Mole concept

Mole concept: A mole is defined as a collection of particles of anything (atoms, ions, electrons, protons, etc) equal in number to the number of atoms present in 1 gm atom of C-12 isotope.

In terms of atoms,

number\ of\ moles = \frac{weight\ in\ gm}{atomic\ weight}

In terms of mass,

number\ of\ moles = \frac{weight\ in\ gm}{molecular\ weight}

In terms of ions,

number\ of\ moles = \frac{weight\ in\ gm}{ionic\ weight}

In terms of gases,

number\ of\ moles = \frac{volume\ at\ NTP}{22.4\ litres}

1 mole of H atoms = 6.023 x 10²³ atoms = 1.008 gm
1 mole of H₂ molecules = 6.023 x 10²³ molecules = 2.016 gm
1 mole of H₂O = 6.023 x 10²³ molecules = 18 gm
1 mole of SO₄^— ions = 6.023 x 10²³ ions = 96 gm

Solved Numericals of mole concept

1. Pure water sample having a mass of 0.36 gm is taken, answer the following questions.

i. How many moles of water are present?

\begin {align*} number\ of\ moles &= \frac{weight\ in\ gm}{molecular\ weight}\\ &=\frac{0.36}{18}= 0.02\ mol \end{align*}

ii. How many molecules of water are present?

1 mol of water = 6.023 x 10²³ molecules
0.02 mol of water = 6.023 x 10²³ x 0.02 = 1.2046 x 10²² molecules

iii. How many atoms of hydrogen are present ?

1 mol of watter = 2 x 6.023 x 10²³ atoms
0.02 mol of water = 2 x 6.023 x 10²³ x 0.02 = 2.4092 x 10²² atoms

iv. How many volume of water is present at NTP ?

1 mol of water at NTP = 22.4 L
0.02 mol of water at NTP = 22.4 x 0.02 = 0.448 L

2. Calculate the weight of 1 molecule of CaCO₃.

\begin {align*} 1\ mol\ of CaCO_{3} &= 100\ gm\\ 6.023 \times 10^{23}\ molecules\ of\ CaCO_{3} &= 100\ gm\\ 1\ molecules\ of\ CaCO_{3} &= \frac{100\ gm}{6.023 \times 10^{23}}\\ &=1.6606 \times 10^{-22}\ gm \end{align*}

3. Calculate the weight of 100 ml of NH3 gas at NTP.

1 mole of NH₃ = 17 gm
22.4 litres of NH₃ at NTP = 17 gm
22400 ml of NH₃ at NTP = 17 gm
100 ml of NH₃ at NTP = (17 x 100) / 22400= 0.0759 gm

4. Which of the following has the largest number of molecules and how? 7 gm of nitrogen or 1 gm of hydrogen.
solution:

→ In 7 gm of nitrogen
Number of moles = 7/28 = 0.25 mol
Number of molecules = 0.25 x 6.023 x 10²³ molecules

→ In 1 gm of hydrogen
Number of moles = 1/2 = 0.5 mol
Number of molecules = 0.5 6.023 10 molecules

So 1 gm of hydrogen has a large number of molecules because its number of moles is large.

What volume of CO₂ is produced when 20 gm of 20% CaCO₃ is heated?
solution:

\underset {1mol(100\ gm)}{CaCO_{3}} \rightarrow \underset {1\ mol}{CaO} + \underset {}{\underset{22.4\ L\ at\ NTP}{\underset {1\ mol}{CO_{2}}}}\\ \begin {align*} Given\ mass\ of\ CaCO_{3} &= 20 \%\ of\ 20gm = 4gm\\ 4\ gm\ of\ CaCO_{3} &= \frac{22.4}{100} \times 4 = 0.896L\ at\ NTP\\ &= 886\ cc\ at\ NTP \end{align*}

5. What mass of 60% pure sulphuric acid is required to decompose 25 gm of CaCO₃?
solution:

\underset {\underset {100\ gm}{1\ mol}}{CaCO_{3}} + \underset {\underset {98\ gm}{1\ mol}}{H_{2}SO_{4}} \rightarrow CaSO_{4} + H_{2}O + CO_{2}

100 gm of CaCO₃ is decomposed by 98 gm of H₂SO₄
25 gm of CaCO₃ is decomposed by 24.5 gm of H₂SO₄
Given H₂SO₄ is 60% pure
let mass of 60% pure H₂SO₄ = x gm

x \times 60 \%\ = 24.5\\ x \times \frac{60}{100} = 24.5\\ \therefore x = 40.833\ gm

Percentage composition

Percent means certain parts present per hundred parts. Percent of an element in a compound is the number of parts by mass of that element per hundred parts by mass of the compound. The percent composition of a compound gives per cent by the mass of elements present in the compound.

\%\ element = \frac{mass\ of\ element}{mass\ of\ the\ compound} \times 100

Solved Numericals

1. Calculate the percentage composition of Al₂O₃.
solution:

Molecular\ mass\ of\ Al_{2}O_{3} = 2 \times 27 + 3 \times 16 = 102\\ \%\ of\ Al = \frac{54}{102} \times 100 = 52.94\ \%\ \\ \%\ of\ O = \frac{48}{102} \times 100 = 47.06\ \%\ 

2. Calculate the percentage composition of the water of crystallization in blue vitriol crystal.
solution:

Molecular weight of CuSO₄.5H₂O = 63.5 + 32 + 4 x 16 + 5 x 18 = 249.5 gm
% of water crystallization = (5 x 18 x 100)/249.5 = 36.07 %

Empirical formula

It is the simplest formula that represents the simplest whole-number ratio of the number of atoms of all elements present in one molecule of the compound. eg. the empirical formula of glucose is CH₂O where its actual formula is C₆H₁₂O₆.

Molecular formula

It is the true formula that represents the actual number of atoms of all the elements present in one molecule of the compound.

Solved Numerical

1. Determine the empirical and molecular formula of a compound having the following percentage composition: 40% carbon, 6.66% hydrogen and 53.44% oxygen. (Mol. mass =60)
solution:

Empirical formula = CH₂O
Empirical formula mass = 12 + 2 + 16 = 30
Molecular mass = 60

Common\ factor\ (n) = \frac{Molecular\ mass}{empirical\ formula\ mass}= \frac{60}{30}=2\\ \begin{align*} Molecular\ formula &= (Empirical\ formula)_{n}\\ &= (CH_{2}O)\\ &= C_{2}H_{4}O_{2} \end{align*}

Limiting reagent

The reagent which is present in the lesser amount gets consumed completely and limits the amount of product formed is limiting reagent.

Solved numericals

5 gm of pure CaCO3 is treated with 5 gm of HCl.
i. Find the limiting reagent.
ii. Calculate the mass of CaCl₂ formed.
iii. How many number of water molecules are produced?
iv. Calculate the volume of CO2 produced at NTP?
Solution:

\underset {\underset {100\ gm}{1\ mol}}{CaCO_{3}} + \underset {\underset {73\ gm}{2\ mol}}{2HCl} \rightarrow \underset {\underset {111\ gm}{1\ mol}}{CaCl_{2}} + \underset {1\ mol}{H_{2}O} + \underset {\underset {22.4\ lit}{1\ mol}}{CO_{2}}

i. 100 gm of CaCO₃ requires 73 gm of HCl
5 gm of CaCO₃ requires 3.65 gm of HCl
The amount of HCl present is 5 gm and the required is 3.65 gm. So HCl is excess and CaCO₃ is the limiting reagent.

ii. 100 gm of CaCO₃ produces 111 gm of CaCl₂
5 gm of CaCO₃ produces 5.55 gm of CaCl₂.

iii. 100 gm of CaCO₃ produces 6.023 x 10²³ water molecules
5 gm of CaCO₃ produces 3.011 x 10²³ water molecules.

iv. 100 gm of CaCO₃ produces 22.4 lit of CO₂ at NTP
5 gm of CaCO₃ produces 1.12 lit of CO₂ at NTP

2. 10 gm of Fe₂O₃ is reacted with 9 gm of CO.
i. Find the limiting reagent.
ii. How many moles of unreacted reagent left over?
iii. Calculate the mole of CO consumed in the reaction.
iv. What mass of NaOH is required to absorb the whole CO₂ produced?
Solution:

\underset {\underset {159.7\ gm}{1\ mol}}{Fe_{2}O_{3}} + \underset {\underset {84\ gm}{3\ mol}}{3CO} \rightarrow \underset {2\ mol}{2Fe} + \underset {3\ mol}{3CO_{2}}

i. 159.7 gm of Fe₂O₃ reacts with 84 gm of CO
10 gm of Fe₂O₃ reacts with 5.25 gm of CO
The amount of CO present is 9 gm and the required is 5.25 gm. So CO is excess and Fe₂O₃ is the limiting reagent.

ii. Mass of unreacted reagent left over = 9 – 5.25 = 3.75 gm
Moles of unreacted reagent left over = 3.75/28 = 0.133 mol

iii. Mole of CO consumed = 5.25/28 = 0.1875 mol

iv. 159.7 gm of Fe₂O₃ produces 3 x 44 gm of CO₂
10 gm of Fe₂O₃ produces 8.26 gm of CO₂

\underset {80\ gm}{2NaOH} + \underset {44\ gm}{CO_{2}} \rightarrow Na_{2}CO_{3} + H_{2}O

To absorb 44 gm of CO₂, 80 gm of NaOH is required.
To absorb 8.26 gm of CO₂, 15 gm of NaOH is required.

Berzelius hypothesis

It states that “Under the similar condition of temperature and pressure, an equal volume of all gases consists of an equal number of atoms”.

Avogadro’s hypothesis (law)

It states that “Under the similar condition of temperature and pressure, an equal volume of all gases consists of an equal number of molecules”.

Applications of Avogadro’s law

1. Determination of atomicity of elementary gases

The number of atoms present in a molecule of elementary gas is called atomicity. One molecule of hydrogen has two atoms. Its atomicity is 2.

Q: Prove that hydrogen is diatomic?

We know that,

1 molecule of Hydrogen chloride = 1/2 molecule of hydrogen
1 molecule of hydrogen chloride = 1 atom of hydrogen
1/2 molecule of hydrogen = 1 atom of hydrogen
1 molecule of hydrogen = 2 atoms of hydrogen

So hydrogen is diatomic molecule.

2. Determination of the relationship between vapour density and molecular mass

Vapour density is the ratio of the weight of a given volume of gas to the same volume of hydrogen under a similar condition of temperature and pressure.

Vapour\ density\ (VD)= \frac{Weight\ of\ certain\ volume\ of\ gas}{Weight\ of\ same\ volume\ of\ hydrogen\ gas}

Let the certain volume of gas contain n molecules. Then applying Avogadro’s law,

Molecular weight is the ratio of the weight of one molecule of gas to the weight of one atom of hydrogen.

\begin {align*} Molecular\ weight &= \frac{Weight\ of\ 1\ molecule\ of\ gas}{Weight\ of\ 1\ atom\ of\ hydrogen\ gas} ----(ii)\\ VD &= \frac{Molecular\ weight}{2}\\ Hence,\ Molecular\ weight &= 2 \times vapour\ density \end {align*}

3. Derivation of the relationship between gram molecular mass and volume of a gas

Molecular weight expressed in gram is called gram molecular mass. Experimentally, one gram molecular mass occupies 22.4 litres of a gas at NTP which is explained by Avogadro’s law.

We have,
Mol. wt. = 2 x V.D

Gram molecular weight = 22.4 x Weight of 1 litre of gas
Gram molecular weight = Weight of 22.4 litres of gas at NTP
Gram molecular weight = 22.4 litres at NTP

4. Derivation of the relationship between gram molecular weight and number of molecules

we know that,

6.023 x 10²³ molecules = gram molecular weight
1 molecule = gram moleculer weight/6.023 x 10²³

Some important questions

Mole Concept

1. A pure water sample having a mass of 0.18 gm is taken.
   a. How many moles of water are present? (0.1)
   b. How many molecules of water are present? (6.023 x 1022)
   c. How many atoms of hydrogen are present? (1.2044 x 1023)
   d. How many gram molecules of water are present? (0.1)
   e. Calculate the volume occupied at NTP? (2.24 L)
2. Which of the following has a larger number of molecules and why? 7 gm of nitrogen or 1 gm of hydrogen. (H₂)
3. How many molecules are contained in 0.35 moles of N₂? (2.1 x 1023)
4. What volume of CO₂ is produced at NTP when 20 gm of 20% pure CaCO₃ is completely heated? (896 cc)
5. How many moles of hydrogen are left when 3×10²¹ molecules of hydrogen are removed from a vessel containing 40 mg of hydrogen? (0.015 mol)
6. What mass of 60% pure sulphuric acid is required to decompose 25 gm of CaCO₃? (40.833 gm)
7. Calculate the mass of 120 cc of nitrogen at NTP. How many molecules are present in it? (0.15 gm, 3.22 x 10²¹)
8. 34.2 gm of sucrose (C₁₂H₂₂O₁₁) is dissolved in 180 gm of water. Calculate the number of oxygen atoms in the solution. (6.684 x 10²⁴)
9. Calculate the number of hydrogen atoms present in 25.6 gm of urea. (1.03 x 10²⁴)
10. Sulphuric acid is manufactured through the following reactions.
    S + O₂ → SO₂
    2SO₂ + O₂ → 2SO₃
    SO₃ + H₂O → H₂SO₄
    What mass of oxygen is needed to produce 4.9 gm of H₂SO₄? (2.4)

Limiting Reagent

1. 5 gm of pure CaCO₃ if treated with 5 gm of HCl to produce CaCl₂, H₂O and CO₂.
   a. Find the limiting reagent. (CaCO₃)
   b. Calculate the mass of CaCl₂ formed. (5.55 gm)
   c. How many water molecules are produced? (3.011×10²²)
   d. Calculate the volume of CO₂ produced at NTP. (1.12 L)
2. 10 gm of pure zinc reacts with excess dilute sulphuric acid to yield zinc sulphate and hydrogen.
   a. Calculate the number of moles of H₂SO₄ consumed. (0.153)
   b. Calculate the mass of ZnSO₄ formed. (24.70 gm)
   c. What volume of hydrogen is evolved at NTP? (3430.32 cc)
3. 10.6 gm of pure Na₂CO₃ if treated with 7.9 gm of HCl to produce NaCl, H₂O and CO₂.
   a. Find the limiting reagent. (Na₂CO₃)
   b. Calculate the mole of unreacted reagent left over. (0.0164)
   c. What volume of CO₂ is produced at NTP? (2.24 L)
   d. Calculate the mass of NaCl formed. (11.7 gm)
4. 200 gm of 90% pure CaCO₃ is completely reacted with excess HCl to produce CaCl₂, H₂O and CO₂.
   a. Calculate the mass of CaCl₂ formed. (199.8 gm)
   b. How many moles of water are formed? (1.8)
   c. What volume of CO₂ is produced if the reaction is carried out at 27oC and 760 mm of Hg? (44.30 L)
5. 17 gm of ammonia is completely reacted with 45 gm of oxygen to produce NO and H₂O.
   a. Which is limiting reagent? (NH₃)
   b. Calculate the number of moles of unreacted reagent left over. (0.156)
   c. What volume of NO is produced at NTP? (22.4 L)
   d. Calculate the mass of water produced. (27 gm)
6. A chemical reaction was carried out by mixing 25 gm of pure CaCO₃ and 0.75 moles of pure HCl.
   a. Which one is limiting reagent? (CaCO₃)
   b. Calculate the mass of CaCl₂ produced. (27.75 gm)
   c. How many water molecules are formed? (1.505×10²³)
   d. What mass of NaOH is required to absorb the whole CO₂ produced in the reaction? (20 gm)

Percentage composition

1. A compound contains 75% carbon and 25% hydrogen. Determine its empirical formula. (CH₄)
2. Butyric acid contains C, H and O. 4.24 gm sample of butyric acid is completely burnt. It gives 8.45 gm of CO₂ and 3.46 gm of H₂O. If molecular mass of butyric acid is 88, determine its molecular formula. (C₄H₈O₂)
3. A sample of pure oxide of sulphur contains 50.1 % sulphur and 49.9% oxygen by mass. What is the simplest formula of compound. (SO₂)
4. 7gm of nitrogen is combined with 12 gm of oxygen. What is the empirical formula. (N₂O₃)
5. In hydrocarbon, the ratio of hydrogen to carbon is 2:1 and the molecular weight is 42. Find its molecular formula. (C₃H₆)
6. During construction of Palpa-Butwal road, a new compound was suspected. It contains 4.6 gm sodium, 6.4 gm sulphur and 4.8 gm oxygen.
   a. Calculate the percentage composition.
   b. What is the empirical formula of compound. (Na₂S₂O₃)