Empirical and molecular formula.

1.The empirical formula of a compound is its simplest formula. It is the simplest whole number ratios in which atoms of elements combine to form the compound. 2.It is mathematically the lowest common multiple (LCM) of the atoms of the elements in the compound
3.Practically the empirical formula of a compound can be determined as in the following examples.

To determine the empirical formula of copper oxide

(a)Method 1:From copper to copper(II)oxide
Procedure.
Weigh a clean dry covered crucible(M1).Put two spatula full of copper powder into the crucible. Weigh again (M2).Heat the crucible on a strong Bunsen flame for five minutes. Lift the lid, and swirl the crucible carefully using a pair of tong. Cover the crucible and continue heating for another five minutes. Remove the lid and stop heating. Allow the crucible to cool. When cool replace the lid and weigh the contents again (M3).

Sample results

Mass of crucible(M1) 15.6g
Mass of crucible + copper before heating(M2) 18.4
Mass of crucible + copper after heating(M3) 19.1

Sample questions

Calculate the mass of copper powder used.
Mass of crucible + copper before heating(M2) = 18.4
Less Mass of crucible(M1) = - 15.6g
Mass of copper 2.8 g
Calculate the mass of Oxygen used to react with copper.
Method I
Mass of crucible + copper after heating(M3) = 19.1g
Mass of crucible + copper before heating(M2) = - 18.4g
Mass of Oxygen = 0.7 g
Method II
Mass of crucible + copper after heating(M3) = 19.1g
Mass of crucible = - 15.6g
Mass of copper(II)Oxide = 3.5 g
Mass of copper(II)Oxide = 3.5 g
Mass of copper = - 2.8 g
Mass of Oxygen = 0.7 g
Calculate the number of moles of:
(i) copper used (Cu = 63.5)
number of moles of copper = mass used => 2.8 = 0.0441moles
Molar mass 63.5
(ii) Oxygen used (O = 16.0)
number of moles of oxygen = mass used => 0.7 = 0.0441moles
Molar mass 16.0
Determine the mole ratio of the reactants
Moles of copper = 0.0441moles = 1 => Mole ratio Cu: O = 1:1
Moles of oxygen 0.0441moles 1
5.What is the empirical, formula of copper oxide formed.
CuO (copper(II)oxide
State and explain the observations made during the experiment.
Observation
Colour change from brown to black
Explanation
Copper powder is brown. On heating it reacts with oxygen from the air to form black copper(II)oxide
Explain why magnesium ribbon/shavings would be unsuitable in a similar experiment as the one above.
Hot magnesium generates enough heat energy to react with both Oxygen and Nitrogen in the air forming a white solid mixture of Magnesuin oxide and magnesium nitride. This causes experimental mass errors.

(b)Method 2:From copper(II)oxide to copper
Procedure.
Weigh a clean dry porcelain boat (M1). Put two spatula full of copper(II)oxide powder into the crucible. Reweigh the porcelain boat (M2).Put the porcelain boat in a glass tube and set up the apparatus as below;

Pass slowly(to prevent copper(II)oxide from being blown away)a stream of either dry Hydrogen /ammonia/laboratory gas/ carbon(II)oxide gas for about two minutes from a suitable generator.
When all the in the apparatus set up is driven out ,heat the copper(II)oxide strongly for about five minutes until there is no further change. Stop heating.
Continue passing the gases until the glass tube is cool.
Turn off the gas generator.
Carefully remove the porcelain boat form the combustion tube.
Reweigh (M3).
Sample results

Mass of boat(M1) 15.6g
Mass of boat before heating(M2) 19.1
Mass of boat after heating(M3) 18.4

Sample questions

Calculate the mass of copper(II)oxide used.
Mass of boat before heating(M2) = 19.1
Mass of empty boat(M1) = - 15.6g
Mass of copper(II)Oxide 3.5 g
Calculate the mass of
(i) Oxygen.
Mass of boat before heating(M2) = 19.1
Mass of boat after heating (M3) = - 18.4g
Mass of oxygen = 0.7 g
(ii)Copper
Mass of copper(II)Oxide = 3.5 g
Mass of oxygen = 0.7 g
Mass of oxygen = 2.8 g
Calculate the number of moles of:
(i) Copper used (Cu = 63.5)
number of moles of copper = mass used => 2.8 = 0.0441moles
Molar mass 63.5
(ii) Oxygen used (O = 16.0)
number of moles of oxygen = mass used => 0.7 = 0.0441moles
Molar mass 16.0
Determine the mole ratio of the reactants
Moles of copper = 0.0441moles = 1 => Mole ratio Cu: O = 1:1
Moles of oxygen 0.0441moles 1
5.What is the empirical, formula of copper oxide formed.
CuO (copper(II)oxide
State and explain the observations made during the experiment.
Observation
Colour change from black to brown
Explanation
Copper(II)oxide powder is black. On heating it is reduced by a suitable reducing agent to brown copper metal.
Explain why magnesium oxide would be unsuitable in a similar experiment as the one above.
Magnesium is high in the reactivity series. None of the above reducing agents is strong enough to reduce the oxide to the metal.
Write the equation for the reaction that would take place when the reducing agent is:

(i) Hydrogen
CuO(s) + H2(g) -> Cu(s) + H2O(l)
(Black) (brown) (colourless liquid form
on cooler parts )
(ii)Carbon(II)oxide
CuO(s) + CO (g) -> Cu(s) + CO2(g)
(Black) (brown) (colourless gas, form
white ppt with lime water )
(iii)Ammonia
3CuO(s) + 2NH3(g) -> 3Cu(s) + N2 (g) + 3H2O(l)
(Black) (brown) (colourless liquid form
on cooler parts )

Explain why the following is necessary during the above experiment;
(i)A stream of dry hydrogen gas should be passed before heating copper (II) Oxide.
Air combine with hydrogen in presence of heat causing an explosion

(ii)A stream of dry hydrogen gas should be passed after heating copper (II) Oxide has been stopped.
Hot metallic copper can be re-oxidized back to copper(II)oxide

(iii) A stream of excess carbon(II)oxide gas should be ignited to burn
Carbon(II)oxide is highly poisonous/toxic. On ignition it burns to form less toxic carbon(IV)oxide gas.
10. State two sources of error in this experiment.
(i)All copper(II)oxide may not be reduced to copper.
(ii)Some copper(II)oxide may be blown out the boat by the reducing agent.

4.Theoreticaly the empirical formula of a compound can be determined as in the following examples.

(a)A oxide of copper contain 80% by mass of copper. Determine its empirical formula. (Cu = 63.5, 16.0)
% of Oxygen = 100% - % of Copper => 100- 80 = 20% of Oxygen

Element Copper Oxygen
Symbol Cu O
Moles present = % composition
Molar mass 80
63.5 20
16
Divide by the smallest value 1.25
1.25 1.25
1.25
Mole ratios 1 1

Empirical formula is CuO

(b)1.60g of an oxide of Magnesium contain 0.84g by mass of Magnesium. Determine its empirical formula(Mg = 24.0, 16.0)

Mass of Oxygen = 1.60 – 0.84 => 0.56 g of Oxygen

Element Magnesium Oxygen
Symbol Mg O
Moles present = % composition
Molar mass 0.84
24 0.56
16
Divide by the smallest value 0.35
0.35 0.35
0.35
Mole ratios 1 1
Empirical formula is MgO

(c)An oxide of Silicon contain 47% by mass of Silicon. What is its empirical formula(Si = 28.0, 16.0)

Mass of Oxygen = 100 – 47 => 53% of Oxygen

Element Silicon Oxygen
Symbol Si O
Moles present = % composition
Molar mass 47
28 53
16
Divide by the smallest value 1.68
1.68 3.31
1.68
Mole ratios 1 1.94 = 2
Empirical formula is SiO2
(d)A compound contain 70% by mass of Iron and 30% Oxygen. What is its empirical formula(Fe = 56.0, 16.0)

Mass of Oxygen = 100 – 47 => 53% of Oxygen

Element Silicon Oxygen
Symbol Si O
Moles present = % composition
Molar mass 47
28 53
16
Divide by the smallest value 1.68
1.68 3.31
1.68
Mole ratios 1 1.94 = 2

Empirical formula is SiO2

The molecular formula is the actual number of each kind of atoms present in a molecule of a compound.

The empirical formula of an ionic compound is the same as the chemical formula but for simple molecular structured compounds, the empirical formula may not be the same as the chemical formula.

The molecular formula is a multiple of empirical formula .It is determined from the relationship:

(i)         n  =    Relative formular mass
        Relative empirical formula
where  n  is a whole number.

(ii) Relative empirical formula x n = Relative formular mass
where n is a whole number.

Practice sample examples

A hydrocarbon was found to contain 92.3% carbon and the remaining Hydrogen.
If the molecular mass of the compound is 78, determine the molecular formula(C=12.0, H =1.0)

Mass of Hydrogen = 100 – 92.3 => 7.7% of Oxygen

Element Carbon Hydrogen
Symbol C H
Moles present = % composition
Molar mass 92.3
12 7.7
1
Divide by the smallest value 7.7
7.7 7.7
7.7
Mole ratios 1 1

Empirical formula is CH
The molecular formular is thus determined :
n = Relative formular mass = 78 = 6
Relative empirical formula 13

The molecular formula is (C H ) x 6 = C6H6

A compound of carbon, hydrogen and oxygen contain 54.55% carbon, 9.09% and remaining36.36% oxygen.
If its relative molecular mass is 88, determine its molecular formula(C=12.0, H =1.0, O= 16.0)

Element Carbon Hydrogen Oxygen
Symbol C H O
Moles present = % composition
Molar mass 54.55
12 9.09
1 36.36
16
Divide by the smallest value 4.5458
2.2725 9.09
2.2725 2.2725
2.2725
Mole ratios 2 4 1

Empirical formula is C2H4O
The molecular formular is thus determined :
n = Relative formular mass = 88 = 2
Relative empirical formula 44

The molecular formula is (C2H4O ) x 2 = C4H8O2.

3.A hydrocarbon burns completely in excess air to form 5.28 g of carbon (IV) oxide and 2,16g of water.
If the molecular mass of the hydrocarbon is 84, what is its molecular.

Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then:
Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 =>
Molar mass of CO2 12 x 5.28 = 1.44g
44
Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O =>
Molar mass of H2O 2 x 2.16 = 0.24g
18

Element Carbon Hydrogen
Symbol C H
Moles present = % composition
Molar mass 1.44g
12 0.24g
1
Divide by the smallest value 0.12
0.12 0.24
0.12

Mole ratios 1 2
Empirical formula is CH2
The molecular formular is thus determined :
n = Relative formular mass = 84 = 6
Relative empirical formula 14

The molecular formula is (CH2 ) x 6 = C6H12.