Chemical Equations and Stoichiometry

New Way Chemistry for Hong Kong A-Level Book 11

Chemical Equations Chemical Equations

and Stoichiometryand Stoichiometry3.13.1 Formulae of CompoundsFormulae of Compounds

3.23.2 Derivation of Empirical FormulaeDerivation of Empirical Formulae

3.33.3 Derivation of Molecular FormulaeDerivation of Molecular Formulae

3.43.4 Chemical EquationsChemical Equations

3.53.5 Calculations Based on Chemical EquationsCalculations Based on Chemical Equations

3.63.6 Simple TitrationsSimple Titrations

33


3.3.11 Formulae of Formulae of

CompoundsCompounds


Formulae of compoundsFormulae of compounds

How can you describe the composition of compound X?

1st way = by chemical formula1st way = by chemical formula

C?H?

ratio of no. of atoms

3.1 Formulae of compounds (SB p.43)


How can you describe the composition of compound X?

Compound X

2nd way = by percentage by mass

2nd way = by percentage by mass

Mass of carbon atoms inside = …. g

Mass of hydrogen atoms inside = …. g

carbon atomshydrogen atoms


Check Point 3-1Check Point 3-1



Compound Empirical formula

Molecular formula

Structural formula

Carbon dioxide

CO2 CO2 O = C =O

Water H2O H2O

Methane CH4 CH4

Glucose CH2O C6H12O6

Sodium fluoride

NaF Not applicable

Na+F-

Different types of formulae of some Different types of formulae of some compoundscompounds

O

H H

C

H

H

HH

O

OH

HH

H

OH

OH

H OH

H

OH


3.3.22 Derivation of Derivation of

Empirical Empirical FormulaeFormulae


From combustion From combustion datadata• During complete combustion, elements in a co

mpound 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)



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

Example 3-2AExample 3-2A Example 3-2BExample 3-2B Check Point 3-2ACheck Point 3-2A


Composition by mass

Empirical formula

3.2 Derivation of Empirical Formulae (SB p.48)

From combustion by massFrom combustion by mass

Example 3-2CExample 3-2C Check Point 3-2BCheck Point 3-2BExample 3-2DExample 3-2D


3.3.33 Derivation of Derivation of

Molecular Molecular FormulaeFormulae


What is molecular What is molecular formulae?formulae?

Molecular formula

= (Empirical formula)n

?

3.3 Derivation of molecular formulae (SB p.49)


Empirical formula Molecular mass

Molecular formula

3.3 Derivation of Molecular Formulae (SB p.49)

From empirical formula and known From empirical formula and known relative molecular massrelative molecular mass

Example 3-3AExample 3-3A

Example 3-3BExample 3-3B



Water of Crystallization Derived from Water of Crystallization Derived from Composition by MassComposition by Mass

Hydrated salt Anhydrous salt

CuSO45H2O

(Blue crystals)

Anhydrous CuSO4

(White powder)

Na2CO310H2O

(Colourless crystals)

Anhydrous Na2CO3

(White powder)

CoCl2 2H2O

(Pink crystals)

Anhydrous CoCl2(Blue crystals)

Example 3-3CExample 3-3C

Check Point 3-3ACheck Point 3-3A


Formula of a compound

Composition by mass


Find composition by mass from Find composition by mass from formulaformula

Example 3-3DExample 3-3D

Example 3-3EExample 3-3E

Check Point 3-3BCheck Point 3-3B


3.3.44 Chemical Chemical

EquationsEquations


Chemical equationsChemical equations

a A + b B c C + d D

mole ratios

(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.53)



3.3.55 Calculations Calculations

Based on Based on Chemical Chemical EquationsEquations


Calculations based on equationsCalculations based on equations

3.5 Calculations Based on Equations (SB p.65)

Calculations involving reacting masses

Example 3-5AExample 3-5A Example 3-5BExample 3-5B


Calculations based on equationsCalculations based on equations

3.5 Calculations Based on Equations (SB p.66)

Calculations involving volumes of gases





3.3.66 Simple TitratiSimple Titrati

onsons


Simple titrationsSimple titrations

Acid-base titrationsAcid-base titrations

Acid-base titrationswith indicators

Acid-base titrationswith indicators

Acid-base titrationswithout indicators

Acid-base titrationswithout indicators

(to be discussed in later chapters)

3.6 Simple titrations (SB p.58)


Finding the concentration of a solutionFinding the concentration of a solution

+solute

solvent

solution

Copper(II) sulphate

Water

Copper(II) sulphate solution




~50 cm3



~50 cm3




~50 cm3




~50 cm3




~50 cm3




~50 cm3




50 cm3

Solution A




~50 cm3




~50 cm3




~50 cm3




~50 cm3




~50 cm3


Finding the concentration of a Finding the concentration of a solutionsolution


~50 cm3




50 cm3

Solution B




~100 cm3




~100 cm3




~100 cm3




~100 cm3





~100 cm3




~100 cm3




Solution C

100 cm3



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

contain the same amount of solute (same concentration)

2 x the amount of solute

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


Concentration is the amount of solute in a unit volume of solution.


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

Concentration is the amount of solute in a unit volume of solution.

no. of spoons mass

no. of moles



MolarityMolarity

Molarity is the number of moles of solute dissolved in 1 dm3 (1000 cm3) of solution.

A way of expressing concentrations

)3dm (in solution of volumesolute of moles of number Molarity

Unit: mol dm-3 (M)



What does this mean?

1 dm3

contains 2 moles of HCl

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


Example 3-6AExample 3-6A Example 3-6BExample 3-6B


Titration without an indicatorTitration without an indicator


By change in pH value



Titration without an indicatorTitration without an indicator


By change in temperature



1. Iodometric titration

I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6

2-(aq)brown colourless

in conical flask

in burette

3.6 Simple Titrations (SB p.65)

Redox titrationsRedox titrations



I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6

2-(aq)

brown colourless

Add starch

During titration : brown yellow

in conical flask in burette





I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6

2-(aq)

brown colourless






I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6

2-(aq)

brown colourless

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



Redox titrationsRedox titrationsExample 3-6EExample 3-6E


2. Titrations involving potassium permanganate

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

purple pale green

In burette

In conical flask



Example 3-6FExample 3-6F Check Point 3-6Check Point 3-6


The END


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

(a) Propene

(b) Nitric acid

(c) Ethanol

(d) Glucose

Back

Answer


C6H12O6C6H12O6(d) Glucose

C2H5OHC2H6O(c) Ethanol

HNO3HNO3(b) Nitric

acid

C3H6CH2(a) Propene

Structural formula

Molecular formula

Empirical

formula

Compound

O N

O

H

O

C

H

H

CH

H

H

OH

O

OH

HH

H

OH

OH

H OH

H

OH

C

H

H

CH C

H H

H


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

0.440.12

0.180.2


Therefore, the empirical formula of the hydrocarbon is CH3.


Back

Carbon Hydrogen

Mass (g) 0.80 0.20

No. of moles (mol)

Relative no. of moles

Simplest mole ratio

1 3

0667.00.12

80.0 20.00.120.0

10667.00667.0 3

0667.020.0


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.

0.440.12

0.180.2


Therefore, the empirical formula of compound X is C2H6O.


Back

0667.00.12

80.0 06.00.106.0

601.006.0 1

01.001.0

Carbon Hydrogen Oxygen

Mass (g) 0.24 0.06 0.16

No. of moles (mol)


Simplest mole ratio

2 6 1

01.00.16

16.0

201.002.0


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


Answer(a) Mass of sulphur = 5 g

Mass of oxygen = (10 – 5) g = 5 g

21Simplest mole ratio


No. of moles (mol)

55Mass (g)

OxygenSulphur

156.01.32

5 313.00.16

5

1156.0156.0 2

156.0313.0


(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(b)

The empirical formula of the oxide is M2O5.



No. of moles (mol)

25.6119.85Mass (g)

OM

64.00.31

85.19 6.10.1661.25

164.064.0 5.2

64.06.1


(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



(c) Mass of Cu = (22.940 – 21.430) g = 1.51 g

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

Therefore, the empirical formula of copper(II) oxide is CuO.

0238.05.63

51.1



No. of moles (mol)

0.3811.51Mass (g)

OxygenCopper

0238.00.16

381.0

10238.00238.0 1

0238.00238.0

Back


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


Answer



Back

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.



No. of moles (mol)

2575Mass (g)

HydrogenCarbon

25.60.12

75 250.1

25

125.625.6 4

25.625


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



Back

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.



No. of moles (mol)

77.4522.55Mass (g)

ChlorinePhosphorus

727.00.3155.22 182.2

5.3545.77

1727.0727.0 3

727.0182.2


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


Answer(a) Let the mass of vitamin C analyzed be 100 g.

The empirical formula of vitamin C is C3H4O3.



No. of moles (mol)

54.54.640.9Mass (g)

OxygenHydrogenCarbon

41.30.129.40 60.4

0.16.4 41.3

0.165.54

141.341.3 35.1

41.36.4 1

41.341.3


(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(b) The masses of the elements are multiplied by 1000 first.

The empirical formula of aspirin is C9H8O4.



No. of moles (mol)

115.414.6195.0Mass (g)


25.160.120.195 6.14

0.16.14 21.7

0.164.115

25.221.725.16 02.2

21.76.14 1

21.721.7

Back


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



No. of moles (mol)

1.04.0Mass (g)

HydrogenCarbon

333.00.120.4 1

0.10.1

1333.0333.0 3

333.01

0.440.12

0.180.2



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.

Back


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.



No. of moles (mol)

49.386.1844.44Mass (g)


70.30.1244.44 18.6

0.118.6 09.3

0.1638.49

2.109.370.3 2

09.318.6 1

09.309.3



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.

Back


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.

10036

x186.159x18

Back


(a)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



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

Therefore, the empirical formula of compound Z is CH2O.


Carbon Hydrogen Oxygen

Mass (g) 40.00 6.67 53.33

No. of moles (mol)


Simplest mole ratio

1 2 1

33.30.1200.40 67.6

0.167.6 33.3

0.1633.53

133.333.3 2

33.367.6 1

33.333.3


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

Answer


(b)

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



No. of moles (mol)

72.7227.28Mass (g)

S(NH4) unit

52.10.1828.27 27.2

1.3272.72

152.152.1 49.1

52.127.2


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

Answer


(c)

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



No. of moles (mol)

51.2248.78Mass (g)

H2OMgSO4

405.04.120

78.48 846.20.1822.51

1405.0405.0 7

405.0846.2

Back


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.

%1000.60

20.12

%1000.6040.1

%1000.60

20.16

Back


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


AnswerRelative 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

%1009.277

8.55

Back


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

Answer


(a) 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 %

%100gmol2.294

gmol)21.39(1

1

%100gmol2.294

gmol)20.52(1

1

%100gmol2.294

gmol)70.16(1

1


(b)Find the mass of metal and water of crystallization in

(i) 100 g of Na2SO4 · 10H2O

(ii) 70 g of Fe2O3 · 8H2OAnswer




(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

g100gmol1.322

gmol)20.23(1

1

g100gmol1.322

gmol)100.18(1

1

g70gmol6.303

gmol)28.55(1

1

g70gmol6.303

gmol)80.18(1

1

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

3.4 Chemical equations (SB p.54)

(a) Zn(s) + H2O(g) ZnO(s) + H2(g)

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

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

Back


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

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

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.

1gmol)0.165.63(g45.12

1mol g 63.5Cu of Mass

Back


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.

13

3

mol cm 24000cm 240

Back


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



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 : C

2H6 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.

Back


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



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.

(g)HC of Volume

(g)CO of Volume

yx

2

1x

3

3

cm 10cm 20

1x

(g)HC of Volume

(g)O of Volume

yx

2

14y

x

3

3

cm 10cm 30

14y

x

4y

x

4y

2

Back


(a)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


(a) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)

No. of moles of H2= No. of moles of Mg

=

Volume of H2 = 2.4 dm3

1-3

2

mol dm 24.0

H of Volume1-mol g 24.3

g 2.43


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

Answer


31

21

(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.

1-

2

mol g 2) (35.5

Cl of Mass

31

1-mol g 3)35.5 1.03(

g 10021

31

21


(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




(c) 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

)4y

x( 2y

4y

x



(c) As x = 3,

= 5

= 2

y = 8

The molecular formula of the hydrocarbon is C3H8.

4y

3

4y


(d)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


(d) 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.

3cm 5x

11

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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



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.

(aq)SOH of moles of NumberNaOH(aq) of moles of Number

42 12

21

33-

-3

dm 10 25.0mol 103.016

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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.

(a)Calculate the molarity of the acid solution.


Answer

(a) Number of moles of acid = = 0.02 mol

Molarity of acid solution = = 0.08 M

1-mol g 26.01g 2.52

33 dm 10250mol 0.02


(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)


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


Answer

(c) 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.

21

21

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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 Na2CO

3 · H2O.

21

211-mol g 18.0n) 316.0 12.0 2 (23.0

g 0.186

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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:


Volume of H2SO4 added

(cm3)

Temperature (oC)

0 20.05 21.8

10 23.415 25.020 26.525 25.230 24.0


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


Answer


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


Answer(b) 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.

33 dm 1025.0mol 0.02


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


Answer(c) Neutralization is an exothermic reaction. When more and more sulphur

ic(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.

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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



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

I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6

2-(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 S2O3

2-(aq)

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

m3

Molarity of IO3-(aq) = 7.33 10-3 M

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

31

21

61

61

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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



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 %.

g 0.22g 0.204

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(a)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



(a) 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

1-mol g 3) 16.0 12.0 2 (23.0g 5

3dm1000100



(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



(b) 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 %%100

g 8.54g 8.26

3dm100020.76

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