Chapter 3 Chemical Formulae and Equations

Post on 15-Jul-2015

486 views 0 download

Transcript of Chapter 3 Chemical Formulae and Equations

Chapter 3 Chemical Formulae

And

Equations

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.

• 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)

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

• 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

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

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

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

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

Relationship between the number of moles and the number of particles

Number of moles, (mol)

Number of particles (atoms, molecules, ions)

× NA

÷ NA

Number of particles

Moles NA

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]

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

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

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

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

Relationship between the number of moles and the mass of a substance

Number of moles, (mol)

Mass (g)

× molar mass

÷ molar mass

Mass (g)

Moles RAM / RMM /

RFM

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]

(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

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]

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

Relationship between the number of moles and the volume of gas

Number of moles, (mol)

Volume of gas (dm3)

× molar volume

÷ molar volume

Volume (dm3)

Moles

22.4 dm3 (STP) / 24 dm3 (room condition)

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

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

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

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

H2

Symbol of hydrogen atom

Shows that there are two hydrogen atom in a hydrogen gas,

H2 molecule

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

• Compound can be represented by two types:

1. Empirical formula

2. Molecular formula

Empirical Formula

• Meaning

Formula that show the simplest whole number ratio of atoms of each element in the compound

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]

Element Al O

Mass of element (g)

Number of mole (mol)

Ratio of moles

Simplest ratio

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

To determine empirical formula of magnesium oxide

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]

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

Result:

Description Mass (g)

Crucible + lid x

Crucible + lid + Mg y

Crucible +lid + MgO z

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

To determine empirical formula copper(II) oxide

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

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

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

Example:

(CH3)n = 30

n [12 + 3(1) ] = 30

15n = 30

n = 30/15

= 2

Molecular formula = (CH3)2

= C2H6

Ionic Formulae

Positive ions

(cation)

Negative ions

(anion)

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+

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

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

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

Example:

Interpreting:

2 mol of hydrogen, H₂ react with 1 mol of oxygen, O₂ to produced 2 mol of water

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]