L.S.T. Leung Chik Wai Memorial School
Ch3: p.1
F.6 Chemistry
Chemical Equations and Stiochiometry
Chapter 3
:
Chemical Equations and Stoichiometry
I.
Chemical Formulae
The empirical formula of a compound is a formula which shows the simplest ratio
of the different atoms or ions present in the compound.
The molecular formula of a compound is a formula which shows the actual
number of each kind of atoms present in one molecule of the compound.
Examples:
II.
Derivation of Empirical Formula using combustion data and Composition by Mass
The empirical formula of any compound can be determined by experiments such as combustion,
or its composition by mass is already known.
(A) An Experiment to find the empirical formula of copper (II) oxide
1.
Experimental set-up
2.
Procedure:
<i> The test tube was weighed. Some black copper(II) oxide was then added and the
test-tube was reweighed.
<ii> Town gas was passed over the oxide and the excess town gas was burnt at the end of the
test tube.
<iii> The black copper (II) oxide was heated and its was reduced to brown copper metal by
the town gas.
<iv> The burner was removed and the tube was allowed to cool in a stream of town gas.
<v> Finally, the town gas was turned off and the test tube with the solid (copper) was
weighed again.
3.
Experimental results
Mass of test tube = 24.76 g
Mass of test tube and copper (II) oxide = 27.94 g
Mass of test tube and copper = 27.30 g
Ch3: p.2
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chemical Equations and Stiochiometry
4.
Calculation :
Mass of copper left behind =
Mass of oxygen removed from copper (II) oxide =
Empirical formula of copper (II) oxide is calculated below:
Copper
Oxygen
Mass (g)
No. of moles of atoms
Simplest ratio of relative
number of moles
5.
Questions:
(a) Explain how town gas changed copper (II) oxide to copper.
(b) (1) Town gas was allowed to pass through the apparatus for sometime before it was
ignited at the end of the tube. Explain.
(2) Explain briefly why copper left inside the test tube was allowed to cool in a stream
of town gas.
Ch3: p.3
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chemical Equations and Stiochiometry
(B) An Experiment to Find the Empirical Formula of Magnesium oxide
1.
Experimental set-up:
2.
Procedure:
A known mass of magnesium was heated in a crucible in contact with air. When cooled,
the mass of magnesium oxide produced was found by reweighing the crucible with its
contents.
3.
Experimental Results:
Mass of Crucible and lid = 22.56 g
Mass of crucible , lid and magnesium = 22.92 g
mass of crucible, lid and magnesium oxide = 23.16 g
4.
Calculation:
Empirical formula of copper (II) oxide is calculated below:
Magnesium
Oxygen
Mass (g)
No. of moles of atoms
Simplest ratio of relative
number of moles
5.
Questions:
(i)
Explain briefly why the crucible should be covered with lid during heating.
(ii)
After the magnesium was ignited, the Bunsen flame was removed and the
crucible lid was raised for a few moments. Explain why this is done.
Ch3: p.4
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chemical Equations and Stiochiometry
III.
Composition by Mass
From the formula of a compound and the relative atomic masses of the elements in it, the
percentage of each element in the compound can be calculated. This is called percentage
composition by mass.
Example: Calculate the percentage mass of nitrogen in urea, NH2CONH2 .
% by mass of N = molar mass of N content / molar mass of the compound
= 14 x 2 / (14x2+1x4+12+16) x 100%
= 46.63 %
x 100%
Other Example,
1. 0.0234 g of hydrocarbon Y produced 0.0792 g carbon dioxide and 0.0162 g of water on
complete combustion. Calculate the percentage of carbon and hydrogen in Y.
L.S.T. Leung Chik Wai Memorial School
Ch3: p.5
F.6 Chemistry
Chemical Equations and Stiochiometry
IV.
Derivation of Molecular Formula from Empirical Formula and Relative Molecular Masses
The molecular formula of any compound can be found if the empirical formula and the relative
molecular mass of the compound are known.
Example:
1. A carboxylic acid P, with relative molecular mass less than 1--, contains C (55.8%), H (7.0%)
and O (37.2%) by mass. Determine the empirical and molecular formulae of the acid.
Ch3: p.6
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chemical Equations and Stiochiometry
Name: _________________ Class no.: _____
Date: __________
Marks: _____________
Exercises:
1. A colourless organic liquid , L, contains 40.0% of carbon, 6.67% of hydrogen and 53.3% of
oxygen. 1.50 g of L, when vaporized, is found to occupy 560 cm3 at s.t.p..
Calculated the empirical and molecular formula of L.
(Given: the molar volume of gas in s.t.p. = 22400 cm3 )
( 4 marks )
2.
1.00 g of a saturated hydrocarbon gave 200.0 cm3 of vapour at 2500 C and 1 atmospheric
pressure. Determine the molecular formula of this saturated hydrocarbon CnH2n+2 .
L.S.T. Leung Chik Wai Memorial School
Ch3: p.7
F.6 Chemistry
Chemical Equations and Stiochiometry
V.
Stoichiometry Relationship between Reactants and Products in a Reaction
Stoichiometry is the relationship between the amounts of reactants and products
in a chemical reaction. A balanced chemical equation is a statement which relates
the number of atoms in the reactants with that in the products.
The following steps are useful in calculation:
<1> Write down the balanced equation for the chemical reactions involved.
<2> Change the amount of relevant substance into mole unit.
<3> Calculate the molar quantities of the required substances using ratios given by the
equation.
<4> Convert the molar quantities back into mass or volume units required by the question.
VI.
Calculations involving Reacting Masses
1. Excess copper (II) oxide is dissolved in 25 cm3 of 3M sulphuric acid. The solution is
then filtered to remove the undissolved copper (II) oxide. The filtrate is concentrated by
evapourating some of the water away and the remaining solution is then left to cool.
On cooling, crystals of hydrated copper (II) sulphate (CuSO4 5H2O) are formed.
Calculate the maximum mass of hydrated copper (II) sulphate crystals that can by
obtained.
CuO(s) + H2SO4(aq)  CuSO4(aq) + H2O(l)
CuSO4 + 5H2O  CuSO4 5H2O
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Ch3: p.8
F.6 Chemistry
Chemical Equations and Stiochiometry
2.
Calculate the volume of carbon dioxide evolved ( at r.t.p.) when 5.3 g of anhydrous
sodium carbonate reacts with 40.0 cm3 of 2M hydrochloric acid.
3.
Ammonium chloride reacts with sodium hydroxide to form ammonia gas:
NH4Cl(s) + NaOH(aq)  NaCl(aq) + NH3(g) + H2O(l)
Ammonia gas reacts with sulphuric acid to form ammonium sulphate, calculate the mass
of ammonium chloride required to produce 2.64 g of ammonium sulphate.
L.S.T. Leung Chik Wai Memorial School
Ch3: p.9
F.6 Chemistry
Chemical Equations and Stiochiometry
VII.
Calculations involving Concentrations and Volumes of Solutions
The concentration of a solution is expressed in mass units of solutes per unit volume of
solution.
A molar (M) solution contains 1 mole of solute dissolved in 1 dm3 of solution.
Molarity of a solution is the number of moles of solute in 1 dm3 of the
solution.
No. of mole of solutes = molarity x volume of solution (dm3)
n
= MxV
Volumetric Analysis
-
-
Volumetric analysis is a method to determine the quantity or concentration of a substance by
measuring the volume of solution that will react with a known volume of another solution.
The essential process called titration involves running a solution from burette into a known
volume of another solution until the two solutions have just reacted together completely.
Volumetric analysis includes titrations of
1. acid against alkali or carbonate
2. oxidizing agent against reducing agent
3. one substance against another, giving precipitate
Before titration is carried out, a standard solution should be prepared first
Standard solution is a solution whose concentration is accurately known and can be
therefore used to
1. determine the concentration of another solution through titration or
2. standardize other solutions.
Example:
Borax (Na2B4O7 . 10H2O) has a relative formula mass of 381.37 . It is often used as a primary
standard in acid-base titrations.
<1> Meaning of “ primary standard”
A primary standard is a pure compound, a known mass of which when dissolved in distilled
water will give a standard solution.
<2> Steps in preparing 250 cm3 of a standard borax solution of approximately 0.1 M
The mass of borax required in this preparation
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Ch3: p.10
F.6 Chemistry
Chemical Equations and Stiochiometry
The steps are outlined as below:
A.
The weighing process
B.
Transferring the solid
1. Add 50 cm3 of distilled water to the beaker and stir the solution with a glass rod until
borax is completely dissolved.
2. Pour the solution into a 250.0 cm3 volumetric flask through a funnel.
3. Rinse the beaker and glass rod with 30 cm3 of distilled water using a wash bottle.
Pour all the washings through the funnel into the volumetric flask. Repeat this steps
several times to ensure complete transference.
C.
Make up to 250.0 cm3 of solution
1. Add distilled water into the volumetric flask until the graduation mark is nearly
reached.
2. Add the last few cm3 of distilled water through a dropper (dropwise) to avoid
overshooting the mark.
(The lowest meniscus is on the mark when seen at the eye level.)
3. Stopper the flask and shake (or invert) it several times to mix the contents
thoroughly.
Ch3: p.11
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chemical Equations and Stiochiometry
VIII.
Titration involving Acid-Base Reactions
A.
Principles of Acid-Base Titrations
A Standard solution of an acid can be used to find the concentration of an alkali. A known
volume of alkali is taken by a pipette. The alkali is then titrated against the standard acid until
the end point is reached. The number of moles of acid can be calculated and the equation
used to the number of moles of alkali neutralized.
1.
Acids and Alkalis
Reactions between acids and alkalis can be represented by:
H+(aq) + OH-(aq) 
H2O(l)
One mole of alkali is needed to neutralize one mole of monobasic acid while two moles of
alkali will be used for dibasic acid.
2.
Acids and Carbonates
The reaction between carbonate and acids can be represented by
2H+(aq) + CO32-(aq) 
H2O(l) + CO2(g)
Note:
The reactions between acid/alkali and acid/carbonate are rapid and essentially
complete. The point at which the reaction is complete can be easily be detected. Therefore,
such reactions are particularly suitable for use in Volumetric Analysis.
B.
The Use of Indicators
1.
-
The choice of indicators
At the end point, when acid and alkali have just reacted completely, the pH changes
remarkably. Therefore, to detect the end point, the indicator should show a sharp colour
change at this point.
The choice of indicator in a given titration depends on the strength of the acid and the alkali
involved. This is summarized in the table below:
-
-
Titration type
pH at the end point
Suitable indicators
Strong acid vs Strong base
7
Methyl orange,
Phenolphthalein
Strong acid vs Weak base
<7
Methyl orange
Weak acid vs Strong base
>7
Phenolphthalein
Weak acid vs Weak base
Depends on the particular acid
and alkali
None
In general the colours of methyl orange and phenolphthalein can be summarized as follows:
Indicator
pH range with which colour
changes
Methyl orange
3.2 – 4.2
Phenolphthalein
8.2 – 10
Colour in acid
Colour in alkali
Red
Yellow
(below pH 3.2) (above pH 4.2)
Colourless
Pink
L.S.T. Leung Chik Wai Memorial School
Ch3: p.12
F.6 Chemistry
Chemical Equations and Stiochiometry
2.
Procedures of Titration
(i)
(ii)
(iii)
(iv)
(v)
In each titration, a fixed volume of the standard solution is measured into a conical
flask with a pipette.
A fixed amount (2-3 drops) of indicator is added in each titration.
Another solution is added from a burette into the conical flask is swirled constantly.
When the colour of the indicator just changes, the end point is reached. The volume of
solution added from the burette is calculated.
Repeat titration to obtain a consistent value.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chemical Equations and Stiochiometry
Ch3: p.13
L.S.T. Leung Chik Wai Memorial School
Ch3: p.14
F.6 Chemistry
Chemical Equations and Stiochiometry
Name: _________________ Class no.: _____
Date: __________
Marks: _____________
Exercises:
1. 5.032 g of dibasic acid of anhydrous relative molecular mass 90, were made up to 1.0 dm3 of
aqueous solution. 25.0 cm3 of this solution required 19.97 cm3 of 0.1 M NaOH for neutralization
with phenolphthalein as indicator.
Calculate the number of moles of water crystallization per mole of the crystalline acid.
2.
1.46 g of a mixture of sodium hydroxide and anhydrous sodium carbonate was dissolved in 100
cm3 of distilled water. This solution required 30.0 cm3 of 1 M hydrochloric acid for complete
reaction.
Find the percentage composition by mass of the mixture.
L.S.T. Leung Chik Wai Memorial School
Ch3: p.15
F.6 Chemistry
Chemical Equations and Stiochiometry
Black Titration
In the technique called back titration, a known excess of one reagent A is allowed to react
with an unknown amount of reagent B. At the end of reaction, the amount of A that remains
unreacted is found by titration. From this, the amount of A that has been used and the amount of B
that has reacted can be calculated.
Example:
1. 1.340 g of a sample of ammonium chloride were boiled with excess sodium hydroxide
solution. The ammonia evolved was absorbed in 50.0 cm3 of 0.5 M sulphuric acid. This
solution is then made up to 250.0 cm3 with deionized water and 25.0 cm3 of it required 25.10
cm3 of 0.1 M sodium hydroxide solution for neutralization.
Calculate the percentage by mass of ammonia available from 1.34 g of the ammonium
chloride sample.
L.S.T. Leung Chik Wai Memorial School
Ch3: p.16
F.6 Chemistry
Chemical Equations and Stiochiometry
D.
Titration without an Indicator
-
Apart from using indicators, the end point of a titration can be determined by following
changes in
electrical conductivity
temperature
pH during the reaction
Example 1:
Conductometric titration
0.1 M sulphuric acid is added by means of a burette in 2 cm3 portions to 25 cm3 of a solution
of barium hydroxide in a conductivity flask as shown in the diagram below:
The mixture is constantly stirred and the reading on the milliammeter is recorded after each
addition of the acid. The results obtained are plotted in a graph as shown below:
a.
Account for the change in the milliammeter reading in the course of the experiment.
b.
Calculate the concentration of barium hydroxide solution.
c.
Explain why an alternating current, but not a direct current , is used in the experiment.
L.S.T. Leung Chik Wai Memorial School
Ch3: p.17
F.6 Chemistry
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Example 2:
Thermometric Titration
50 cm3 of 1M solution of sodium hydroxide is transferred by a pipette into an expended
polystyrene cup. A molar solution of a strong acid HnA is added from a burette, 5 cm3 at a time.
The solution is stirred thoroughly and the experiment of the solution is recorded after each
addition of the acid.
The temperature change in the course of the experiment is shown in the graph below:
a.
Describe and account for the change in the temperature of the solution.
b.
Calculate the value of n in the acid HnA.
Write a full equation for the reaction between the acid and sodium hydroxide.
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Ch3: p.18
F.6 Chemistry
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Example 3 : Following the pH change during titration
25.0 cm3 of 0.1 M hydrochloric acid is placed in a conical flask. Sodium hydroxide is added from
a burette and the pH value of the solution is measured by a pH meter. The results are recorded and
shown in the following graph.
a.
Explain the shape of the curve.
b.
Calculate the concentration of the sodium hydroxide solution.
L.S.T. Leung Chik Wai Memorial School
Ch3: p.19
F.6 Chemistry
Chemical Equations and Stiochiometry
IX.
A.
Titrations involving Redox Reactions
Redox Reactions:
Oxidation and Reduction always occur simultaneously in the same reaction. Such reactions are
known as redox reactions.
-
B.
Oxidation involves a loss of electrons.
Reduction involves a gain of electrons.
A reducing agent is a substance which undergoes oxidation.
A oxidizing agent is a substance which undergoes reduction.
Common Redox Titrations
The main types of redox titrations are based on the use of the following agents:
1.
2.
3.
Potassium permanganate (KMnO4 )
Potassium dichromate (K2Cr2O7 )
Sodium thiosulphate (Na2S2O3)
<1> Potassium Permanganate is a strong oxidizing agent in acidic solution.
MnO4-(aq) + 8H+(aq) + 5e 
(purple)
Mn2+(aq) + 4H2O(l)
(pale pink)
-
The colour suddenly changes form purple to pale pink when the reaction completes.
This suggests that no indicator is needed to show the end-point.
-
As an example , MnO4-(aq) oxidized Fe2+ in acidic solution:
MnO4- + 8H+ + 5Fe2+ 
Mn2+ + 4H2O + 5 Fe3+
<2> Potassium Dichromate is a strong oxidizing agent in acidic solution.
Cr2O72-(aq) + 14H+(aq) + 6e  2Cr3+(aq) + 7 H2O(l)
-
The colour change from orange to green at the end-point is not sharp. Therefore, an indicator
is needed for the detection of end-point.
As an example, acidified potassium dichromate oxidizes ethanedioate to carbon dioxide:
Cr2O72-(aq) + 14H+(aq) + 3C2O42-(aq) 
2Cr3+(aq) + 7H2O(l) + 6CO2(g)
L.S.T. Leung Chik Wai Memorial School
Ch3: p.20
F.6 Chemistry
Chemical Equations and Stiochiometry
<3> Sodium thiosulphate, Na2S2O3, is a reducing agent.
2S2O32-(aq)  S4O62-(aq) + 2e
-
Standard sodium thiosulphate solution is used to estimate the quantity of iodine.
2S2O32-(aq) + I2(aq) 
-
C.
S4O62-(aq) + 2I-(aq)
It reduces iodine to iodide ions, itself being oxidized in the process.
When the brown colour of iodine fades as the end point approaches, a little starch solution
(about 5 ml) is added. This gives an intense blue black colour with even a trace of iodine.
At the end point, when all iodine has been reduced to iodide ions, the blue colour disapper
completely.
Sodium thiosulphate is more frequently used to quantity any oxidizing agent which liberates
iodine from potassium iodide.
Calculations in Redox Titration / Reactions
In doing calculations involving redox reactions, the oxidation half-equation and reduction
half-equation are combined to give a balanced equation. Calculations are again based on the mole
ratio of reactants in the balanced equation.
Ch3: p.21
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chemical Equations and Stiochiometry
Name: _________________ Class no.: _____
Date: __________
Marks: _____________
Exercises
1. Potassium dichromate reacts with sulphur dioxide according to the equation:
3SO2 + Cr2O72- + 2H+ 
3SO42- + 2Cr3+ + 2H2O
a certain solution contain 0.1 M of potassium dichromate . Calculate the volume of this solution
which is required to remove sulphur dioxide from 2 dm3 of polluted air, which contains 3.6% by
volume of sulphur dioxide.
(Assume that all measurement are made at room temperature and pressure.)
( 4 marks )
2.
2.50 g of a sample of bleaching powder were ground with water, transferred to a volumetric flask
and made up to 250.0 cm3 of this solution, when added to excess potassium iodide solution and
acidified, liberated iodine, which requires 24.20 cm3 of 0.1M sodium thiosulphate solution for
reaction. The reaction involved are:
Cl2(g)
+
2I-(aq)  I2(aq)
+ 2Cl-(aq)
2S2O32- +
I2

S4O62-
+
2I-
Calculate the percentage of “available chlorine” in the bleaching powder.
( 4 marks )