O level Chemistry Notes

O level Chemistry Notes (Page 3)

O level Chemistry Notes (Page 3)

O level Chemistry Notes covering Atomic and molecular masses, Percentage composition of elements in compounds, The mole and molar mass, Amount of substance, Empirical formula and Molecular formula .These notes only show the key points in O’level chemistry curriculum.

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Atomic and molecular masses

The relative atomic mass (Ar) of an atom is found of the periodic table.

For example:

ElementSymbolRelative Atomic Mass
HydrogenH1
CarbonC12
Nitrogen N14
Oxygen O16
Sodium Na23
Aluminium Al27
Sulphur S32
Chlorine Cl35.5
Copper Cu64

The relative molecular mass or relative formula mass (Mr) of a compound is obtained by adding up the masses of all the atoms in the formula:

CompoundFormulaMr
Carbon dioxideCO212 + 2 ×16 = 44
Copper (ii) nitrateCu(NO3)264 + 2 × (14 + 3×16) = 188

Percentage composition of elements in compounds

The percentage by mass is useful in finding formulae. If the formula is already known it is simply a question of finding the fraction of the formula mass made up by the element in question.

% mass = Mass of the element in the compound/Mr of the compound x 100

Example 1

What is the percentage by mass of aluminium in bauxite, Al2O3 ?

Solution 1

Mr of Al2O3 = 2×27 + 3×16 = 102

Mass of aluminium in Al2O3 = 2×27 = 54

% by mass of Al = 54/102 × 100 = 52.9%

Example 2

What is the percentage by mass of oxygen in CuSO4.5H2O?

Solution 2

Mr of CuSO4.5H2O

= 64 + 32 + 4×16 + 5× (2+16) = 250

Mass of oxygen in CuSO4.5H2O
= 4×16 + 5×16 = 144

% by mass of Oxygen = 144/250 × 100 = 57.6%.

The mole and molar mass

One mole is the amount of substance which contains 6 × 1023 particles. This means that one mole of any substance contains the same number of atoms (or molecules, or ions).

6 × 1023 is called the Avogadro Constant.

Mathematically, the mass of one mole of a substance is equal to the relative formula mass of a substance. The mass of one mole of a substance is called its molar mass.

e.g. What is the mass of one mole of

  • CO2
  • Cu(NO3)2 ?
[C=12, O=16, Cu=64, N=14]

Solution 1

Mr of CO2

= 12 + 2×16 = 44

The molar mass is the Mr expressed in grams

Therefore Molar mass of CO2 = 44 g

Solution 2

Mr of Cu(NO3)2

= 64 + 2 × (14+ 3×16) = 188

Molar mass of Cu(NO3)2 = 188 g

Amount of substance

The “amount of substance” is a special name for the number of moles. If we know the mass of a substance and its molar mass, we can find the number of moles:

Number of moles = mass/molar mass

N = m/Mr

Which also means:

mass = number of moles × molar mass

Empirical formula

The simplest ratio of atoms in a compound is called the empirical formula. Empirical formulae is calculated as follows:

  1. Find the masses of all the different elements which combine with each other.
  2. Convert each mass to amount of substance: i.e. by dividing the mass of each element by its relative atomic mass.
  3. Find the molar ratio by dividing all the amounts by the amount of the element with the smallest amount.
  4. Then try multiplying by small whole numbers (2, 3 or 4) or rounding off to get a whole-number ratio.

For example:

2.88 g of magnesium is heated in nitrogen, and forms 4.00 g of magnesium nitride. Find the empirical formula of magnesium nitride. [Mg=24, N=14]

MagnesiumNitrogen
Mass2.884.00 – 2.88 = 1.12 g
Moles2.88/241.12/14
= 0.12 mol= 0.08 mol
Molar Ratio0.12/0.080.08/0.08
= 1.5= 1
32

Molecular formula

The molecular formula is the formula showing the actual number of each type of atom in one molecule.

For example, butene, like all alkenes has the empirical formula CH2 but its molecular formula is C4H8.

Molecular formula = formula units × the empirical formula.

We find the formula units using the Mr and the mass of the empirical formula.

Formula units = Mr/empirical mass

Example 1:

A compound of carbon, hydrogen and oxygen is found to be 40.0% carbon and 6.7% hydrogen by mass. Its relative molecular mass is 120.

Find:

  • its empirical formula
  • its molecular formula.

Solution 1:

First of all we treat the percentages as masses out of 100g. Which means in 100 g of the compound there is:

  • 40.0 g of Carbon
  • 6.7 g of Hydrogen
  • and (100 – 40.0 – 6.7) = 53.3 g of Oxygen.
CarbonHydrogenOxygen
Mass40.06.753.3
Moles40/126.7/153.3/16
= 3.33 mol= 6.7 mol= 3.33
Molar Ratio3.33/3.336.7/3.333.33/3.33
= 1= 2.01= 1
121

Empirical formula = CH2O

Empirical mass = 12 + 2 + 16 = 30
Mr(given in question) = 120

Formula units = 120/30
= 4

Molecular formula = 4 × CH2O
= C4H8O4

Finding the formulae of metal oxides

Example 2:

A crucible was weighed with its lid and the mass was found to be 6.20g. A coiled piece of magnesium ribbon was added, and it was weighed again resulting in a mass of 6.68g. The crucible was heated until the magnesium started to burn, then the lid was raised for brief periods until it had ceased to burn. It was heated to constant mass with the lid off. The final mass of crucible, lid and magnesium oxide was found to be 7.00g. Find the empirical formula of the magnesium oxide. [Mg=24; O=16]

Solution 2

  • mass of magnesium used = 6.68 – 6.20 = 0.48g
  • mass of oxygen gained = 7.00 – 6.68 = 0.32g

amount of magnesium : amount of oxygen

= 0.48/24 : 0.32/16

= 0.020 mol : 0.020 mol

= 1 : 1

Empirical formula = MgO