB p.61)
Answer
(c) Number of moles of H2X = Number of moles of NaOH
0.08 mol dm-3 25.0 10-3 dm3
= Molarity of NaOH 28.5 10-3 dm3
Molarity of NaOH = 0.14 M
Therefore, the molarity of the sodium hydroxide solution was 0.14 M.
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New Way Chemistry for Hong Kong A-Level Book 1113
0.186 g of a sample of hydrated sodium carbonate, Na2CO3 · nH2O, was dissolved in 100 cm3 of distilled water in a conical flask. 0.10 M hydrochloric acid was added from a burette, 2 cm3 at a time. The pH value of the reaction mixture was measured with a pH meter. The results were recorded and shown in the following figure. Calculate the value of n in Na2CO3 · nH2O.
3.6 Simple titrations (SB p.62)
Answer
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New Way Chemistry for Hong Kong A-Level Book 1114
3.6 Simple titrations (SB p.63)
There is a sudden drop in the pH value of the solution (from pH 8 to pH 3) with the equivalence point at 30.0 cm3.
Na2CO3 · nH2O(s) + 2HCl(aq) 2NaCl(aq) + CO2(g) + (n + 1)H2O(l)
Number of moles of Na2CO3 · nH2O = Number of moles of HCl
= 0.10 mol dm-3 30.0 10-3 dm3
106.0 + 18.0n = 124.0
n = 1
Therefore, the chemical formula of the hydrated sodium carbonate is Na2CO
3 · H2O.
21
211-mol g 18.0n) 316.0 12.0 2 (23.0
g 0.186
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New Way Chemistry for Hong Kong A-Level Book 1115
5 cm3 of 0.5 M sulphuric(VI) acid was added to 25.0 cm3 of potassium hydroxide solution. The mixture was then stirred and the highest temperature was recorded. The experiment was repeated with different volumes of the sulphuric(VI) acid. The laboratory set-up and the results were as follows:
3.6 Simple titrations (SB p.63)
Volume of H2SO4 added
(cm3)
Temperature (oC)
0 20.05 21.8
10 23.415 25.020 26.525 25.230 24.0
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New Way Chemistry for Hong Kong A-Level Book 1116
(a) Plot a graph of temperature against volume of sulphuric(VI) acid added.
3.6 Simple titrations (SB p.63)
Answer
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New Way Chemistry for Hong Kong A-Level Book 1117
(b) Calculate the molarity of the potassium hydroxide solution.
3.6 Simple titrations (SB p.63)
Answer(b) From the graph, it is found that the equivalence point of the titration is reached when 20 cm3 of H2SO4 is added.
Number of moles of H2SO4 = 0.5 mol dm-3 20 10-3 dm3
= 0.01 mol
2KOH(aq) + H2SO4(aq) K2SO4(aq) + 2H2O(l)
2 mol : 1 mol
From the equation,
mole ratio of KOH(aq) : H2SO4(aq) = 2 : 1
Number of moles of KOH(aq) = 2 0.01 mol = 0.02 mol
Molarity of KOH(aq) = = 0.8 M
Therefore, the molarity of potassium hydroxide solution was 0.8 M.
33 dm 1025.0mol 0.02
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New Way Chemistry for Hong Kong A-Level Book 1118
(c) Explain why the temperature rose to a maximum and then fell.
3.6 Simple titrations (SB p.63)
Answer(c) Neutralization is an exothermic reaction. When more and more sulphur
ic(VI) acid was added and reacted with potassium hydroxide, the temperature rose. The temperature rose to a maximum value at which the equivalence point of the reaction was reached. After that, any excess sulphuric(VI) acid added would cool down the reacting solution, causing the temperature to drop.
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New Way Chemistry for Hong Kong A-Level Book 1119
When excess potassium iodide solution (KI) is added to 25.0 cm3 of acidified potassium iodate solution (KIO3) of unknown concentration, the solution turns brown. This brown solution requires 22.0 cm3 of 0.05 M sodium thiosulphate solution to react completely with the iodine formed, using starch solution as an indicator. Find the molarity of the acidified potassium iodate solution.
3.6 Simple titrations (SB p.66)
Answer
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New Way Chemistry for Hong Kong A-Level Book 1120
3.6 Simple titrations (SB p.66)
IO3-(aq) + 5I-(aq) + 6H+(aq) 3I2(aq) + 3H2O(l) … … (1)
I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6
2-(aq) … … (2)
From (1), Number of moles of IO3-(aq) = Number of moles of I2(aq)
From (2), Number of moles of I2(aq) = Number of moles of S2O32-(aq)
Number of moles of IO3-(aq) = Number of moles of S2O3
2-(aq)
Molarity of IO3-(aq) 25.0 10-3 dm3 = 0.05 mol dm-3 22.0 10-3 d
m3
Molarity of IO3-(aq) = 7.33 10-3 M
Therefore, the molarity of the acidified potassium iodate solution is 7.33 10-3 M.
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New Way Chemistry for Hong Kong A-Level Book 1121
3.6 Simple titrations (SB p.67)
A piece of impure iron wire weighs 0.22 g. When it is dissolved in hydrochloric acid, it is oxidized to iron(II) ions. The solution requires 36.5 cm3 of 0.02 M acidified potassium manganate(VII) solution for complete reaction to form iron(III) ions. What is the percentage purity of the iron wire?
Answer
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New Way Chemistry for Hong Kong A-Level Book 1122
3.6 Simple titrations (SB p.67)
MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)
Number of moles of MnO4-(aq) : Number of moles of Fe2+(aq) = 1 : 5
Number of moles of Fe2+(aq) = 5 Number of moles of MnO4-(aq)
= 5 0.02 mol dm-3 36.5 10-3 dm3
= 3.65 10-3 mol
Number of moles of Fe dissolved = Number of moles of Fe2+ formed
= 3.65 10-3 mol
Mass of Fe = 3.65 10-3 mol 55.8 g mol-1 = 0.204 g
Percentage purity of Fe = 100 % = 92.73 %
Therefore, the percentage purity of the iron wire is 92.73 %.
g 0.22g 0.204
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New Way Chemistry for Hong Kong A-Level Book 1123
3.6 Simple titrations (SB p.67)
(a)5 g of anhydrous sodium carbonate is added to 100 cm3 of 2 M hydrochloric acid. What is the volume of gas evolved at room temperature and pressure?
(Molar volume of gas at R.T.P. = 24.0 dm3 mol-
1)Answer
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New Way Chemistry for Hong Kong A-Level Book 1124
3.6 Simple titrations (SB p.67)
(a) Na2CO3(s) + 2HCl(aq) 2NaCl(aq) + H2O(l) + CO2(g)
No. of moles of Na2CO3 used =
= 0.0472 mol
No. of moles of HCl used = 2 M
= 0.2 mol
Since HCl is in excess, Na2CO3 is the limiting agent.
No. of moles of CO2 produced = No. of moles of Na2CO3 used
= 0.0472 mol
Volume of CO2 produced = 0.0472 mol 24.0 dm3 mol-1
= 1.133 dm3
1-mol g 3) 16.0 12.0 2 (23.0g 5
3dm1000100
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New Way Chemistry for Hong Kong A-Level Book 1125
3.6 Simple titrations (SB p.67)
(b) 8.54 g of impure hydrated iron(II) sulphate (formula mass of 392.14) was dissolved in water and made up to 250.0 cm3. 25.0 cm3 of this solution required 20.76 cm3 of 0.020 3 M acidified potassium manganate(VII) solution for complete reaction. Determine the percentage purity of the hydrated iron(II) sulphate.
Answer
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New Way Chemistry for Hong Kong A-Level Book 1126
3.6 Simple titrations (SB p.67)
(b) MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)
No. of moles of MnO4- ions = 0.0203 M
= 4.214 10-4 mol
No. of moles of Fe2+ ions = 5 No. of moles of MnO4- ions
= 2.107 10-3 mol
No. of moles of Fe2+ ions in 25.0 cm3 solution = 2.107 10-3 mol
No. of moles of Fe2+ ions in 250.0 cm3 solution = 0.02107 mol
Molar mass of hydrated FeSO4 = 392.14 g mol-1
Mass of hydrated FeSO4 = 0.02107 mol 392.14 g mol-1 = 8.26 g
% purity of FeSO4 = = 96.72 %%100
g 8.54g 8.26
3dm100020.76
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