Chemistry Form 6 Chap 01

5/23/2018 Chemistry Form 6 Chap 01

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PRE UNIVERSITYCHEMISTRY

SEMESTER 1

HAPTER 1 : MATTER


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1.1 The Atom

1.1.1 Historical development of atomic theory When atom was first discovered by John Dalton

(1808), he claim that atom are the simplest unitin a substance.

Later, physicist J.J. Thomson (1897) found outthat atom are made up of even smaller particleswith negative charge electron, Daltons theorywas being rejected.

After that, subatom with a positive charge proton was discovered by Rutherford in thecenter of the atom.

At the same decade, Neils Bohr discover that

electrons surrounding nucleus similar as planetsurrounding the Sun, and electrons move aboutin a rich electron region called orbital.

Few years later, Chadwick discovered that not

only proton exist in the center of an atom butalso a non-charge subatom neutron.


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Electron

Nucleus

Proton

Neutron

Particle Symbol Mass Relative Charge Relative

g assProton p or 11H 1.67 x

10-271 a.m.u +1.6 x

10-19+1

Neutron n or 10n 1.67 x

10-27

1 a.m.u 0 0

Electron e or 0-1e 9.11 x10-31

_1_ amu1834

-1.6 x10-19

-1


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The path of moving protons, electrons and neutrons is deflected by both

electric field and magnetic field

electron proton

electr

on

Northneutron

ne

+

<

proto

n

Path of proton, electron and neutron

in an electrical field


in a magnetic field


in an electrical field


in a magnetic field

South

tron


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1.1.2 Structure of atoms

In atom, proton & neutron are located at the centre of

the atom, which is also known as nucleus whileelectrons are surrounding it. The mass of an atom areconcentrated at the center of the atom, as the mass ofelectron is very small.

In periodic table, the symbol of elements is usually writtenin such way

nuc eon num er c arge

proton number no. of same element

Ion is formed when electron is donated or received

PZ x


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Positive charged ion (cation) Negative charged ion (anion)

Formed Formed when an atom donate electrons Formed when an atom receive electrons

Dot and

cross

dia ra

2-

+ 2 e-

ms

Ionic

equatio

n

Na Na+ + e- O + 2 e- O2-

No of

e- 10 10

+ e-

oxide ion, O2-


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Particles Number of protons Number of neutrons Number of electrons

2010Ne

168 O

2-

3517 Cl

-

4020Ca

2+

52 3+

10 10 10

8 8 10

17 18 1820 20 18

24

12251Sb

3-

OH-

CO32-

21D3O

+

[168O-N-188O]-

51 71 54

9 8 10

30 30 32

11 11 10

23 25 24


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1.2 Isotopes

Isotopes are atoms which have the same proton number

but different nucleon number

In the previous table, ____________ are isotopes

Other examples of isotopes

Element Isotopes No of proton No of neutron % ofabundance

Protium, 11H 99.0

HYDROGEN & DEUTERIUM

1 0y rogen euter um, : 1 .

Tritium, T : 31H 0.01

Oxygen

Oxygen-16 : 168O 98.9

Oxygen-17 :17

8O 1.00

Oxygen-18 : 188O 0.01

Chlorine

Chlorine 35 ; 3517Cl 75

Chlorine 37 ; 3717Cl 25

1 2

8 8

8 98 10

17 18

17 20


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Isotope 54Fe 56Fe 57Fe 58Fe

% composition 5.8 91.6 2.2 0.4

No. of protons 26 26 26 26

No. of neutron 28 30 31 32

The relative abundance (% composition) of the isotopes in the sample

of an element is not the same for all the isotopes present.

Same DifferentProton number

No. of electron (neutralatom)

Electronic configuration

Chemical properties

No. of neutron in nucleus

Density

Mass

Rate of diffusion


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If compounds are formed from different isotopes, themolecular mass of that particular compound is thesummation of all the isotopic mass involved

Example

C16O2 and C18O2

1 C + 2 (16O) = 44 1 C + 2 (18O) = 48 H2O and T2O

2 H + 1 O = 18 2 T + 1 O = 22

N35Cl3 and N37Cl31 N + 3 (35Cl) = 119 1 N + 3 (37Cl) = 125

79Br79Br and 79Br81Br

79 + 79 = 158 79 + 81 = 160


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1.2.4 Uses of Radioisotopes

Widely used as tracers in biological processes. For exampletracing the uptake of phosphorus by plant using 32P

Carbon-14 is used in carbon dating, which is used todetermine the age of archeological artifacts

Gamma radiation from

60

Co is used in radiotherapy todestroy malignant tissues in cancer patients.

Energy released by nuclear fission is used to generateelectricity in nuclear plants.

To sterilise food or surgical instruments. Used in leak management. Underground leakage, especially

in water or fuel pipeline leakage. Sudden increase inradioactivity mean that theres a leakage.


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1.3 Relative Mass

Mass of an atom is approximately equal to the sumof the mass of all the sub-atomic particles present.Example

In this method, the mass of the atom is comparedto the mass of another atom which is used asreference. Initially, hydrogen was used as standardbecause it is the lightest. Subsequently, the oxygen

atom was used to replace hydrogen as standarddue to a few reason.

In 1961, carbon-12 was chosen as the standard forcomparing relative atomic masses because it iseasily available and its solid in room temp. It isknown as 12C scale. On this scale, an atom of 12Chas the mass exactly 12 atomic mass unit (a.m.u).


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1.3.1 Relative Isotopic Mass (RIM)

The relative isotopic mass of an isotope is the massof 1 atom of the isotope relative to 1/12 times themass of one atom of 12C.

RIM = mass of 1 atom of the isotope

1/12 x mass of 1 atom of C-12 The relative isotopic mass of an isotope is

approximately equal to its nucleon number. Forexam le,

Isotope Relative isotopic mass40Ca 40.08019F 18.999

127

I 126.910 Thus for most calculation involving atomic mass, the

nucleon number can be use as a substitute for theactual RIM.


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1.3.2 Relative Atomic Mass (RAM) Most elements consist of a mixture of isotopes with

different abundance as mention in slide 7. Therefore,the relative abundance of the abundance of theisotopes has to be taken into consideration whencalculating the average mass of an atom of the

element. The relative atomic mass (Ar) of an element isdefined as the average mass of 1 atom of theelement relative to 1/12 times the mass of 1 atom of12 .

RAM = average mass of 1 atom of the element1/12 x mass of 1 atom of C-12

Example : The element oxygen consist of 3 isotopes, 16O,17O and 18O in the ratio of 99.76 : 0.04 : 0.20. Calculate

RAM of oxygen.RAM=(16 x 99.76) + (17 x 0.04) + (18 x 0.20) = 16.004499.76 + 0.04 + 0.20


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1.3.3 Relative Molecular Mass (RMM)

RMM of a molecular substance is the mass of 1 molecule of

the substance relative to 1/12 times the mass of 1 atom of 12CRMM = average mass of 1 molecule of substance

1/12 x mass of 1 atom C-12

~ is equal to the sum of the relative mass of all the atoms

shown in the molecular formula.

For ionic compound the terms relative formula mass (RFM) isused because it do not exist as discrete molecules but consist

.

formula unit of the compound relative to 1/12 times the massof 1 atom of carbon-12.


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1.4 Introduction to Mass Spectroscopy (MS)

Relative mass of atom can be determine by usinganalytical instrument. One of the instrument that isfrequently used nowadays is MASS SPECTROSCOPY

Mass Spec can be used to determine :

Relative isotopic mass

Relative abundance of the isotopes

Relative atomic mass

Relative molecular mass

Structural formula of compound

Figure on the next slide shows a simplified diagram

of a mass spec.


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Mass Spectrometer


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This machine is used to find out the relative atomic mass ofan element. There are 4 main stages in the process:

1. Ionisation - after a vapourised sample is put into themass spectrometer, it is ionised - electrons are removed -usually one electron is removed but sometimes two. Thepositively charged species then go into the accelerating

chamber.One electron is removed : A (g) A+ (g) + e

2nd electron is removed : A+ (g) A2+ (g) + e

If molecule involved e.g. : ABCABC (ABC)+ + e-

or A+ + (BC)+ + 2 e-/ C+ + (AB)+ + 2 e-

or A+ + B+ + C+ + 3 e

2. Acceleration - the ions are subjected to a negativelycharged electric plate in order to accelerate the ion


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2. Acceleration - the ions are subjected to a negativelycharged electric plate in order to accelerate the ion.

3. Deflection - the heavy ions (ones with the larger atomicmass) are deflected less than the the lighterions. Therefore the ions are separated according to theiratomic masses and travel a different path in the mass

spectrometer (shown by the dotted lines in the diagrambelow).56Fe+ ;59Fe+

56Fe+ ;56Fe2+

4. Detection - only ions of a certain mass actually end up at

this point (the ones taking the green path). To make surethat all of the ions are detected, you have to vary thestrength of the accelerating field. The detector recordseach species as a peak on a trace.


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Ratio of 79Br to 81Br is 50 : 50 or 1 : 1, so RAM of Br = 80

For peak m/e 158, it is due to the existence of (79Br79Br)+ ; so

the probability of peak P(79

Br79

Br) = (1/2)(1/2) = For peak m/e 160, it is due to the existence of (79Br81Br)+ ; or,

it may also be (81Br79Br)+ , so the probability of peak is

P(79Br81Br) = (1/2)(1/2) + P(81Br79Br) = (1/2)(1/2) =

For peak m/e 162, it is due to the existence of (81Br81Br)+ ; sothe probability of peak P(81Br81Br) = (1/2)(1/2) =


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RAM = 206(15) + 207(72) +208(8) + 209(5)___15 + 72 + 8 + 5

= 207.03


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N+

O+

NO+

N2O+

NO2+

N2O2+

N2O3+

N2O4+


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CH2+

CH3+

C2H2+

C2H3+

C3H5+

C3H6+

C4H8+


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CH3

+

O+

C2H3+

C2H5+

COH+

C3H6+

C2

H3

O+

C3H6O+


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1.5 Mole

One mole of substance is the amount of substance that

contains the same number of particles as the number ofatom in exactly 12 g of the C-12 isotope.

The number of particles in one mole of any substance is aconstant known as the Avogadro constant (L)

Avogadro Constant = 6.023 x 1023 mol-1

1 mole of any substance is the same as the relative atomic/molecular/formula mass of that substance expressed in gram

e.g. : 3 mole of boron

= 3 x 6.023 x 1023 atoms of boron


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1.5.1 Moles of Gases

In reactions involving gases, the volume of the gases that

take part in the reaction is usually more important than themass of the gases involved.

The relationship between the amount of gas (in moles) andthe volume of gas is given in Avogadros Law

Avogadros Law state that under the same conditions oftemperature and pressure, an equal volume of gasescontains equal number of moles where ;

volume number of moles

For examp e, un er room con ition, 5 cm3 o C 2 wi containthe same number of mole/molecules as5 cm3 NH3

At standard temperature and pressure, (s.t.p ; 273 K and101 kPa), 1 mole of all gases will occupy a volume of 22.4dm3 . This volume is known as molar volume (Vm). Under

room condition, 1 mole of gas occupies 24.4 dm3.


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1.5.2 Moles and Solutions

The concentration of a solution is usually expressed as the

mass solute per 1.0 dm3 of solution (g dm-3) or mole ofsolute in 1.0 dm3 of solution (mol dm-3).

Concentration in unit mol dm-3 is also known as molarity, M.

The relationship between molarity and concentration is givenby expression :

Molarity (M) = concentration (g dm-3)relative molecular mass of solution (g mol-1)

The number of moles of solute present in a given volume ofsolution (of molarity M) is

mole = molarity x volume = MV


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1.6 Empirical Formula and Molecular Formula

Empirical Formula of a compound shows the simplest whole

number ratio for atom of all the different elements present inone molecule of the compound

Molecular Formula of a compound shows the actual numberof atoms of different elements in one molecule of compound

Example

Ethene C2H4 CH2

Phosphorous (V) oxide P4O10 P2O5

Hydrogen peroxide H2O2 HO

Ethanoic acid CH3COOH CH2O


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1. A saturated hydrocarbon (hydrocarbon which only have carbonand hydrogen in it) contains 82.66% of carbon.

i. What is its empirical formula?Element

C H

Mass 82.66 17.34

Mol 82.66

12= 6.89 mol

17.34

1=17.34 mol

Ratio 6.89/6.89= 1

17.34/6.89= 2.5

Empirical formulaC2H5

ii. What is its molecular formula if given the relative molecularmass of the hydrocarbon is 58.0

Empirical formula = C2H5(C2H5)n = 58

(12(2) + 5(1))n = 58n = 2

Molecular formula = (C2H5)2= C4H10


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2. Ester is the main chemical substance applied in perfume. Inan analysis of determining the molecular formulae of the

ester, it is say that this ester contains 54.5% of carbon,9.10% of hydrogen and 36.4% of oxygen. Given themolecular mass of the ester is 88.0, determine themolecular formulae of this ester.

Element C H OMass 54.5 9.10 36.4

Mol54.5

129.101

36.416

=4.542 =9.10 = 2.275

Ratio4.542 / 2.275= 2

9.10 / 2.275= 4

2.275 / 2.275= 1

Empirical formula = C2H4O(C2H4O)n = 88

(12(2) + 4(1) + 16(1))n = 88n = 2

Molecular formula = (C2H4O)2= C4H8O2

3 A i id h th f ll i iti b C 40 0%


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3 An organic acid has the following composition by mass: C, 40.0%;H, 6.7%; O, 53.3%. Its mass spectrum shows major peaks(including the molecular ion) at the following m/e (mass) values:15, 17, 43, 45, 60.

(a) Calculate the empirical formula of the acid, and use the massspectrum to suggest its molecular formula and its structuralformula. [3]

Element C H O

Mass 40.0 6.7 53.3

Mol40.0 / 12

6.7 / 1

53.3 / 16

Empirical = CH2O(CH2O)n = 60(12(1) + 1(2) + 16(1))n= 60

(b) By suggesting their molecular formulae, identify the variousspecies responsible for the peaks in the mass spectrum. [3]

m/e = 15 CH3+

m/e = 17 OH+

m/e = 43 CH3CO+

m/e = 45 COOH+

m/e = 60 CH3COOH+

. . .

Ratio 3.33 / 3.33= 1

6.7 / 3.33= 2

3.33 / 3.33= 1

n = 2

Molecular = (CH2O)2= C2H4O2

1 6 St i hi t


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1.6 Stoichiometry

The term Stoichiometric comes from a balanced chemical

equation where the amount of mole required for reactants toform a certain amount of mole of products.

The moles of reactants required to form how many moles ofproducts are referred from the chemical equation

Example 1 : In the reaction of

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

Stoichiometrically : 1 mole of sodium hydroxide is required toform 1 mole of sodium chloride

1 mol NaOH 1 mol NaClExample 2 : In the reaction of

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

Stoichiometrically : 2 mole of potassium hydroxide is required to

form 1 mole of potassium sulphate2 mol KOH 1 mol of K2SO4

Students are required to understand thus balanced the chemicalequation before knowing the stoichiometry between the reactants

and products.


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1. The composition of an organic compound is 76.6 % C,6.38 % H and 17.02 % O. Its relative molecular mass is

94. What are the empirical and molecular formulas of thecompound?

Element C H O

Mass 76.60 6.38 17.02

Mol

76.6012

=

6.381

=

17.0216

=

Ratio6.38 / 1.06= 6

6.38 / 1.06= 6

1.06 / 1.06= 1

Empirical formula = C6H6O(C

6

H6

O)n = 94[(12(6) + 1(6) + 16(1)]n = 94

n = 1Molecular formula = (C6H6O)1

= C6H6O

2 What volume of oxygen (at STP) is required to burn


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2. What volume of oxygen (at STP) is required to burnexactly

(a) 100 cm3

methane, CH4?

(b) 200 cm3 ethanol, C2H5OH?

CH4 + 2 O2 CO2 + 2 H2OSince 1 CH4 = 2 O2Volume of oxygen gas = 2 (100)

= 200 cm3

C2H5OH + 3 O2 2 CO2 + 3 H2O

Since 1 C2H5OH = 3 O2

(c) 2.2 dm3 propanone, C3H6O?

(d) 3.0 dm3 octane, C8H18?

o ume o oxygen gas =

= 600 cm3

C3H6O + 4 O2 3 CO2 + 3 H2O

Since 1 C3H6O = 4 O2

Volume of oxygen gas = 4 (2.2)= 8.8 dm3

C8H18 + 25/2 O2 8CO2 + 9H2OSince 1 CH4 = 25 / 2 O2

Volume of oxygen gas = 25/2 (3.0)= 37.5 dm3

3 When 1 25 g of a mixture of ethane C H and propene


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3. When 1.25 g of a mixture of ethane C2H6 and propene,C3H6 was burned in excess oxygen, 3.78 g of CO2 was

obtained. What is the percentage by mass of C2H6 in themixture?

Since gas is mixture of C2H6 and C3H6If the total mass of gas = 1.25 g

Then, lets assume mass of C2H6 = x ; while C3H6 = 1.25 xThe mole of each gas are

Given the mass of CO formed from mixture is 3.78

30

xHCofmol 62 =

42

x25.1HCofmol 63

=

mol of CO2 = 3.78 / 44 ; mol = 0.0859 molThe amount of CO2 released from each gas is known from equationC2H6 + 7/2 O2 2 CO2 + 3 H2O (2 mol of CO2 is given by 1 C2H6)C3H6 + 9/2 O2 3 CO2 + 3 H2O (3 mol of CO2 is given by 1 C3H6)Hence the total mol of CO2,

x = 0.69 g% C2H6 = 0.69 / 1.25 x 100%

= 55.2 %

0859.042

x25.13

30

x2 =

+

4 20 0 cm3 of a gaseous hydrocarbon X mixed with 150 cm3


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4. 20.0 cm3 of a gaseous hydrocarbon X mixed with 150 cm3

of oxygen and bum completely. When the mixture is cooled

the total volume of gas is 110 cm3

. When the gaseousmixture is passed through concentrated potassiumhydroxide solution, 30.0 cm3 of gas remains. Determine themolecular formula of X.Given the chemical equation of combustion for hydrocarbon

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

Initial 20 cm3 150 cm3

After 0 cm3 30 cm3 80 cm3 ?

The volume of O2

used for water = 150 110 = 40 cm3

According to Avogadros Law

Since 20 cm3 of CXHY formed 80 cm3 of CO2

So, the mol of CO2 = 80 / 20 = 4 ; Hence x = 4

As for H, since the 20 cm3 of CXHY react with 40 cm3 oxygen for water

So, mol of O for water = 40 / 20 = 2

Since y / 4 = 2 ; so y = 8

As a conclusion, the formula of hydrocarbon is C4H8

5 Q is more electropositive than Y When 1 92 g of metal Q is


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5 Q is more electropositive than Y. When 1.92 g of metal Q isadded to an aqueous solution containing Y2+ ions, 12.4 g

of metal Y is obtained. In this reaction, Q3+

ions areproduced.

(a) Write an ionic equation to represent the reaction above.

2 Q + 3 Y2+ 3 Y + 2 Q3+

(b) What is the relative atomic mass of Q if the relativeatomic mass of Y is 207?

Mol of Y = mass RAM

Y = 12.42 / 207Y = 0.060 molSince from equation 3 Y = 2 QMol of Q = 0.060 x 2 / 3

Q = 0.040 mol

RAM of Q = 1.92 / 0.040RAM of Q = 48

6 1 0 dm3 sample of air containing carbon dioxide is passed


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6. 1.0 dm sample of air containing carbon dioxide is passedthrough aqueous calcium hydroxide. If 0.080 g of calcium

carbonate is formed, determine(a) the number of moles of CO2 present.

The equation between CO2 and Ca(OH)2CO2 + Ca(OH)2 CaCO3 + H2O

Since 1 mol of CO2 = 1 mol of CaCO3= 0.080 g / [40 + 12 + 3 (16)]= 8.0 x 10-4 mol

(b) the percentage by volume of CO2 in the sample at STP.

Since mol of CO2 = 8.0 x 10-4 mol

So volume of CO2 in sample = 8.0 x 10-4 (22.4 dm3)

V = 0.018 dm3

% by V of CO2 = 0.018 dm3 / 1.0 dm3 x 100%= 1.8 %

7. When 6.70 g of iron is burned in 3.64 g oxygen, Fe2O3 is


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7. When 6.70 g of iron is burned in 3.64 g oxygen, Fe2O3 isformed. [Ar O = 16; Fe = 55.8]

(a) Write a balanced equation for the action.4 Fe + 3 O2 2 Fe2O3

(b) What mass of Fe2O3 will be produceMol of Fe = mass / RAM ; mol of Fe = 6.70 / 55.8

mol of Fe = 0.120 molSince 4 Fe = 2 Fe2O3

Mol of Fe2O3 = 0.120 / 2 = 0.060 molMass of Fe O = 0.060 x 2 55.8 + 3 16

(c) What mass of oxygen will be left over at the end of thereaction?

= 9. 58 g

From equation above, since 4 Fe = 3 O2

Mol of O2 = 0.120 x 3 / 4 = 0.090 molMass of O2 = 0.090 x [2(16)]= 2.88 g

Mass of O2 unreacted = 3.64 2.88= 0.96 g

8. Calculate the concentration in mol dm-3 of the resulting


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8. Calculate the concentration in mol dm of the resultingsolution when 300 cm3 of 0.40 mol dm-3 Na2SO4 is mixed

with 200 cm

3

of 1.2 mol dm

-3

Na2SO4. What are the molarconcentration of Na+ and SO42- ions in the resulting solution?

When mixing both Na2SO4Mol of Na2SO4 in A mol of Na2SO4 in B

mol = 0.12 mol mol = 0.24 mol

1000)300)(40.0(mol;

1000MVmol ==

1000)200)(2.1(mol;

1000MVmol ==

Tota mo = 0.12 + 0.24 = 0.36 mo

When mixed, total volume, V = 300 + 200V = 500 cm3

Concentration after mixture

M = 0.72 mol dm-3

Since Na2SO4 2 Na+ + SO42-

So, [Na+] = 2 (0.72) = 1.44 mol dm-3

[SO42-] = 1 (0.72) = 0.72 mol dm-3

500

)1000)(36.0(

M;V

1000mol

M tot=

=

9. Manganate (VII) ions react with oxalate ions according to


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g ( ) gthe reaction below: ;

2 MnO4-

+ 5 C2O42-

+ 16 H+

8 H2O + 2 Mn2+

+ 10 CO2What is the volume of 0.200 mol dm-3 KMnO4

- required tocompletely oxidize 28.5 cm3 of 0.500 mol dm-3 Na2C2O4?

[Ar H = 1.0; C = 12.0; O = 16.0; K = 39.1]

5

2

)50.28)(500.0(

V)200.0(;

b

a

VM

VM a

bb

aa ==

a = . cm

10. Brass is an alloy of copper, containiq 90.0% copper and


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y pp , q pp10.0 % zinc by mass. When nitric acid, HNO3 is added to tin

alloy, the following reactions occur:Cu + 4 H+ + 2 NO3- Cu2+ + 2 NO2 + 2 H2O

4 Zn + 10 H+ + NO3- 4 Zn2+ + NH4

+ + 3 H2O

(a) What volume of 2.00 M nitric acid is required to react

completely with 10.0 g of brass?

In 10.0 g of brass ; 9.0 g is Cu and 1.0 g ZnMol of Cu = 9.0 / 63.5 = 0.1417 mol

o o n = . . = . mo

Based on the mol of H+ in both equation ;Total mol of HNO3 = 4 (0.1417) + 10/4 (0.0153)

= 0.605 molVHNO3 = mol x 1000 / M

= 0.605 x 1000 / 2.00= 302.5 cm3

(b) What volume of NO2 gas will be produced at 25C and


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(b) What volume of NO2 gas will be produced at 25 C and1.01 x 105 Pa? [Ar Cu = 63.5: Zn = 65.4]

Since NO2 is only produced from reaction with CuSo, mol of NO2 = 2 (0.1417)

= 0.2834 molUnder room condition,V = mol x Vm

= 0.2834 x 22.4 dm3

= 6.91 dm3

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Chemistry Form 6 Chap 01