The Mole | Chemistry Form 3

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The Mole - Chemistry Form 3

The background knowledge required in this topic is obtained from the periodic table. This includes the chemical symbols of elements and their relative atomic masses.



Elements in the periodic table.


In everyday life two common methods for specifying the quantity of material in a substance exist. These methods are in terms of units of amount. Both units of mass and units of amount are used in daily basis. When buying potatoes at the market, we decide on other quantity in mass units (50 kg, 10 kg, 2 kg) or amount units (10 potatoes, 15 potatoes or 20 potatoes). When buying eggs, amount units are used exclusively-12 eggs (1 dozen), 24 eggs (2 dozens). On the other hand sugar and rice are always purchased in weighed quantities.

Potatoes

Eggs

In everyday life two common methods for specifying the quantity of material in a substance exist. These methods are in terms of units of amount. Both units of mass and units of amount are used in daily basis. When buying potatoes at the market, we decide on other quantity in mass units (50 kg, 10 kg, 2 kg) or amount units (10 potatoes, 15 potatoes or 20 potatoes). When buying eggs, amount units are used exclusively-12 eggs (1 dozen), 24 eggs (2 dozens). On the other hand sugar and rice are always purchased in weighed quantities.

Potatoes

Eggs

By the end of the lesson you should be able to:-
a) Define the mole.
b) Relate the mole to the relative atomic mass.
c) Convert mass to mole.


From the activity, it's clear that a pair represents two items. E.g. a pair of socks means we have two socks. Similarly a dozen represents twelve items e.g dozen of book means that we have twelve books. This means a pair and a dozen are counting units representing two items and twelve items respectively.
The mass of ions, atoms and molecules is very small and therefore is difficult to weigh them. The counting unit in chemistry is the mole. The mole contains 6.023x1023 particles. This counting unit is extremely large due to the small size of the particles.
This particles include; atoms, molecules, ions and electrons. The number 6.023x1023 is reffered to as avogadro's number or constant unit is represented by letter L.

The table below shows the R.A.M and atomic mass of some elements.

Atomic mass of an element in grams is equal to 1 mole of the element.

For example the R.A.M of 1 Carbon atom =12.00 a.m.u (Atomic Mass Unit)

The mass of 1 Mole of Carbon atom = 12.00g

Therefore the mass of one mole of a substance is reffered to as Molar mass.

Relative Atomic Mass

This is the amount of any substance that contains 6.023 x1023 (Avogrado's number) of particles.
In form II we defined the R.A.M of an element as mass of an atom compared with C-12 atom. That is:-
R.A.M=average mass of one atom of the element/one-twelve(1/12) of the mass of one atom of carbon-12


R.A.M is a ratio therefore has no units.
The R.A.M of one carbon atom=12.00 a.m.u (atomic mass unit).
The mass of one mole of carbon atom=12.00g
Therefore the mass of one mole of substance is referred to as molar mass.

In order to convert mass of an element to number of moles, we use the following formula
[Number of moles=mass in grams/mass of one mole]


EXAMPLE I: -
How many moles of carbon are present in 6g of carbon?
Step I
The mass of one mole of carbon=12g
Step II
Substitute the values
Mole of carbon=6/12
=0.5 moles

What is the mass of 0.25 moles of magnesium (Relative atomic mass of Mg=24)
Step I
Using the formula
[Number of moles of Mg =mass in grams/mass of one mole]
It follows that

Mass in grams=number of moles X Mass of 1 mole
Step II
Substitute the values
Mass in gram of mg=0.25x24
=6 g

Relative molecular mass is obtained by adding up the relative atomic masses of all the atoms present in a molecule. For example
1. In a Nitrogen molecule, the relative R.M.M N2 is equal to 14+14=28
2. CO2 molecule, has a R.M.M equal to 12+(16x2)=44 where 12 is R.A.M of carbon, 16 is the R.A.M of Oxygen.
Relative formula mass is obtained by adding up the relative atomic masses of all the atoms present in a compound.

Relative Formula mass of Sodium Chloride


Relative formula mass is obtained by adding up the relative
atomic masses of all the atoms present in a compound.

For example the R.F.M of NaCl is: -
Relative atomic mass of Na=23
Relative atomic mass of Cl=35.5

=23+35.5
=58.5


R.F.M of Calcium Carbonate (CaCO3)
Relative atomic mass of Ca=40
Relative atomic mass of C=12
Relative atomic mass of O=16
=40+12+(16x3)
=100

To convert the mass of compounds/molecules to moles

The formulae Number of moles in a compound/molecules=mass of compound/molecule /RMM/RFM
Example I


Click on the ENTER button to follow a worked out example.

By the end of the lesson you should be able to:
(i) Explain the term concentration, molarity and dilution of a solution.
(ii) Define molar solutions.
(iii) Prepare molar solution


In this lesson we will discuss Molar solutions.



Examples of molar solutions


Consider three boiling tubes with equal amounts of water. In the first boiling tube one spoonful of salt is dissolved, in the second two spoonfuls of salt are dissolved while in the third three spoonfuls of salt are dissolved as shown in the following animation.

Test tube C contains the highest amount of salt and therefore is the most concentrated solution.

A contains the least amount of salt and therefore the least concentrated or the most dilute of the three solutions.

Click on the ENTER button to follow up on a worked example.



The term concentration refers to the amount of solute dissolved in a specific volume of solvent.
When mole of a solvent is dissolved in water and the volume of a solution is made up to 1 dm3,the solution is said to be a molar solution.
A concentration of one mole per dm3 is written as 1 mole dm-3 or 1 M (1 molar). This concentration is referred to as molarity of a solution, abbreviated as M.
The concentration/molarity can also be expressed as in g/dm3.

Converting to 1g/ dm3 or g/litre



Click on the ENTER button to follow up on a worked example.


Click to play the video to observe how a solution of 1M of sodium hydroxide is prepared in the laboratory.

Click on the ENTER button to follow a worked out example.


Click on the ENTER button to follow up on a worked example.

Dilution is the process by which the concentration of a solution is lowered by adding more solvent.Only the volume of the solvent is increased but the number of moles of the solute in final solution remains the same.



Diluting a solution. Number of particles remain the same.

The concentration of the original soluton can be expressed as M1 and the original volume of solution is expressed as V1.
If we take the concentration of the dilute solution as M2, then the relationship below is applicable .
M1V1=M2V2
Where
M
1............... Concentration of the original solution
V
1................Volume of the original solution
M
2 ............... concentration of the dilute solution
V
2..................volume of dilute solution

Click on the ENTER button to follow a worked example.


In this lesson we will discuss Volumetric analysis.



Some apparatus and reagents used in volumetric analysis.

In our everyday life, most processes in industry involves analysis of product to ascertain their composition. Such industries include smithkline Becham, East African Breweries limited and Nairobi Bottlers.Others carry out analysis for quality certification and safety of the products in the market. These include Kenya bureau of standards and Nairobi water and Sewerage Company .



Kenya Bureau of Standards laboratories

By the end of the lesson you should be able to: -
a) Carry out experiments involving titrations
b) Work out problems involving molar solution


Titration is a process in which a solution of known concentration (molarity) is added to a solution of unknown concentration (molarity) until reaction is complete. The end- point of reaction is indicated by the color change of the indicator used. Titration is commonly used in neutralization reaction between acids and bases.

The following apparatus are used in titration experiments.
i) Burette
ii) Pipette
iii) Pipette filler
iv) Conical flask
v) Stand
vi) Clamp
vii) White tile
viii) Filter funnel

In order to determine the end- point of neutralization reaction, indicators are used. Indicators are substances that show different colors in acids and bases.

Click to play the video to observe the colour changes when different indicators are used.



Click to play the video to observe how a titration experiment is carried out.


The following table shows sample results obtained from an experiment.

From the table given the average volume can be calculated.The average volume is =21.00+21.00+21.00/3
=21.03 cm3
=21.00 cm3


Click on the ENTER button to follow a worked out example.


Redox titration does not require indicator because solutions change colours when oxidation states change. This titration involves both reduction and oxidation simultaneously.

Click on the ENTER button to follow a worked out example.

MOLAR GAS VOLUMES

In this lesson we will discuss Molar gas volume.



Molar gas volume


By the end of the lesson you should be able to:-
a) Define atomicity of a gas.
b) Define molar gas volume
c) Carry out calculations involving molar gas volume.


The atomicity of a gas refers to the number of atoms that make one molecule of the gas. The atomicity of the noble gases (He, Ne, Ar) is one. They are monoatomic.
The atomicity of oxygen gas, O2, Nitrogen N2 and hydrogen H2 is two. They are diatomic.
Ozone is a triatomic


The atomicity of a gas refers to the number of atoms that make one molecule of the gas. The atomicity of the noble gases (He, Ne, Ar) is one. They are monoatomic.
The atomicity of oxygen gas, o2, Nitrogen N2 and hydrogen H2 is two. They are diatomic.
Ozone is a triatomic .


Find the volume of 1 mole of hydrogen at s.t.p. (H=1.0008, density of hydrogen is 0.09g/dm3)
Solution
Step I
[Write the formulae]
Volume=mass/density
Step II
[Work out the RMM of hydrogen]
1 mol of H2=1.0082=2.016g
Step III
Substitute variables in the formula
Volume=2.016/0.09=22.4dm3

Find the volume of one mole o nitrogen gas at s.t.p given that its desity is 1.25/dm3 and R.A.M of nitrogen is 14.
Solution
Step I
Nitrogen gas is diatomic. Hence its R.M.M is
214=28

Step II
Volume=mass/density
Step III
Mass of one mole of N2 gas=28g
Density of N2 gas=1.25g/dm3
Step IV

Substitute variables in the formula
Volume=28g/1.25g/dm3




The volume of one mole of a gas is called molar gas volume. At standard temperature and pressure this molar gas volume is taken to be 22.4dm3 for all gases.
At room temperature and pressure, the molar gas volume is 24.0 dm3.

Remember
1. Gases have mass and volume
2. 1 mole of any gas occupies 22.4dm3 at s.t.p and 24dm3 at r.t.p.
Click on each of the steps given to follow a worked out example.




Equal volumes of gases though with different molar masses, contain same number of particles. This law was developed by Amedeo avogardos' law which states that equal volume of all gases at same temperature and pressure contain the same number of particles. It's usually represented by the letter L, where L=6.023 x1023 particles.

Click on the ENTER button to follow up on a worked example.


Click on the ENTER button to follow up on a worked example.

The expression Gay-Lussac's law is used for each of the two relationships named after the French chemist Joseph Louis Gay-Lussac and which concern the properties of gases, though it is more usually applied to his law of combining volumes. One law relates to volumes before and after a chemical reaction while the other concerns the pressure and temperature relationship for a sample of gas.

The law of combining volumes states that, when gases react together to form other gases, and all volumes are measured at the same temperature and pressure:
The ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.

The results obtained conform to a law which describe the behavior of gases and is referred as Gay-lussac's law. "when gases combine they do so in volumes which bear a simple ratio to one another and to the volume of the product if gaseous".

Click on the ENTER button to follow a worked out example.


Click on the ENTER button to follow a worked out example.

By the end of the learner should be able to:-
i) Write the correct formula.
ii) Write the correct stochiometric equations.
iii) Write correct ionic equations.
iv) Calculate reacting masses and volumes.


In this lesson we will discuss reacting masses and volumes.

Ionic equation represents a chemical change by means of ions.



An ionic equation

Remember ions are charged particles i.e. charged atoms, radicals.



Formation of Sodium ion



Consider the reaction between lead (II) nitrate solution and potassium iodide solution. At the start, all the ions K+, I-, Pb2+ and NO3- are free to move.On mixing potassium iodide solution and lead (II) nitrate solution, lead ions (Pb2+) combine with iodide ions (I-) to form a yellow precipitate of lead (II) iodide.
Potassium ions (K+ ) and nitrate ions (NO3- ) remain in solution.
 

Such ions that do not take part in reaction are called spectator ions.
i) Stoichiometric equation

Pb (NO3)2(aq)+ 2KI(aq) PbI2(s) + 2KNO3(aq)

ii) Ionic equation
 

Step I
Check for substances which remain in the same state. Where ions are free.

Pb2+(aq) +2NO3- (aq) + 2K+(aq) + 2I-(aq) PbI2(s) + 2K+(aq) + 2NO3-(aq)

Step II
Cross out the common ions which appear on both sides of the equation
 

Pb2+(aq) + 2I-(aq) PbI2(s)

Equations can be used to calculate the exact masses and volumes of reactants and products involved.



Example of a chemical equation

Click on the ENTER button to follow up on a worked example.


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