O level Chemistry Notes

O level Chemistry Notes (Page 4)

O level Chemistry Notes (Page 4)

O level Chemistry Notes covering Water of crystallisation, Calculations from chemical reactions, Percentage Yield, Calculations of reacting gas volumes, Avogadro’s Law of Gases.These notes only show the key points in O’level chemistry curriculum.

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Water of crystallisation

Some salts contain water molecules that form part of their crystal structure. This water is called water of crystallisation and is written in the formula of the salt. e.g. MgSO4.H2O refers to hydrated magnesium sulphate, Na2SO4.6H2O refers to hydrated sodium sulphate.

The % water of crystallisation and the formula of the salt are calculated as follows:

  • A known mass of hydrated salt is heated gently in a crucible until it reaches constant mass.
  • The mass of anhydrous salt remaining and the mass of water lost are then calculated.
  • These are converted to moles and the formula of the hydrated salt can be found from the molar ratio.

Example 1.

6.25g of blue hydrated copper(II) sulphate, CuSO4.xH2O was gently heated in a crucible until the mass remaining was 4.00g of white anhydrous copper(II) sulphate CuSO4. Find the formula of hydrated copper sulphate.

Solution 1

The mass of anhydrous CuSO4 = 4.00g

Mr of CuSO4 = 64 + 32 + (4×18)
= 160

Therefore: Moles of CuSO4 = mass/Mr = 4 / 160 = 0.025

Mass of water of crystallisation driven off = 6.25 – 4.00 = 2.25g

Mr of H2O = 1+1+16 = 18

Therefore: Moles of H2O = 2.25 / 18 = 0.125

The molar ratio of CuSO4 : H2O
= 0.025 : 0.125
= 1 : 5

Therefore the formula of the hydrated salt is CuSO4.5H2O.

Calculations from chemical reactions

The steps involved in a calculation are as follows :

  1. Write down the balanced chemical equation with reactants on the left hand side and products on the right hand side.
  2. Convert the information given to moles of one substance.
  3. Use the chemical equation to find moles of other substance needed.
  4. Convert back from moles to mass, concentration or volume according to what has been asked by the question.

Example 1

What mass of oxygen is needed to burn 3.00 kg of propane, C3H8?

Solution 1

In this case the chemical equation is not given, so we must start by writing it down:

C3H8 + 5O2 → 3CO2 + 4H2O

Since we are given the mass of propane, we convert it to amount in moles.

Mr of C3H8

= 3 × 12 + 8 × 1 = 44

So its molar mass = 44 g

Therefore Number of moles of C3H8

= 3000/44

= 68.2 mol

According to the chemical equation molar ratio C3H8 : O2 = 1:5.

This means that 1 mol of propane reacts with 5 mol of oxygen molecules.

Therefore 68.2 mol of propane reacts with 5 × 68.2 = 341 mol of oxygen molecules

Finally convert back the moles of oxygen molecules to mass because the question requires mass.

Molar mass of O2 molecules = 2 × 16 = 32 g

So mass of oxygen needed = 341 × 32 = 10912 g

Example 2

In the reaction Fe3O4 + 4CO → 3Fe + 4CO2, what mass of the iron oxide is needed to form 2.50g of carbon dioxide?

Molar mass of CO2

= 12 + (2 × 16)

= 44 g/mol

Amount of CO2

= 2.5/44

= 0.0568 mol

From equation, 4 mol of CO2 is formed from 1 mol of Fe3O4

This means the molar ratio is of CO2: Fe3O4 = 4:1.

4:1 = 0.0568:x

x = 0.0568/4

= 0.0142

Therefore 0.0568 mol of CO2 is formed from 0.0142 mol of Fe3O4.

Finally lets convert the number of moles of Fe3O4 to mass.

Molar mass of Fe3O4 = 3×56 + 4×16 = 232 g

mass of 0.0142 mol of Fe3O4 = 0.0142 mol × 232 g/mol = 3.29g (3 s.f.)

Percentage Yield

The percentage yield of a reaction is the percentage of the product obtained compared to the theoretical maximum as calculated from the balanced equation.

However in any chemical process it is almost impossible to get 100% of the product because:

  • the reaction may not go to completion especially if it is reversible
  • there may be side-reactions which use up some of the starting material
  • it may not be possible to separate all of the product from the mixture of the products.

Percentage yield

Example 1

2.8g of iron was heated with excess sulphur to form iron sulphide as follows

Fe + S → FeS

The excess sulphur was dissolved in a solvent and the FeS filtered, washed and dried. 4.1g of purified iron sulphide was finally obtained, what was the % yield?

Solution 1

First calculate the maximum theoretical amount of moles to be produced.

2.8g of iron = mass/Mr = 2.8/56 = 0.05 moles.

According to the chemical equation, the molar ratio is 1:1 which means 0.05 moles of Fe theoretically makes 0.05 moles of FeS.

Therefore the theoretical mass of FeS = 0.05 x (56+32) = 0.05 x 88 = 4.4g

% yield = (4.1/4.4) x 100 = 93.2%.

Calculations of reacting gas volumes

At room temperature and pressure (r.t.p.), one mole of any gas occupies a volume of 24 dm3 or (24000 cm3). This is called the molar gas volume.

Since volume of any gas increases with temperature, the hotter the place the higher the molar volume of the gas. For Cambridge students, the molar gas volume = 24 dm3, which is the average volume of one mole of any gas at r.t.p in U.K. For Zimsec students, the molar gas volume = 28 dm3, which is the average volume of one mole of any gas at r.t.p in Zimbabwe.

Number of moles

Example 1

What total volume of gas at r.t.p. can be obtained from heating 99.3g of lead(II) nitrate? [Pb=207, N=14, O=16]

Solution 1

2Pb(NO3)2 → 2PbO + 4NO2+ O2

Molar mass of lead(II) nitrate = 207 + 2 × (14 + 3×16) = 331g

Amount of lead(II) nitrate = 99.3/331 = 0.3 mol

According to the equation, 2 mol of lead(II) nitrate give 4 moles of Nitrogen dioxide and 1 mole of Oxygen which gives a total of 5 moles of gas.

Therefore 0.3 mol gives 0.3× 2.5 = 0.75 mol of gas

Volume of 0.75 mol of gas = 0.75 × 28 = 21 dm3 at r.t.p

Example 2

What mass of sodium hydroxide is needed to absorb 1.00 dm3 of carbon dioxide gas at r.t.p.?

Solution 2

Equation: 2NaOH + CO2 → Na2CO3 + H2O

Amount of carbon dioxide volume/molar volume = 1/28 = 0.0375 mol

According to the equation 1 mol of carbon dioxide reacts with 2 mol of NaOH which means 0.0417 mol of CO reacts with 2 × 0.0417 = 0.0833 mol of NaOH

Molar mass of NaOH = 23 + 16 + 1 = 40 g

mass of NaOH needed = 0.0833 × 40 g = 3.33 g

Avogadro’s Law of Gases

When two gases react, the ratio of volumes is the same as the ratio of moles. This is because equal volumes of all gases, under the same conditions, contain the same number of molecules.

Example 1: When propane gas is burnt in excess air the equation is:

C3H8 + 5O2 → 3CO2 + 4H2O

What volume of oxygen will react with 2.5 dm3 of propane, at r.t.p.?

According to the equation: 1 mol of C3H8 reacts with 5 mol of Oxygen gas.

Therefore 1 dm3 of C3H8 reacts with 5 dm3 of Oxygen gas.

Which means 2.5 dm3 of C3H8 reacts with 2.5 × 5 = 12.5 dm3 of Oxygen gas.