ICSE Revision Notes for Mole Concept and Stoichiometry Class 10 Chemistry
Chapter Name | Mole Concept and Stoichiometry |
Topics Covered |
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Related Study |
Boyle's Law and Charle's Law
Boyle’s Law
Relation between pressure (p) and volume (V)
Statement: At constant temperature, the pressure of a fixed amount (number of moles, n) of a gas is inversely proportional to its volume.
Explanation: Based on kinetic theory,
Number of particles and their average kinetic energy is constant for a given mass of gas.
When volume of a certain mass of gas is reduced to half, the particles have lesser space to move around.
The number of collision of the particles with the walls of the container doubles, thus increasing the pressure to twice the original value.
Mathematically,
p ∝ 1/V (at constant T and n)
⇒ p = k1 1/V, where k1 = Proportionally constant
⇒ pV = k1
From the above equation, it is found that at constant temperature, the product of pressure and volume of a fixed amount of a gas is constant.
The value of k1 depends upon
- amount of the gas
- temperature of the gas
- units of p and V
Graphical representation of Boyle's law
V1 = Volume of a gas at pressure p1
V2 = Volume of the same gas at pressure p2
Then,
p1V1 = p2V2 = Constant
⇒ p1/p2 = V2/V1
Relationship between density (d) and pressure (p):
We know that,
Where, m = Mass of a gas
V = Volume of the gas
⇒ d = (m/k1)p (since p = k1. 1/V)
⇒ d = k’p
⇒ d ∝ p
From the above equation, it is known that density is proportional to the pressure of a fixed amount of a gas.
Significance of Boyle's law
Mountaineers carry oxygen cylinders with them as at higher altitudes as the pressure is low.
Example: Rita has two cylinders. One is empty and the other contains compressed nitrogen at 25 atm. She wants to distribute the gas in the two cylinders. To do so, she connects the two cylinders. If the volume of the cylinder containing the gas is 50 L and that of the empty one is 80 L, then what will be the pressure inside the two cylinders?
Answer
According to Boyle’s law,
p1V1 = p2V2
Given, p1 = 25 atm
V1 = 50 L
V2 = (50 + 80)L = 130 L
Now, 25 atm × 50 L = p2 × 130 L
⇒ p2 = (25 × 50)/130 atm
= 9.62 atm
Hence, the pressure inside the cylinders is 9.62 atm.
Charles’ Law
Relation between temperature (T) and volume (V)
Statement: At constant pressure, the volume of a fixed amount of a gas is directly proportional to its absolute temperature.
Explanation: On the basis of kinetic theory
Average kinetic energy of the particles of a gas is directly proportional to the absolute temperature of the gas
When temperature is increased at constant pressure, the kinetic energy of the particles increases.
The number and intensity of collisions with the walls of the container increase, thereby increasing the volume at constant pressure.
Mathematically,
V ∝ T
⇒ V = k2T where,
k2 = Proportionality constant
T = Proportionality constant
The value of k2 depends upon
- pressure of the gas
- amount of the gas
- unit of volume
Graphical representation of Charles' Law
- Straight line
- Interception on zero volume at 273.15°C
- Each line is called isobar (constant pressure plot).
Derivation
For each degree rise in temperature, volume of a gas increases by (1/273.15) of the original volume of the gas at 0°C.
Suppose, V0 = Volume of a gas at 0°C
Vt = Volume of the same gas at t°C
Then,
V1 = V0 + (t/273.15).V0
⇒ V1 = V0 (1 + t/273.15)
⇒ V1 = V0 (273.15 + t)/273.15 ....(i)
According to Kelvin temperature scale (also called absolute temperature scale or thermodynamic scale)
T = 273.15 + t
T0 = 273.15
From equation (i), we obtain
V1 = V0 (T1/T0)
⇒ V1/V0 = T1/T0
Or, we can write
V2/V1 = T2/T1
⇒ V1/T1 = V2/T2
⇒ V/T = constant = k2
⇒ V = k2T
Significance of Charle's law
Hot air is filled in the balloons used for meteorological purposes.
Example
It is desired to increase the volume of 5 L of a gas by 40% without changing the pressure. To what temperature should the gas be heated if its initial temperature is 298 K?
Answer:
Desired increase in the volume of gas = 40% of 5 L
= 40/100 × 5 L
= 2 L
Therefore, final volume of the gas = (5 + 2 ) L = 7 L
Applying Charles’ law,
V1/T1 = V2/T2
Now, V1 = 5 L
T1 = 298 K
V2 = 7 L
Therefore,
5L/298 K = 7L/T2
⇒ T2 = (7L × 298 K)/5L
= 417. 2 K
Standard Temperature and Pressure(STP)
The pressure and temperature of the gas keeps varying frequently. Hence, we choose a standard value for temperature and pressure to which the gas volumes can be referred. The standard value chosen are 0°0°C or 273K for temperature and 1 atm or 760 mm of Hg for pressure and are commonly known as S.T.P.
Diffusion
Diffusion is defined as the random movement of gaseous molecules from regions of higher concentration to regions of lower concentration. It is a physical process and can only occur if the gases do not react with each other.
Graham's Law of Diffusion
It states that the rate of diffusion of gas is inversely proportional to the square root of its density at the given temperature and pressure.
r ∝ 1/√d
r = K/√d or K = r/√d
r = rate of diffusion
d = density of gas
K = proportionality constant
Relationship between diffusion and mass
r = K/√d
It means that the rate of diffusion is inversely proportional to the square root of mass of the gas.
The Mole Concept
The Mole Concept: A Brief Overview
Mole defines the quantity of a substance.
One mole of any substance will always contain 6.022 × 1023 particles, no matter what that substance is.
Therefore, we can say:
- 1 mole of sodium atoms (Na) contains 6.022 × 1023 sodium atoms.
- 1 mole of sodium ions (Na+) contains 6.022 × 1023 sodium ions.
- 1 mole of hydrogen atoms (H) contains 6.022 × 1023 hydrogen atoms.
- 1 mole of hydrogen molecules (H2) contains 6.022 × 1023 hydrogen molecules.
The word ‘mole’ is derived from the Latin word ‘moles’ which means ‘heap’ or ‘pile’. It was first used by the German chemist Wilhelm Ostwald in 1896. It was accepted universally much later, in 1967, as a way of indicating the number of atoms or molecules in a sample.
Thus, mole can be defined as a unit of measurement used for determining the number of atoms or molecules or ions in a given sample. It is also used to express the number of reactants and products in a chemical reaction.
Consider the formation of water by the combination of hydrogen and oxygen.
2H2 + O2 → 2H2O
This reaction implies that 2 moles of hydrogen molecules combine with 1 mole oxygen molecules to form 2 moles of water molecules.
When carbon (C) reacts with oxygen (O), carbon dioxide is produced. Can you write the chemical equation for the same?
The chemical equation for the reaction is:
C + O2 → CO2
In this reaction, one atom of carbon combines with one molecule (or two atoms) of oxygen to form one molecule of carbon dioxide. We can also say that in this chemical reaction, 12 u of carbon combines with 32 u of oxygen to give 44 u of carbon dioxide. Clearly, we can represent the quantities of substances in terms of their masses.
However, a chemical equation only indicates the numbers of atoms or molecules taking part in the chemical reaction. Therefore, it is easier to represent the quantities of substances involved in a chemical reaction by the numbers of their atoms or molecules rather than their masses. In order to do the same, the concept of mole is used.
Mole Concept
In 1909, the French physicist Jean Perrin found that one gram atom of any element contains the same number of atoms and one gram molecule of any substance contains the same number of molecules, which is equal to 6.022 × 1023.
He proposed naming this number in honour of the Italian physicist Amedeo Avogadro. Hence, 6.022 × 1023 is known as Avogadro’s number (or Avogadro’s constant) and the amount of a substance containing 6.022 × 1023 atoms/molecules/ions is called a mole.
Mole is a counting unit in chemistry as it is used to express large numbers of atoms or molecules. One mole of any substance can be defined as the amount of a substance that contains as many particles (atoms, molecules or ions) as there are atoms in 12 g of carbon-12 isotope. So,
1 mole of oxygen atoms (O) = 6.022 × 1023 oxygen atoms
1 mole of oxygen molecules (O2) = 6.022 × 1023 oxygen molecules
Know Your Scientist
In 1908, he studied Brownian motion using an ultramicroscope and gave experimental confirmation to the hypothesis that the random motion of suspended particles is due to the particulate nature of matter and the inter-particle interactions. He is also credited with estimating the size of a water molecule and the number of molecules of water present in a given amount of water.
Molar Mass
The molar mass of a substance can be defined as the mass of one mole of a substance in grams. It is numerically equal to atomic/molecular/formula unit mass in u.
Molar mass = (Mass of substance)/(Number of moles)
The mass of one atom is called atomic mass and its unit is unified mass (u), while the mass of one mole of atoms is called molar mass of atoms and its unit is gram (g). Molar mass of atoms is also called gram atomic mass.
For example, the atomic mass of nitrogen (N) is 14 u, while its gram atomic mass is 14 g. So, while 14 u of nitrogen contains only 1 atom of nitrogen, 14 g of nitrogen contains 1 mole of nitrogen atoms, i.e., 6.022 × 1023 nitrogen atoms.
The mass of one molecule is called molecular mass and its unit is unified mass (u), while the mass of one mole of molecules is called molecular mass and its unit is gram (g). When molecular mass is expressed in grams, it is called gram molecular mass or gram molecule.
For example, the molecular mass of oxygen (O2) is 32 u, while its gram molecular mass is 32 g. So, while 32 u of oxygen contains only 1 molecule of oxygen, 32 g of oxygen contains 1 mole of oxygen molecules, i.e., 6.022 × 1023 oxygen molecules.
The volume of one mole of any substance is called its molar volume.
The molar volume of a gas at STP is numerically equal to 22.4 L.
Solved Examples
Example 1: Calculate the mass of 3.3 moles of ammonia molecule.
Answer:
Molar mass of ammonia molecule (NH3) = 17 g
Number of moles of ammonia molecule = 3.3
We know that:
Molar mass = (Mass of substance)/(Number of moles)
So,
17g = (mass of ammonia molecule)/3.3
⇒ Mass of ammonia molecule = (17 × 3.3) = 56.1 g
Example 2: Calculate the volume of 14 g of nitrogen gas at STP.
Answer:
Mass of nitrogen gas (N2) = 14 g
Molar mass of nitrogen gas = 28 g
We know that:
Molar mass = Mass of substance/Number of moles
So,
28 g = 14g/(Number of moles)
Number of moles = 14g/28 g = 0.5
The volume of 1 mole of a gas at STP is 22.4 L.
Therefore,
Volume of 0.5 mole of nitrogen gas at STP = (22.4L)/1 × 0.5 = 11.2 L
Avogadro’s Law
In 1811, Avogadro hypothesized that under the same conditions of temperature and pressure, equal volumes of all gases contain an equal number of moles. For example, at the same temperature and pressure, the two gases, oxygen and nitrogen possessing the same volume contain the same number of molecules. This hypothesis is named Avogadro’s law. The mole concept provides the following information.
If one mole of a substance (atoms, molecules or ions) is present, then the number of elementary particles present in that substance is equal to 6.022 × 1023.
The mass of one mole of a substance (atoms, molecules or ions) is equal to its molar mass.
While carrying out reactions, scientists require the number of atoms and molecules. This requirement is fulfilled by the use of the mole concept as follows:
1 mole = 6.022 × 1023 = Relative mass in grams.
Relationship between Mole, Avogadro’s Number and Mass
The relationship between mole, Avogadro’s number and mass is summarized in the given figure.
Applications of Avogadro's Law
It provides an explanation for Gay-Lussac's law.
The volumes of different combining gases bear a simple ratio to one another because according to Avogadro's law at constant temperature and pressure equal volumes of gases contain the same number of molecules.
It helps in determination of atomicity of gases.
Consider the formation of hydrogen chloride gas by the direct combination of hydrogen and chlorine gases:
H2 + Cl2 → 2 HCl
According to Avogadro's law:
1 molecule of H2 + 1 molecule of Cl2 → 2 molecules of HCl
Or, 1/2 molecule of H2 + 1/2 molecule of Cl2 → 1 molecule of HCl
As atoms are indivisible, therefore, half a molecule of H2 and Cl2 indicate they both contain two atoms per molecule.
It helps in determination of molecular formula of a gas.
Consider the formation of hydrogen chloride gas by the direct combination of hydrogen and chlorine gases:
H2 + Cl2 → 2 HCl
According to Avogadro's law:
1 molecule of H2 + 1 molecule of Cl2 → 2 molecules of HCl
Or, 1/2 molecule of H2 + 1/2 molecule of Cl2 → 1 molecule of HCl
Or, 1 atom of H + 1 atom of Cl → 1 molecule of HCl
Hydrogen chloride gas has one atom of hydrogen and one atom of chlorine. Therefore, its molecular formula is HCl.
It helps in the establishment of the relationship between molecular mass and vapour density (VD).
2 × VD = (Mass of 1 molecule of a gas)/(mass of 1 atom of hydrogen)
It provides the relation between gram molecular mass and gram molecular volume.
Molar volume of a gas = 22.4 L
Gram molecular mass of a gas occupies 22.4 L and contains 6.02 × 1023 molecules/atoms of the gas.
Gay-Lussac’s Law
Mathematically,
p ∝ T
⇒ p/T = constant = k3
If at constant volume,
p1 = Pressure of a gas at T1
p2 = Pressure of the same gas at T2
Then,
p1/T1 = p2/T2
Mole Concept and Molar Masses
1 mole of any substance can be defined as:
Amount of a substance that contains as many particles (atoms, molecules or ions) as there are atoms in 12 g of the 12C isotope
Avogadro number or Avogadro constant (NA); it is equal to 6.022 × 1023 particles
Example: 1 mole of oxygen atoms = 6.022 × 1023 atoms of oxygen 1 mole of carbon dioxide molecules = 6.022 × 1023 molecules of carbon dioxide 1 mole of sodium chloride = 6.022 × 1023 formula units of sodium chloride
Relative atomic mass: Relative atomic mass of an element is the ratio of the average mass of one atom of an element to one-twelfth of the mass of an atom of carbon-12.
Relative atomic mass = (Average mass of atom)/(1/12 × mass of carbon(C12))
Molar mass of a substance can be defined as:
Mass of one mole of a substance in grams
Numerically equal to atomic/molecular/formula mass in u.
Example − Molar mass of CO2 = 44.011 g mol−1
Relative molecular mass: It is defined as the ratio of the mass of a molecule to the atomic mass unit of the molecule. It is a unitless quantity.
Relative molecular mass = (Average mass of Molecule/Compound)/(1/12 × Mass of carbon (C12))
Molar mass of NaCl = 58.5 g mol−1
Examples
1. What number of moles contains 3.011 × 1023 molecules of glucose?
Answer
1 mole of glucose is equivalent to 6.022 × 1023 molecules of glucose. Hence, 3.011 × 1023 molecules of glucose will be present in
= (1 × 3.011 × 1023)/(6.022 × 1023) mol
= 0.5 mol (of glucose)
mol = 0.5 mol (of glucose)
Thus, 0.5 mole of glucose contains 3.011 × 1023 molecules of glucose.
2. What is the mass of a fluorine molecule?
Answer
1 mole of fluorine molecule contains 6.022 × 1023 molecules and weighs 38 g.
Therefore, mass of a fluorine molecule = 38/(6.022 × 1023) g
= 6.31 × 10−23 g
Atomicity
It is defined as the total number of atoms of constituent elements which combine to form a molecule.
One molecule of hydrogen combines with one molecule of chlorine to form two molecules of hydrogen chloride.
One molecule of hydrogen or chlorine contains two atoms of each.
Percentage Composition
Mass percent of an element = (Mass of that element in compound × 100%)/(Molar mass of the compound)
Example: What is the mass percent of oxygen in potassium nitrate? (Atomic mass of K = 39.10 u, atomic mass of N = 14.007 u, atomic mass of O = 16.00 u)
Answer:
Atomic mass of K = 39.10 u (Given)
Atomic mass of N = 14.007 u (Given)
Atomic mass of O = 16.00 u (Given)
Therefore, molar mass of potassium nitrate (KNO3)
= 39.10 + 14.007 + 3(16.00)
= 101.107 g
Therefore, mass percent of oxygen in KNO3
= (Mass of oxygen in KNO3 × 100%)/(Molar mass of KNO3)
= (3 × 16.00)/101.107 × 100%
= 47.47% (approx)
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Empirical formula |
Molecular formula |
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Represents the simplest whole number ratio of various atoms present in a compound |
Represents the exact number of different types of atoms present in a molecule of a compound |
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Empirical formula is determined if mass % of various elements are known. |
Molecular formula is determined from empirical formula if molar mass is known. |
Example: A compound contains 92.26% carbon and 7.74% hydrogen. If the molar mass of the compound is 26.038 g mol−1, then what are its empirical and molecular formulae?
Answer:
Mass percent of carbon (C) = 92.26% (Given)
Mass percent of hydrogen (H) = 7.74% (Given)
Therefore, 100 g of the compound contains 92.26 g and 7.74 g of of hydrogen
Number of moles of carbon present in the compound = 92.26/12.011
= 7.68 mol
Number of moles of hydrogen present in the compound = 7.74/1.008
= 7.68 mol
Thus, in the given compound, carbon and hydrogen are present in the ratio C : H = 7.68 : 7.68
= 1 : 1
Therefore, the empirical formula of the compound is CH.
Empirical formula mass of CH = (12.011 + 1.008)g
= 13.019 g
Molar mass of the compound = 26.038 g (Given)
Therefore, n = Molar mass/Empirical formula mass
= 26. 038 g/13.019 g
= 2
Hence, the molecular mass of the compound is (CH)n, i.e., (CH)2 or C2H2
Interconversion among Number of Moles, Mass and Number of Molecules
From the above balanced chemical equation, the following information is obtained:
One mole of C3H8(g) reacts with five moles of O2(g) to give three moles of CO2(g) and four moles of H2O(l).
One molecule of C3H8(g) reacts with five molecules of O2(g) to give three molecules of CO2(g) and four molecules of H2O(l).
44 g of C3H8(g) reacts with (5 × 32 = 160) g of O2(g) to give (3 × 44 = 132) g of CO2(g) and (4 × 18 = 72) g of H2O(l).
22.4 L of C3H8(g) reacts with (5 × 22.4 = 112) L of O2(g) to give (3 × 22.4 = 67.2) L of CO2(g) and (4 × 22.4 = 89.6) L of H2O(l).
Example: Nitric acid (HNO3) is commercially manufactured by reacting nitrogen dioxide (NO2) with water (H2O). The balanced chemical equation is represented as follows:
3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g)
Calculate the mass of NO2 required for producing 5 moles of HNO3.
Answer:
According to the given balanced chemical equation, 3 moles of NO2 will produce 2 moles of HNO3.
Therefore, 2 moles of HNO3 require 3 moles of NO2.
Hence, 5 moles of HNO3 require = 3/2 × 5 moles of NO2
= 7.5 moles of NO2
Molar mass of NO2 = (14 + 2 × 16) g mol-1
= 46 g mol-1
Thus, required mass of NO2 = (7.5 × 46) g mol-1
= 345 g
Example: Lead nitrate reacts with sodium iodide to give lead iodide and sodium nitrate in the following manner:
Pb(NO3)2 + 2NaI → PbI2 + 2NaNO3
What amount of sodium nitrate is obtained when 30 g of lead nitrate reacts with 30 g of sodium iodide?
Answer:
Molar mass of Pb(NO3)2 = 207 + [{14 + (16 × 3)} × 2]
= 331 g mol-1
Molar mass of Nal = (23 + 127) = 150 g mol-1
According to the given equation, 1 mole of Pb(NO3)2 reacts with 2 moles of NaI, i.e.,
331 g of Pb(NO3)2 reacts with 300 g of NaI to give PbI2 and NaNO3
Thus, 30 g of Pb(NO3)2 will react with (30 × 300) / 331 g of NaI = 27.19g of NaI
However, we have 30 g of NaI. So, NaI is present in excess and Pb(NO3)2 is the limiting reagent.
Now, number of moles in 30 g of Pb(NO3)2 30 g = 30/331 = 0.09 mole
According to the equation, 1 mole of Pb(NO3)2 gives 2 moles of NaNO3.So
0.09 moles of Pb(NO3)2 will give (2 × 0.09) moles of NaNO3 = 0.18 moles of NaNO3.
Ways for expressing the concentration of a solution:
Mass per cent or weight per cent (w/w%)
Mass per cent = Mass of solute/Mass of solution × 100%
Example: 4.4 g of oxalic acid is dissolved in 200 mL of a solution. What is the mass per cent of oxalic acid in the solution? (Density of the solution = 1.1 g mL−1)
Answer:
Density of the solution = 1.1 g mL−1
So the mass of the solution = (200 mL) × (1.1 g mL−1)
= 220 g
Mass of oxalic acid = 4.4 g
Therefore, mass per cent of oxalic acid in the solution
= (Mass of Oxalic acid)/(Mass of the solution) × 100%
= 4.4g/220 g × 100%
= 2%
If a substance ‘A’ dissolves in a substance ‘B’, then mole fraction of
A = (Number of moles of A)/(Number of moles of the solution)
= nA/(nA + nB)
Mole fraction of B = (Number of moles of B)/(Number of moles of the solution)
= nB/(nA + nB)
nA − Number of moles of A
nB − Number of moles of B
Example: A solution is prepared by dissolving 45 g of a substance X (molar mass = 25 g mol−1) in 235 g of a substance Y (molar mass = 18 g mol−1). Calculate the mole fractions of X and Y.
Answer:
Moles of X, nX = 45/25
= 1.8 mol
Moles of Y, nY = 235/18
= 13.06 mol
Therefore, mole fraction of X, nX
= 1.8/(1.8 + 13.06)
= 0.121
And, mole fraction of Y, nY = 1 − nX
= 1 − 0.121
= 0.879
Number of moles of solute in 1 L of solution
Molarity (M) = (Number of moles of solute)/(Volume of solution in litres)
For a given solution, the molarity equation is as follows:
M1V1 = M2V2
M1 = Molarity of a solution when its volume is V1
M2 = Molarity of the same solution when its volume is V2
Example 1: 10g of HCl is dissolved in enough water to form 500 mL of the solution. Calculate the molarity of the solution.
Answer:
Molar mass of HCl = 36.5 g mol−1
So the moles of HCl = 10/36.5 mol
= 0.274 mol
Volume of the solution = 500 mL = 0.5 L
Therefore, molarity = (Number of moles of HCl)/(Volume of solution in litres)
= 0.274 mol/0.5 L
= 0.548 M
Example 2: Commercially available concentrated HCl contains 38% HCl by mass. What volume of concentrated HCl is required to make 2.5 L of 0.2 M HCl? (Density of the solution = 1.19 g mL−1)
Answer:
38% HCl by mass means that 38g of HCl is present in 100 g of the solution.
Moles of HCl = 38/36.5 = 1.04 mol
Volume of the solution = Mass/Density
= 100g/1.19g mL−1
= 84.03 mL
= 0.08403L
Therefore, molarity of the solution = 1.04/0.08403 L
= 12.38 M
According to molarity equation,
M1V1 = M2V2
Here,
M1 = 12.38 M
M2 = 0.2 M
V2 = 2.5 L
Now, M1V1 = M2V2
⇒ V1 = M2V2/M1
= (0.2 × 2.5)/12.38
= 0.0404 L (approx)
Hence, required volume of HCl = 0.0404 L
Number of moles of solute present in 1 kg of solvent
Molality (m) = (Number of moles of solute)/(Mass of solvent in kg)
Example: What is the molality of a solution of glucose in water, which is labelled as 15% (w/w)?
Answer:
15% (w/w) solution means that 15 g of glucose is present in 100 g of the solution, i.e. (100 − 15) g = 85 g of water = 0.085 kg of water
Moles of glucose = 15g /180g mol-1
= 0.083 mol
Therefore, molality of the solution = 0.083 mol/0.085 kg
= 0.976 m