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Chapter 3

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Chapter 3

Chemical Equations

and Stoichiometry

3.1Formulae of Compounds

3.2Derivation of Empirical Formulae

3.3Derivation of Molecular Formulae

3.4 Chemical Equations

3.5 Calculations Based on Equations

3.6 Simple Titrations

3.1 Formulae of Compounds (SB p.54)

ratio of no. of atoms

How can you describe the composition of compound X?

1st way = by chemical formula

C?H?

3.1 Formulae of Compounds (SB p.54)

carbonatoms

hydrogen atoms

How can you describe the composition of compound X?

Compound X

2nd way = by percentage by mass

Mass of carbon atoms inside = …. g

Mass of hydrogen atoms inside = …. g

3.1 Formulae of Compounds (SB p.54)

Compound

Empirical

formula

Molecular formula

Structural formula

(a) Propene

CH2

C3H6

(b) Nitric

acid

HNO3

HNO3

(c) Ethanol

C2H6O

C2H5OH

(d) Glucose

C6H12O6

C6H12O6

Check Point 3-1

Give the empirical molecular and structural formula for the following compounds

Answer

3.1 Formulae of Compounds (SB p.55)

The different types of formulae of some compounds

3.2 Derivation of Empirical Formulae (SB p.56)

Solution:

The relative molecular mass of CO2

= 12.0 + 2 x 16.0 = 44.0

Mass of carbon in 2.93 g of CO2

= 2.93 g x 12.0/44.0 = 0.80 g

The relative molecular mass of H2O

= 2 x 1.0 + 16.0 = 18.0

Mass of hydrogen in 1.80 g of H2O

= 1.80 g x 2.0/18.0 = 0.20 g

Example 3-1

A hydrogen was burnt completely in excess oxygen. It was found that 1.00 g of the hydrocarbon gives 2.93 g of carbon dioxide and 1.8 g of water. Find the empirical formula of the hydrocarbon.? (R.a.m.* : H = 1.0, C = 12.0, O = 16.0)

Answer

3.2 Derivation of Empirical Formulae (SB p.57)

Solution: (cont’d)

Let the empirical formula of the hydrocarbon be CxHy.

Mass of carbon in CxHy = Mass of carbon in CO2

Mass of hydrogen in CxHy = Mass of hydrogen in H2O

The simplest whole number ratio of x and y can be determined by the following the steps in the below table.

3.2 Derivation of Empirical Formulae (SB p.57)

3.2 Derivation of Empirical Formulae (SB p.57)

Solution:

Mass of compound X = 0.46g

Mass of carbon in compound X

= 0.88 g x 12.0/44.0 = 0.24 g

Mass of hydrogen in compound X

= 0.54 g x 2.0/18.0 = 0..06g

Mass of oxygen in compound X

= 0.46 g – 0.24 g – 0.06 g = 0.16 g

Example 3-2

Compound X is known to contain carbon, hydrogen and oxygen only. When it is burnt completely in excess oxygen, carbon dioxide and water are given out as the only products. It is found that 0.46 g of compound X gives 0.88 g of carbon dioxide and 0.54 g of water. Find the empirical formula of compound X. (R.a.m.* : H = 1.0, C = 12.0, O = 16.0)

Answer

3.2 Derivation of Empirical Formulae (SB p.57)

Solution: (cont’d)

Let the empirical formula of compound X be CxHyOz.

Therefore, the empirical formula of compound X is C2H6O.

3.1 Formulae of Compounds (SB p.58)

Check Point 3-2

(a) 5 g of sulphur forms 10 g of an oxide on burning.What is the empirical formula of the oxide?

(R.a.m. : O = 16.0, S = 32.1)

(b) 19.85 f of element M combines with 25.61 g of oxygen to form an oxide. If the relative atomic mass of M is 331.0, find the empirical formula of the oxide.

(R.a.m. : O =16.0)

(c) Determine the empirical formula of copper(II) oxide using the following results.

Experimental results:

Mass of test tube = 21.430 g

Mass of test tube + Mass of copper(II) oxide = 23.321g

Mass of test tube + Mass of copper = 22.940g

(R.a.m. : Cu = 63.5, O = 16.0)

(a)Mass of sulphur = 5 g

Mass of oxygen = (10 – 5) g

The empirical formula of the sulphur oxide is SO2.

Answer

3.1 Formulae of Compounds (SB p.58)

(b)

The empirical formula of the oxide is M2O5.

3.1 Formulae of Compounds (SB p.58)

(c)

Mass of Cu = (22.940 - 21.430) g = 1.51g

Mass of O = (23.321 - 22.940) = 0.381 g

The empirical formula of the oxide is CuO.

3.2 Derivation of Empirical Formulae (SB p.58)

From Combustion by Mass

Composition by mass

Empirical formula

3.2 Derivation of Empirical Formulae (SB p.58)

Solution:

Let the empirical formula of the hydrocarbon be CxHy, and the mass of the compound be 100 g.

Mass of carbon in the compound = 75 g

Mass of hydrogen in the compound=(100 –75) g = 25 g

Therefore, the empirical formula of the hydrocarbon is CH4.

Example3-3Compound A contains carbon and hydrogen only. It is found that the compound contains 75% carbon by mass. Determine its empirical formula. (Relative atomic masses: C=12, H=1 )

Answer

3.2 Derivation of Empirical Formulae (SB p.59)

Solution:

Let the mass of phosphorus chloride be 100g. Then,

Mass of phosphorus in the compound = 22.55g

Mass of chloride in the compound = 77.45g

Therefore, the empirical formula of the phosphorus chloride is PCl3.

Example 3-4

The percentage by mass of phosphorus and chlorine in a sample of a phosphorus chloride are 22.55% and 77.45% respectively. Find the empirical formula of the chloride. (R.a.m. : P = 31.0, Cl = 35.5)

Answer

3.2 Derivation of Empirical Formulae (SB p.59)

- Let the mass of vitamin C analyzed be 100g.
- The empirical formula of vitamin C is C3H4O3.

- Check Point 3-3
- Find the empirical formula of vitamin C if it consists of 40.9% caarbon, 54.5% oxygen and 4.6% hydrogen by mass. ( R.a.m.: C = 12.0, H = 1.0, O = 16.0)
- Each 325 mg tablet of aspirin consists of 195.0 mg carbon 14.6 mg hydrogen and 115.4mg oxygen. Determine the empirical formula of aspirin. (R.a.m. : C= 12.0, H = 1.0, O = 16.0)

Answer

3.2 Derivation of Empirical Formulae (SB p.59)

(b) In order to facilitate calculation, the masses of the elements are multiplied by 1000 first.

The empirical formula of aspirin is C9H8O4.

3.3 Derivation of Molecular Formulae (SB p.60)

Molecular formula

?

= (Empirical formula)n

3.3 Derivation of Molecular Formulae (SB p.60)

Empirical formula

Molecular mass

Molecular formula

3.3 Derivation of Molecular Formulae (SB p.60)

Solution:

Let the empirical formula of the hydrocarbon be CxHy.

Mass of carbon in the hydrocarbon

= 14.6g x 12.0/44.0 = 4.0g

Mass of hydrogen in the hydrocarbon

= 9.0g x 2.0/18.0 = 1.0g

Example 3-5

A hydrogen was burnt completely in excess oxygen. It was found that 5.00 g of the hydrocarbon gives 14.6 g of carbon dioxide and 9.0 g of water. Given that the relative molecular mass of the hydrocarbon is 30.0, determine its molecular formula. hydrocarbon.? (R.a.m.* : H = 1.0, C = 12.0, O = 16.0)

Answer

3.3 Derivation of Molecular Formulae (SB p.60)

Solution: (cont’d)

Therefore, the empirical formula of the hydrocarbon is CH3. The molecular formula of the hydrocarbon is (CH3)n.

Relative molecular mass of (CH3)n = 30.0

n x (12.0 + 1.0 x 3) = 30.0

n= 2

Therefore, the molecular formula of the hydrocarbon is C2H6.

3.3 Derivation of Molecular Formulae (SB p.61)

Solution:

Let the empirical formula of the hydrocarbon be CxHyOz. Mass of carbon in the compound = 44.44g

Mass of hydrogen in the compound = 6.18g

Mass of oxygen in the compound = 49.38g

Example 3-6

Compound X is known to contain 44.44% carbon, 6.18% hydrogen and 49.38% oxygen by mass. A typical analysis shows that it has a relative molecular mass of 162.0. Find its molecular formula(R.a.m.* : H = 1.0, C = 12.0, O = 16.0)

Answer

3.3 Derivation of Molecular Formulae (SB p.61)

Solution(cont’d)

The empirical formula of compound X is C6H10O5.

The molecular formula of compound X is (C6H10O5)n.

Relative molecular mass of (C6H10O5)n = 162.0

n x (12.0 x 6 + 1.0 x 10 + 16.0 x 5) = 162.0

n = 1

Therefore, the molecular formula of compound is C6H10O5.

3.3 Derivation of Molecular Formulae (SB p.61)

Water of Crystallization Derived from

Composition by Mass

3.3 Derivation of Molecular Formulae (SB p.61)

Example 3-7

The chemical formula of hydrated copper(II) sulphate is known to be CuSO4.xH2O. It is found that the percentage of water by mass in the compound is 36%. Find x.(R.a.m. : H=1.0, O=16.0, S=32.1, Cu=63.5)

Solution:

Let Relative molecular mass of CuSO4xH2O

= 63.5 + 32.1 + 16.0 x 4 + (1.0x2 = 16.0)x

= 159.6 + 18x

Relative molecular mass of water of crystallization =18x

18x/(159.6 + 18x) = 36/100

1800x=5745.6 + 648 x

1152x= 5745.6

x = 4.99 5

Therefore, the chemical formula of hydrated copper(II)

sulphate is CuSO4 5H2O

Answer

3.3 Derivation of Molecular Formulae (SB p.63)

- (i) Let the mass of compound Z be 100g.
- The empirical formula of compound Z is CH2O.

- Check Point 3-4
- Find Compound Z is the major component of a healthy drink. It contains 40.00% carbon, 6.67% hydrogen and 53.33% oxygen.
- (i) Find the empirical formula of compound Z.
- (ii) If the relative molecular mass of compound Z is 180, finds its molecular formula.(R.a.m. : C= 12.0, H = 1.0, O = 16.0)

Answer

3.3 Derivation of Molecular Formulae (SB p.63)

(a) (ii) Let the molecular formula of

compound Z be (CH2O)n.

n x (12.0 = 1.0 x 2 = 16.0) = 180

30n = 180

n = 6

The molecular formula of Z is C6H12O6.

3.3 Derivation of Molecular Formulae (SB p.63)

- (b)
- Since the chemical formula of (NH4)2Sx is (NH4)2S3, the value of x is 3.

Check Point 3-4

(b) (NH4)2Sx contains 72.72% sulphur by mass is water. Find the value of x.

(R.a.m.: H = 1.0, N = 14.0, O = 16.0)

(c) In the compound MgSO4nH2O, 51.22% by mass is water. Find the value of n.

(R.a.m.: H = 1.0, O = 16.0, Mg = 24.3, S = 32.1)

Answer

3.3 Derivation of Molecular Formulae (SB p.63)

- (c)
- Since the chemical formula of MgSO4nH2O
- is MgSO47H2O , the value of x is 7.

3.3 Derivation of Molecular Formulae (SB p.63)

Example 3-8

The chemical formula of ethanoic acid is CH3COOH. Calculate the percentages by mass of carbon, hydrogen and oxygen by mass respectively. (R.a.m. : C=12.0, H=1.0, O=16.0 )

Solution:

Relative molecular mass of CH3COOH

= 12.0 x 2 + 1.0 x 4 + 16.0 x 2 = 60.0

% by mass of C = 12.0 x 2/ 60.0 x 100%= 40.00%

% by mass of H = 1.0 x 4 /60.0 x 100% = 6.67%

% by mass of O = 16.0 x 2/60.0 x 100% = 53.33%

The percentage by mass of carbon, hydrogen and oxygen are 40.00%, 6.67% and 53.33% respectively.

Answer

3.3 Derivation of Molecular Formulae (SB p.63)

Example 3-9

Calculate the mass of iron metal in a sample of 20g of hydrated iron (II) sulphate, FeSO47H2O. (R.a.m. : Fe = 55.8 , H=1.0, O=16.0 )

Solution:

Relative molecular mass of FeSO4·7H2O

= 55.8 + 32.1 + 16.0 x 4 + (1.0x2+16.0) x 7=277.9

% by mass of Fe = 55.8/277.9 x 100% = 20.08%

Mass of Fe = 20g x 20.08% = 4.02g

Answer

3.3 Derivation of Molecular Formulae (SB p.63)

- Molar mass of K2Cr2O7
- = (39.1x2+52.0+16.0x7) g mol-1 = 294.2 g mol-1
- % by mass of K
- = 39.1 x 2 g mol-1/294.2 g mol-1 x 100% = 26.58%
- % by mass of Cr
- = 52.0 x 2 g mol-1/294.2g mol-1 x 100% =35.25%
- % by mass of O
- = 16.0 x 7 g mol-1/294.2g mol-1 x 100% = 38.07%

- Check Point 3-5
- (a) Calculate percentages by mass of potassium, chromium and oxygen in potassium chromate (VI), K2Cr2O7.(R.a.m. : K = 39.1 . Cr = 52.0, O = 16.0)
- (b) Find the mass of metal and water of crystallization in
- 100 g of Na2SO4·10H2O;
- 70g of Fe2O3·8H2O.
- (R.a.m.: H = 1.0, O = 16.0, Na = 23, S = 32.1, Fe = 55.8)

Answer

3.3 Derivation of Molecular Formulae (SB p.63)

(b)( i) Molar mass of Na2SO4·10H2O= 322.1 g mol-1

Mass of Na = 23.0 x 2 g mol-1/ 322.1 g mol-1 x 100g

= 14.28 g

Mass of H2O = 18.0 x 10 g mol-1/ 322.1 g mol-1 x 100g = 14.28 g

(ii) Molar mass of Fe2O3·8H2O= 303.6 g mol-1

Mass of Fe = 55.8 x 2 g mol-1/303.6g mol-1 x 70g

= 25.73 g

Mass of H2O = 18.0 x 8 g mol-1/303.6g mol-1 x 70g

= 33.20 g

3.4 Chemical Equations (SB p.64)

mole ratios

Chemical Equations

a A + b B c C + d D

(can also be volume ratios for gases)

Stoichiometry

= relative no. of moles of substances involved

in a chemical reaction.

3.4 Chemical Equations (SB p.64)

Check Point 3-6

Give the chemical equations for the following reactions:

(a) Zinc + steam zinc oxide + hydrogen

(b) Magnesium + silver nitrate

silver + magnesium nitrate

(c) Butane + oxygen carbon dioxide + water

- Zn(s) + H2O(g) ZnO(s) + H2(g)
- (b) Mg(s) + 2 AgNO3(aq) 2Ag(s) + Mg(NO3)2(aq)
- (c) 2C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(l)

Answer

3.5 Calculations Based on Equations (SB p.65)

Calculations Based on Equations

Calculations involving Reacting Masses

3.5 Calculations Based on Equations (SB p.65)

Solution:

CuO(s) + H2(g) Cu(s) + H2O(l)

As the mole ratio of Cu : CuO is 1 : 1, the number of moles of Cu formed is the same as the number of moles of CuO reduced.

Number of moles of CuO reduced

= 12.45/ (63.5 + 13.0) g mol-1 = 0.157 mol

Number of mole of Cu formed = 0.157 mol

Mass of Cu / 63.5 g mol-1 = 0.157

Mass of Cu = 0.157 mol x 63.5 g mol-1= 9.97g

Therefore, the mass of copper formed in the reaction is 9.97g.

Example 3-10

Calculate the mass of copper formed when 12.45g of copper(II) oxide is completely reduced by hydrogen. (R.a.m. : H=1.0, O=16.0, Cu = 63.5 )

Answer

3.5 Calculations Based on Equations (SB p.65)

Example 3-11

Sodium hydrogencarbonate decomposes according to the following equation.

2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(l)

In order to obtain 240 cm3 of CO2 at room temperature and pressure, what is the minimum amount of sodium hydrogencarbonate required?

(R.a.m. : H = 1.0, C =12.0, O = 16.0, Na = 23.0;

molar volume of gas at R.T.P. = 24.0 dm3mol-1)

Solution:

Number of moles of CO2 formed

= 240cm3/ 24000cm3 mol-1 = 0.01 mol

From the equation, 2 moles of NaHCO3(s) will form 1 mole of CO2(g).

Number of moles of NaHCO3 required

= 0.01 x 2 = 0.02 mol

Mass of NaHCO3 required

= 0.02 mol x(23.0 + 1.0 + 16.0 x 3) g mol-1

= 0.02 mol x 84.0g mol-1

= 1.68 g

Therefore, the minimum amount of sodium hydrogencarbonate required is 1.68g.

Answer

3.5 Calculations Based on Equations (SB p.66)

Calculations Based on Equations

Calculations involving Volumes of Gases

3.5 Calculations Based on Equations (SB p.66)

Solution:

Number of moles of CO2 formed

C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l)

2 mol : 7 mol : 4 mol : 6 mol

(from equation)

2 volumes: 7 volumes : 4 volumes : -

(by Avogadro’s law)

It can be judged from the equation that the mole ratio of CO2 : C2H6 is 4 :2, and the volume ratio of CO2 : C2H6 should also be 4:2.

Let x be the volume of CO2(g) formed

x /20cm3 = 4/2

x = 40 cm3

Therefore, the volume of CO2 formed is 40 cm3.

Example 3-12

Calculate the volume of carbon dioxide formed when 20 cm3 of ethane and 70 cm3 of oxygen are exploded, assuming all volumes are measured at room temperature and pressure.

Answer

3.5 Calculations Based on Equations (SB p.67)

Solution:

Let the molecular formula of the hydrocarbon be CxHy.

Volume of hydrogen reacted = 10 cm3

Volume of O2(g) unreacted = 50 cm3

Volume of O2(g) reacted = 30 cm3

Volume of CO2(g) formed = 20 cm3

CxHy + (x + y/4) O2 CO2 + y/2 H2O

1 volume : (x + y/4) volumes : x volumes

Volume of CO2 (g)/ volume of CxHy(g)

= 20 cm3/ 10cm3 = 2 X =2

Volume of O2(g) / volume of CxHy(g)

=(x + y/4) / 1= 30/ 10

(x + y/4)= 3 Y = 2

Molecular formula is C2H4

Example 3-13

10 cm3 of a gaseous hydrocarbon was mixed with 80cm3 of oxygen which was in excess. The mixture was exploded and then cooled. The volume left was 70cm3. Upon passing the resulting gaseous mixture through concentrated sodium hydroxide solution ( to absorb carbon dioxide), the volume of the residual gas became 50 cm3. Find the molecular formula of the hydrocarbon.

Answer

3.5 Calculations Based on Equations (SB p.68)

- No. of moles of H2 = No. of moles of Mg
- Volume of H2/ 24.0 dm3 mol-1 = 2.43 g / 24.3 g mol-1
- Volume of H2 =2.4 dm3
- (b) 1/3 x no. of moles of Cl2 =1/2x no. of moles of PCl3
- 1/3 x mass of Cl2 / (35.5 x 2) g mol-1
- = 1/2 x 100g / (31.0 + 35.5 + 3 ) g mol-1
- Mass of Cl2 =77.45g

- Check Point 3-7
- Find the volume of hydrogen produced at R.T.P. when 2.43g of magnesium reacts with excess hydrochloric acid. (R.a.m. : Mg = 24.3; molar volume of gas at R.T.P. = 24.0 dm3mol-1.
- Find the minimum mass of chlorine required to produced 100 g of phosphorus trichloride ( PCl3).
- 20 cm3 of a gaseous hydrocarbon and 150 cm3 of oxygen were exploded in a closed vessel. After cooling, 110 cm3 of gases remained. After passing through a solution of concentrated sodium hydroxide, the volume left was 50 cm3 .Determine the molecular of the hydrocarbon.
- Calculate the volume of carbon dioxide formed when 5cm3 of methane burns in excess oxygen, assuming all volumes are measured at room temperature and pressure.

Answer

3.5 Calculations Based on Equations (SB p.68)

(c)Volume of CxHy used = 20 cm3

Volume of CO2 formed = 60 cm3

Volume of O2 used = 100 cm3

Volume of CxHy : volume of CO2 = 1 : x = 20 : 60

x = 3

Volume of CxHy : volume of O2 = 1 : x + y/4 = 20 : 100

x + y/4 = 5

3 + y/4 = 5

y = 8

3.5 Calculations Based on Equations (SB p.68)

(d)Volume of CxHy used = 20 cm3

It can be judged from the equation that the mole ratio of CO2 : CH4 is 1:1, the volume ratio of CO2 : CH4 should also be 1:1.

x / 5 = 1/1

x = 5

The volume of carbon dioxide gas is 5 cm3.

3.6 Simple Titrations (SB p.68)

Acid-Base Titrations

Acid-Base Titrationswith Indicators

Acid-Base Titrationswithout Indicators

(to be discussed in later chapters)

3.6 Simple Titrations (SB p.69)

Copper(II) sulphate solution

Copper(II) sulphate

+

solute

Water

solvent

solution

3.6 Simple Titrations (SB p.69)

~50 cm3

3.6 Simple Titrations (SB p.69)

~50 cm3

3.6 Simple Titrations (SB p.69)

~50 cm3

3.6 Simple Titrations (SB p.69)

~50 cm3

3.6 Simple Titrations (SB p.69)

~50 cm3

3.6 Simple Titrations (SB p.69)

~50 cm3

3.6 Simple Titrations (SB p.69)

50 cm3

Solution A

3.6 Simple Titrations (SB p.69)

~50 cm3

3.6 Simple Titrations (SB p.69)

~50 cm3

3.6 Simple Titrations (SB p.69)

~50 cm3

3.6 Simple Titrations (SB p.69)

~50 cm3

3.6 Simple Titrations (SB p.69)

~50 cm3

3.6 Simple Titrations (SB p.69)

~50 cm3

3.6 Simple Titrations (SB p.69)

50 cm3

Solution B

3.6 Simple Titrations (SB p.69)

~100 cm3

3.6 Simple Titrations (SB p.69)

~100 cm3

3.6 Simple Titrations (SB p.69)

~100 cm3

3.6 Simple Titrations (SB p.69)

~100 cm3

3.6 Simple Titrations (SB p.69)

~100 cm3

3.6 Simple Titrations (SB p.69)

~100 cm3

3.6 Simple Titrations (SB p.69)

100 cm3

Solution C

3.6 Simple Titrations (SB p.69)

2 x the amount of solute

Concentration of solution B is 2 times that of the concentrations of solutions A & B.

contain the same amount of solute (same concentration)

Concentrationis theamount of solutein aunit volume of solution.

3.6 Simple Titrations (SB p.69)

Class Practice

Suppose the right-handed side figure shows the number of solute particles in solution D. Draw similar particle models for Solutions A, B and C.

3.6 Simple Titrations (SB p.69)

Class Practice Answers

3.6 Simple Titrations (SB p.69)

no. of spoons

no. of moles

mass

Concentrationis theamount of solutein aunit volume of solution.

3.6 Simple Titrations (SB p.69)

Unit: moles/dm3

A way of expressing concentrations

Molarityis thenumber of molesof solutedissolved in 1 dm3 (1000 cm3) of solution.

(M)

3.6 Simple Titrations (SB p.69)

What does this mean?

1 dm3

contains 2 moles of HCl

“In every 1 dm3 of the solution, 2 moles of HCl is dissolved.”

3.6 Simple Titrations (SB p.71)

Solution:

Number of moles of NaOH(aq) 2

Number of moles of H2SO4(aq) = 1

½ x Number of moles of NaOH(aq)

= Number of moles of H2SO4 (aq)

= 0.067 mol dm-3 x 22.5 x 10-3 dm3 = 1.508 x 10-3 mol

Number of moles of NaOH(aq) = 2 x 1.508 x 10-3 mol

= 3.016 x 10-3 mol

Molarity of NaOH(aq)

= 3.012 x 10-3 mol / 25.0 x 10-3 mol

= 0.1221 mol dm-3

The molarity of NaOH is 0.121M

Example 3-14

25.0cm3 of sodium hydroxide solution was titrated against 0.067 M of sulphuric(VI) acid using methyl orange as indicator. The indicator changed colour from yellow to red when 22.5 cm3 of sulphuric(VI) acid had benn added. Calculate the molarity of the sodium hydroxide solution.

Answer

3.6 Simple Titrations (SB p.71)

- Solution:
- Number of moles of acid = 2.52 g/ 126.0 g mol-1
- = 0.02mol
- Molarity of acid solution = 0.02 mol / 250 x 10-3 dm3
- = 0.08M
- (b) H2X(aq) + 2NaOH(aq) Na2X(aq) + 2NaOH(l)
- (c) Number of moles of H2X
- =½ x number of moles of NaOH
- Molarity of NaOH = 0.14M

- Example 3-15
- 2.52 g of a pure dibasic acid with formula mass of 126.0 was dissolved in water and made up to 250.0 cm3 in a volumetric flask 25.0 cm3 of this solution was found to neutralize 28.5 cm3 of sodium hydroxide solution.
- Calculate the molarity of the acid solution.
- If the dibasic acid is represented by H2X, write an equation for the reaction between the acid and sodium hydroxide.
- Calculate the molarity of the sodium hydroxide solution.

Answer

3.6 Simple Titrations (SB p.72)

- Solution:
- Number of moles of acid = 2.52 g / 126.0 g mol-1
- There is a sudden drop in the pH value of the solution (from pH 3 to pH 8) with the end point at 30.0 cm3.
- Na2CO3·nH2O(s) + 2 HCl(aq)
- 2NaCl(aq) + CO2(g) + (n+1)H2O(l)
- Number of moles of Na2CO3·nH2O
- = ½ x 0.1 mol dm-3 x 30 x 10-3 dm3
- 106.0 + 18.0n = 124.0
- n = 1
- The formula is Na2CO3·H2O

Example 3-16

0.186g of sample of hydrate sodium carbonate, NaCO2·nH2O, was dissolved in 100 cm3 of distilled water in conical flask. 0.10 M by hydrochloric acid was added from a burette, 2 cm3 at a time. The pH value of the solution was measured by a pH meter. The result was recorded and shown in the following figure.

Calculate the value of n in NaCO2·nH2O.

Answer

3.6 Simple Titrations (SB p.73)

Solution:

(a)

- Example 3-17
- 5 cm3 of 0.5M sulphuric(VI) acid was added to 25.0 cm3 of potassium hydroxide solution. The mixture was then stirred and the highest temperature was recorded. The experiment was repeated with different volumes of the sulphuric(VI) acid. The laboratory set-up and the results were as follows:
- Plot the graph of temperature against volume of sulphuric(VI) acid added.
- Calculate the molarity of the potassium hydroxide solution.
- Explain why the temperature rose to a maximum and the fell.

Answer

3.6 Simple Titrations (SB p.74)

Solution: (cont’d)

(b) From the graph, it is found that the end point of the titration is reached when 20 cm3 of H2SO4 is added.

Number of moles of = 0.5 mol dm-3 x 20/1000 dm3

= 0.01 mol

2KOH(aq) + H2SO4(aq) K2SO4(aq) + 2H2O(l)

2 mol 1 mol

Mole of KOH(aq) : H2SO4 = 2 : 1

Number of moles of KOH(aq)

= 2 x 0.01 mol = 0.02 mol

Molarity of KOH(aq)

= 0.02 mol / (25 x 10-3 dm3) = 0.8M

3.6 Simple Titrations (SB p.74)

Solution: (cont’d)

(c) Neutralization is an exothermic reaction. When more and more sulphuric(VI) acid was added and reacted with potassium hydroxide, the temperature rose. The temperature rose to a maximum value at which the equivalence point of the reaction was reached. After that, any excess sulphuric (VI) acid added cooled down the reacting solution, causing the temperature to drop.

3.6 Simple Titrations (SB p.76)

in burette

in conical flask

Redox Titrations

Some Examples

Iodometric Titrations

I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O62-(aq)

brown

colourless

3.6 Simple Titrations (SB p.76)

Add starch

in burette

in conical flask

Redox Titrations

Some Examples

Iodometric Titrations

I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O62-(aq)

brown

colourless

During titration : brown yellow

3.6 Simple Titrations (SB p.76)

Redox Titrations

Some Examples

Iodometric Titrations

I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O62-(aq)

brown

colourless

During titration : brown yellow

3.6 Simple Titrations (SB p.76)

Redox Titrations

Some Examples

Iodometric Titrations

I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O62-(aq)

brown

colourless

During titration : brown yellow

End point : blue black colourless(after addition of starch indicator)

3.6 Simple Titrations (SB p.76)

Solution:

IO3-(aq) + 5I- + 6H+(aq) 3I2(aq) + 3H2O(l) …(1)

I2(aq) + 2S2O3 2-(aq) 2I-(aq) + S4O6 2-(aq) …(2)

From (1), Number of moles of IO3-(aq)

=1/3 x number of moles of I2(aq)

From(2), Number of moles of I2(aq)

=1/2 x number of moles of S2O32-(aq)

Number of moles of IO3-(aq)

= 1/6 x number of moles of S2O3 2-(aq)

Molority of IO3-(aq) x 25.0/1000 dm3

=1/6 x 0.05 mol dm-3 x 22.0/1000 dm3

Molarity ofIO3-(aq) = 7.33 x 10-3 M

Example 3-18

When excess potassium iodide solution (KI) is added to 25.0 cm3 of acidified potassium iodate solution (KIO3) of unknown concentration, the solution turns brown. This brown solution requires 22.0 cm3 of 0.05 M sodium thiosulphate solution to react completely with the iodine formed, using starch solution as indicator. Find the molarity of the potassium iodate solution.

Answer

3.6 Simple Titrations (SB p.76)

In burette

In conical flask

Redox Titrations

Some Examples

Titrations Involving Potassium Permanganate

MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + H2O(aq) + 5Fe3+(aq)

purple

colourless

3.6 Simple Titrations (SB p.76)

In burette

In conical flask

Redox Titrations

Some Examples

Titrations Involving Potassium Permanganate

MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + H2O(aq) + 5Fe3+(aq)

purple

colourless

During titration : pale green yellow

3.6 Simple Titrations (SB p.76)

In burette

In conical flask

Redox Titrations

Some Examples

Titrations Involving Potassium Permanganate

MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + H2O(aq) + 5Fe3+(aq)

purple

colourless

During titration : pale green yellow

End point : yellow light purple

3.6 Simple Titrations (SB p.77)

Solution:

MnO4-(aq) + 8H+(aq) + 5Fe2+(aq)

Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)

Number of moles of Fe2+(aq)

= 5 x number of moles of MnO4-(aq)

= 5 x 0.02 mol dm-3 x 36.5 x 10-3 dm3

= 3.65 x 10-3 mol

Number of moles of Fe dissolved

= number of mole of Fe 2+ formed = 3.65 x 10-3 mol

Mass of Fe = 3.65 x 10-3 mol x 55.8 g mol-1= 0.204g

Percentage purity of Fe

= 0.204g/0.22g x 100% = 92.73%

Example 3-19

A piece of impure iron wire weighs 0.22g. When it is dissolved in hydrochloric acid, it is oxidized to iron(II) ions. The solution requires 36.5 cm3 of 0.02 M acidified potassium manganate(VII) for complete reaction to form iron(III) ions. What is the percentage purity of the iron wire?

Answer

3.6 Simple Titrations (SB p.78)

- Na2CO3 (s)+ 2 HCl (aq)
- 2NaCl(aq) + H2O(l) + CO2(g)
- No. of moles of Na2CO3 used
- = 5g / (23.0x 2 +12 +16.0 x 3)g mol-1 = 0.0472 mol
- No. of moles of HCl used
- = 2M x 100/ 1000 dm3 = 0.2 mol
- Since HCl is in excess, Na2CO3 is the limiting agent.
- No. of moles of CO2 produced
- =no. of moles of Na2CO3 used = 0.0472 mol
- Volume of CO2 produced
- = 0.0472 mol x 24.0 dm3 mol-1= 1.133 dm3

- Check Point 3-8
- 5g of anhydrous sodium carbonate is added to 100 cm3 of 2 M hydrochloric acid. What is the volume of gas evolved at room temperature and pressure?
- (R.a.m. : C = 12.0, O = 16.0, Na = 23.0; molar volume of gas at R.T.P. = 24.0 dm3 mol-1)
- (b) 8.54g of impure hydrated iron(II) sulphate (formula mass of 392.14) was dissolved in water and made up to 250 cm3. 25cm3 of this solution required 20.76cm3 of 0.0203M acidified potassium manganate(VII) solution for complete reaction. Determine the percentage purity of the hydrated iron(II) sulphate.

Answer

3.6 Simple Titrations (SB p.78)

(b)No. of moles of MnO4-

= 0.0203M x 20.76/1000 dm3 = 4.214 x 10-4 mol

No. of moles of Fe2+

= 5 x no. of moles of MnO4- = 2.107 x 10-3 mol

No. of mole of Fe2+ in 25.0 cm3 solution = 2.107 x 10-3 mol

No. of mole of Fe2+ in 250.0 cm3 solution = 0.021 07 mol

Molar mass of hydrated FeSO4 =392.14 g mol –1

Mass of hydrated FeSO4

= 0.021 07 mol x 392.14 g mol –1= 8.26g

5 purity of FeSO4 = 8.26g/8.54g x 100% = 96.72%

The END