Since it is not possible to calculate the weight of particles individually, a collection of such particles called mole is taken for all practical purposes. Avogadro discovered that under standard conditions of temperature and pressure, (1atm and 273K) a sample of gas occupies a volume of 22.4 L.
It was discovered that the number of atoms present in 12g of carbon is 6.023 x 1023 atoms. This is referred to as Avogadro number.
A mole of a gas is the amount of a substance containing 6.023 x 1023 particles.
Avogadro’s number is the number of atoms present in C12 isotope = 12g of C.
One mole of any gas at STP will have a volume of 22.4 L.
Mole in term of mass, volume number and ions
In chemistry, the term mole represents a pile or mass of atoms, molecules, ions or electrons. Just as common man measures quantity in terms of kilograms or dozens, a chemical scientist deals with a ‘mole’ of atoms, molecules, ions or electrons.
Mole is defined as the amount of a substance, which contains the same number of chemical units (atoms, molecules, ions or electrons) as there are atoms in exactly 12 grams of pure carbon-12. A mole represents a collection of 6.023 x 1023 ( Avogadro’s number) chemical units. Thus a mole represents the quantity of material, which contains one Avogadro’s number of chemical units of any substance. The unit of mole is denoted as ‘mol’.
For example
One mole of hydrogen atoms = 6.023 x 1023 atoms of hydrogen
One mole of hydrogen molecules = 6.023 x 1023 molecule of hydrogen
One mole of electrons = 6.023 x 1023 electrons
One mole of sodium ions (Na+) = 6.023 x 1023 Na+ ions
It can be thus concluded that,
One mole of atoms = 6.023 x 1023 atoms = Gram atomic mass of the element.
One mole of molecules = 6.023 x 1023 molecules = Gram molecular mass.
Molar Volume
The volume occupied by one mole of any substance is called its molar volume. It is denoted by Vm. Molar volume of the substance depends on temperature and pressure. One mole of all gaseous substances at 273 K and 1 atm pressure occupies a volume equal to 22.4 litre or 22,400 mL. The unit of molar volume is litre per mol or millilitre per mol.
The various relationships of mole can be summarized as follows:
- A sample of gaseous substance weighing 0.5 g occupies a volume of 1.12 litre under NTP conditions. Calculate the molar mass of the substance.
Solution
1 mole of any gaseous substance at NTP occupies 22.4 L.
1.12 L of gaseous substance = 0.5 g
The molar mass of the substance therefore is 10 g/mol.
Firstly, let’s calculate the molar masses:
Molar mass of Na = 1* Relative Atomic Mass of Na
= 1* 23 = 23
Thus, 1mole of Na weighs 23 g/mol.
Molar mass of Cl2 = 2* Relative Atomic Mass of Cl
= 2 * 35.5 = 71
Hence, 1mole of Cl2 molecule weighs 71 g/mol.
Similarly, Molar mass of NaCl = (1* Relative Atomic Mass of Na) + (1* Relative Atomic Mass of Cl)
= (1* 23) + (1 * 35.5)
= 23 + 35.5 = 58.5
Therefore, 1mole of NaCl molecule weighs 58.5 g/mol.
Now, let’s fly:
2Na(s) + Cl2(g) → 2NaCl(s)
Mole ratio 2 : 1 : 2
Molar mass 23 71 58.5
Molar vol – 22.4dm^3 –
Rxting mass 2*23 1*71 2*58.5
46g 71g 117g
(*Rxting Mass = Reacting Mass)
From the above, it means that if we were to prepare sodium chloride in the lab, we would need 46g of sodium metal and 71g of chlorine gas, which would give us 117g of sodium chloride.
Also, remember that chlorine is a gas at room temperature and therefore, possesses molar volume. This implies that 46g of sodium metal can also react with 22.4dm^3 of chlorine gas to form 117g of sodium chloride at s.t.p.
The question now is what if we do not have up to this amounts of reagents in the lab, does it mean we will not be able to prepare sodium chloride? Of course, we will. Let’s say, we have:
(i) 35g of sodium, and
(ii) 50g of chlorine in the lab, how much sodium chloride can we prepare from each?
According to the equation:
2Na(s) + Cl2(g) → 2NaCl(s)
(i) If 46g of Na produce 117g of NaCl
Then, 35g of Na will produce xg of NaCl
Solve for x by cross multiplying:
46 * x = 117 * 35
46x = 4095
x = 4095/46
= 89.02g of NaCl
(ii) If 71g of Cl produce 117g of NaCl
Then, 50g of Cl will produce xg of NaCl
71 * x = 117 * 50
71x = 5850
x = 5850/71
= 82.39g of NaCl
We can see from the above that we can achieve whatever we desire in the lab, irrespective of the amount of reagents available. All we need is the stoichiometry of the reaction, and every other thing will fall into place.
There is something l want us to note in our calculations above:
In (i), we ignored Cl, while in (ii), we ignored Na. I like calling species like these, ‘distractions‘, because though, they are part of the equation, the question did not provide any information about them in each case. In other words, the questions were just silent about them.
Also, we made use of simple proportions in our calculations, by using the relationship between the two species involved in each question, i.e, Na and NaCl in question (i), and Cl and NaCl in question (ii)
Beginners Steps to Take When Solving Problems Involving Mole Concept
1) Write out the balanced chemical equation.
2) Get the stoichiometry or mole ratio of the reaction.
3) Calculate the molar masses of the substances involved.
4) Calculate the molar volumes of the gaseous substances involved (if any).
5) Calculate the reacting masses/volumes of the substances as the case may be.
6) Identity the ‘distractions’ in the equation and ignore them, based on the question(s).
7) Apply simple proportion in solving the problem using the reacting masses/volumes/no. of particles.
Examples
Question 1
Calculate the number of moles present in 8g of oxygen gas (O = 16)
Answer
Oxygen gas, O2 is a diatomic molecule, i.e, it is made up of two atoms.
Reacting mass = 8g
Molar mass = 2 x Relative Atomic Mass of O
= 2 x 16 = 32g/mol
but,
No of moles = Reacting mass/Molar mass
= 8/32 = 0.25mole
Question 2
Calculate the number of
(i) atoms (ii) molecules
present in 35g of nitrogen gas [N = 14, Avogadro No = 6.02 x 10^23/mol]
Answer
Nitrogen gas, N2 is also a diatomic molecule. Before calculating the number of particles in a substance, first calculate the molar mass.
Molar mass of N2 = 2 x 14
= 28g/mol
Reacting mass = 35g
No of moles = Reacting mass/Molar mass
= 35/28 = 1.25moles
remember,
1 mole of any substance = 6.02 x 10^23 entities or particles
(i) N2 contains 2 moles of atoms (diatomic)
Thus, 1mole of N2 = 2 x 6.02 x 10^23 atoms
= 1.204 x 10^24 atoms
Therefore,
If 1mole of N2 = 1.204 x 10^24 atoms
Then, 1.25 moles of N2 = x atoms
x = 1.25 x 1.204 x 10^24
= 1.505 x 10^24 atoms
(ii)N2 is 1 mole of a molecule of nitrogen, and 1 mole of N2 = 6.02 x 10^23 molecules. Therefore,
If 1mole of N2 = 6.02 x 10^23 molecules
Then, 1.25 moles of N2 = x molecules
x = 1.25 x 6.02 x 10^23
= 7.525 x 10^23 molecules
Question 3
What volume in cm^3 will be occupied by 45g of ozone at s.t.p? [O = 16, G.M.V = 22.4dm^3]
Answer
Ozone, O3 is a triatomic molecule, i.e, it is made up of three atoms.
Reacting mass = 45g
Molar mass = 3 x Relative Atomic Mass of O
= 3 x 16 = 48g/mol
but,
No of moles = Reacting mass/ Molar mass
= 45/48 = 0.938mole
1 mole = 22.4dm^3 or 22400cm^3
(Since the answer is expected in cm^3, then we will use 22400cm^3)
So,
If 1mole of O3 gas = 22400cm^3
Then, 0.938mole of O3 = x cm^3
x = 0.938 x 22400
= 21011.2cm^3
Therefore, 0.938mole of ozone will occupy 21011.2cm^3.
Question 4
C3H8(g) + 5O2(g) → 4H2O(g) + 3CO2(g)
From the equation above, calculate the volume of oxygen at s.t.p. required to burn 50cm^3 of propane. [C = 12, H = 1, O = 16, Molar gas volume at s.t.p. = 22.4dm^3]
Answer
From the equation and the accompanying question, all the species are in the gaseous phase, and as such, we will work with their volumes. In other words, we do not need to waste our time calculating their molar masses, so we make use of the relationship between mole and volume.
C3H8(g) + 5O2(g) → 4H2O(g) + 3CO2(g)
Mole ratio 1 : 5 : 4 : 3
Molar vol 22.4 22.4 22.4 22.4
Rxting vol (1×22400) (5×22400) (4×22400) (3×22400)
= 22400 112000 89600 67200
Therefore, from the stoichiometry,
22400cm^3 of propane = 112000cm^3 of oxygen
Then, 50cm^3 of propane = x cm^3 of oxygen
Solve for x by cross multiplying
22400 * x = 50 * 112000
22400x = 5600000
x = 5600000/22400
= 250cm^3
Therefore, 250cm^3 of oxygen will be required to burn 50cm^3 of propane.
Question 5
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
From the equation above, calculate the mass of sodium hydroxide produced by 2.3g of sodium [H = 1, O = 16, Na = 23].
Answer
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
Mole ratio 2 : 2 : 2 : 1
Molar mass 23 18 40 2
Rxting mass 46 32 80 2
From the stoichiometry,
46g of Na produce 80g of NaOH
Then, 2.3g of Na will produce xg of NaOH
Solving for x gives,
46 * x = 2.3 * 80
46x = 184
x = 184/46
= 4g
Therefore, 4g of sodium hydroxide will be produced by 2.3g of sodium.
Question 6
16.8g of sodium hydrogentrioxocarbonate (IV) is completely decomposed by heat. Calculate the volume of carbon (IV) oxide given off at s.t.p. [Na = 23, C = 12, O = 16, H = 1, Molar volume of a gas at s.t.p = 22.4dm^3]
Answer
Equation of reaction:
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
(Note that when a hydrogentrioxocarbonate (IV) salt, [HCO3]– is heated, it decomposes to give a trioxocarbonate (IV) salt, [CO3–], water and carbon (IV) oxide)
Molar masses,
NaHCO3 = (1*Na) + (1*H) + (1*C) + (3*O)
= (1*23) + (1*1) + (1*12) + (3*16)
= 23 + 1 + 12 + 48
= 84g/mol
CO2 = (1*C) + (2*O)
= (1*12) + (2*16)
= 12 + 32
= 44g/mol
(We will focus only on the two compounds, whose molar masses are calculated above. The others [Na2CO3 and H2O] are considered as ‘distractions’, because nothing was mentioned about them in the question)
Mole ratio 2 : 1 : 1 : 1
Molar mass 84 – – 44
Molar vol – – – 22.4
Rxting mass 168 – – 44
Rxting vol – – – 22.4
(Since there is a relationship between mass and volume using the mole, based on the question, we can directly compare the reacting mass of NaHCO3 to the reacting volume of CO2)
Hence, from the stoichiometry;
168g of NaHCO3 = 22.4dm^3 of CO2
16.8g of NaHCO3 = x dm^3 of CO2
168 * x = 16.8 * 22.4
168x = 376.32
x = 376.32/168
= 2.24dm^3
Therefore, 2.24dm^3 of CO2 is liberated at s.t.p. when 16.8g of NaHCO3 decomposes on heating.
Application of Mole Concept
The following are some of the areas where mole concept is applied:
1) Gay-Lussac’s Law
2) Avogadro’s Law
3) Water of Crystallization
4) Law of Conservation of Mass
5) Calculation of Masses from Chemical Equations
6) Redox Reactions
7) Acid-Base Reactions
8) Organic Chemistry
Do These
Question 1
How many moles of limestone, CaCO3 will be required to produce 5.6g of quicklime, CaO?
CaCO3(s) → CaO(s) + CO2(g)
[Ca = 40, C = 12, O = 16]
A. 0.20 B. 0.10 C. 1.12 D. 0.56
Question 2
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
What volume of gas is evolved at s.t.p. if 2g of calcium trioxocarbonate (IV) is added to a solution of hydrochloric acid? [Ca = 40, C = 12, O = 16, Cl = 35.5, H = 1, Molar volume of a gas at s.t.p. = 22.4dm^3]
A. 112cm^3 B. 224cm^3 C. 448cm^3 D. 2240cm^3
Question 3
If a solution contains 4.9g of tetraoxosulphate (VI) acid, calculate the amount of copper (II) oxide that will react with it. [Cu = 64, O = 16, S = 32, H = 1]
A. 0.8g B. 4.0g C. 40.0g D.
80g
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