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3. Chemical Equations and Stoichiometry. 3.1 Formulae of Compounds 3.2 Derivation of Empirical Formulae 3.3 Derivation of Molecular Formulae 3.4 Chemical Equations 3.5 Calculations Based on Chemical Equations 3.6 Simple Titrations. 3.1. Formulae of Compounds.

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3

Chemical Equations

and Stoichiometry

3.1 Formulae of Compounds

3.2 Derivation of Empirical Formulae

3.3 Derivation of Molecular Formulae

3.4 Chemical Equations

3.5 Calculations Based on Chemical Equations

3.6 Simple Titrations


3.1

Formulae of Compounds


Formulae of compounds

3.1 Formulae of compounds (SB p.43)

ratio of no. of atoms

Formulae of compounds

How can you describe the composition of compound X?

1st way = by chemical formula

C?H?


3.1 Formulae of compounds (SB p.43)

carbonatoms

Check Point 3-1

hydrogen atoms

How can you describe the composition of compound X?

2nd way = by percentage by mass

Compound X

Mass of carbon atoms inside = …. g

Mass of hydrogen atoms inside = …. g


3.1 Formulae of compounds (SB p.44)

Different types of formulae of some compounds


3.2

Derivation of Empirical Formulae


3.2 Derivation of empirical formulae (SB p.45)

From combustion data

  • During complete combustion, elements in a compound are oxidized.

  • e.g. carbon to carbon dioxide, hydrogen to water, sulphur to sulphur dioxide

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


3.2 Derivation of empirical formulae (SB p.45)

Example 3-2A

Example 3-2B

Check Point 3-2A

The laboratory set-up used for determining the empirical formula of a gaseous hydrocarbon


3.2 Derivation of Empirical Formulae (SB p.48)

Example 3-2D

Example 3-2C

Check Point 3-2B

From combustion by mass

Composition by mass

Empirical formula


3.3

Derivation of Molecular Formulae


What is molecular formulae

3.3 Derivation of molecular formulae (SB p.49)

What is molecular formulae?

Molecular formula

?

= (Empirical formula)n


From empirical formula and known relative molecular mass

3.3 Derivation of Molecular Formulae (SB p.49)

Example 3-3A

Example 3-3B

Molecular formula

From empirical formula and known relative molecular mass

Empirical formula

Molecular mass


3.3 Derivation of Molecular Formulae (SB p.51)

Example 3-3C

Check Point 3-3A

Water of Crystallization Derived from Composition by Mass


3.3 Derivation of Molecular Formulae (SB p.52)

Example 3-3D

Example 3-3E

Check Point 3-3B

Composition by mass

Find composition by mass from formula

Formula of a compound


3.4

Chemical Equations


mole ratios

Check Point 3-4

3.4 Chemical equations (SB p.53)

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


3.5

Calculations Based on Chemical Equations


3.5 Calculations Based on Equations (SB p.65)

Example 3-5A

Example 3-5B

Calculations based on equations

Calculations involving reacting masses


3.5 Calculations Based on Equations (SB p.66)

Example 3-5C

Check Point 3-5

Example 3-5D

Calculations based on equations

Calculations involving volumes of gases


3.6

Simple Titrations


Simple titrations

3.6 Simple titrations (SB p.58)

Simple titrations

Acid-base titrations

Acid-base titrationswith indicators

Acid-base titrationswithout indicators

(to be discussed in later chapters)


Finding the concentration of a solution

3.6 Simple titrations (SB p.59)

Copper(II) sulphate solution

Finding the concentration of a solution

Copper(II) sulphate

+

solute

Water

solvent

solution


Finding the concentration of a solution1

3.6 Simple titrations (SB p.59)

Finding the concentration of a solution

~50 cm3


Finding the concentration of a solution2

3.6 Simple titrations (SB p.59)

Finding the concentration of a solution

~50 cm3


Finding the concentration of a solution3

3.6 Simple titrations (SB p.59)

Finding the concentration of a solution

~50 cm3


Finding the concentration of a solution4

3.6 Simple titrations (SB p.59)

Finding the concentration of a solution

~50 cm3


Finding the concentration of a solution5

3.6 Simple titrations (SB p.59)

Finding the concentration of a solution

~50 cm3


Finding the concentration of a solution6

3.6 Simple titrations (SB p.59)

Finding the concentration of a solution

~50 cm3


Finding the concentration of a solution7

3.6 Simple titrations (SB p.59)

Finding the concentration of a solution

50 cm3

Solution A


Finding the concentration of a solution8

3.6 Simple titrations (SB p.59)

Finding the concentration of a solution

~50 cm3


Finding the concentration of a solution9

3.6 Simple titrations (SB p.59)

Finding the concentration of a solution

~50 cm3


Finding the concentration of a solution10

3.6 Simple titrations (SB p.59)

Finding the concentration of a solution

~50 cm3


Finding the concentration of a solution11

3.6 Simple titrations (SB p.59)

Finding the concentration of a solution

~50 cm3


Finding the concentration of a solution12

3.6 Simple titrations (SB p.59)

Finding the concentration of a solution

~50 cm3


Finding the concentration of a solution13

3.6 Simple titrations (SB p.59)

Finding the concentration of a solution

~50 cm3


Finding the concentration of a solution14

3.6 Simple titrations (SB p.59)

Finding the concentration of a solution

50 cm3

Solution B


Finding the concentration of a solution15

3.6 Simple titrations (SB p.59)

Finding the concentration of a solution

~100 cm3


Finding the concentration of a solution16

3.6 Simple titrations (SB p.59)

Finding the concentration of a solution

~100 cm3


Finding the concentration of a solution17

3.6 Simple titrations (SB p.59)

Finding the concentration of a solution

~100 cm3


Finding the concentration of a solution18

3.6 Simple titrations (SB p.69)

Finding the concentration of a solution

~100 cm3


Finding the concentration of a solution19

3.6 Simple titrations (SB p.59)

Finding the concentration of a solution

~100 cm3


Finding the concentration of a solution20

3.6 Simple titrations (SB p.59)

Finding the concentration of a solution

~100 cm3


Finding the concentration of a solution21

3.6 Simple titrations (SB p.59)

Finding the concentration of a solution

100 cm3

Solution C


Comment on the concentrations of solutions a b and c

3.6 Simple titrations (SB p.59)

2 x the amount of solute

Comment on the concentrations of solutions A, B and C !

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.


Comment on the concentrations of solutions a b and c1

3.6 Simple titrations (SB p.59)

Comment on the concentrations of solutions A, B and C !

no. of spoons

no. of moles

mass

Concentrationis theamount of solutein aunit volume of solution.


Molarity

3.6 Simple titrations (SB p.59)

Unit: mol dm-3

Molarity

A way of expressing concentrations

Molarity is the number of molesof solute dissolved in1 dm3 (1000 cm3) of solution.

(M)


3.6 Simple titrations (SB p.59)

What does this mean?

1 dm3

contains 2 moles of HCl

Example 3-6A

Example 3-6B

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


3.6 Simple titrations (SB p.62)

Example 3-6C

Titration without an indicator

By change in pH value


3.6 Simple titrations (SB p.62)

Example 3-6D

Titration without an indicator

By change in temperature


3.6 Simple Titrations (SB p.65)

in burette

in conical flask

Redox titrations

1. Iodometric titration

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

brown

colourless


3.6 Simple Titrations (SB p.65)

Add starch

in burette

in conical flask

Redox titrations

1. Iodometric titration

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

brown

colourless

During titration : brown  yellow


3.6 Simple Titrations (SB p.65)

Redox titrations

1. Iodometric titration

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

brown

colourless

During titration : brown  yellow


3.6 Simple Titrations (SB p.65)

Example 3-6E

Redox titrations

1. Iodometric titration

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

colourless

brown

During titration : brown  yellow

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


3.6 Simple Titrations (SB p.66)

In conical flask

In burette

Example 3-6F

Check Point 3-6

Redox titrations

2. Titrations involving potassium permanganate

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

purple

pale green



3.1 Formulae of compounds (SB p.45)

Compound

Empirical

formula

Molecular formula

Structural formula

(a) Propene

CH2

C3H6

(b) Nitric

acid

HNO3

HNO3

(c) Ethanol

C2H6O

C2H5OH

(d) Glucose

C6H12O6

C6H12O6

Back

Check Point 3-1

Give the empirical, molecular and structural formulae for the following compounds:

(a) Propene

(b) Nitric acid

(c) Ethanol

(d) Glucose

Answer


3.2 Derivation of empirical formulae (SB p.46)

Example 3-2A

A hydrocarbon was burnt completely in excess oxygen. It was found that 1.00 g of the hydrocarbon gives 2.93 g of carbon dioxide and 1.80 g of water. Find the empirical formula of the hydrocarbon.

Answer


The relative molecular mass of CO2 = 12.0 + 16.0  2 = 44.0

Mass of carbon in 2.93 g of CO2 = 2.93 g  = 0.80 g

The relative molecular mass of H2O = 1.0  2 + 16.0 = 18.0

Mass of hydrogen in 1.80 g of H2O = 1.80 g  = 0.20 g

Let the empirical formula of the hydrocarbon be CxHy.

Mass of carbon in CxHy = Mass of carbon in CO2

Mass of hydrogen in CxHy = Mass of hydrogen in H2O

3.2 Derivation of empirical formulae (SB p.46)

Example 3-2A


3.2 Derivation of empirical formulae (SB p.46)

Back

Example 3-2A

Therefore, the empirical formula of the hydrocarbon is CH3.


3.2 Derivation of empirical formulae (SB p.46)

Example 3-2B

Compound X is known to contain carbon, hdyrogen and oxygen only. When it is burnt completely in excess oxygen, carbon dioxide and water are given out as the only products. It is found that 0.46 g of compound X gives 0.88 g of carbon dioxide and 0.54 g of water. Find the empirical formula of compound X.

Answer


Mass of compound X = 0.46 g

Mass of carbon in compound X = 0.88 g  = 0.24 g

Mass of hydrogen in compound X = 0.54 g  = 0.06 g

Mass of oxygen in compound X = 0.46 g – 0.24 g – 0.06 g = 0.16 g

Let the empirical formula of compound X be CxHyOz.

3.2 Derivation of empirical formulae (SB p.47)

Example 3-2B


3.2 Derivation of empirical formulae (SB p.47)

Back

Example 3-2B

Therefore, the empirical formula of compound X is C2H6O.


Sulphur

Oxygen

Mass (g)

5

5

No. of moles (mol)

Relative no. of moles

Simplest mole ratio

1

2

3.2 Derivation of empirical formulae (SB p.47)

Check Point 3-2A

  • 5 g of sulphur forms 10 g of an oxide on complete combustion. What is the empirical formula of the oxide?

Answer


  • (b)

  • The empirical formula of the oxide is M2O5.

M

O

Mass (g)

19.85

25.61

No. of moles (mol)

Relative no. of moles

Simplest mole ratio

2

5

3.2 Derivation of empirical formulae (SB p.47)

Check Point 3-2A

  • (b) 19.85 g of element M combines with 25.61 g of oxygen to form an oxide. If the relative atomic mass of M is 31.0, find the empirical formula of the oxide.

Answer


3.2 Derivation of empirical formulae (SB p.47)

Check Point 3-2A

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

Experimental results:

Mass of test tube = 21.430 g

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

Mass of test tube + Mass of copper = 22.940 g

Answer


Copper

Oxygen

Mass (g)

1.51

0.381

No. of moles (mol)

Relative no. of moles

Simplest mole ratio

1

1

3.2 Derivation of empirical formulae (SB p.47)

Back

Check Point 3-2A


3.2 Derivation of empirical formulae (SB p.48)

Example 3-2C

Compound A contains carbon and hydrogen atoms only. It is found that the compound contains 75 % carbon by mass. Determine its empirical formula.

Answer


Let the empirical formula of compound A be CxHy, and the mass of the compound be 100 g. Then, mass of carbon in the compound = 75 g

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

Therefore, the empirical formula of compound A is CH4.

Carbon

Hydrogen

Mass (g)

75

25

No. of moles (mol)

Relative no. of moles

Simplest mole ratio

1

4

3.2 Derivation of empirical formulae (SB p.48)

Back

Example 3-2C


3.2 Derivation of empirical formulae (SB p.48)

Example 3-2D

The percentages by mass of phosphorus and chlorine in a sample of phosphorus chloride are 22.55 % and 77.45 % respectively. Find the empirical formula of the phosphorus chloride.

Answer


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

Mass of phosphorus in the compound = 22.55 g

Mass of chlorine in the compound = 77.45 g

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

Phosphorus

Chlorine

Mass (g)

22.55

77.45

No. of moles (mol)

Relative no. of moles

Simplest mole ratio

1

3

3.2 Derivation of empirical formulae (SB p.48)

Back

Example 3-2D


Carbon

Hydrogen

Oxygen

Mass (g)

40.9

4.6

54.5

No. of moles (mol)

Relative no. of moles

Simplest mole ratio

3

4

3

3.2 Derivation of empirical formulae (SB p.49)

Check Point 3-2B

  • Find the empirical formula of vitamin C if it consists of 40.9 % carbon, 54.5 % oxygen and 4.6 % hydrogen by mass.

Answer


Carbon

Hydrogen

Oxygen

Mass (g)

195.0

14.6

115.4

No. of moles (mol)

Relative no. of moles

Simplest mole ratio

9

8

4

3.2 Derivation of empirical formulae (SB p.49)

Back

Check Point 3-2B

(b) Each 325 mg tablet of aspirin consists of 195.0 mg carbon, 14.6 mg hydrogen and 115.4 mg oxygen. Determine the empirical formula of aspirin.

Answer


3.3 Derivation of molecular formulae (SB p.50)

Example 3-3A

A hydrocarbon was burnt completely in excess oxygen. It was found that 5.0 g of the hydrocarbon gave 14.6 g of carbon dioxide and 9.0 g of water. Given that the relative molecular mass of the hydrocarbon is 30.0, determine its molecular formula.

Answer


Let the empirical formula of the hydrocarbon be CxHy.

Mass of carbon in the hydrocarbon = 14.6 g  = 4.0 g

Mass of hydrogen in the hydrocarbon = 9.0 g  = 1.0 g

Carbon

Hydrogen

Mass (g)

4.0

1.0

No. of moles (mol)

Relative no. of moles

Simplest mole ratio

1

3

3.3 Derivation of molecular formulae (SB p.50)

Example 3-3A


3.3 Derivation of molecular formulae (SB p.50)

Back

Example 3-3A

Therefore, the empirical formula of the hydrocarbon is CH3.

Let the molecular formula of the hydrocarbon be (CH3)n.

Relative molecular mass of (CH3)n = 30.0

n  (12.0 + 1.0  3) = 30.0

n = 2

Therefore, the molecular formula of the hydrocarbon is C2H6.


3.3 Derivation of molecular formulae (SB p.50)

Example 3-3B

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.

Answer


Let the empirical formula of compound X is CxHyOz, and the mass of the compound be 100 g. Then,

Mass of carbon in the compound = 44.44 g

Mass of hydrogen in the compound = 6.18 g

Mass of oxygen in the compound = 49.38 g

The empirical formula of compound X is C6H10O5.

Carbon

Hydrogen

Oxygen

Mass (g)

44.44

6.18

49.38

No. of moles (mol)

Relative no. of moles

Simplest mole ratio

6

10

5

3.3 Derivation of molecular formulae (SB p.50)

Example 3-3B


3.3 Derivation of molecular formulae (SB p.50)

Back

Example 3-3B

Let the molecular formula of compound X be (C6H10O5)n.

Relative molecular mass of (C6H10O5)n = 162.0

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

n = 1

Therefore, the molecular formula of compound X is C6H10O5.


3.3 Derivation of molecular formulae (SB p.51)

Example 3-3C

The chemical formula of hydrated copper(II) sulphate is known to be CuSO4 · xH2O. It is found that the percentage of water of crystallization by mass in the compound is 36 %. Find the value of x.

Answer


Relative formula mass of CuSO4· xH2O

= 63.5 + 32.1 + 16.0  4 + (1.0  2 + 16.0)x

= 159.6 + 18x

Relative molecular mass of water of crystallization = 18x

1800x = 5745.6 + 648x

1152x = 5745.6

x = 4.99

 5

Therefore, the chemical formula of the hydrated copper(II) sulphate is CuSO4 · 5H2O.

3.3 Derivation of molecular formulae (SB p.51)

Example 3-3C

Back


3.3 Derivation of molecular formulae (SB p.52)

Check Point 3-3A

  • Compound Z is the major ingredient of a healthy drink. It contains 40.00 % carbon, 6.67 % hydrogen and 53.33 % oxygen.

  • (i) Find the empirical formula of compound Z.

  • (ii) If the relative molecular mass of compound Z is 180, find its molecular formula.

Answer


3.3 Derivation of molecular formulae (SB p.52)

Check Point 3-3A

  • (i) Let the mass of compound Z be 100 g.

  • Therefore, the empirical formula of compound Z is CH2O.


3.3 Derivation of molecular formulae (SB p.52)

Check Point 3-3A

(ii) Let the empirical formula of compound Z be (CH2O)n.

n  (12.0 + 1.0  2 + 16.0) = 180

30n = 180

n = 6

Therefore, the molecular formula of compound Z is C6H12O6.


(b)

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

(NH4) unit

S

Mass (g)

27.28

72.72

No. of moles (mol)

Relative no. of moles

Simplest mole ratio

2

3

3.3 Derivation of molecular formulae (SB p.52)

Check Point 3-3A

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

Answer


(c)

Since the chemical formula of MgSO4· nH2O is MgSO4 · 7H2O, the value of n is 7.

MgSO4

H2O

Mass (g)

48.78

51.22

No. of moles (mol)

Relative no. of moles

Simplest mole ratio

1

7

3.3 Derivation of molecular formulae (SB p.52)

Back

Check Point 3-3A

(c) In the compound MgSO4 · nH2O, 51.22 % by mass is water. Find the value of n.

Answer


3.3 Derivation of molecular formulae (SB p.52)

Example 3-3D

The chemical formula of ethanoic acid is CH3COOH. Calculate the percentage of mass of carbon, hydrogen and oxygen respectively.

Answer


Relative molecular mass of CH3COOH

= 12.0  2 + 1.0  4 + 16.0  2

= 60.0

% by mass of C =

= 40.00 %

% by mass of H =

= 6.67 %

% by mass of O =

= 53.33 %

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

3.3 Derivation of molecular formulae (SB p.52)

Example 3-3D

Back


Relative formula mass of FeSO4· 7H2O

= 55.8 + 32.1 + 16.0  4 + (1.0  2 + 16.0)  7 = 277.9

% by mass of Fe =

= 20.08 %

Mass of Fe = 20 g  20.08 % = 4.02 g

3.3 Derivation of molecular formulae (SB p.53)

Back

Example 3-3E

Calculate the mass of iron in a sample of 20 g of hydrated iron(II) sulphate, FeSO4 · 7H2O.

Answer


  • Molar mass of K2Cr2O7 = (39.1  2 + 52.0  2 + 16.0  7) g mol-1

  • = 294.2 g mol-1

  • % by mass of K =

  • = 26.58 %

  • % by mass of Cr =

  • = 35.35 %

  • % by mass of O =

  • = 38.07 %

3.3 Derivation of molecular formulae (SB p.53)

Check Point 3-3B

  • Calculate the percentages by mass of potassium , chromium and oxygen in potassium chromate(VI), K2Cr2O7.

Answer


3.3 Derivation of molecular formulae (SB p.53)

Check Point 3-3B

  • Find the mass of metal and water of crystallization in

  • (i) 100 g of Na2SO4 · 10H2O

  • (ii) 70 g of Fe2O3 · 8H2O

Answer


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

Mass of Na =

= 14.28 g

Mass of H2O =

= 55.88 g

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

Mass of Fe =

= 25.73 g

Mass of H2O =

= 33.20 g

3.3 Derivation of molecular formulae (SB p.53)

Check Point 3-3B

Back


(b) Mg(s) + 2AgNO3(aq)  2Ag(s) + Mg(NO3)2(aq)

(c) 2C4H10(g) + 13O2(g)  8CO2(g) + 10H2O(l)

3.4 Chemical equations (SB p.54)

Check Point 3-4

Back

  • Give the chemical equations for the following reactions:

  • Zinc + steam  zinc oxide + hydrogen

  • (b) Magnesium + silver nitrate  silver + magnesium nitrate

  • (c) Butane + oxygen  carbon dioxide + water

Answer


3.5 Calculations based on chemical equations (SB p.55)

Example 3-5A

Calculate the mass of copper formed when 12.45 g of copper(II) oxide is completely reduced by hydrogen.

Answer


CuO(s) + H2(g)  Cu(s) + H2O(l)

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

Number of moles of CuO reduced =

= 0.157 mol

Number of moles of Cu formed = 0.157 mol

= 0.157 mol

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

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

3.5 Calculations based on chemical equations (SB p.55)

Back

Example 3-5A


3.5 Calculations based on chemical equations (SB p.55)

Example 3-5B

Sodium hydrogencarbonate decomposes according to the following chemial equation:

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

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

(Molar volume of gas at R.T.P. = 24.0 dm3 mol-1)

Answer


Number of moles of CO2 required

= = 0.01 mol

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

Number of moles of NaHCO3 required = 0.01  2 = 0.02 mol

Mass of NaHCO3 required

= 0.02 mol  (23.0 + 1.0 + 12.0 + 16.0  3) g mol-1

= 0.02 mol  84.0 g mol-1

= 1.68 g

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

3.5 Calculations based on chemical equations (SB p.55)

Back

Example 3-5B


3.5 Calculations based on chemical equations (SB p.56)

Example 3-5C

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

Answer


3.5 Calculations based on chemical equations (SB p.56)

Back

Example 3-5C

2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(l)

2 mol : 7 mol : 4 mol : 6 mol (from equation)

2 volumes : 7 volumes : 4 volumes : - (by Avogadro’s law)

It can be judged from the chemical equation that the mole ratio of CO2 : C2H6 is 4 : 2, and the volume ratio of CO2 : C2H6 should also be 4 : 2 according to the Avogadro’s law.

Let x be the volume of CO2(g) formed.

Number of moles of CO2(g) formed : number of moles of C2H6(g) used = 4 : 2

Volume of CO2(g) : volume of C2H6(g) = 4 : 2

x : 20 cm3 = 4 : 2

x = 40 cm3

Therefore, the volume of CO2(g) formed is 40 cm3.


3.5 Calculations based on chemical equations (SB p.57)

Example 3-5D

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

Answer


3.5 Calculations based on chemical equations (SB p.57)

Example 3-5D

Let the molecular formula of the hydrocarbon be CxHy.

Volume of hydrocarbon reacted = 10 cm3

Volume of O2(g) unreacted = 50 cm3 (the residual gas after reaction)

Volume of O2(g) reacted = (80 – 50) cm3 = 30 cm3

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

CxHy + O2  xCO2 + H2O

1 mol : mol : x mol

1 volume : volumes : x volumes


=

=

x = 2

=

=

= 3

As x = 2, = 3

y = 4

Therefore, the molecular formula of the hydrocarbon is C2H4.

3.5 Calculations based on chemical equations (SB p.57)

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Example 3-5D


3.5 Calculations based on chemical equations (SB p.58)

Check Point 3-5

  • Find the volume of hydrogen produced at R.T.P. when 2.43 g of magnesium reacts with excess hydrochloric acid.

  • (Molar volume of gas at R.T.P. = 24.0 dm3 mol-1)

Answer


(b) 2P(s) + 3Cl2(g)  2PCl3(l)

 No. of moles of Cl2 =  No. of moles of PCl3

=

Mass of Cl2 = 77.45 g

The minimum mass of chlorine required is 77.45 g.

3.5 Calculations based on chemical equations (SB p.58)

Check Point 3-5

(b) Find the minimum mass of chlorine required to produce 100 g of phosphorus trichloride (PCl3).

Answer


3.5 Calculations based on chemical equations (SB p.58)

Check Point 3-5

(c) 20 cm3 of a gaseous hydrocarbon and 150 cm3 of oxygen (which was in excess) were exploded in a closed vessel. After cooling, 110 cm3 of gases remained. After passing the resulting gaseous mixture through concentrated sodium hydroxide solution, the volume of the residual gas became 50 cm3. Determine the molecular formula of the hydrocarbon.

Answer


  • CxHy(g) + O2(g)  xCO2(g) + H2O(l)

  • Volume of CxHy used = 20 cm3

  • Volume of CO2 formed = (110 – 50) cm3 = 60 cm3

  • Volume of O2 used = (150 – 50) cm3 = 100 cm3

  • Volume of CxHy : Volume of CO2 = 1 : x

  • = 20 : 60

  • x = 3

  • Volume of CxHy : Volume of O2 = 1 :

  • = 20 : 100

  • = 5

3.5 Calculations based on chemical equations (SB p.58)

Check Point 3-5


  • As x = 3,

  • = 5

  • = 2

  • y = 8

  • The molecular formula of the hydrocarbon is C3H8.

3.5 Calculations based on chemical equations (SB p.58)

Check Point 3-5


  • CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)

  • 1 mol : 2 mol : 1 mol : 2 mol (from equation)

  • 1 volume : 2 volumes : 1 volume : - (from Avogadro’s law)

  • 5 cm3 x cm3

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

  • =

  • x = 5 cm3

  • The volume of CO2(g) formed is 5 cm3.

3.5 Calculations based on chemical equations (SB p.58)

Check Point 3-5

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  • Calculate the volume of carbon dioxide formed when 5 cm3 of methane was burnt completely in excess oxygen, assuming all volumes of gases are measured at room temperature and pressure.

Answer


3.6 Simple titrations (SB p.61)

Example 3-6A

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

Answer


3.6 Simple titrations (SB p.61)

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Example 3-6A

2NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2H2O(l)

=

 Number of moles of NaOH(aq) = Number of moles of H2SO4(aq)

Number of moles of H2SO4(aq) = 0.067 mol dm-3  22.5  10-3 dm3

= 1.508  10-3 mol

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

= 3.016  10-3 mol

Molarity of NaOH(aq) =

= 0.121 mol dm-3

Therefore, the molarity of the sodium hydroxide solution was 0.121 M.


3.6 Simple titrations (SB p.61)

Example 3-6B

  • 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.

Answer


3.6 Simple titrations (SB p.61)

Example 3-6B

  • (b) If the dibasic acid is represented by H2X, write an equation for the reaction between the acid and sodium hydroxide.

Answer

(b) H2X(aq) + 2NaOH(aq)  Na2X(aq) + 2H2O(l)


  • Number of moles of H2X =  Number of moles of NaOH

  • 0.08 mol dm-3  25.0  10-3 dm3

  • =  Molarity of NaOH  28.5  10-3 dm3

  • Molarity of NaOH = 0.14 M

  • Therefore, the molarity of the sodium hydroxide solution was 0.14 M.

3.6 Simple titrations (SB p.61)

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Example 3-6B

  • (c) Calculate the molarity of the sodium hydroxide solution.

Answer


3.6 Simple titrations (SB p.62)

Example 3-6C

0.186 g of a sample of hydrated sodium carbonate, Na2CO3 · nH2O, was dissolved in 100 cm3 of distilled water in a conical flask. 0.10 M hydrochloric acid was added from a burette, 2 cm3 at a time. The pH value of the reaction mixture was measured with a pH meter. The results were recorded and shown in the following figure. Calculate the value of n in Na2CO3 · nH2O.

Answer


There is a sudden drop in the pH value of the solution (from pH 8 to pH 3) with the equivalence point at 30.0 cm3.

Na2CO3· nH2O(s) + 2HCl(aq)  2NaCl(aq) + CO2(g) + (n + 1)H2O(l)

Number of moles of Na2CO3 · nH2O =  Number of moles of HCl

=  0.10 mol dm-3  30.0  10-3 dm3

106.0 + 18.0n = 124.0

n = 1

Therefore, the chemical formula of the hydrated sodium carbonate is Na2CO3· H2O.

3.6 Simple titrations (SB p.63)

Example 3-6C

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3.6 pH 8 to pH 3) with the equivalence point at 30.0 cm Simple titrations (SB p.63)

Example 3-6D

5 cm3 of 0.5 M sulphuric(VI) acid was added to 25.0 cm3 of potassium hydroxide solution. The mixture was then stirred and the highest temperature was recorded. The experiment was repeated with different volumes of the sulphuric(VI) acid. The laboratory set-up and the results were as follows:


3.6 pH 8 to pH 3) with the equivalence point at 30.0 cm Simple titrations (SB p.63)

Example 3-6D

  • Plot a graph of temperature against volume of sulphuric(VI) acid added.

Answer


  • From the graph, it is found that the equivalence point of the titration is reached when 20 cm3 of H2SO4 is added.

  • Number of moles of H2SO4 = 0.5 mol dm-3 20  10-3 dm3

  • = 0.01 mol

  • 2KOH(aq) + H2SO4(aq)  K2SO4(aq) + 2H2O(l)

  • 2 mol : 1 mol

  • From the equation,

  • mole ratio of KOH(aq) : H2SO4(aq) = 2 : 1

  • Number of moles of KOH(aq) = 2  0.01 mol = 0.02 mol

  • Molarity of KOH(aq) = = 0.8 M

  • Therefore, the molarity of potassium hydroxide solution was 0.8 M.

3.6 Simple titrations (SB p.63)

Example 3-6D

(b) Calculate the molarity of the potassium hydroxide solution.

Answer


3.6 the titration is reached when 20 cm Simple titrations (SB p.63)

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Example 3-6D

(c) Explain why the temperature rose to a maximum and then fell.

Answer

(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 would cool down the reacting solution, causing the temperature to drop.


3.6 the titration is reached when 20 cm Simple titrations (SB p.66)

Example 3-6E

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

Answer


3.6 the titration is reached when 20 cm Simple titrations (SB p.66)

Example 3-6E

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

I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) … … (2)

From (1), Number of moles of IO3-(aq) =  Number of moles of I2(aq)

From (2), Number of moles of I2(aq) =  Number of moles of S2O32-(aq)

Number of moles of IO3-(aq) =  Number of moles of S2O32-(aq)

Molarity of IO3-(aq)  25.0  10-3 dm3 =  0.05 mol dm-3  22.0  10-3 dm3

Molarity of IO3-(aq) = 7.33  10-3 M

Therefore, the molarity of the acidified potassium iodate solution is 7.33  10-3 M.


3.6 the titration is reached when 20 cm Simple titrations (SB p.67)

Example 3-6F

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

Answer


3.6 the titration is reached when 20 cm Simple titrations (SB p.67)

Example 3-6F

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MnO4-(aq) + 8H+(aq) + 5Fe2+(aq)  Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)

Number of moles of MnO4-(aq) : Number of moles of Fe2+(aq) = 1 : 5

Number of moles of Fe2+(aq) = 5  Number of moles of MnO4-(aq)

= 5  0.02 mol dm-3  36.5  10-3 dm3

= 3.65  10-3 mol

Number of moles of Fe dissolved = Number of moles of Fe2+ formed

= 3.65  10-3 mol

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

Percentage purity of Fe =  100 % = 92.73 %

Therefore, the percentage purity of the iron wire is 92.73 %.


3.6 the titration is reached when 20 cm Simple titrations (SB p.67)

Check Point 3-6

  • 5 g of anhydrous sodium carbonate is added to 100 cm3 of 2 M hydrochloric acid. What is the volume of gas evolved at room temperature and pressure?

  • (Molar volume of gas at R.T.P. = 24.0 dm3 mol-1)

Answer


3.6 the titration is reached when 20 cm Simple titrations (SB p.67)

Check Point 3-6

  • Na2CO3(s) + 2HCl(aq)  2NaCl(aq) + H2O(l) + CO2(g)

  • No. of moles of Na2CO3 used =

  • = 0.0472 mol

  • No. of moles of HCl used = 2 M 

  • = 0.2 mol

  • Since HCl is in excess, Na2CO3 is the limiting agent.

  • No. of moles of CO2 produced = No. of moles of Na2CO3 used

  • = 0.0472 mol

  • Volume of CO2 produced = 0.0472 mol  24.0 dm3 mol-1

  • = 1.133 dm3


3.6 the titration is reached when 20 cm Simple titrations (SB p.67)

Check Point 3-6

(b) 8.54 g of impure hydrated iron(II) sulphate (formula mass of 392.14) was dissolved in water and made up to 250.0 cm3. 25.0 cm3 of this solution required 20.76 cm3 of 0.020 3 M acidified potassium manganate(VII) solution for complete reaction. Determine the percentage purity of the hydrated iron(II) sulphate.

Answer


3.6 the titration is reached when 20 cm Simple titrations (SB p.67)

Check Point 3-6

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  • MnO4-(aq) + 8H+(aq) + 5Fe2+(aq)  Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)

  • No. of moles of MnO4- ions = 0.0203 M 

  • = 4.214  10-4 mol

  • No. of moles of Fe2+ ions = 5  No. of moles of MnO4- ions

  • = 2.107  10-3 mol

  • No. of moles of Fe2+ ions in 25.0 cm3 solution = 2.107  10-3 mol

  • No. of moles of Fe2+ ions in 250.0 cm3 solution = 0.02107 mol

  • Molar mass of hydrated FeSO4 = 392.14 g mol-1

  • Mass of hydrated FeSO4 = 0.02107 mol  392.14 g mol-1 = 8.26 g

  • % purity of FeSO4 = = 96.72 %


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