Saturday, July 14, 2012

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 massAr 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 scalestandard 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 massMr 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, Na2CO3·10H2O = 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 dmmol-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
So, do come back for more Berry Essential Notes. If it is not simplified to its essense, then it is not from Berry Berry Easy. Let us do the summarising while you understand the core before you master the whole topic. Do share out this site with your friends.
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)reacts with 2 moles of KI formed 1 mole PbIof and2 moles of KNO3.
Step 2: Convert to moles.
§  No. of moles of Pb(NO3)2
= Mass of Pb(NO3)/ 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)
= 2/1
§  Number of moles of KNO
= (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



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