Chapter 3 Chemical Formulae and Equations
Transcript of Chapter 3 Chemical Formulae and Equations
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Chapter 3 Chemical Formulae
And
Equations
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A. Relative Atomic Mass and Relative Molecular Mass
• Based on the theory of particles:
particles are very small and discrete. A single atom is too small and light and cannot be weighed directly
• Thus, the mass of an atom is obtained by comparing it with another atom which is taken as a standard.
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• 3 types of scale to determine the mass of the particles
a) Compared with a hydrogen-1 scale
b) Compared with an oxygen-16
c) Compared with carbon-12 (modern comparison UNTILL TODAY)
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Relative atomic mass, RAM
• Meaning;
The average mass of one atom of the element when compared with 1/12 of the mass of an atom of
carbon-12.
Relative Atomic Mass, RAM
= Average mass of one atom of the element
1/12 x the mass of an atom of carbon-12
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• Example:
RAM of magnesium
= 24 = 24
1/12 x 12
= magnesium is 24 times larger than carbon-12
** THE VALUE OF NUCLEON NUMBER IN THE PERIODIC TABLE OF ELEMENT
= RELATIVE ATOMIC MASS, RAM
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Relative molecular mass, RMM
• Meaning; The average mass of one molecule when compared
with 1/12 of the mass of an atom of carbon-12. Relative Molecular Mass, RMM = Average mass of one molecule 1/12 x the mass of an atom of carbon-12 • Calculate RMM/RFM by adding up the relative atomic
mass of all the atoms that present in the molecule/ionic compound
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B. The Mole and the Number of Particles
• The number of particles in matter is measured in mole.
• Definition:
The amount of substance that contains as many particles as the number of atoms in exactly 12 g of carbon-12
• Symbol of mole: mol
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How many atoms are there in 12 g of carbon-12?
= 6.02 × 1023
• The value of 6.02 × 1023 is called the Avogadro constant or Avogadro number
• Avogadro constant, NA
The number of particles in one mole of a substance
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Point to note:
One mole of any substance contains 6.02 × 1023 particles
1 mol of atomic substance contains 6.02 × 1023 atoms
1 mol of molecular substance contains 6.02 × 1023 molecules
1 mol of ionic substance contains 6.02 × 1023 ions
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Relationship between the number of moles and the number of particles
Number of moles, (mol)
Number of particles (atoms, molecules, ions)
× NA
÷ NA
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Number of particles
Moles NA
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Example 1:
A closed glass bottle contains 0.5 mol of oxygen gas, O2
(a) How many oxygen molecules, O2 are there in the bottle?
(b) How many oxygen atoms are there in the bottle?
[Avogadro constant: 6.02 × 1023 mol-1]
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a) The number of oxygen molecules, O2
= 0.5 mol × 6.02 × 1023 mol-1
= 3.01 × 1023 molecules
b) The number of oxygen atoms
= 0.5 mol × 6.02 × 1023 mol-1 × 2
= 6.02 × 1023 atoms
Number of particles
Moles NA
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Example 2:
Find the number of moles of 9.03 × 1023 molecules in a sample containing molecules of carbon dioxide, CO2 [Avogadro constant: 6.02 × 1023 mol-1]
The number of moles carbon dioxide
= 9.03 × 1023
6.02 × 1023 mol-1
= 1.5 mol
Number of particles
Moles NA
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C. The Mole and the Mass of Substances
• The mass of one mole of any substance is called molar mass
• Units: g mol-1
• The molar mass of substances are numerically equal to relative mass
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Element Relative mass Mass of
1 mol Molar mass
Helium 4 4 4 g mol-1
Sodium 23 23 23 g mol-1
Water, H2O 2(1) + 16 = 18 18 18 g mol-1
Ammomia, NH3 14 + 3(1) = 17 17 17 g mol-1
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Relationship between the number of moles and the mass of a substance
Number of moles, (mol)
Mass (g)
× molar mass
÷ molar mass
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Mass (g)
Moles RAM / RMM /
RFM
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Example 1:
What is the mass of
(a) 0.1 mol of magnesium?
(b) 2.408 × 1023 atoms of magnesium?
[Relative atomic mass: Mg=24; Avogadro constant: 6.02 × 1023 mol-1]
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(b) The number of moles Mg atoms
= 2.408 × 1023
6.02 × 1023 mol-1
= 0.4 mol
Mass of Mg atoms
= 0.4 mol × 24 g mol-1 = 9.6 g
(a) Molar mass of Mg = 24 g mol-1
Mass of Mg = 0.1 mol × 24 g mol-1
= 2.4 g
Mass (g)
Moles
RAM / RMM /
RFM
Number of particles
Moles NA
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Example 2:
RMM of SO2
= 32 + 2(16) = 64
Molar mass of SO2 = 64 g mol-1
The number of moles = 16 g 64 g mol-1
= 0.25 mol
Mass (g)
Moles
RAM / RMM /
RFM
How many moles of molecules are there in 16 g of sulphur dioxide gas, SO2?
[Relative atomic mass: O=16, S=32]
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D. The Mole and the Volume of Gas
• The volume occupied by one mole of the gas is called molar volume
• One mole of any gas always has the same volume under the same temperature and pressure
• The molar volume of any gas is 22.4 dm3 at STP or 24 dm3 at room condition
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Relationship between the number of moles and the volume of gas
Number of moles, (mol)
Volume of gas (dm3)
× molar volume
÷ molar volume
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Volume (dm3)
Moles
22.4 dm3 (STP) / 24 dm3 (room condition)
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What is the volume of 1.2 mol of ammonia gas, NH3 at STP?
[Molar volume: 22.4 dm3 mol-1 at STP]
Example 1:
Volume (dm3)
Moles
22.4 dm3 (STP) / 24 dm3 (RC)
The volume of ammonia gas, NH3
= 1.2 mol × 22.4 dm3 mol-1
= 26.88 dm3
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How many moles of ammonia gas, NH3 are present in 600 cm3 of the gas measured at room conditions?
[Molar volume: 24 dm3 mol-1 at room condition]
Example 2:
Volume (dm3)
Moles
22.4 dm3 (STP) / 24 dm3 (RC)
The number of moles of ammonia gas, NH3
= 600 cm3 1000 = 0.6 dm3
= 0.6 dm3
24 dm3 mol-1
= 0.025 mol
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Relationship between the number of moles, number of particles, mass and the volume of gas
Number of moles, (mol)
Mass (g)
× molar volume ÷ molar volume
Number of particles
Volume of gas (dm3)
÷ NA
× NA × molar mass
÷ molar mass
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E. Chemical Formulae
• A chemical formulae
A representation of a chemical substance using letters for atom and subscript numbers to show the numbers of each type of atoms that are present in the substance
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H2
Symbol of hydrogen atom
Shows that there are two hydrogen atom in a hydrogen gas,
H2 molecule
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H2O
Symbol of hydrogen atom
Shows that there are two hydrogen atom in a water molecule
Symbol of oxygen atom
Shows that there are one oxygen atom in
a water molecule
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• Compound can be represented by two types:
1. Empirical formula
2. Molecular formula
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Empirical Formula
• Meaning
Formula that show the simplest whole number ratio of atoms of each element in the compound
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Example
A sample of aluminium oxide contains 1.08 g of aluminium and 0.96 g of oxygen. What is the empirical formula of this compound?
[Relative atomic mass: O = 16; Al = 27]
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Element Al O
Mass of element (g)
Number of mole (mol)
Ratio of moles
Simplest ratio
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To determine empirical formula of magnesium oxide
Burn magnesium with oxygen
To determine empirical formula copper(II) oxide
Use hydrogen to removed oxygen from copper(II) oxide
Weigh mass of copper
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To determine empirical formula of magnesium oxide
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Experiment question:
Describe how you can carry out an experiment to determine the empirical formula of magnesium oxide. Your description should include
• Procedure of experiment
• Tabulation of result
• Calculation of the results obtained
[Relative atomic mass: O = 16; Mg = 24]
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Procedure:
1. Clean (5-15 cm) magnesium ribbon with sandpaper and coil it
2. Weigh an empty crucible with its lid
3. Place the magnesium in the crucible and weigh again
4. Record the reading
5. Heat the crucible strongly without its lid
6. When magnesium start burning close the crucible. Open and close the lid very quickly interval time
7. When burning is complete, stop the heating
8. Let the crucible cool and then weigh it again
9. The heating, cooling and weighing process is repeated until a constant mass is recorded
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Result:
Description Mass (g)
Crucible + lid x
Crucible + lid + Mg y
Crucible +lid + MgO z
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Calculation:
Mg O
Mass (g) y-x z-y
No. of mole
(mol) (y-x)/24 (z-y)/16
Ratio 1 1
Empirical formula = MgO
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To determine empirical formula copper(II) oxide
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Discussion
1. H2 gas must be flowed through the apparatus to remove all the air
2. H2 gas must be flowed throughout the experiment to prevent the air from outside mixing with the H2 gas
3. H2 gas flowed through the apparatus during cooling to prevent copper being oxidised by air into copper(II) oxide
4. Repeat heating, cooling & weighing process to ensure all the copper(II) oxide changed into copper
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5. This method is to determine empirical formula of oxide of metals which are less reactive than H2 in the reactivity series
6. Other example: Lead(II) oxide, Iron(II) oxide
7. Function anhydrous calcium chloride – to dry the H2 gas
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Molecular Formula
• Meaning
Formula that show the actual number of atoms of each element that are present in a molecule of the compound
Molecular formula = (Empirical formula)n
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Example:
(CH3)n = 30
n [12 + 3(1) ] = 30
15n = 30
n = 30/15
= 2
Molecular formula = (CH3)2
= C2H6
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Ionic Formulae
Positive ions
(cation)
Negative ions
(anion)
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Formulae of cations & anions
Cation Formula Anion Formula
Sodium ion Na+ Chloride ion Cl-
Potassium ion K+ Bromide ion Br-
Zinc ion Zn2+ Iodide ion I-
Magnesium ion Mg2+ Oxide ion O2-
Calcium ion Ca2+ Hydroxide ion OH-
Aluminium ion Al3+ Sulphate ion SO42-
Iron(II) ion Fe2+ Carbonate ion CO32-
Iron(III) ion Fe3+ Nitrate ion NO3-
Copper(II) ion Cu2+ Phosphate ion PO43-
Ammonium ion NH4+
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Try this..
• Iron(II) hydroxide • Lithium oxide • Silver chloride • Calcium carbonate • Lead(II) oxide • Sulphuric acid • Hydrochloric acid • Nitric acid • Phosphoric acid
• Zinc sulphate
• Ammonium nitrate
• Copper(II) nitrate
• Ammonium carbonate
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F. Chemical Equation
• A chemical equation
Satu cara penulisan untuk menghuraikan sesuatu tindak balas kimia
• In qualitative aspect, equation shows:
Reactant produces products
Reactant → Product
A + B → C + D
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In quantitative aspect:
• Stoichiometry : A study of quantitative composition of a substances involved in chemical reaction
• The coefficients in a balanced chemical equation tell the exact proportions of reactants and products in equation
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Example:
Interpreting:
2 mol of hydrogen, H₂ react with 1 mol of oxygen, O₂ to produced 2 mol of water
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Numerical Problems Involving Chemical Equations
Copper(II) oxide, CuO reacts with aluminium according to the following equations.
3CuO + 2Al → Al2O3 + 3Cu
Calculate the mass of aluminium required to react completely with 12 g of copper(II) oxide, CuO
[Relative atomic mass: O, 16; Al, 27; Cu, 64]