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SPM CHEMISTRY Chapter 3: Chemical Formulae & Equations Prepared by: Ms. Brintha Ganapathy BSc. MSc. 3.1 Relative atomic mass and Relative molecular mass 1. Relative atomic mass, Ar is the __________________________ of an atom when compared to a __________________________. 2. Standard atom: a. Hydrogen scale: hydrogen is the lightest atom of all and the mass of one hydrogen atom was assigned 1 unit . Weakness of Hydrogen scale: not too many elements can react readily with hydrogen, hydrogen exists as a gas at room temperature and has a number of isotopes with different masses. b. Helium scale: the second lightest atom of all and the mass of one helium atom were assigned 1 unit . Weakness of Helium scale: 1

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SPM CHEMISTRY CHAPTER 3: CHEMICAL FORMULAE AND EQUATIONS

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Page 1: Spm Chemistry Chapter 3

SPM CHEMISTRY

Chapter 3: Chemical Formulae & Equations

Prepared by: Ms. Brintha Ganapathy BSc. MSc.

3.1 Relative atomic mass and Relative molecular mass

1. Relative atomic mass, Ar is the __________________________ of an atom

when compared to a __________________________.

2. Standard atom:

a. Hydrogen scale: hydrogen is the lightest atom of all and the mass

of one hydrogen atom was assigned 1 unit.

Weakness of Hydrogen scale:

not too many elements can react readily with hydrogen,

hydrogen exists as a gas at room temperature and

has a number of isotopes with different masses.

b. Helium scale: the second lightest atom of all and the mass of one

helium atom were assigned 1 unit.

Weakness of Helium scale:

helium exists as a gas at room temperature and

helium is an inert gas.

c. Oxygen scale: chose as the standard atom to compare the masses

of atoms

Weakness of Oxygen scale:

the existence of three isotopes of oxygen were discovered

natural oxygen (containing all the three isotopes) as the

standard

used the isotopes oxygen-16 as the standard

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Page 2: Spm Chemistry Chapter 3

d. Carbon scale: standard atom of comparison internationally.

a carbon-12 atom is __________________________than an atom of

hydrogen

exists as a solid at room temperature

most abundant carbon isotope, happening about 98.89%

3. Relative atomic mass, Ar of an atom is

4. Relative molecular mass, Mr of a substance is

5. Example:

Relative atomic mass, Ar of helium = 4

Relative molecular mass, Mr of H O ₂

Relative formula mass, Na2CO3·10H2O

=

2

Page 3: Spm Chemistry Chapter 3

Practice 3.1

Calculate the RMM of the following substances.

[RMM: H, 1; C, 12; O, 16; Cl, 35.5, Na, 23; Mg, 24; Al, 27; S, 32; Ca, 40]

a. Ethanol, C H OH₂ ₅ b. Oxygen gas, O₂ c. Carbon dioxide,

CO₂

d. Water molecule,

H O₂e. Tetrachloromethane,

CCl₄f. Sodium chloride,

NaCl

g. Carbon monoxide,

CO

h. Aluminium

carbonate, Al (CO )₂ ₃ ₃i. Magnesium

chloride, MgCl₂

j. Calcium carbonate,

CaCO₃k. Calcium sulphate,

CaSO₄l. Hydrochloric acid,

HCl

3

Page 4: Spm Chemistry Chapter 3

3.2, 3.3, 3.4 Relationship between the number of moles and the

number of particles, mass and volume of gas

1. Avogadro constant / Avogadro’s number is 6.02 x 1023

2. Room temperature and pressure (r.t.p.) = 24 dm3 mol-1

(25°C and 1 atm)

3. Standard temperature and pressure (s.t.p) = 22.4 dm3 mol-1

Mass (g) = Number of moles x Molar mass

Number of particles = Number of moles x Avogadro constant

Volume (dm3) = Number of moles x Molar volume

Question!

State three formulae to find number of moles

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Page 5: Spm Chemistry Chapter 3

3.5 Empirical and Molecular Formulae

1. Empirical formulae definition

Simplest ratio of atoms of each element that present in the compound

2. Molecular formulae definition

Actual number of atoms of each element that is present in one

molecule of the compound

3. Molecular formula = (empirical formula)n,

Calculation of Empirical Formula

Example 1 Example 2

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Page 6: Spm Chemistry Chapter 3

List of cations = positively-charged ions

Charge Cations Formula

+ 1 Ammonium ion NH4+

+ 1 Copper(I) ion Cu+

+ 1 Hydrogen ion H+

+ 1 Lithium ion Li+

+ 1 Nickel(I) ion Ni+

+ 1 Potassium ion K+

+ 1 Silver ion Ag+

+ 1 Sodium ion Na+

+ 2 Barium ion Ba2+

+ 2 Calcium ion Ca2+

+ 2 Copper(II) ion Cu2+

+ 2 Iron(II) ion Fe2+

+ 2 Lead(II) ion Pb2+

+ 2 Magnesium ion Mg2+

+ 2 Manganese(II) ion Mn2+

+ 2 Nickel(II) ion Ni2+

+ 2 Tin(II) ion Sn2+

+ 2 Zinc ion Zn2+

+ 3 Aluminium ion Al3+

+ 3 Chromium(III) ion Cr3+

+ 3 Iron(III) ion Fe3+

+ 4 Lead(IV) ion Pb4+

+ 4 Tin(IV) ion Sn4+

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PLEASE MEMORISE!

Page 7: Spm Chemistry Chapter 3

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Page 8: Spm Chemistry Chapter 3

List of Anions = negatively-charged ions

Charge Anions Formula

- 1 Bromide ion Br-

- 1 Chloride ion Cl-

- 1 Chlorate(V) ion ClO3-

- 1 Ethanoate ion CH3COO-

- 1 Fluoride ion F-

- 1 Hydride ion H-

- 1 Hydroxide ion OH-

- 1 Iodide ion I-

- 1 Manganate(VII) ion MnO4-

- 1 Nitrate ion NO3-

- 1 Nitrite ion NO2-

- 2 Oxide ion O2-

- 2 Carbonate ion CO32-

- 2 Chromate(VI) ion CrO42-

- 2 Dichromate(VI) ion Cr2O72-

- 2 Sulphide ion S2-

- 2 Sulphate ion SO42-

- 2 Sulphite ion SO32-

- 2 Thiosulphate ion S2O32-

- 3 Nitride ion N3-

- 3 Phosphate ion PO43-

- 3 Phosphite ion PO33-

8

PLEASE MEMORISE!

Page 9: Spm Chemistry Chapter 3

Chemical formulae for ionic compounds

Name Chemical

formula

Number of

cation

Number of

anion

Zinc chloride ZnCl2

Copper(II)

sulphate

CuSO4

Aluminium

sulphate

Al2(SO4)3

Prefixes

Prefix Meaning

Mono- 1

Di- 2

Tri- 3

Tetra- 4

Penta- 5

Hexa- 6

Hepta- 7

Octa- 8

Nona- 9

Deca- 10

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Page 10: Spm Chemistry Chapter 3

Chemical Equation

nA + nB nC + nD

Example 1:

Word equation:

Sodium hydroxide + sulphuric acid sodium sulphate + water

Chemical equation: NaOH + H2SO4 Na2SO4 + H2O

Balancing equation: 2NaOH + H2SO4 Na2SO4 + 2H2O

Complete chemical equation: 2NaOH + H2SO4 Na2SO4 + 2H2O

Example 2:

Word equation:

Aluminium + copper(II) oxide aluminium(III) oxide + copper

Chemical equation: Al + CuO Al2O3 + Cu

Balancing equation: __________________________________________________________

Complete chemical equation: _______________________________________________

Example 3:

Word equation: Nitrogen + hydrogen ammonia

Chemical equation: N2 + H2 NH3

Balancing equation: __________________________________________________________

Complete chemical equation: _______________________________________________

Information obtainable from chemical equations

i) mass of reactants

ii) volume of reacting gas

iii) mass of products formed

iv) volume of gas produced

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Page 11: Spm Chemistry Chapter 3

Balancing Chemical Equations

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TIPS!Always balance the chemical equations

starting from the gases

Page 12: Spm Chemistry Chapter 3

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Page 13: Spm Chemistry Chapter 3

Example:

2 cm3 of lead (II) nitrate solution is added to excess of potassium iodide

solution. How many molecules of potassium nitrate will be formed?

[Relative atomic mass: N, 14; O, 16; K, 39; I, 127; Pb, 207; Avogadro's constant:

6.02 x 1023 mol-1]

Step 1: Write a complete chemical equation.

Pb(NO3)2(aq) + 2KI(aq) –> PbI2(s) + 2KNO3(aq)

From the equation, 1 mole of Pb(NO3)2 reacts with 2 moles of KI formed

1 mole PbI2 of and 2 moles of KNO3.

Step 2: Convert to moles.

No. of moles of Pb(NO3)2

= Mass of Pb(NO3)2 / Relative molecular mass

= 2 / [207 + 2(14 + 3 x 16)]

= 6.04 x 10-3 mol

Step 3: Ratio of moles.

Number of moles of KNO3/ Number of moles of Pb(NO3)2

= 2/1

Number of moles of KNO3

= (2 x 6.04 x 10-3) / 1

= 12.08 x 10-3 mol

Step 4: Convert to the number of molecules of potassium nitrate.

Number of molecules of KNO3

= 12.08 x 10-3 x 6.02 x 1023

= 7.27 x 1021

13

This part of the chapter is the most challenging, but

once you hit the technique it will be easy forever.