Section 1 Atoms, molecules and Stoichiometry

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Modern College F.6 Chemistry (2009 – 10) Section 1

1.1 The Atomic Structure

A. Dalton’s atomic theory

Atoms are the smallest particles of an element that can take part in a chemical change.

All elements are made up of atoms

Atoms of a given element are identical in chemical behaviour

Chemical compounds are formed when different atoms combine chemically;

The atoms themselves are not changed in a chemical reaction, they just reorganized.

Constituents of atoms

Definition of key terms:

Atomic number

= Number of protons in an atom; or

= Number of electrons in the neutral atom;

= Ordinal number of that element in the Periodic Table;

It determines the chemical properties of an element.

Mass Number

Number of protons + number of neutrons

~ Relative atomic mass of an atom

Isotopes

Atoms with the same number of protons but different number of neutrons.

Atoms with the same atomic number but different mass number.

Mass number (A)

11

23

Na Element symbol

Atomic number (Z)

Prepared by Mr. Chau Chi Keung, Richard Page 1Notation of an element




1.2 Relative Isotopic, Atomic and Molecular Masses

A. Mass spectrometer

It measures the mass to charge ratio of a particle.

Atomic masses can be determined by mass spectrometry.

Components of a mass spectrometer and their functions:

1. Vaporization chamber: To vaporize the sample

2. Ionization chamber: The vaporized sample is then bombarded by fast moving electrons to

give positive ions.

M(g) + e – → M+(g) + 2e –

3. Accelerating electric field: The ions are accelerated by an electric field.

4. Deflecting magnetic field: It deflects the moving ions along a circular path.

As a result, the ions are separated as different ions have different deflection in the

magnetic field. The lighter (lower mass/charge ratio) the positive ion, the greater the

deflection will be.

By varying the strength of the deflecting magnetic field, ions of a particular mass/charge

ratio are brought to the ion detector

5. Ion detector: It detects the signals generated by the ions and passes them to the recorder.

6. Recorder: A mass spectrum is recorded using the recorder.

7. Vacuum pump: The mass spectrometer is operated at low pressure to prevent ions from

colliding with other particles.

B. Uses of mass spectrometer

Example 1

The mass spectrum of naturally occurring chlorine atom is as follows:

75.77%

Relative abundance

24.23%

Prepared by Mr. Chau Chi Keung, Richard Page 2




35 37

(a) Why there are two peaks of chlorine?

There are two isotopes of chlorine atoms in nature. The two isotopes are

35

Cl and

37

Cl.(b) What is the relative atomic mass of the naturally occurring chlorine atom?

Relative atomic mass of chlorine = 35 x 0.7577 + 37 x 0.2423 =

In general, only ions with +1 charge are detected. It is because +2 ions are very difficult to

generate, as large amount of energy is required to remove the second electron.

The highest m/e ratio of a sample most likely represents the molecular ion.

When the sample is NOT monoatomic, many peaks may appear. We should consider different

isotopes in a molecule:

e.g. Peaks of HCl in a mass spectrum

m/e ratio peaks: fragments:

36 1H – 35Cl+

37 2H – 35Cl+

38

39

Peaks of 37 and 39 may be very small due to low % abundance of 2H.

Example 2

The following mass spectrum was obtained for gaseous chlorine molecules. (Chlorine molecules are

diatomic)

Relative intensities 27

18

3

m/e ratio

70 72 74(a) How many types of chlorine molecules are there? What are they?

(b) Does chlorine has an isotope with atomic mass 36, i.e. 36Cl?

(c) Calculate the relative atomic mass of chlorine in this sample.

(d) What is the molar mass of chlorine gas (i.e. the relative molecular mass for chlorine molecules)?

(e) Two very small peaks appear at 35 and 37. Explain why.





The covalent bond within chlorine molecules may be broken by the bombardment of fast moving

electrons. Thus the ions 35Cl+ and 37Cl+ may be formed.

Example 3

The mass spectrum of an element, which is diatomic in the vapor state, gives peaks which corresponded

to masses of 158, 160 and 162.

(a) What deduction may be made about this element if there were no other peaks in the vicinity of

the above numbers?

(b) The heights of the peaks in (a) were in the ratio of 1:2:1.

(i) What is the relative abundance of each of the isotopes you have stated in (a)?

(ii) What is the relative atomic mass of the element?

(c) What is the molar mass (i.e. molecular mass) of the diatomic molecule?

Example 4

The mass spectrum of neon consists 3 lines corresponding to relative m/e ratio of 20,21and 22 with

relative intensities of 0.91:0.0026:0.088 respectively.

Relative intensities

m/e ratio20 21 22

How many isotopes are there for neon?

Calculate the relative atomic mass of neon.

1.3 The mole concept

A. Mole concept revisited (P.18 – 23)

The mole is defined as the amount of a substance containing the same number of particles as the





number of carbon atoms in 12 g of carbon-12. The particles may be atoms, molecules, ions or

electrons.

1 mole = 6.02×1023 particles.

L = 6.02×1023 is called the Avogadro constant

B. Molar volume of gases (P.24 – 26)

From the Avogadro’s Law, equal volumes of gases, measured at the same temperature and pressure,

contains the same number of molecules (i.e. same no. of mole).

The gas molar volume measured at R.T.P. (room temperature and pressure) = 24.056 dm3 and

the gas molar volume measured at S.T.P. (standard temperature and pressure) = 22.414 dm 3.

Gas Molar mass Density at 25°C and 1 atm Molar volume at 25°C and 1 atm.

= molar mass/density

Hydrogen 2.0 0.083 24.1

Oxygen 32.0 1.333 24.0

Chlorine 71.0 2.994 23.7

Ammonia 17.0 0.706 24.1

Carbondioxide

44.0 1.811 24.3

Exercise

Calculate:

1. Volume of 2.50 moles of hydrogen at R.T.P.

2. Volume of 1.3 moles of carbon dioxide at S.T.P.

3. The number of atoms in 120 cm3 of chlorine gas at R.T.P.


Important equations!1. Number of particles in a substance = Number of moles of the substance × L

2. massMolar

massmoleof Number =

3.)(dmsolutionof Volume

(mol)soluteof molesof Number )(moldmsolutionaof Molarity

3

3- =

4. Dilution: M1V1 = M2V2




C. Ideal gas equation (P. 27 – 31)

Foundation: quantitative properties of gas (volume, pressure, temperature and amount of gas) and

their inter-relationship.

(i) Boyle’s Law: At constant amount and temperature, P

V 1

∝ or t cons PV tan=

(ii) Charles’ Law: At constant amount and pressure, T V ∝ or T t consV ×= tan

(iii) Avogadro’s Law: At constant temperature and pressure, nV ∝ or nt consV ×= tan

Gases approach ideal behaviors at low pressures and high temperatures.

Molecules occupy no volume

Molecules are in constant random motion

No intermolecular forces between the particles

The collisions involving the gas molecules are totally elastic

Avogadro’s, Boyle’s and Charles’s laws (see textbook P.36 – 37 for details) are now known to be

three special cases of a more general equation, the ideal gas equation:

Where P – Pressure (Nm –2, atm. or mmHg)

V – Volume (dm3, cm3)

n – No. of moles (mol)

R – Universal gas constant, 8.3140JK –1mol –1/ 0.0821 atm. dm3 K –1 mol –1

T – Absolute temperature (K)

Example 1

Calculate the pressure inside a television picture tube which has a volume of 5.0 dm3 at a temperature of

25°C, and it contains 0.010 mg of nitrogen.

Gas constant = 0.08206 atm dm3

K –1

mol –1

Relative atomic mass of nitrogen = 14.01

Example 2

What is the volume of 20 g of CO2 at 300°C and 2 atm?

(Given: Relative atomic mass: C = 12; O = 16; R = 8.314 JK –1mol –1; 1 atm. = 1.0130×105 Nm –2)


PV = nRT

Units P V T R

SI units Nm-2 (Pa) m3 K 8.314 JK –1mol –1

Non SI units atm dm3 K 0.0821 atmdm3K –1mol –1




Determination of the relative molecular mass of volatile liquid by using the ideal

gas law

1. A gas syringe containing a small but known volume of air is fitted with a self sealing rubber cap.

A small volume of air is needed to provide some space for the volatile liquid to expand

smoothly).

2. Steam is passed through the steam jacket, until the thermometer reading and the volume of air in

the syringe become steady. Steady temperature (T1) and initial volume of air (V1) are recorded.

3. The hypodermic syringe is filled with about 1 cm3 of the volatile liquid (e.g. propanone).

4. Hypodermic syringe with the volatile liquid is weighed (m1) (Air bubbles must be expelled from

the hypodermic syringe, to ensure that it is completely filled.)

5. Small amount (e.g. 0.2 cm3) is then injected into the air space of the gas syringe through the

rubber cap.

6. After injection, small hypodermic syringe with small amount of volatile liquid is weighed (m2).

7. After the volatile is completely vaporized and final volume of the gas syringe (V2) is recorded.

8. The atmospheric pressure (P1) is finally recorded.

9. Calculations:

Mass of liquid injected: 21mmm −=

Volume of vapour formed by the injected liquid: 12 VVV −=

By using the equation nRT PV = , we have RT M m PV = , where M is the molar mass of the

gas/volatile liquid

P)VV(

RT)mm(

PV

mRTM

12

21

−

−==∴ (*)


Hint: If the density of that volatile liquid (ρ) is known, we can also determine the value of M by modifying (*) as follows:

V

m= ρ

From (*),

=

P RT

V mM

P

RT M

ρ =∴




Example 3

In order to determine the relative molecular mass of an unknown compound, an experiment was taken at

363 K by using the above method, the following data was obtained:

Volume and mass of the gas syringe:

Mass of the gas syringe/g Volume of the gas syringe/cm3

Before injection 20.120 V1

After injection 20.255 69.8

What is the relative molecular mass of the compound?

Possible sources of errors:

1. The assumption that the vapour obeys ideal gas behaviour may not be valid.

2. Errors in obtaining the mass of volatile liquid since volatile liquid may evaporate during weighing.

3. Errors in recording the temperature. (T of the steam jacket ≠ T of vapour inside the gas syringe).

4. The volatile liquid does not vapourize completely

5. Error in measuring pressure.

Exercise

Using ideal gas equation to calculate:

1. The volume of 1.50 g of hydrogen. H2 at 15°C and a pressure of 750 mmHg (1 atm. = 760 mmHg).

2. The temperature at which 4.71 g of nitrogen will occupy 12.0 dm3 at 760 mmHg?





D. Partial pressure of a gas and its relationship with mole fraction (P. 35 – 39)

According to Dalton’s law of partial pressure:

In a mixture of gases, the total pressure exerted is the sum of the pressurethat each gas would exert if it were present alone under the same condition.

That is,

Ptotal = P A + PB + …PN

Where P total is the total pressure and P A + P B + …P N are the partial pressure of gas A, B and

N respectively

The partial pressure of gas of each species can be calculated by using the concept of mole fraction.

From the ideal gas law,

V

RT n P A

A = ,

V

RT n P B

B=

V

RT n P C

C = ……,

Where nA, nB, nC……are the amount of the corresponding components in the mixture

Then, ......+++= C B Atotal P P P P

= ......+++V

RT n

V

RT n

V

RT n C B A

= )...)((V

RT nnn C B A +++

= )(V RT ntotal

Consider the partial pressure of A and the total pressure. By dividing PA by Ptotal, we have

total

A

total

A

total

A

n

n

V

RT n

V

RT n

P

P ==

Rearrangement gives

total A A

total

total

A

A

P P

P n

n

P

χ =

=

The quantity χ A is called the mole fraction of A; it is the ratio of the number of moles of A to the

total number of moles of gases present

Note: PA + PB + PC + … = Ptotal

∴(χ A + χ B + χ C +…)Ptotal = Ptotal

∴χ A + χ B + χ C +… = 1

Exercise

1. 0.25 mole of nitrogen and 0.30 mole of oxygen are introduced into a vessel of 12 dm3 at 50°C.

Calculate the partial pressures of nitrogen and oxygen and hence the total pressure exerted by the





gases.

2. 46 dm3 of O2 at 25°C and 1.0 atmosphere was pumped along with 12 dm3 (or litres) of He at 25°C and

1.0 atmosphere into a tank with a total volume of 5.0 dm3. Calculate the partial pressure of each gas

and the total pressure in the tank at 25°C. (Given R = 0.0821 atm dm3 K -1 mol-1)

3. A containing vessel holds a gaseous mixture of nitrogen and butane. The pressure in the vessel at

126.9°C is 3.0 atm. At 0°C the butane condenses completely and the pressure drops to 1.0 atm.

Calculate the mole fraction of nitrogen in the original gaseous mixture.

1.4 Formulae of compounds (P.43 – 53) The empirical formula is the formula representing the simplest integral ratio of atoms of elements





making up the compound or ion. It does not represent the actual number of atoms present in a

compound or ion.

The molecular formula is the formula showing the actual number of atoms of each element making

up the ion or compound. It is often a simple multiple of the empirical formula of a compound, that is:

Molecular formula = n x Empirical formula, where n is a positive integer. The structural formula is used to show how the constituent atoms are joined up within the

compound or ion.

Example: Butene

Empirical formula Molecular formula Structural formula

CH2 C4H8

A. Relationship between Empirical formula and molecular formula

1. Compound X has an empirical formula CH2 and relative molecular mass 84.0, what is the molecular

formula of X?

Let the molecular formula of compound X be (CH 2 )n , where n is a positive integer.

Relative molecular mass of (CH 2 )n = 84.0

∴(12.0 + 1.0 x 2) n = 84.0

∴n = 6

∴The molecular formula of compound X is C 6 H 12.

B. Derivation of Empirical formula using composition by mass

2. A sample of Mg of mass 0.450 g reacts with excess nitrogen to form 0.623 g of magnesium nitride.

Determine the empirical formula of magnesium nitride.

Given: relative molecular masses: Mg = 24.31; N = 14.01

Original mass of Mg = Mass of Mg in magnesium nitride = 0.450 g

No. of mole of Mg present = 0185.0

31.24

450.0=

Mass of N in magnesium nitride = 0.623 – mass of Mg

= 0.623 – 0.450

= 0.173 g

∴

No. of mole of N present = 0123.001.14

173.0=

∴Mg : N = 0.0185 : 0.0123 = 1.5 : 1

= 3 : 2 (simplest integral ratio)

∴

The empirical formula of magnesium nitride should be Mg 3 N 2.





C. Derivation of Empirical formula using combustion data

During complete combustion (i.e. O2 is in excess), elements in a compound are oxidized (e.g. H to

H2O, C to CO2, S to SO2, etc).

From the masses of the products formed, the number of moles of these atoms originally present can

be found.

Exercise

3. (a) Vitamin C is an organic compound known to contain the elements carbon, hydrogen and oxygen

only. Complete combustion of a 0.2000g sample of this compound yields 0.2998 g CO2 and 0.0819

g H2O. What is the molecular formula vitamin C?

(b) Vitamin C is found by mass spectrometry to have a relative molecular mass 176, what is its

molecular formula?

(a) Mass of C in vitamin C = Mass of C in CO 2 collected

=01.44

0.122998.0 ×

= 0.0818 g

Mass of H in vitamin C = Mass of H in H 2O collected

=02.18

0.20819.0 ×

= 0.0092 g

Mass of O in vitamin C = Mass of vitamin C – Mass of C – Mass of H

= 0.2000 – 0.0818 – 0.0092

= 0.1090 g ∴Mole ratio of atoms

C : H : O =0.16

1090.0:

00.1

0092.0:

01.12

0818.0

= 1 : 1.33 :1

= 3 : 4 : 3

∴The empirical formula of vitamin C should be C 3 H 4O3.

(b) Let the molecular formula of vitamin C be (C 3 H 4O3 )n , where n is a positive integer.

Relative molecular mass of (C 3 H 4O3 )n = 176

∴(3 x 12.0 + 4 x 1.0 + 3 x 16) n = 176

n = 2





∴The molecular formula of vitamin C should be C 6 H 8O6 .

*Hint: To deal with this type of calculation more effectively, you may consider tabulating all the things above as follows:

Carbon Hydrogen Oxygen

Mass (g) 0.0818 0.0092 0.1090 No. of moles 31082.6

0.12

0818.0−

×= 0092.0

0.1

0092.0=

31081.6

0.16

1090.0−

×=

Relative no. of moles1

1081.6

1082.63

3

≈

×

×

−

−

35.11081.6

0092.03≈

×−

11081.6

1081.63

3

=

×

×

−

−

Simplest mole ratio 3 4 3

∴The empirical formula of vitamin C should be C 3 H 4O3.

D. Determination of chemical formulae – A summary

To determine the chemical formula of a compound, there are many aspects of analysis to be

considered.

Qualitative analysis: e.g. observing chemical reactions with other substances, to identify the species

present in the compound.

Quantitative analysis: e.g. using combustion data, to find the empirical formula.

Instrumental analysis

Determination of molecular formula: e.g. using mass spectrometer to find the relative molecular

mass and hence the molecular formula.

Determination of structural formula: using advanced instruments like infra-red (IR)

spectroscopy, NMR (nuclear magnetic resonance), etc.

1.5 Chemical Equations and Stoichiometry

Stoichiometry is the calculation of quantitative relationships of the reactants and products in chemical

reactions

For a chemical reaction as shown below:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

The relative amounts (mole ratio) of reactants and products, are indicated by the coefficients in the

balanced equation.

These coefficients are called the stoichiometric coefficients. They reflect the mole ratio of all the

species involved in the chemical equation.

To deal with stoichiometric calculations, writing a balanced equation for the reaction and working out

the appropriate mole ratios are often necessary.

A. Stoichiometric calculations involving reacting masses

Example

Calculate the mass of magnesium oxide formed when 2.43 g of magnesium are burnt with

(a) excess oxygen(b) 1.28g of oxygen





Given: relative molecular masses: Mg = 24.30; O = 15.99

The balanced equation of the reaction: 2Mg(s) + O2(g) → 2MgO(s)

No. of mole of Mg = 1.03.24

43.2=

(a) From the equation, 2 moles of Mg give 2 moles of MgO after reaction.

Since O2 is in excess, no. of mole of MgO = no. of mole of Mg = 0.1

∴Mass of MgO formed = 0.1 x 40.3

= 4.03 g

(b) No. of mole of 1.28 g O2 = 04.00.162

28.1=

×

From the equation, 1 moles of O2 can react with 2 moles of Mg.

∴0.04 moles of O2 can react with 0.08 moles of Mg (i.e. O2 is the limiting reactant while Mg is in

excess).

∴ No. of mole of MgO formed = 0.08

∴Mass of MgO formed = 0.08 x 40.3

= 3.22 g

B. Stoichiometric calculations involving volume of gases

Consider a reaction: a A(g) + b B(g)→ c C(g) + d D(g)

Mole ratio:

By the Avogadro’s Law, at fixed T and P, nV ∝

In the reactions involving gaseous reactants and products, mole ratio = volume ratio.

Example 1

Calcium oxide (quicklime) is produced by thermal decomposition of calcium carbonate. Calculate the

volume of CO2 at R.T.P produced from the decomposition of 152 g of CaCO3, according to the reaction

CaCO3(s)→CaO(s) + CO2(g)

(Given the relative atomic mass: C = 12.0, O = 16.0 and Ca = 40.0)

No. of mole of 152 g of CaCO3 = 52.1100

152=

According to the given equation, 1.52 mol of CaCO3 gives 1.52 mol of CO2 after decomposition.

∴Volume of CO2 produced at R.T.P = 1.52 × 24.056 = 37.53 dm3

Example 2

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 residual gas

became 50 cm3. Find the molecular formula of that hydrocarbon.


d:c: b:an:n:n:ndc ba=

d:c: b:aV:V:V:V dc ba =∴




Let the molecular formula of the hydrocarbon be C x H y.

Volume of hydrocarbon reacted = 10 cm3.

Residual gas ⇒ Unreacted O2(g)

Volume of residual gas = 50 cm3

Volume of O2

(g) reacted = 80 – 50 = 30 cm

3

Volume of CO2(g) formed = 70 – 50 = 20 cm3

C x H y(g) + )4

(y

x + O2(g) → xCO2(g) +2

y H 2O(l)

For mole ratio, C x H y : O2 : CO2 = x y

x :)4

(:1 +

∴ For volume ratio, C x H y : O2 : CO2 = x y

x :)4

(:1 +

1

x

)(gHCof Volume

(g)COof Volume

yx

2

=

∴ x = 2

1

4

yx

)(gHCof Volume

(g)Oof Volume

yx

2

+=

∴ 34=+

y x

As x = 2, we have 34

2 =+ y

y = 4

∴The molecular formula of the hydrocarbon is C 2 H 4.

Exercise

1. What is the volume of oxygen needed for the complete combustion of 2 dm3 of propane?

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

Mole ratio, 1 : 5 : 3 : 4

∴Volume ratio, 1 : 5 : 3 : 4∴Volume of oxygen needed = 2 x 5 = 10 dm3

2. 10 cm3 of a gaseous hydrocarbon was mixed with 33 cm3 of oxygen which was in excess. The mixture

was exploded and then cooled to room temperature. The remaining volume of gas occupied 28 cm3.

On adding concentrated potassium hydroxide the volume decreased to 8 cm3. Find the molecular of

that hydrocarbon.


Volume of hydrocarbon reacted = 10 cm3

Volume of CO2(g) formed = 28 – 8 = 20 cm3





Volume of O2(g) reacted = 33 – 8 = 25 cm3

C x H y(g) + )4

(y

x + O2(g) → xCO2(g) +2

y H 2O(l)

For mole ratio, 1 : )4

(y

x + : x

For volume ratio, 10 : 25 : 20

20

101=

x

∴x = 2

10

25

4=+

y x

10

25

42 =+

y

y = 2

The molecular formula of the hydrocarbon is C2H2.

3. 10 cm3 of a hydrocarbon X was exploded with excess oxygen. The mixture was exploded and then

cooled to room temperature. There was a contraction in volume of 35 cm3 (all volumes were measured

at R.T.P. After treatment with concentrated sodium hydroxide, There was another contraction of 40

cm3. Deduce the molecular formula of X.


Volume of hydrocarbon reacted = 10 cm3.

Volume of CO2(g) formed = 40 cm3

For mole ratio, 1 : )4

(y

x + : x

For volume ratio, 1 : )4

(y

x + : x

10 : 10 )

4

(y

x + : 10x

10 x = 40

x = 4

By contraction (i.e. decrease) in volume of 35 cm3,

3540)]4

yx(1010[ =−++

y = 10

The molecular formula of the hydrocarbon is C4H10.

C. Stoichiometric calculations involving concentrations and volumes of solutions.





Always remember: Molarity =)dm(insolutionof Volume

soluteof moleof No.3

Example

(a) 11.5 g of solid NaOH was dissolved in distilled water to make 1.50 dm 3 solution. What is the

concentration of NaOH in mol-1dm3?

(b) What mass of hydrogen chloride, HCl, would be in 100cm3 of 2M HCl(aq)?

(c) To neutralize 50 cm3 of solution NaOH in (a), what volume of 2M HCl is used?

(a) No. of mole of NaOH = 288.00.40

5.11=

Concentration of NaOH solution = 192.05.1

288.0= M

(b) 100 cm3 = 0.1 dm3

No. of mole of hydrogen chloride in this acid = 0.2

Mass of HCl = 0.2 x (1.0 + 35.5) = 7.3 g

(c) Let V be the volume of 2M HCl required.

The balanced equation of the reaction: NaOH(aq) + HCl(aq) → NaCl(aq) + H 2O(l)

No. of mole of HCl required = No. of mole of NaOH

0.192 x 0.05 = 2V

V = 0.0048 dm3 = 4.8cm3

1.6 Titrations

Titration is the procedure for finding the molarity of an unknown solution by controlled addition of

a known volume of standard solution to another solution with unknown concentration, until

complete reaction.

A. Standard solution (標準溶液)(P. 68-69)

Standard solution: A solution of known concentration.

Primary standard: A pure compound from which a standard solution of accurately known

concentration can be prepared directly.

Criteria to be met for a compound to be used as a primary standard:

must be available in pure form;

should be stable in air over long periods of time;

should not absorb CO2 and H2O from the atmosphere;

should dissolve in water easily;

should not decompose in solution;

should have a larger molar mass (usually M > 100 gmol -1), so that the weighing error in using a

given number of moles of the solid is reduced.





Some commonly used primary standards: sodium carbonate (Na2CO3, Mr = 106), sodium

ethanedioate (i.e. sodium oxalate, Na2C2O4, Mr = 134).

B. Back Titration

In some cases, direct titration may not be easily done for the reaction between reagent X andreagent Y, because:

1. One of the reagents is a solid and thus the reaction is very slow. (i.e. we do not know whether

the reaction has completed.) For example, we cannot titrate solid Mg(s) or CaCO3(s) with

standard HCl(aq) although they react with each other.

2. No suitable indicator for the detection of end-point. For example, titration between NH4Cl(aq)

with KOH(aq) would have NO suitable acid-base indicator.

In this case, we would use back titration.

An excess but known amount of reagent A is added to the reagent B. When the reaction is

completed, the excess reagent A can be titrated with another reagent C . (A suitable indicator

can be found for the titration between reagent A and C).

For case 1:

If we want to find out the amount of CaCO3 in a sample, first, add excess with known concentration

of HCl into the CaCO3(s) sample. After complete reaction, the excess amount of acid can be ‘back

titrated’ with standard NaOH solution.

CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) No. of moles of HCl = 2(no. of moles of CaCO3(s)) + no. of moles of NaOH

For case 2:

The amount of ammonium salt (e.g. NH4 NO3) in a sample can be found by adding a known volume

of standard alkali, e.g. 50.0 cm3 of 2 M NaOH(aq). The reaction mixture is then warmed to drive off

the product NH3(g). When the reaction is completed, excess NaOH can be determined by titration

with standard acid, e.g. 2 M HCl.

NH4 NO3(aq) + NaOH(aq) → NaNO3(aq) + NH3(g) + H2O(l)

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

No. of moles of NaOH = no. of moles of NH 4 NO3 + no. of moles of HCl

C. Types of titrations

1. Acid-Base titrations ( 酸鹼滴定)

Example: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

Equivalence point (當量點): The point at which the amount of standard solution added is equivalent,

or equal, to the amount of substance to be analyzed present in the sample.

When performing a manual titration, it may be difficult or impossible to detect when the equivalence





point is reached. An acid-base indicator is often used, which would give a sharp colour change at the

end point (終點) that is supposed to be very close to the equivalence point.

End point: The point which the indicator changes colour.

Different acid-base indicator may be used according to different strength of acid and base involved.

For choices of indicator, refer to textbook P. 59 – 60 and they will be further discussed in Section 6.

The end point can also be detected by:

A change in temperature (thermometric titration). Since neutralization is exothermic, the

temperature of the reaction mixture is expected to rise to a maximum at the equivalence point.

A change in electrical conductivity. The equivalence point of a neutralization reaction should be

with electrical conductivity.

Change in pH value (using pH meter)

2. Acid-carbonate titrations

Similar to acid-base titration. For example: Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(l)

For soluble carbonate (e.g. Na2CO3), the end-point can be detected by acid-base indicator.

For insoluble carbonate (e.g. CaCO3), back titration is used.

A known amount of excess acid is added to the insoluble carbonate so that all the carbonate has





been completed reacted.

The amount of excess acid is determined by titration with standard alkali.

3. Redox titrations( 氧化還原滴定 )

Redox titration means titration of two solutions which undergo redox reaction.

Some applications: determination of heavy metal content in drinking water, vitamin C content in

fruits and vegetables, etc.

Case 1: Determination of thiosulphate ( 硫代硫酸根離子 ) or iodine

A standard iodine solution is used to titrate a thiosulphate solution to determine its concentration (i.e.

to standardize it).

The standardized thiosulphate solution can be used to titrate another iodine solution of unknown

concentration.

Standardization of thiosulphate solution:

1. A standard solution of iodine is prepared by dissolving a known amount of pure potassium

iodate(V) (KIO3) in an acidic medium, containing excess potassium iodide.

IO3 – (aq) + 5I – (aq) + 6H+(aq) → 3I2(aq) + 3H2O(l)

Note: No. of mole of I2(aq) formed = No. of mole of IO3-(aq) added × 3

Hint 1: Standard iodine solution cannot be prepared by dissolving iodine in water because:

iodine is only slightly soluble in water

iodine may easily lost through sublimation

Hint 2: The solution of iodine is brown in colour and should be used immediately ( iodine

may easily lost through sublimation).

2. The standard iodine solution is then used to titrate with a thiosulphate solution of unknown

concentration. Here iodine oxidizes thiosulphate ions to tetrathionate, and forms iodide ions.

I2(aq) + 2S2O32– (aq) → S4O6

2– (aq) + 2I – (aq)

thiosulphate tetrathionate

(brown) (colourless) (colourless) (colourless)

Hint: Standardization of thiosulphate with standard I 2(aq) is necessary because it is unstable:

in acidic medium,

in the presence of microorganism,

in the presence of Cu(II) ions (Cu2+ ),

under sunlight

3. Since the colour of iodine in iodide solution cannot be used to accurately detect the end point ( colour change: brown →pale yellow→colorless, very difficult to observe). Thus, starch is

used as the indicator:

Starch + iodine→ blue complex (reversible)

4. Since starch would react with iodine irreversibly at high concentration of I2, so starch solution

cannot be added at the beginning of the titration as it would hinder the reaction between iodineand thiosulphate.





5. Starch solution should be added in a later stage of the titration (when the brown colour of

iodine changes from brown to pale yellow). After the adding of starch solution, the solution

turns deep blue.

6. The end point is shown by the complete decolorization of the blue colour. (Iodine is

completely reacted).

7. After standardization, that thiosulphate solution can be used to titrate another solution of iodine

of unknown molarity.

Example 1 – Thiosulphate determination

2.015 g of pure potassium iodate(V)(碘酸鉀) was dissolved and made up to 250cm3. Excess potassium

iodide and dilute sulphuric acid were also added to a 25.0 cm3 portion of this standard solution. The

solution was then titrated with a solution of sodium thiosulphate, starch solution was added near the

end-point. In the end, 29.8 cm3 of thiosulphate solution were required. Calculate the concentration of

thiosulphate solution. (Given Mr of KIO3 = 214)

No. of mole of IO3-(aq) added =

10

1

214

015.2× = 9.42

×10-4

∴ No. of mole of I 2 (aq) formed = 3×9.42×10-3

= 2.83×10-3

From the equation I 2(aq) + 2S 2O32-(aq) → S 4O6

2-(aq) + 2I -(aq)

No. of mole of S 2O32-(aq) used = 2.83×10-3×2

= 5.66 ×10-3

29.8 cm3 = 0.0298 dm3

∴S 2O32-(aq)] =

0298.0

1066.5 3−×

= 0.190 M

Example 2 – Estimation of copper(II) ions

It is known that copper(II) ions oxidize iodide ions to iodine:

2Cu2+(aq) + 4I-(aq) → I2(aq) + 2CuI(s)

The iodine produced can be titrated with standard thiosulphate solution, the amount of copper(II) ions

can be deduced from the amount of iodine produced. A sample of 4.256 g of copper(II) sulphate-5-

water was dissolved and made up to 250 cm3. A 25.0 cm3 portion is added to an excess of potassium

iodide. The iodine formed required 18.0 cm3 of a 0.0950M sodium thiosulphate solution for reaction.

Calculate the percentage of copper in the crystal. (Relative atomic mass of Cu = 63.5)

I 2(aq) + 2S 2O32–-(aq) → S 4O6

2–-(aq) + 2I – (aq) -------------------- (1)

2Cu2+(aq) + 4I – (aq) → I 2(aq) + 2CuI(s) -------------------------(2)

From (1), No. of mole of S 2O32-(aq) = No. of mole of I 2 (aq) formed ×2

∴ No. of mole of I 2 (aq) formed = )1000

0.18(095.0

2

1×× =8.55

×10-4

∴ No. of mole of Cu2+(aq) in 25.0 cm3 solution = 2×8.55×10-4

= 1.71×10-3





∴ No. of mole of Cu2+(aq) in 250 cm3 solution = 1.71×10-2

∴Mass of copper in the sample = 1.71×10-2 ×63.5 = 1.086 g

% by mass of copper in the crystals = %5.25%100256.4

086.1=×

Case 2 Potassium manganate(VII )titrations Acidified potassium manganate(VII) solution is a strong oxidizing agent

Equation:

Once it has been standardized, it can be used to determine the amount of various reducing agents

such as iron(II) salts.

No indicator is needed as the oxidant changes from purple to colourless during titration. At the end

point, a light but permanent purple colour (persists for at least 30 seconds) appears due to the

addition of a drop of excess KMnO4(aq).

It should be noted that the redox reaction involved occurs slowly at room temperature. Thus the

reaction mixture should be heated to about 60°C so that the reaction takes places at a suitable rate.

Example 1 – Standardization of potassium manganate(VII) solution by using sodium ethanedioate*

For example, a 25.0 cm3 of sodium ethanedioate solution of concentration 0.200 moldm -3 was titrated

against a solution of acidified potassium manganate(VII). If 17.2 cm3

of potassium manganate(VII)were used, what is the concentration of the solution?

From the equation: 2MnO4-(aq) + 5C 2O4

2-(aq) + 16H +(aq) → 2Mn2+(aq) + 10CO2(g) + 8H 2O(l)

Mole ratio, MnO4-(aq) : C 2O4

2-(aq) = 2 : 5

No. of mole of MnO4-(aq) = ×

5

2 No. of mole of C 2O4

2-(aq)

No. of mole of C 2O42-(aq) = 0.200 ×25×10-3 = 5.00 ×10-3





∴ No. of mole of MnO4-(aq) = ×

5

25.00×

10-3 = 2.00×

10-3

∴[MnO4-(aq)] = =

×

×

−

−

3

3

102.17

1000.20.116 moldm-3

*Note 1: ethanedioate ions are also known as oxalate ions (草酸根離子).

Note 2: Standardized potassium permanganate solution can be used in a wide variety of redox

reaction for the determination of many different species, for example:

oxalate content in some vegetables (e.g. spinach)

hydrogen peroxide content in commercially available hydrogen peroxide solution

iron content of iron tablets that can be obtained from any pharmacy

metal content (e.g. copper, iron and tin) in some alloys or impure metal samples

Example 2 – Iron determination

An impure sample of iron of mass 7.50 g was dissolved in dilute sulphuric(VI) acid and the solution

was made up to 250 cm3. The solution contained iron(II) ions together with the impurities. 25.0 cm3 of

this solution was titrated with potassium manganate(VII) solution mentioned in the Example 1. The

average volume of potassium manganate(VII) solution used was 20.00 cm3. Determine the percentage

of purity of iron in the sample (Given relative atomic mass of iron = 56.0).

From the equation: MnO4-(aq) + 5Fe2+(aq) + 8H +(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H 2O(l)

Mole ratio, MnO4-(aq) : Fe2+-(aq) = 1 : 5

No. of mole of Fe2+-(aq) in 25.0 cm3 solution = No. of mole of MnO4-(aq) ×5

= (0.116 ×20×10-3 )×5

= 0.0116

∴ No. of mole of Fe2+(aq) in the original solution = 0.116

∴Mass of iron in the sample = 0.116 ×56.0 = 6.50 g

∴% purity = %10050.7

50.6× = 86.7%

Supplementary Notes on Titrations

On preparation of standard solutions





On techniques of performing acid-base titration













Documents

Section 1 Atoms, molecules and Stoichiometry