Chemistry Form 4 Notes – Chapter 3: Chemical Formulae & Equation [PART1]

PART 1

RAM - relative atomic mass
RMM - relative molecular mass

Pertama anda perlu faham konsep ini. Anda perlu tahu cara mengiranya. Ok, sebelum itu,kita faham ini dahulu:

1. Relative atomic mass, Ar is the atomic mass of an atom when compared to a standardatom

2. Standard atom:

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,
• the reactive masses of some elements were not accurate,
• hydrogen exists as a gas at room temperature and
• has a number of isotopes with different masses.

Helium scale: the second lightest atom of all and the mass of one helium atom was assigned 1 unit.

Weakness of Helium scale:

• Mass of 1 helium atom = 4 times the mass of a hydrogen atom
• So, mass of 1 helium atom = 4 times 1/12 mass of a carbon atom
• helium exists as a gas at room temperature and
• helium is an inert gas.

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 (Chemist) and
• used the isotopes oxygen-16 as the standard (Physicists).

Carbon scale: standard atom of comparison internationally.

• a carbon-12 atom is 12 times heavier than an atom of hydrogen,
• used as the reference standard in mass spectrometers,
• exists as a solid at room temperature,
• most abundant carbon isotope, happening about 98.89% and
• carbon-12 is close to the agreement based on oxygen.

3. Relative molecular mass, Mr of a substances is the average mass of a molecule (two or more atoms) of the substances when compared 1/12 with of the mass of a carbon-12 atom.

4. Relative formula mass, Fr is for ionic compound which is calculated by adding up therelative atomic masses of all the atoms.

5. Example:

• Relative atomic mass, Ar of helium = 4
• Relative molecular mass, Mr of CO₂ = 12 + 2(16) = 44
• Relative formula mass, Fr of NaCl = 23 + 35.5 = 58.5
• Relative formula mass, Na₂CO_3·10H₂O = 2(23) + 12 + 3(16) + 10 [2(1) + 16] = 286

Contoh di atas adalah cara untuk mengira jisim sesuatu sebatian. Ini sangat penting bila kita nak kira bilangan mol nanti.

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

2. Atomic substances

§  Elements – all the particles are atoms.

§  Example: zinc (Zn), sodium (Na), aluminium (Al) and all noble gases, argon (Ar), helium (He) and neon (Ne).

§  RAM (Relative Atomic Mass) of  Na = 23

3. Molecular substances

§  Covalent compounds – the particles are molecules.

§  Example: carbon dioxide (CO₂), water (H₂O) and non-metal elements, iodine (I₂), nitrogen (N₂) and oxygen (O₂).

§  RMM (Relative Molecular Mass) of I₂ = 127 + 127 = 254

4. Ionic substances

§  Ionic compounds – the particles are ions.

§  Example: sodium chloride (NaCl), hydrochloric acid (HCl) and potassium iodide (KI).

§  RFM (Relative Formula Mass) of HCl = 1 + 35.5 = 36.5

5. Avogadro’s Law / Gas Law states that equal volumes of all gases contain the same numberof molecules under the same temperature and pressure.

§  Example: equal volumes of molecular hydrogen and nitrogen would contain the same number of molecules under the same temperature and pressure.

6. Volume of gas (dm³) = Number of moles of gas x Molar volume

7. Room temperature and pressure (r.t.p.) = 24 dm³ mol^-1 (25°C and 1 atm)

§  Example: What is the volume of 5.0 mol helium gas at s.t.p.?

§  Volume of gas = Number of moles x Molar gas volume
= 5.0 mol x 24 dm³ mol^-1
= 120 dm³

8. Standard temperature and pressure (s.t.p.) = 22.4 dm³ mol^-1 (0°C and 1 atm)

§  Example: What is the volume of 5.0 mol helium gas at s.t.p.?

§  Volume of gas = Number of moles x Molar gas volume
= 5.0 mol x 22.4 dm³ mol^-1
= 112 dm³

9. Mass (g) = Number of moles x Molar mass

10. Number of particles = Number of moles x Avogadro constant

11. Volume (dm³) = Number of moles x Molar volume

Be sure to copy down all these formulae a few times on paper so that you will have a better chance recalling it in the future. Copying them onto a card to bring around will be very helpful.

Empirical and Molecular Formulae

1. Empirical (simplest ratio of atoms of each element that present in the compound) and molecular formulae (actual number of atoms of each element that are present in one molecule of the compound) indicate:

§  the types of the elements

§  the symbols of the elements and the ratio of atoms or

§  moles of atoms of each element in a compound.

2. Molecular formula = (empirical formula)_n

n is a positive number

Compound	Molecular formula	n	Empirical formula
Carbon dioxide	CO2	1	(CO2) = CO2
Ethane	CH3	2	(CH3)2 = C2H6
Propene	CH2	3	(CH2)3 = C3H6
Glucose	CH2O	6	(CH2O)6 = C6H12O6
Quinine	C10H12NO	2	C20H24N2O2

3. Chemical formulae for covalent compounds.

Name	Chemical formula	Number of each element
Nitrogen gas	N2	2 nitrogen atoms
Oxygen gas	O2	2 oxygen atoms
Ammonia	NH3	1 nitrogen atom and 3 hydrogen atoms
Water	H2O	2 hydrogen atoms and 1 oxygen atom

4. Cations are 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+

* refer to the Roman numerals

5. Anions are 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-

6. Chemical formulae for ionic compounds

Name	Chemical formula	Number of cation	Number of anion
Zinc chloride	ZnCl2	1 Zn2+	2 Cl-
Copper(II) sulphate	CuSO4	2 Cu2+	2 SO42-
Aluminium sulphate	Al2(SO4)3	2 Al3+	3 SO42-

7. Meaning of prefixes

Prefix	Meaning
Mono-	1
Di-	2
Tri-	3
Tetra-	4
Penta-	5
Hexa-	6
Hepta-	7
Octa-	8
Nona-	9
Deca-	10

8. Naming of chemical (non-metal) compounds with Greek numerical prefixes.

Non-metal compound	Chemical formula
Carbon monoxide	CO
Carbon dioxide	CO2
Sulphur dioxide	SO2
Sulphur trioxide	SO3
Carbon tetrachloride (tetrachloromethane)	CCl4

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Chemical Equation

1. Importance of chemical equation:
The types of reactants; the physical conditions; the quantity of reactants and products and stated in moles.
nA + nB –> pC + pD

2. Reactants are written in the left side of the reaction and products are written in the right side of the reaction.

§  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: 2Al + 3CuO –> Al₂O₃ + 3Cu
Complete chemical equation: 2Al + 3CuO –> Al₂O₃ + 3Cu

§  Example 3:
Word equation: Nitrogen + hydrogen <–> ammonia
Chemical equation: N₂ + H₂ <–> NH₃
Balancing equation: N₂ + 3H₂ <–> 2NH₃
Complete chemical equation: N₂ + 3H₂ <–> 2NH₃

3. Information obtainable from chemical equations.

§  i) mass of reactants

§  ii) volume of reacting gas

§  iii) mass of products formed

§  iv) volume of gas produced

§  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 10²³mol^-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 and2 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