Int Chem Chap 3

8/14/2019 Int Chem Chap 3

1/70

3 CHEMICAL FORMULAEAND EQUATIONS


2/70

Relative atomic mass, Ar

1. The mass of an atom is very small andtherefore difficult to be determined accurately.

2. However, we can determine the mass of anatom relative to a standard atom.

3. The atomic mass of an atom found by

comparing it with a standard atom, is calledrelative atomic mass, or A

r, for short.


3/70


Using Hydrogen4. This method had several weaknesses.(i) Relative masses of certain elements

cannot be determined because these elements

do not combine readily with hydrogen or are,unable to displace hydrogen;

(ii) It was found that the relative massesof some elements were not accurate.

(iii) Since hydrogen exists as a gas atroom temperature, it is difficult to determineits mass accurately.


4/70


5. (a) Oxygen was then used asa standard to compare the masses

of atoms,(b) However, problems arose

when the existence ofthree

isotopes of oxygen werediscovered.


5/70

Relative atomic mass based oncarbon-12 scale

1. Carbon-12 was chosen for a few reasons.

(a) Carbon-12 is used as a reference standard inthe mass spectrometer.

(b) Many elements can combine with carbon-12.

(c) Carbon-12 exists as a solid at roomtemperature and thus can be handled easily.

(d) Carbon-12 is the most abundant carbonisotope, occurring about 98.89%.

(e) The table of relative atomic masses based oncarbon-12 is in close agreement with the tablesbased on oxygen.


6/70


2. Carbon-12 isotope is assigned a massof exactly 12 units.

Relative atomic mass of an element=

112

12

theaveragemassof oneatomof theelement

of themassof anatomof carbon


7/70


3. For example, the average mass of onenitrogen atom is 14 times larger than 1/12 ofthe mass of carbon-12 atom. Therefore, therelative atomic mass of nitrogen is 14.

[Relative atomic mass is not the actualmass of the atom. It is only a comparisonvalue. Therefore, it has no unit.]


8/70

Mr

1.The relative molecular mass of asubstance is the average mass of a

molecule of the substance whencompared with of the mass of an atomof carbon-12.

Relative molecular mass of a

substance tan1

1212

the average mass of one molecule of the subs ce

of themassof anatomof carbon

=


9/70

Mr

2. The relative molecular mass of a substancecan be calculated by adding up therelative atomic masses of all theatoms present in a molecule of thesubstance.

3. For this reason, it is essential to know the

molecular formula of the substance first.


10/70

Mr

4. Table 3.1 shows how tocalculate the relativemolecular masses of somesubstances.

Substance Molecular formula

Relativemolecular

mass,Mr

Hydrogen

gas

H2

2(1)=2

Oxygengas O2 2(16)=32

Water H2O 2(1)+16=18

Ammonia NH3

14+3(1)=17

Propane C3H

83(12)+8(1)=44

Ethanol C2H

5OH 2(12)+5(1)+16+1

=46

[Relative atomic mass:

H,1; C,12; N,14; O,16]


11/70


12/70

Mr

6. Table 3.2shows thecalculation of

the relativeformula massesof some ioniccompounds.

Substance Molecular formula

Relative formulamass, M

r

Sodium

chloride

NaCl 23+35.5=58.5

Potassium

oxide

K2O 2(39)+16=94

Aluminium

sulphate

Al2(SO

4)

32(27)+3[32+4(16

)]=342

Hydrated

copper(II)

sulphate

CuSO4.5H

2O

64+32+4(16)+5[

2(1)+16]=250[Relative atomic mass: H,1;

O,16; Na,23; Al,27; S,32;

Cl,35.5; K,39; Cu,64]


13/70

3.2 The Mole and the Number ofParticles

1. In daily life, we count objects in units such as dozen andpair. In chemistry, we measure substances in, `mole'. Thesymbol of mole is mol.

2. One mole is an amount of substance that contains as

many particles as the number of atoms in exactly 12 g ofcarbon-12, which is 6.02 x 1023 particles.

3. The number of particles per mole (6.02 x 1023 mol-') isdetermined experimentally and is known as the Avogadroconstant or Avogadro number.

The Avogadro constant (NA) is defined as the number of

particles in one mole of a substance.


14/70


4.In simple words, 1 mole of substances contains

6.02 x 1023 particles, 2 moles ofsubstance contain 2 x 6.02 x 1023

particles, and soon.

Atom, molecules,ions


15/70


5. We can convert the number ofmoles of any substance to the

number of particles in it and viceversa using the followingrelationship.

A

x

N

AN

Numberof mol Numberof particles


16/70

3.3 The Mole and the Mass ofSubstances

1. The molar mass of asubstance is the mass ofone

mole of the substance. It has aunit of grams per mole (g mol-1). 2. In other words, the molar mass

of a substance is the mass of 6.02x 1023 particles of the substance.


17/70


18/70


4. We can easily calculate the mass ofany number of moles of substance or viceversa using the following relationship.

molar mass

molar massnumber of moles mass in grams


19/70


5. Equal numbers ofmoles ofsubstances always contain the samenumber ofparticles.

6. For this, reason, we can compare

the number of .particles insubstances just by comparing thenumber of moles of the substances.


20/70


21/70

3.4 The Mole and the Volume ofGas

5. The molar volume of any gas is 22.4 dm3mol-1 at STP or 24 dm3 mol-1 at room conditions.

For example

1 mole of oxygen gas occupies 22.4 dm3 atSTP.

1 mole of hydrogen gas occupies 24 dm3 atroom temperature.


22/70


6. The following relationship shows the coneof the volume of a gas to the number of molesand vice versa.

molar volume

molar volumenumber of moles volume of gas


23/70


7. The following shows the relationshipsbetween the number of moles, number ofparticles, mass and volume of gases.


24/70


25/70

3.5Chemical Formulae

1. Each chemical substance it given a name and achemical formula.

2. A chemical formula is a representation of a chemicalsubstance using letters for atoms and subscriptnumbers to show the numbers of each type o f atomsthat are present in the substance.

3. For example, the chemical formula of water is H2O.

The chemical formula of carbon dioxide is CO2.


26/70

3.5Chemical Formulae

4. Based on the chemical formula of asubstance, we can gather the following

information. (a) The elements that make up the

substance. (b) The ratio or number of atoms of

each element in the substance.


27/70

Chemical formulae of

elements 1. For elements that exist as atoms, their chemical

formulae represent their atoms. For example, theformulae for carbon and copper are C and Cu

respectively.

2. For elements that exist as molecules, their chemicalformulae represent their molecules. The subscriptnumber at the bottom right shows the number of atoms

in each molecule. 3. For example, the chemical formula of oxygen gas is

as follows:


28/70


compounds (a) Glucose consists of three elements

namely, carbon, hydrogen and oxygen. Sixcarbon atoms combine with twelve hydrogen

atom and six oxygen atoms.


29/70


compounds (b) Magnesium nitrate consists of the

elements magnesium, nitrogen and oxygen.Each magnesium atom combines with two

nitrogen atoms and six oxygen atoms.

E i i l f l d


30/70

Empirical formulae andmolecular formulae

1. There are two types of chemicalformulae - empirical formulae and

molecular formulae 2. The empirical formula of a

compound gives the simplest whole

number ratio of atoms of eachelement present in the compound.


31/70

Determining empirical


32/70

Determining empiricalformulae

1. The empirical formula of acompound can be determined

experimentally by finding the simplestratio of moles of atoms of each elementin the compound.



33/70


2.The following shows the steps in determiningthe empirical formula of a compound.

(a) Find the mass of each element in thecompound experimentally.

(b) Convert the masses to the numbers ofmoles of atoms.

(c) Find the simplest ratioof moles of theelements



34/70


Example

2.24 g of iron combines chemically with0.96 g of oxygen to form an oxide. Whatis the empirical formula of the oxide?[Relative atomic mass: 0, 16; Fe, 56]



35/70


Solution:

Since 2moles of ironatomscombinewith 3 molesof oxygenatoms, theempiricalformula ofthe oxide is

Fe2O3.

2.240.04

56=

0.960.06

16=

0.04

10.04=

0.06

1.50.04=

Element Iron, Fe Oxygen, O

Mass(g) 2.24 0.96

No. of moles of

atoms

Ratio of moles

Simplest ratio of

moles

2 3



36/70


Example

A potassium compound has a percentagecomposition as the following.

K, 31.84%; Cl, 28.98%; O, 39.18%

What is the empirical formula of thepotassium compound?

[Relative atomic mass: 0, 16; Cl, 35.5; K,39]



37/70


Example 3.23

Solution:

From the percentage composition, we knowthat every 100 g of the compoundcontains 31.84 g of potassium, 28.98 g ofchlorine and 39.18 g of oxygen. So, by

taking 100 g of the compound:



38/70


Example

Solution:

1 mole of potassiumatoms combineswith 1 mole ofchlorine atoms and

3 moles of oxygenatoms. Therefore,the, empiricalformula. of thepotassiumcompound is KClO3

31.840.816

39=

28.980.816

35.5=

39.182.449

16=

0.816

10.816 =

0.816

10.816=

2.449

30.816 =

Element Potassium, K

Chlorine, Cl

Oxygen,O

Mass(g) 31.84 28.98 39.18

No. of moles

of atoms

Ratio of

moles

Simplest

ratio of

moles

1 1 3

Activity: Determining the


39/70

Activity: Determining theempirical formula of copper (II)

oxide Apparatus: Combustion tube with

a small hole at the end, Bunsen

burner, stoppers, glass tube,retort stand; and clamp, balance,U tube, spatula, porcelain dish.


40/70

Activity: determining the


41/70

Activity: determining theempirical formula of copper (II)

oxide

Activity: determining the


42/70

Activity: determining theempirical formula of copper (II)

oxide 1. This method can also be used to

determine the empirical formula of

oxides of low reactivity metals such astin(II) oxide and lead(II) oxide. 2. The empirical formula of copper(II)

oxide cannot be determined by heating

copper(II) oxide with reactive metalssuch as magnesium or calcium.



43/70

Activity: Determining theempirical formula of magnesium

oxide



44/70

Activity: Determining theempirical formula of magnesium

oxide1. The following are the precautions taken in this activity. (a) The crucible is covered with its lid to prevent the white

fumes of magnesium oxide from escaping. This wouldaffect the accuracy coif the mass obtained.

(b) The lid is removed at intervals to allow oxygen to enterthe crucible and reacts with the magnesium ribbon.

(c) Heating, cooling and weighing are repeated until aconstant mass is obtained to ensure that, the magnesiumribbon reacts completely to form magnesium oxide.


45/70

Finding molecular

formulae 1. Actually, the molecular formula of a

compound is a multiple of its empirical Molecular formula=(Empirical formula) x n whereby n is a positive integer.


46/70

Finding molecular

formulae 2. Table 3.5 shows the molecular and empirical

formulae of some compounds.

Compound Empirical

formula

Molecular

Formula

n

Water H2O (H

2O)

1= H

2O 1

Ethane CH3 (CH3)2= C2H6 2

Propene CH2

(CH2)

3= C

3H

63


47/70

Finding molecular

formulae Example A hydrocarbon compound has an

empirical formula of CH2 and a relativemolecular mass of 70. Find themolecular formula of the compound.

[Relative atomic mass: H, 1; C, 12]


48/70

Finding molecular

formulae Solution: Let the molecular formula of the compound to be (CH2) n, Based on the formula (CH2) n, the relative molecular mass = n[12 + 2(1)] = 14n However, it is given that the relative molecular mass = 70 Therefore, 14n = 70

n =

n =5

So, the molecular formula of the compound is (CH2)5 ,whichis C5H10 .

70

14


49/70

Chemical formulae of ionic

compounds 1. Ionic compounds are compounds

consisting of cations and anions.

Cations are positively charged ionswhereas anions are negatively chargedions.

2. It is important that you know the

formulae of cations and anions beforeconstructing the chemical formulae ofionic compounds.


50/70


compounds 3. Table 3.6 and Table 3.7 show the formulae ofsome common cations and anions

Cation(positive ion)

Formula of cation Charge of cation

Sodium ion Na+ +1

Potassium ion K + +1

Silver ion Ag+ +1

Hydrogen ion H+ +1


51/70


compounds



Ammonium ion NH4

+ +1

Copper(I) ion Cu+ +1

Calcium ion Ca2+ +2

Magnesium ion Mg2+ +2

Zinc ion Zn2+ +2


52/70


compounds



Barium ion Ba2+ +2

Iron(II) ion Fe2+ +2

Copper(II) ion Cu2+ +2

Tin(II) ion Sn2+ +2


53/70


compounds



Lead(II) ion Pb2+ +2

Aluminium ion Al3+ +3

Iron(III) ion Fe3+ +3

Chromium(III) ion Cr 3+ +3

Tin(IV) ion Sn4+ +4

Lead(IV) ion Pb4+ +4


54/70


55/70


compounds

Anion

(negative ion)

Formula of anion Charge of anion

Ethanoate ion CH3COO- -1

Manganate(IV) ion MnO4

- -1

Oxide ion O2- -2

Carbonate ion CO3

2- -2

Sulphide ion S2- -2


56/70


compounds 3. Table 3.6 and Table 3.7 show the formulae of some commoncations and anions

Anion

(negative ion)

Formula of anion Charge of anion

Sulphate ion SO4

2- -2

Sulphite ion SO3

2- -2

Thiosulphate ion S2O

32- -2

Chromate(VI) ion CrO4

2- -2

Dichromate(VI) ion Cr 2O

72- -2

phosphate ion PO4

3- -3


57/70


compounds 4. Even though ionic compounds contain

charged particles, their chemical formulaeare electrically neutral. This is because the

total of positive charges are equal tothe total of negative.


58/70

Chemical formulae of ioniccompounds 5. The chemical formula of an ionic compound

can be constructed as the following. (a) From its name, identify and write down the

formula of its cation and anion. (b) Determine the number of cations and anions

by balancing the positive and negative charges. (c) Write the formula of the compound. The

number of cations and anions are written assubscript numbers.


59/70

Chemical formulae of ioniccompounds Example


60/70

Chemical formulae of ioniccompounds Example


61/70

Naming of chemicalcompounds

1. Chemical compounds arenamed systematic according to the

guidelines given by InternationalUnion of Pure and App Chemistry(IUPAC).


62/70


63/70


3. Certain elements such as transitionmetals, can form more than one type ofions. Roman numerals (such as I, II and III)

are used in their naming to differentiate theions.

(a) For example, iron can form two cation: Fe 2+ - named as iron(II) ion Fe 3+ - named as iron(III) ion


64/70


3. Certain elements such as transition metals,can form more than one type of ions. Romannumerals (such as I, II and III) are used in their

naming to differentiate the ions. (a) For example, iron can form two cation: Fe 2+ - named as iron(II) ion Fe 3+ - named as iron(III) ion

(b) Therefore, the names of oxides with ironare, iron(II) oxide and iron(III) oxiderespectively.


65/70


4. For simple molecular compounds,the name of the first element is

maintained as it is. However, thename of the second element is addedwith an ide. For example, asmolecular; compound consisting of

hydrogen and chlorine is given thename hydrogen chloride.


66/70


5. Greek prefixes are used to show the number ofatoms of each element in a compound.

(a) Here are some examples.

CO- Carbon monoxide CO2 Carbon dioxide

SO3 Sulphur trioxide

CCl4 Carbon tetrachloride

PCl5 Phosphorus pentachloride

N2O4 Dinitrogen tetroxide

Cl2O7 Dichlorine heptoxide


67/70

3 CHEMICAL FORMULAEAND EQUATIONS

3.6 Chemical Equations


68/70

3.6Chemical Equations

1. A chemical reaction is said to occurwhen a few starting substances react toproduce new substances.

2. The starting substances are calledreactants. The new substances formedare called products.

productsreactants producing


69/70


70/70

Documents

Int Chem Chap 3