Section 1 Atoms, molecules and Stoichiometry

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) 23 11

Na

Element symbol

Atomic number (Z) Prepared by Mr. Chau Chi Keung, Richard

Notation of an element Page 1

Modern College F.6 Chemistry (2009 – 10)

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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: Vaporization chamber: To vaporize the sample 1. Ionization chamber: The vaporized sample is then bombarded by fast moving electrons to 2. give positive ions. M(g) + e– → M+(g) + 2e– 3. Accelerating electric field: The ions are accelerated by an electric field. Deflecting magnetic field: It deflects the moving ions along a circular path. 4. 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 Ion detector: It detects the signals generated by the ions and passes them to the recorder. 5. Recorder: A mass spectrum is recorded using the recorder. 6. Vacuum pump: The mass spectrometer is operated at low pressure to prevent ions from 7. 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

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Modern College F.6 Chemistry (2009 – 10)

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35 37 (a) Why there are two peaks of chlorine? There are two isotopes of chlorine atoms in nature. The two isotopes are 35Cl and 37Cl. (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: 1 35 36 H – Cl+ 2 37 H – 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 18

27 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. Prepared by Mr. Chau Chi Keung, Richard

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Modern College F.6 Chemistry (2009 – 10)

Section 1

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 ratio 20 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  Prepared by Mr. Chau Chi Keung, Richard

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Modern College F.6 Chemistry (2009 – 10)

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Section 1

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

Important equations!

1. Number of particles in a substance = Number of moles of the substance × L 2. Number of mole =

mass Molar mass

-3 3. Molarity of a solution (moldm ) =

Number of moles of solute (mol) Volume of solution (dm 3 )

4. Dilution: M1V1 = M2V2 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 dm3. Gas Hydrogen Oxygen Chlorine Ammonia Carbon dioxide

Molar mass Density at 25°C and 1 atm Molar volume at 25°C and 1 atm. = molar mass/density 2.0 0.083 24.1 32.0 1.333 24.0 71.0 2.994 23.7 17.0 0.706 24.1 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.

Prepared by Mr. Chau Chi Keung, Richard

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Modern College F.6 Chemistry (2009 – 10)

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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, V ∝





1 or PV = cons tan t P

(ii) Charles’ Law: At constant amount and pressure, V ∝ T or V = cons tan t × T (iii) Avogadro’s Law: At constant temperature and pressure, V ∝ n or V = cons tan t × n 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:

PV = nRT 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) 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

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 × 105Nm–2)

Prepared by Mr. Chau Chi Keung, Richard

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Modern College F.6 Chemistry (2009 – 10)

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

1.

2. 3. 4. 5. 6. 7. 8. 9.

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). 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. The hypodermic syringe is filled with about 1 cm3 of the volatile liquid (e.g. propanone). 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.) Small amount (e.g. 0.2 cm3) is then injected into the air space of the gas syringe through the rubber cap. After injection, small hypodermic syringe with small amount of volatile liquid is weighed (m2). After the volatile is completely vaporized and final volume of the gas syringe (V2) is recorded. The atmospheric pressure (P1) is finally recorded. Calculations: Mass of liquid injected: m = m1 − m 2 Volume of vapour formed by the injected liquid: V = V2 −V 1 By using the equation PV = nRT , we have PV =

m RT , where M is the molar mass of the M

gas/volatile liquid mRT (m1 − m 2 )RT ∴M = = (*) PV (V2 − V1 )P Hint: If the density of that volatile liquid (ρ) is known, we can also determine the value of M by modifying (*) as follows: ρ=

m V  m  RT     V  P 

From (*), M =  ∴M =

ρRT P

Prepared by Mr. Chau Chi Keung, Richard

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Modern College F.6 Chemistry (2009 – 10)

Section 1

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 Before injection 20.120 After injection 20.255 What is the relative molecular mass of the compound?

Volume of the gas syringe/cm3 V1 69.8

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?

Prepared by Mr. Chau Chi Keung, Richard

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Modern College F.6 Chemistry (2009 – 10)

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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 pressure that each gas would exert if it were present alone under the same condition. That is,

Ptotal = PA + PB + …PN Where Ptotal is the total pressure and PA + PB + …PN 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, n RT n B RT n RT PA = A PC = C , PB= ……, V V V Where nA, nB, nC……are the amount of the corresponding components in the mixture Then, Ptotal = PA + PB + PC +......

=

n A RT n RT nC RT + B + + ...... V V V

= (n A + n B = ntotal

(

+ nC + ...)(

RT ) V

RT ) V

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

RT V = nA = RT ntotal ntotal V nA

Rearrangement gives n PA = A Ptotal ntotal PA =χ APtotal 



The quantity χA 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

∴(χA + χB + χC +…)Ptotal = Ptotal ∴χA + χB + χC +… = 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 Prepared by Mr. Chau Chi Keung, Richard

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Modern College F.6 Chemistry (2009 – 10)

Section 1

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

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Modern College F.6 Chemistry (2009 – 10)

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



Section 1

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 (CH2)n, where n is a positive integer. Relative molecular mass of (CH2)n = 84.0 ∴(12.0 + 1.0 x 2) n = 84.0 ∴n = 6 ∴The molecular formula of compound X is C6H12. 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 =

0.450 = 0.0185 24 .31

Mass of N in magnesium nitride = 0.623 – mass of Mg = 0.623 – 0.450 = 0.173 g 0.173

= 0.0123

∴No. of mole of N present = 14 .01 ∴Mg : N = 0.0185 : 0.0123 = 1.5 : 1

= 3 : 2 (simplest integral ratio) ∴The empirical formula of magnesium nitride should be Mg3N2.

Prepared by Mr. Chau Chi Keung, Richard

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Modern College F.6 Chemistry (2009 – 10)

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C. Derivation of Empirical formula using combustion data





(a)

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? Mass of C in vitamin C = Mass of C in CO2 collected = 0.2998 ×

12 .0 44 .01

= 0.0818 g Mass of H in vitamin C = Mass of H in H2O collected 2.0 18 .02

= 0.0819 ×

= 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=

(b)

0.0818 0.0092 0.1090 : : 12 .01 1.00 16 .0

= 1 : 1.33 :1 =3:4:3 ∴The empirical formula of vitamin C should be C3H4O3. Let the molecular formula of vitamin C be (C3H4O3)n, where n is a positive integer. Relative molecular mass of (C3H4O3)n = 176 ∴(3 x 12.0 + 4 x 1.0 + 3 x 16) n = 176 n=2 Prepared by Mr. Chau Chi Keung, Richard

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Modern College F.6 Chemistry (2009 – 10)

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∴The molecular formula of vitamin C should be C6H8O6. *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 0.0818 0 . 0092 0 . 1090 No. of moles −3 12 .0

= 6.82 ×10

1.0

= 0.0092

16 .0

Relative no. of moles

0.0092 6.82 ×10 −3 ≈ 1.35 ≈1 −3 6.81 ×10 −3 6.81 ×10 Simplest mole ratio 3 4 ∴The empirical formula of vitamin C should be C3H4O3.

= 6.81 ×10 −3

6.81 ×10 −3 =1 6.81 ×10 −3 3

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 Prepared by Mr. Chau Chi Keung, Richard

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Modern College F.6 Chemistry (2009 – 10)

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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 =

2.43 = 0.1 24 .3

(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 =

1.28 = 0.04 2 ×16 .0

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: n a : n b : n c : n d = a : b : c : d  By the Avogadro’s Law, at fixed T and P, V ∝ n 

Va : Vb : involving Vc : Vd = a gaseous : b : c : d reactants and products, mole ratio = volume ratio. In the ∴ reactions

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 =

152 = 1.52 100

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. Prepared by Mr. Chau Chi Keung, Richard

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Modern College F.6 Chemistry (2009 – 10)

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Let the molecular formula of the hydrocarbon be CxHy. 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 cm3 Volume of CO2(g) formed = 70 – 50 = 20 cm3 y 4

CxHy(g) + ( x + ) O2(g) → xCO2(g) +

y H2O(l) 2 y 4

For mole ratio, CxHy : O2 : CO2 = 1 : ( x + ) : x

∴For volume ratio, CxHy : O2 : CO2 = 1 : ( x + 4y ) : x Volume of CO 2 (g) x = Volume of C x H y (g ) 1

∴x = 2 Volume of O 2 (g) = Volume of C x H y (g )

x+

y 4

1

∴x + 4y = 3 As x = 2, we have 2 +

y =3 4

y=4 ∴The molecular formula of the hydrocarbon is C2H4. 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 cm 3. On adding concentrated potassium hydroxide the volume decreased to 8 cm3. Find the molecular of that hydrocarbon. Let the molecular formula of the hydrocarbon be CxHy. Volume of hydrocarbon reacted = 10 cm3 Volume of CO2(g) formed = 28 – 8 = 20 cm3 Prepared by Mr. Chau Chi Keung, Richard

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Modern College F.6 Chemistry (2009 – 10)

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Volume of O2(g) reacted = 33 – 8 = 25 cm3 y 4

CxHy(g) + ( x + ) O2(g) → xCO2(g) + For mole ratio,

y H2O(l) 2

y 4

: (x + ) : x

1

For volume ratio, 10

:

25

: 20

1 10 = x 20

∴x = 2 x+

y 25 = 4 10

25 ∴2 + 4y = 10

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. Let the molecular formula of the hydrocarbon be CxHy. Volume of hydrocarbon reacted = 10 cm3. Volume of CO2(g) formed = 40 cm3 For mole ratio, For volume ratio,

y 4

: (x + ) : x

1 1

y 4

: (x + ) : x

10

y 4

: 10 ( x + ) : 10x

10 x = 40 x=4 By contraction (i.e. decrease) in volume of 35 cm3, y [10 +10 ( x + )] − 40 = 35 4

y = 10 ∴The molecular formula of the hydrocarbon is C4H10.

C. Stoichiometric calculations involving concentrations and volumes of solutions. Prepared by Mr. Chau Chi Keung, Richard

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Modern College F.6 Chemistry (2009 – 10) 

Always remember: Molarity =

Section 1

No. of mole of solute Volume of solution (in dm 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 =

11 .5 = 0.288 40 .0

Concentration of NaOH solution =

0.288 = 0.192 M 1 .5

(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) + H2O(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.



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.

A.

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Modern College F.6 Chemistry (2009 – 10) 

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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 and  reagent Y, because: One of the reagents is a solid and thus the reaction is very slow. (i.e. we do not know whether 1. 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. No suitable indicator for the detection of end-point. For example, titration between NH 4Cl(aq) 2. 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. NH4NO3) 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. NH4NO3(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 NH4NO3 + 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  Prepared by Mr. Chau Chi Keung, Richard

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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

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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:  A standard solution of iodine is prepared by dissolving a known amount of pure potassium 1. 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) → S4O62–(aq) + 2I–(aq) thiosulphate tetrathionate (brown) (colourless) (colourless) (colourless)

3.

Hint: Standardization of thiosulphate with standard I2(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 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 I 2, so starch solution cannot be added at the beginning of the titration as it would hinder the reaction between iodine and thiosulphate.

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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. The end point is shown by the complete decolorization of the blue colour. (Iodine is completely reacted). 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 =

2.015 1 × = 9.42 10-4 × 214 10

∴No. of mole of I2 (aq) formed = 3 × 9.42 × 10-3 = 2.83 × 10-3

From the equation I2(aq) + 2S2O32-(aq) → S4O62-(aq) + 2I-(aq) No. of mole of S2O32-(aq) used = 2.83 × 10-3 × 2 = 5.66 × 10-3 29.8 cm3 = 0.0298 dm3

∴S2O32-(aq)] =

5.66 ×10 −3 = 0.190 M 0.0298

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-5water 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) I2(aq) + 2S2O32–-(aq) → S4O62–-(aq) + 2I–(aq) -------------------- (1) 2Cu2+(aq) + 4I–(aq) → I2(aq) + 2CuI(s) -------------------------(2) From (1), No. of mole of S2O32-(aq) = No. of mole of I2 (aq) formed × 2

∴No. of mole of I2 (aq) formed =

1 18 .0 × 0.095 × ( ) =8.55 10-4 × 2 1000

∴No. of mole of Cu2+(aq) in 25.0 cm3 solution = 2 × 8.55 × 10-4 = 1.71 × 10-3 Prepared by Mr. Chau Chi Keung, Richard

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∴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 =

1.086 × 100% = 25.5% 4.256

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 cm 3 of potassium manganate(VII) were used, what is the concentration of the solution? From the equation: 2MnO4-(aq) + 5C2O42-(aq) + 16H+(aq) → 2Mn2+(aq) + 10CO2(g) + 8H2O(l) Mole ratio, MnO4-(aq) : C2O42-(aq) = 2 : 5

2 × No. of mole of C2O42-(aq) 5 No. of mole of C2O42-(aq) = 0.200 × 25 × 10-3 = 5.00 × 10-3 No. of mole of MnO4-(aq) =

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2 × 5.00 10-3 = 2.00 10-3 5 × ×

2.00 ×10 −3 = 0.116 moldm-3 ∴[MnO4 (aq)] = 17 .2 ×10 −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 cm 3. 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) + 4H2O(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 =

6.50 ×100 % = 86.7% 7.50

Supplementary Notes on Titrations On preparation of standard solutions

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 On techniques of performing acid-base titration Prepared by Mr. Chau Chi Keung, Richard

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