General
- Chemistry S2 SB File Modified 24/01/22, 13:38
Unit 7: The Mole Concept and Gas Laws
LEARNING OBJECTIVES
After reading this unit, you will be able to:
•explain the mole concept.
•explain the concepts of: relative atomic mass, relative formula mass, relative molecular mass, molar mass, limiting reactant, empirical and molecular formulae.
•state the gas laws: Gay-Lussac, Charles’ law, Boyle’s law and the ideal gas law, Grahams’ law of diffusion
KNOWLEDGE GAIN
Amedeo Avogardo Cerreto was anItalian scientist. He is most noted for his contributions to molecular theory, including what is known as Avogadro’s law.
In tribute to him, the number of elementary entities (atoms, molecules, ions or other particles) in 1 mole of a substance, 6.022×1023, is known as the Avogadro constant.
7.1 AVOGADRO NUMBER AND THE MOLE CONCEPT
ACTIVITY 7.1: Illustrating Link between Number of Particles and Mass of Particles
Materials Required
A paper bag, 5 handfuls peanuts, balance Can you tell how many peanuts are there? (without counting)
•Using balance, measure the mass of the paper bag (say 50 g)•Now measure the mass of a single peanut (say 0.5 g)
•Put 5 handfuls peanuts in a paper bag
Note: Size of each peanut should be same.
•Now, weigh all the peanuts. Let it be 250 g.
Calculation
•Paper bag + 5 handfuls peanuts
= 250 g5 handfuls peanuts
= 250 g – 50 g (weight of paper bag)
= 200 g
•Number of peanuts
Total number of peanuts = 400
In this way, you can count the number of objects by weighing the object.In the same manner, bank people count a large number of coins. In chemistry also, scientists link the mass of an element or compound to the number of atoms or molecules present in them. This is done through the mole. Thus, mole is a link between the mass of atoms (or molecules) and the number of atoms (or molecules).
The word mole was introduced around 1896 by a German Physical Chemist Wilhem Ostwald who derived the term from the Latin word, ‘mole’ which means a pile or a heap. A substance may be considered as a heap of atoms or molecules.
In everyday life, we use counting units like a dozen (12 objects) and a gross (144 objects) to deal with a large quantities. In Chemistry, the unit for dealing with the number of atoms, ions or molecules is mole. A mole is a group of 6.022×1023 particles (atoms, molecules or ions) of a substance. The SI symbol of mole is mol.
A mole is defined as the amount of matter that contains as many objects (atoms, molecules, ions or whatever objects we are considering) as the number of atoms in exactly 12 g of pure 12C (carbon-12 isotope).
From numerous experiments, scientists have determined the number of atoms in 12 g of 12C (carbon-12) is found to be 6.022×1023. This number is called Avogadro’s number, in honor of an Italian scientist Amedeo Avogadro. In this book, we will use 6.02 × 1023 or 6.022 × 1023 for Avogadro’s number. Avogadro’s number is represented by NA.Take an example of the reaction of hydrogen and oxygen to form water:
2H2 + O2→ 2H2O.
The above reaction indicates that two molecules of hydrogen combine with one molecule of oxygen to form two molecules of water.
We can infer from the above equation that the quantity of a substance can be characterized by its mass or the number of molecules. But, a chemical reaction equation indicates directly the number of atoms or molecules taking part in the reaction. Therefore, it is more convenient to refer to the quantity of a substance in terms of moles (number of its molecules or atoms) rather than their masses. One mole of any particles (atoms, molecules, ions or particles) is that quantity in number having a mass equal to its atomic or molecular mass in grams.
The number of particles present in 1 mole of any substance is 6.022×1023.The mass of 1 mole of a substance is equal to its relative atomic or molecular mass in grams. The atomic mass of an element gives us the mass of one atom of that element in atomic mass units (u). To get the mass of 1 mole of atom of that element, we have to take the same numerical value but change the units from ‘u’ to ‘g’. The mass of atoms in grams is also known as gram atomic mass. For example, atomic mass of hydrogen = 1 u. So, gram atomic mass of hydrogen = 1 g.
1 u hydrogen has only 1 atom of hydrogen, 1 g hydrogen has 1 mole atoms, that is, 6.022×1023 atoms of hydrogen.
Similarly,
16 u oxygen has only 1 atom of oxygen,
16 g oxygen has 1 mole atoms, that is,
6.022×1023 atoms of oxygen.
EXERCISE 7.1
1. Avogadro number is the number of atoms in 1 gram-atom of an element.(True or False)
2. A group of 6.022 × 1023 particles (atoms, ions, or molecules) of a substance is called a ______ of that substance.
3. One mole of CO2 contains _______ atoms of carbon.
4. The mass of one Avogadro number of nitrogen atoms is equal to:
(a) 14 amu (b) 14 g (c) 28 g (d) 14 kg
5. Mole is a link between the _________ of atoms (or molecules) and the _____ of atoms (or molecules).
ACTIVITY 7.2: Illustrating Concept of Moles
In groups, discuss the concept of moles as a way of expressing the amount of substance and make a presentation.
7.2 CALCULATION OF THE NUMBER OF MOLES
Example 1
Calculate the number of moles of 12.044 × 1023helium atoms.
Solution
Example 2
Calculate the number of particles in 0.1 mole of carbon atoms.
Solution
Example 3
Calculate the number of moles of 3.011 × 1023hydrogen atoms.
Solution
Number of moles
Example 4
(a) Distinguish between N2 and 2N.Calculate the mass of particles in the following:
(b) 0.5 mole of nitrogen atom
(c) 0.5 mole of nitrogen molecule
(Given: Atomic mass of nitrogen atom = 14 g Molecular mass of nitrogen molecule = 28 g)
Solution
(a) N2 stands for 1 mole of nitrogen molecules and 2N stands for 2 moles of nitrogen atoms.
(b) Mass of nitrogen atoms = Atomic mass of nitrogen atom × number of moles
Mass of nitrogen atoms = ?
Atomic mass of nitrogen = 14 g\
Molecular mass of N2 = 28 g
No. of mole = 0.5 = 14 × 0.5 = 7 grams
The mass of 0.5 mole of nitrogen atom is 7 grams.
(c) Mass of nitrogen molecules (N2)
= Molecular mass of molecules × no. of moles= 28 × 0.5
= 14 grams
Example 5
Calculate the no. of moles in
(i) 0.478 g of magnesium
(ii) 12 g oxygen gas
Given: Atomic mass of Mg = 24
Atomic mass of O = 16.
Solution
(i) No. of mole
(ii) In Oxygen gas, there are two atoms of Oxygen (O2)
Molecular mass of O2= 16 × 2 = 32 g
Example 6
Calculate the mass of 6.022 × 1023 nitrogen molecules.
Solution
The mass of nitrogen molecules (N2) = 14 × 2 = 28 g
1 mole of N2 molecules = 6.022 × 102 molecules of nitrogen.
1 mole of N2 molecule weighs = 28 g
Therefore, 6.022 × 1023 molecules will weigh 28 grams.
EXERCISE 7.2
1. Convert 12 g of oxygen gas into moles.
2. What is the mass of
(i) 0.5 mole of water molecules? (ii) 0.2 mole of oxygen atoms?
7.3 DEFINITION OF RELATIVE ATOMIC MASS
Just like mole, relative atomic mass is based on Carbon-12 isotope as a standard measure.The relative atomic mass (Ar) of an element is the average relative mass of an atom of an element as compared with an atom of 12C taken as 12 atomic mass unit. Hence,Relative atomic mass of an element (Ar)
The relative atomic mass (Ar) of an element is a pure number and it does not have a unit. The relative atomic masses of some elements taking 12C ( = 12.000 u). The relative atomic masses of some elements are given in table 7.1.
The common values are used in numerical problems.
7.4 DEFINITION AND CALCULATION OF RELATIVE MOLECULAR MASS
Relative Molecular Mass (Mr)
Like the atomic masses, the molecular masses of compounds are also very small and these cannot be measured directly. The molecular masses of compounds are also expressed as relative molecular masses (Mr). The relative molecular mass of a compound is the average relative mass of its molecule as compared with the mass of one 12C atom taken as 12 u.Relative molecular mass (Mr) =
The relative molecular mass (Mr) of a compound is a pure number and it does not have a unit.Molecular Mass (M)The average mass of one molecule of a compound in atomic mass unit is called molecular mass (M). Hence,
Molecular mass (M) = Mr × 1 u = Mr u
The molecular mass (M) has the unit of mass, i.e., g, kg or u. Note that the magnitudes of molecular mass (M) and relative molecular mass (Mr) are equal. They differ only in their units.
Calculation of molecular mass from atomic masses: The molecular mass of a compound is calculated by adding the atomic masses of all the atoms present in one molecule of the compound. This is illustrated on next page.
(a) Molecular mass of water: The molecular formula of water is H2O.
Hence,Molecular mass of water = (2 × Atomic mass of hydrogen) + (1 × Atomic mass of oxygen)
= (2 × 1) + (1 × 16) = 18 u.
(b) Molecular mass of nitric acid (HNO3): The molecular formula of nitric acid is HNO3.
Hence, Molecular mass of nitric acid
= (1 × Atomic mass of hydrogen) + (1 × Atomic mass of nitrogen)+ (3 × Atomic mass of oxygen)
= (1 × 1) + (1 × 14) + (3 × 16) = 1 + 14 + 48 = 63 u.
Example 7
Calculate the molecular mass of glucose (C6H12O6).
Solution
Molecular mass of glucose = (6 × atomic mass of C) + (12 × atomic mass of H) + (6 × atomic mass of O)
= 6 × 12 u + 12 × 1 u + 6 × 16 u= 72 u + 12 u + 96 u = 180 u
EXERCISE 7.3
1. Calculate the molecular mass of
(a) Phosphorus molecule, P4
(b) Sulphur molecule, S8
Given: Atomic masses S = 32 u and P = 31 u
7.5 DEFINITION AND CALCULATION OF RELATIVE FORMULA MASS
The sum of the atomic masses of all the atoms in a formula unit of a compound is called the formula unit mass of the compound. Always remember that the concept of the formula unit mass is used in case of ionic compounds. This is because in case of ionic compounds there are no separate individual molecules but they exist as aggregates, i.e., as cluster of ions.
For example, solid sodium chloride is represented as (Na+Cl–)n and in the simplified form it is written as NaCl.The formula unit mass of ionic compounds is calculated in the same manner as in the case of the molecular mass described earlier. For example, the formula unit mass of NaCl
= Atomic mass of Na + Atomic mass of Cl= 23 u + 35.5 u = 58.5 u.
Example 8
Calculate the formula unit masses of ZnO, Na2O and K2CO3. (Given atomic mass: Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u and O = 16 u)
Solution
Formula unit mass of ZnO = atomic mass of Zn + atomic mass of O = 65 u + 16 u = 81 u.
Formula unit mass of Na2O = (2 × atomic mass of Na) + atomic mass of O= 2 × 23 u + 16 u
= 46 u + 16 u = 62 u
Formula unit mass of K2CO3= (2 × atomic mass of K) + atomic mass of C + (3 × atomic mass of O)
= 2 × 39 u + 12 u + 3 × 16 u = 78 u + 12 u + 48 u
= 138 u
EXERCISE 7.4
1. Calculate the formula unit mass of
(i) ZnO
(ii) Na2CO3
(iii) C6H12O6
Given:Atomic mass : Zn = 65 u, Na = 23 u, C = 12 u, H = 1u, O = 16 u
7.6 CALCULATION OF MOLAR MASS
ACTIVITY 7.3: Understanding Molar Mass
We know that one methane molecule contains one carbon atom and four hydrogen atoms. 1 mole of methane molecules consists of 1 mole of carbon atoms and 4 moles of hydrogen atoms. The mass of 1 mole of methane can be found by adding the masses of carbon and hydrogen present.Mass of 1 mole carbon = 1 × 12.01 g = 12.01 g mol–1
Mass of 4 moles hydrogen = 4 × 1.00 g = 4.00 g mol–1
Mass of 1 mole methane = 12.01 + 4.00 = 16.01 g mol–1
The quantity 16.01 g mol–1 is called the molar mass of methane.1 mole methane has 6.022×1023 molecules so the mass of 6.022×1023 molecules is also 16.01 g. Calculate the mass of 12 moles of methane yourself.Mass of one mole of any substance is called its molar mass. Alternatively, the average mass of one mole of any substance is called its molar mass. Molar mass is represented by M and its unit is g mol–1. It is given by,Molar mass (M)
Molar mass of a substance = Mass of 6.023×1023 chemical unit of that substance
Molar mass of one hydrogen atom = Mass of 6.023×1023 atoms of hydrogen
Molar mass of one hydrogen molecule = Mass of 6.023×1023 molecules of hydrogen
Example 9
Calculate the molar mass of sulphur dioxide.
Solution
Mass of 1 mol S = 1 × 32.07 = 32.07 g mol–1
Mass of 2 mol O = 2 × 16.00 = 32.00 g mol–1
Mass of 1 mol SO2 = 64.07 g mol–1
Molar mass = 64.07
The molar mass of SO2 is 64.07 g mol–1. It represents the mass of 1 mole of SO2 molecules.
EXERCISE 7.5
1. Calculate the molar masses of the following substances :
(a) Ethyne, C2H2 (b) Hydrochloric acid, HCI (c) Nitric acid, HNO3(Atomic masses :
C = 12 u, H = 1 u, Cl = 35.5 u, N = 14 u, O = 16 u)
7.7 RELATIONSHIP BETWEEN NUMBER OF MOLES, MASS AND MOLAR MASS
How many moles are there in a certain mass of a substance?
The number of moles in a certain mass of a substance is calculated using the following formula:No. of moles of a substance, X
where w is the mass of the substance and M is the molar mass of the substance.
How many molecules are there in a certain mass of a substance?
The number of molecules present in a certain mass of a substance is calculated using the following formula:
No. of molecules of a substance
where m is the mass of the substance and M is the molar mass of the substance.
Example 10
How many atoms and S8 molecules are present in 50 g of sulphur? The relative atomic mass of sulphur is 32.
Solution
Example 11
The antibiotic penicillin has the molecular formula C16H18N2SO4. One injection of penicillin contains 500 mg of penicillin. When one intra-muscular injection of penicillin is administered:
(a) How many moles of penicillin are administered?
(b) How many molecules of penicillin are administered?
(c) How many atoms of nitrogen are injected?
Solution
(a) Molar mass of penicillin (C16H18N2SO4) = (16 × 12 + 18 × 1 + 2 × 14 + 1 × 32 + 4 × 16) g mol–1
= 334 g mol–1
334 g of penicillin = 1 mol
(b) 1 mole of penicillin = 6.023 × 1023molecules of penicillin.1.497 × 10–3 moles of penicillin contain 6.023 × 1023 × 1.497 × 10–3molecules = 9.016 × 1020 molecules of penicillin.
(c) 1 molecule of penicillin contains 2 atoms of nitrogen.9.016 × 1020 molecules of penicillin contain 2 × 9.016 × 1020 = 1.8032 × 1021 molecules of nitrogen.
Example 12
Calculate the number of molecules of chloroform (CHCl3) weighing 0.0239 g (H = 1, C = 12, Cl = 35.5).
Solution
Molar mass of CHCl3= 12 + 1 + 3 × 35.5 = 119.5 g mol–1
119.5 g of CHCl3 contain 1 mole of molecules of chloroform = 6.023×1023 molecules. 0.0239 g of CHCl3 contains
Example 13
Find the number of atoms in the following:
(a) 52 mol of Ne
(b) 52 g of Ne.
Solution
(a) 1 mole of Ne contains 6.023×1023atoms of Ne.52 moles of Ne contain 6.023 × 1023× 52 atoms of Ne = 3.132×1025 atoms of Ne.
(b)1 mole of Ne weighs 20 g and has 6.023 × 1023 atoms of Ne,
i.e., 20 g of Ne contain 6.023 × 1023 atoms of Ne.52 g of Ne contain 602310522023.×atoms of Ne = 1.57 × 1024 atoms of Ne
Example 14
Calculate the mass of
(a)1 atom of nitrogen
(b) 1 atom of silver
(c) 1 molecule of benzene (C6H6)
Solution
(a) Molecular mass of nitrogen = 14 u
Molar mass of nitrogen = 14.007 g mol–1Number of nitrogen atoms in 1 mole = 6.023 × 1023Mass of 1 atom of nitrogen
(b) Molecular mass of silver = 107.87 u
Molar mass of silver = 107.87 g mol–1Mass of 1 atom of silver
(c)Molecular mass of benzene (C6H6)= 12 × 6 + 1 × 6 = 72 + 6 = 78 u
Mass of 1 molecule of benzene
Example 15
Arrange the following in the decreasing order of masses:
(a) one atom of gold
(b) one gram-atom of nitrogen
(c) one mole of calcium
(d) one gram of iron.Given:
Molar mass of gold = 196.97 u
Molar mass of calcium = 40 u
Solution
(a) Molecular mass of gold = 196.97 u
Molar mass of gold = 196.97 g mol–1Mass of one atom of gold
(b) Mass of one gram-atom of nitrogen = 14.0 g
(c) Mass of one mole of calcium = Molar mass of calcium in gram = 40 g
(d) Given: mass of iron = 1.0 g
Hence the order of decreasing mass is:one mole of calcium > one gram-atom of nitrogen > one gram of iron > one atom of gold.
Example 16
Calculate
(a) the number of molecules of sulphur (S8) in 16 g of solid sulphur
(b)the number of aluminium ions in 0.051 g of aluminium oxide.
Solution
(a) No. of moles of S8 in 16 g of sulphur
No. of S8 molecules= No. of moles × Avogadro’s number
(b) Molar mass of Al2O3= (2 × 27 + 3 × 16) g mol–1
= 102 g mol–1
No. of moles of Al2O3
1 mol of Al2O3 contains 2 × 6.023 × 1023ions of aluminium.
5.00 × 10–4 mol of Al2O3 contains 2 × 6.023 × 1023 × 5.00 × 10–4 ions of aluminium = 6.0 × 1020 ions
Hence, 0.051 g of Al2O3 contains6.0 × 1020 ions of aluminium.
Example 17
Chlorophyll, the green coloring matter in plants involved in photosynthesis, contains 2.7% magnesium by mass. Calculate the number of magnesium atoms in 1.0 g of chlorophyll.
Solution
100 g of chlorophyll contains 2.7 g of magnesium.1 g of chlorophyll contains 0.027 g of magnesium.
1 mole of Mg = 24 g of magnesium = 6.023 × 1023 atoms of magnesium.2.7 × 10–2 g of magnesium contains
Example 18
Calculate the mass of carbon dioxide which contains the same number of molecules as are contained in 40 g of oxygen.
Solution
Molar mass of O2 = 32 g mol–1
32 g of O2 contain 6.023 × 1023 molecules.
40 g of O2 contain
Mass of 6.023 × 1023 molecules of CO2= 44 g
Mass of 7.529 × 1023 molecules of CO2
Example 19
Calculate the number of oxygen atoms present in 88 g of CO2.
What would be the mass of CO having the same number of oxygen atoms?
Solution
Number of moles of CO2 in 88 g of CO2
Since one mole of CO2 contains two moles of oxygen atoms, two moles of CO2 contain four moles of oxygen atoms. Hence,
1 mole of oxygen atoms contains 6.023 × 1023 oxygen atoms.4 moles of oxygen atoms contain 6.023 × 1023 × 4 = 2.4092 × 1024 oxygen atoms.
Since 1 mole of oxygen atoms is present in 1 mole of CO, 4 moles of oxygen atoms are present in 4 moles of CO.
Mass of 4 moles of CO = Number of moles of CO × gram-molecular mass of CO= 4 × (12 + 16) g = 112 g
Example 20
How many molecules are present in
(a) 9 g of water
(b) 17 g of ammonia?
Solution
(a) Given: mass of water = 9 g
Molar mass of water (H2O)= (2 × 1 + 1 × 16) g mol–1 = 18 g mol–1
18 g of water contain 6.023 × 1023molecules
(b) Given: mass of ammonia = 17 g
Molar mass of ammonia (NH3)= (1 × 14 + 3 × 1) g mol–1= 17 g mol–1
17 g of ammonia contain 6.023 × 1023molecules.
Example 21
What is the mass of
(a) 0.2 mol of oxygen atoms
(b) 0.5 mol of water molecules?
Solution
(a) 0.2 mole of oxygen atoms = 0.2 mol × molar mass of oxygen atom = 0.2 mol × 16 g mol–1 = 3.2 g (b)0.5 mole of water molecules = 0.5 mol × molar mass of water = 0.5 mol × 18 g mol–1 = 9 g
Example 22
What is the mass of
(a) 1 mole of N atoms
(b) 4 moles of Al atoms
(c) 1.50 mole of Na+ ions
(d)10 moles of Na2S2O3?
Solution
(a)Mass of 1 mole of N atoms = Mass of 6.023 × 1023 atoms of N
= Gram-atomic mass of N = 14 g
(b)Mass of 4 moles of Al atoms = Mass of 4 × 6.023 × 1023 atoms of Al
= 4 × gram-atomic mass of Al = 4 × 27 g = 108 g
(c) Mass of 1 mole of Na+ ions = Mass of 6.023 × 1023 ions of Na
= Gram-atomic mass of Na = 23 g
(d) Mass of 10 moles of Na2S2O3= 10 mol × molar mass of Na2S2O3
= 10 mol × (2 × 23 g mol–1+ 1 × 32 g mol–1 + 3 × 16 g mol–1)
= 10 mol × (46 g mol–1 + 32 g mol–1+ 48 g mol–1) = 10 mol × 126 g mol–1 = 1260 g
Example 23
Convert into moles
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide.
Solution
(a)Given mass of oxygen gas = 12 g
Molar mass of oxygen gas= 2 × 16 g mol–1 = 32 g mol–1
No. of moles of oxygen gas
Hence, 12 g of oxygen gas is equal to 0.375 mol of oxygen gas.
(b) Given: mass of water = 20 g
Molar mass of water (H2O) = (2 × 1 g mol–1 + 1 × 16 g mol–1) = 18 g mol–1
No. of moles of water
Hence, 20 g of water is equal to 1.11 mol of water.
(c) Given: mass of carbon dioxide = 22 g
Molar mass of carbon dioxide (CO2)= (1 × 12 g mol–1 + 2 × 16 g mol–1) = 44 g mol–1
No. of moles of carbon dioxide
Hence, 22 g of carbon dioxide is equal to 0.5 mole of carbon dioxide.
Example 24
Calculate the number of moles in the following masses
(a) 7.85 g Fe (at. mass 56)
(b) 65.5 μg of Carbon (at. mass 12)
(c) 4.68 mg of Si (at. mass 28)
(d) 1.46 metric tons of Al (at. mass 27)
(e) 7.9 mg of Ca (at. mass 40).
(Note: 1 metric ton = 103 kg).
Solution
Number of moles
Example 25
(a) Calculate the mass of 2.5 moles of calcium. Atomic mass of calcium is 40.
(b) Calculate the mass of 1.5 mole of water (H2O).
Solution
(a) 1 mole of calcium= molar mass of calcium atoms= 40 g mol–1
2.5 moles of calcium= 40 × 2.5 = 100 g
(b) Molecular mass of water (H2O)= 1 × 2 + 16 = 18 u= Molar mass of H2O = 18 g mol–1.
1.5 mole of H2O= 18 × 1.5 = 27 g
Example 26
Calculate the number of gram-atom and gram-mole in 25.4 mg of iodine (I2).
Atomic mass of I = 127 u.
Solution
Example 27
Calculate the molar mass of glucose (C6H12O6) and the number of atoms of each kind in it.
Solution
Molecular mass of glucose (C6H12O6)= 6 × (12.011 u) + 12 × (1.008 u) + 6 × (16.00 u)
= 72.066 u + 12.096 u + 96.00 u= 180.162 u
Calculation of number of atoms of each kind1 mole of glucose (C6H12O6)
≡ 6 moles of carbon + 12 moles of hydrogen+ 6 moles of oxygen
Hence, Atoms of carbon= 6 × 6.02 × 1023= 36.12 × 1023Atoms of hydrogen
= 12 × 6.02 × 1023= 72.24 × 1023
Atoms of oxygen= 6 × 6.02 × 1023= 36.12 × 1023
Example 28
Calculate mass of the following:
(i) one atom of calcium
(ii) one molecule of sulphur dioxide (SO2).
Solution
(i) Mass of 6.022 × 1023 atoms of calcium= gram atomic mass of calcium = 40 g
∴ Mass of 1 atom of calcium
(ii) Mass of 6.022 × 1023 molecules of SO2= molar mass of SO2= 64 g
∴ Mass of 1 molecule of SO2
Example 29
(a) Calculate number of atoms in each of the following:
(i) 0.5 mole of nitrogen atoms
(ii) 0.2 mole of nitrogen molecules
(iii) 3.2 g of sulphur.
(b) Calculate number of molecules in each of the following:
(i) 14 g of nitrogen
(ii) 3.4 g of hydrogen sulphide (H2S).
Solution
(a) (i) 1 mole of nitrogen atom= 6.022 × 1023 atoms
∴ 0.5 mole of nitrogen atoms= 6.022 × 1023 × 0.5 = 3.011 × 1023 atoms
(ii) 1 mole of nitrogen molecule= 6.022 × 1023 molecules
∴ 0.2 mole of nitrogen molecule= 6.022 × 1023 × 0.2= 1.2044 × 1023 molecules.
1 molecule of nitrogen= 2 atoms1.2044 × 1023 molecules of nitrogen
= 1.2044 × 1023 × 2 =2.409 × 1023 atoms
(iii) 32 g sulphur contain= 6.022 × 1023 atoms
∴ 3.2 g of sulphur contain
(b) (i) Molar mass of nitrogen = 28 g mol–1
28 g of nitrogen contain= 6.022 × 1023 molecules
∴ 14 g of nitrogen contain
(ii) Molar mass of H2S= (2 + 32) g = 34 g mol–1
34 g of H2S contain = 6.022 × 1023 molecules
3.4 g of H2S contain
Example 30
How many atoms of oxygen are present in 300 g of CaCO3?
Solution
Gram formula mass of CaCO3 = 100 g
Now, 1 mole of CaCO3 contains= 3 mole of O atoms.
or 100 g of CaCO3 contain = 3 × 6.022 × 1023 O atoms
∴ 300 g of CaCO3 contain O atoms
Example 31
How many molecules of water of hydration are present in 252 mg of oxalic acid,
(H2C2O4 .2H2O)?
Solution
Molar mass of H2C2O4 .2 H2O = 126 g mol–1
Now, water molecules in 1 mol of oxalic acid = 2 mol.or water molecules in 126 g of oxalic acid = 2 × 6.022 × 1023
∴ Water molecules in 252 × 10–3 g of oxalic acid
Example 32
Calculate mass of sodium which contains same number of atoms as are present in 4 g of calcium. Atomic masses of sodium and calcium are 23 and 40 respectively.
Solution
40 g of calcium contain = 6.023 × 1023 atoms
∴ 4 g of calcium contain
Now, 6.022 × 1023 atoms of sodium have mass= 23 g
∴ 6.022 × 1022 atoms of sodium have mass
Example 33
Chlorophyll, the green coloring matter of plants contains 2.68% of magnesium by mass. Calculate the number of magnesium atoms in 5.00 g of this complex.
Solution
Mass of magnesium in 5.00 g of complex
Gram atomic mass of magnesium= 24 g
24 g of magnesium contain= 6.022 × 1023 atom
0.134 g of magnesium would contain
Therefore, 5.00 g of the given complex would contain 3.36 × 1021 atoms of magnesium.
Example 34
Calculate number of atoms of each type in 3.42 g of sucrose (C12H22O11).
Solution
Molecular mass of sucrose = (12 × 12 + 1 × 22 + 16 × 11) = 342 u
342 g of sucrose contain = 6.022 × 1023 molecules
∴ 3.42 g of sucrose contain
Number of atoms of carbon in 3.42 g of sucrose
1 molecule of sucrose contains= 12 atoms of carbon6.022 × 1021 molecules of sucrose contain
= 12 × 6.022 × 1021 atoms of carbon= 7.226 × 1022 atoms of carbon
Number of atoms of hydrogen in 3.42 g of sucrose
1 molecule of sucrose contains
= 22 atoms of hydrogen6.022 × 1021 molecules of sucrose contain= 22 × 6.022 × 1021 atoms of hydrogen= 1.324 × 1023 atoms of hydrogen
Number of atoms of oxygen in 3.42 g of sucrose
1 molecule of sucrose contains= 11 atoms of oxygen
6.022 × 1021 molecules of sucrose contain= 11 × 6.022 × 1021 atoms of oxygen.
= 6.624 × 1022 atoms of oxygen
7.8 CALCULATION OF MASS PER CENT COMPOSITION OF AN ELEMENT IN A COMPOUND
A compound contains two or more elements combined in a certain fixed ratio. The percentage composition of a compound is the mass of each element of the compound, present in 100 g of that compound, i.e., the mass percentage of each element present in the compound. The mass percentage of each element in a compound can be calculated using either of the following two equations.
When the masses of compound and each element are given
Mass percentage of an element can be obtained if we know the mass of that element in a known mass of the compound.Mass percentage of an element X
When the formula of the compound and the atomic masses of the elements are given, the molecular mass of the compound can be calculated by adding the masses of all the elements present in the compound, and then the mass percentage of each element can be calculated using the following formula:
Mass percentage of an element
The above two methods of calculating the percentage composition of a compound are given in the following examples:
Example 35
0.24 g sample of a compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by mass.
Solution
Percentage of boron
Hence, the mass percentages of boron and oxygen in the given compound are 40.0% and 60.0% respectively.
Example 36
Calculate the mass percentage of each element present in water.
Solution
Molecular mass of water (H2O)= 2 × 1 u + 16 u = 18 u
Mass percentage of hydrogen (H) in water
Mass percentage of oxygen (O) in water
Hence, the composition of water by mass is H = 11.11% and O = 88.89%.
Example 37
Find the percentage composition of glucose whose formula is C6H12O6.
Solution
Molecular mass of glucose (C6H12O6)= 6 × 12 u + 12 × 1 u + 6 × 16 u
= 72 u + 12 u + 96 u = 180 u
Mass percentage of carbon (C) in glucose
Mass percentage of hydrogen (H) in glucose
Mass percentage of oxygen (O) in glucose
Hence, glucose contains 40.0% carbon, 6.67% hydrogen and 53.3% oxygen.
Example 38
Calculate the percentage of water of crystallization in washing soda whose formula is Na2CO3.10H2O.
Solution
Molecular mass of Na2CO3.10H2O = 2 × 23 u + 12 u + 3 × 16 u + 10(2 × 1 u + 1 × 16 u)
= 46 u + 12 u + 48 u + 180 u
= 286 u286 g of washing soda contain 180 g of water of crystallization.
Example 39
Find the percentage composition of potassium permanganate.
Solution
Molecular mass of KMnO4= 39 + 55 + 16 × 4 = 39 + 55 + 64 = 158 u
Percentage of potassium
The percentage composition of potassium permanganate is as follows:
Potassium = 24.68%
Manganese = 34.81%
Oxygen = 40.51%
Example 40
Ferric sulphate is a crystalline compound of iron. It is used in water and sewage treatment to help the removal of suspended impurities. Calculate the mass percentage of iron, sulphur and oxygen in the compound.
Solution
The formula of the compound is Fe2(SO4)3.The formula mass = 2 × 56 + 3(32 + 64)= 400 u
Example 41
Calculate the mass per cent of different elements present in ethyl alcohol (C2H5OH).
Solution
Molar mass of ethanol = 2 × 12.01 + 6 × 1.008 + 16.00= 46.068 g
Example 42
Calculate the percentage of hydration, total oxygen and copper in copper (II) sulphate pentahydrate.
Solution
The formula of the compound is CuSO4.5H2O
Formula mass = 63.5 + 32 + 64 + 5 × 18 = 249.5 u
EXPERIMENT 1
Aim
Determining the Per cent Composition of a magnesium in magnesium oxide
Safety
•The burning of magnesium generates an intense white light. Mixed in with the white light there is ultraviolet light as well.
•Do not look at the magnesium as it is burning.
Apparatus and Material
Stand, ceramic tile hot plate, ring clamp, pipe clay triangle, crucible and lid, crucible tongs, Bunsen burner, 25 cm of Magnesium ribbon
Procedure
1. Obtain a 25 cm piece of magnesium ribbon. The exact length is not important.
2.Find the mass of the magnesium ribbon and record it in the data table.
3.Set up the apparatus with the stand, ring clamp, Bunsen burner sitting on the ceramic tile hot plate.
4.Find the mass of the empty crucible and record it in the data table.
5.Roll the ribbon of magnesium into a spiral or crush it into a compacted mass and place it in the crucible.
6.Cover the crucible and heat strongly for 5 minutes or until red hot.
7.With the crucible tongs gently lift the crucible lid to let air into the crucible.
8.With the lid left slightly open discontinue heating.
9.Remove the crucible lid and let the crucible cool before massing it again.
10.Mass the crucible and contents. The contents are your new compound of magnesium oxide. Record the mass on the data table.
Data Table
Mass of magnesium ribbon= __________
Mass of crucible= __________
Mass of crucible and contents= __________
Calculations
1.Find the mass of the contents of the crucible. This is your new compound of mag-nesium oxide.
2.The percentage of magnesium in the compound can be found like this:
3.Find the percentage of oxygen in the compound.
Conclusions
The percentage composition of magnesium oxide is ......
Error Analysis
The above answers are your actual values.If the formula for magnesium oxide is MgO use your periodic table to calculate the theoretical percentage composition of both the magnesium and the oxygen.
Calculate the percentage composition for MgO: __________
Calculate the % error for O: __________
Calculate the % error for Mg:
EXERCISE 7.6
1. Calculate the mass percentage of each element present in water.
2. Find the percentage composition of glucose whose formula is C6H12O6.
3. Calculate the percentage of water of crystallization in washing soda whose formula is Na2CO3.10H2O.
4. Find the percentage composition of potassium permanganate.
7.9 EMPIRICAL FORMULA AND MOLECULAR FORMULA
An empirical formula represents the simplest whole number ratio of various atoms present in a compound whereas the molecular formula shows the exact number of different types of atoms present in a molecule of a compound.
Relation between the Two Formulae
Molecular formula is whole number multiple of empirical formula. Thus,Molecular formula = Empirical formula × n where n = 1, 2, 3...
* Empirical formula mass of a substance is equal to the sum of atomic masses of all the atoms.
Note: In many compounds; empirical formula is same as molecular formula. For example, NH3, H2O, CH4, etc.
If the mass per cent of various elements present in a compound is known, its empirical formula can be determined. Molecular formula can further be obtained if the molar mass is known. The following example illustrates this sequence.
EXAMPLE
A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine by mass. Its molar mass is 98.96 g. What are its empirical and molecular formulas?
Solution
Step 1: Conversion of mass per cent to grams.
Since we are having mass per cent, it is convenient to use 100 g of the compound as the starting material. Thus, in the 100 g sample of the above compound, 4.07 g hydrogen is present, 24.27 g carbon is present and 71.65 g chlorine is present.
Step 2: Convert into number moles of each element.
Divide the masses obtained above by respective atomic masses of various elements.
Step 3: Divide the mole value obtained above by the smallest number
Since 2.021 is smallest value, division by it gives a ratio of 2 : 1 : 1 for H : C : C : Cl.In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable co-efficient.
Step 4: Write empirical formula by mentioning the numbers after writing the symbols of respective elements.
CH2Cl is, thus, the empirical formula of the above compound.
Step 5: Writing molecular formula
(a) Determine empirical formula mass
Add the atomic masses of various atoms present in the empirical formula.For CH2Cl, empirical formula mass is 12.01 + 2 × 1.008 + 35.453 = 49.48 g
(b) Divide Molar mass by empirical formula mass
(c) Multiply empirical formula by n obtained above to get the molecular formula
Empirical formula = CH2Cl, n = 2. Hence molecular formula is C2H4Cl2.
Steps for Writing the Empirical Formula
The percentage of the elements in the compound is determined by suitable methods and from the data collected, the empirical formula is determined by the following steps:
Step 1: Divide the percentage of each element by its atomic mass. This gives the relative number of moles of various elements present in the compound.
Step 2: Divide the quotients obtained in the above step by the smallest of them so as to get a simple ratio of moles of various elements.
Step 3: Multiply the figures, so obtained by a suitable integer, if necessary, in order to obtain whole number ratio.
Step 4: Finally, write down the symbols of the various elements side by side and put the above numbers as the subscripts to the lower right hand corner of each symbol. This will represent the empirical formula of the compound.
Steps for Writing the Molecular Formula
Step 1: Calculate the empirical formula as described above.
Step 2: Find out the empirical formula mass by adding the atomic masses of all the atoms present in the empirical formula of the compound.
Step 3: Divide the molecular mass (determined experimentally by some suitable method) by the empirical formula mass and find out the value of n.
Step 4: Multiply the empirical formula of the compound with n so as to find out the molecular formula of the compound.
Example 43
Write the empirical formula of the compounds having molecular formulae:
Solution
Empirical formula is the simplest whole number ratio of atoms in the molecule, therefore the empirical formula of given compounds are:
Example 44
A substance, on analysis, gave the following percentage composition: Na = 43.4%, C = 11.3%, O = 45.3%. Calculate its empirical formula. [Na = 23, C = 12, O = 16]
Solution
Therefore, the empirical formula is Na2CO3.
Example 45
A compound has the following composition: Mg = 9.76%, S = 13.01%, O = 26.01%, H2O = 51.22%. What is its empirical formula? [Mg = 24, S = 32, O = 16, H = 1]
Solution
Example 46
What is the simplest formula of the compound which has the following percentage composition: Carbon 80%, Hydrogen 20%? If the molecular mass is 30, calculate its molecular formula.
Solution
Calculation of empirical formula:
∴ Empirical formula is CH3
Calculation of molecular formula:
Empirical formula mass = 12 × 1 + 1 × 3 = 15
Molecular formula
Example 47
Butyric acid contains C, H, O elements. A 4.24 mg sample of butyric acid is completely burnt in oxygen. It gives 8.45 mg of carbon dioxide and 3.46 mg of water. What is the mass percentage of each element?
Determine the empirical and molecular formula of butyric acid if molecular mass of butyric acid is determined to be 88 u.
Solution
Calculation of empirical formula:
∴ Empirical formula is C2H4O.
Calculation of molecular formula:
Empirical formula mass = 12 × 2 + 1 × 4 + 16 × 1 = 44
Molecular mass = 88 u
Molecular formula = Empirical formula × n
= C2H4O × 2 = C4H8O2
Example 48
An organic compound on analysis gave the following data : C = 57.82%, H = 3.6%, and the rest is oxygen. Its vapor density is 83. Find its empirical and molecular formula.
Solution
Calculation of empirical formula:
Calculation of molecular formula:
Empirical formula mass = 12 × 4 + 1 × 3 + 2 × 16 = 83
Molecular mass = 2 × V.D. = 2 × 83 = 166
Molecular formula = Empirical formula × n= C4H3O2 × 2 = C8H6O4
Example 49
2.746 g of a compound gave on analysis 1.94 g of silver, 0.268 g of sulphur and 0.538 g of oxygen. Find the empirical formula of the compound. (At. masses: Ag = 108, S = 32, O = 16)
Solution
To calculate empirical formula:
Example 50
A compound on analysis gave the following percentage composition: Na = 14.31%, S = 9.97%, H = 6.22%, O = 69.5%.
Calculate the molecular formula of the compound on the assumption that all the hydrogen in the compound is present in combination with oxygen as water of crystallization. Molecular mass of the compound is 322. [Na = 23, S = 32, N = 1 and O = 16]
Solution
Calculation of empirical formula:
∴ The empirical formula is Na2SH20O14.
Calculation of molecular formula:
Empirical formula mass = 23 × 2 + 32 + 20 × 1 + 16 × 14 = 322
Hence, molecular formula = Na2SH20O14.
Since all hydrogen is present as H2O in the compound, it means 20 hydrogen atoms must have combined with 10 atoms of oxygen to form 10 molecules of water of crystallization. The remaining (14 – 10 = 4) atoms of oxygen should be present with the rest of the compound.
Hence, molecular formula = Na2SO4.10H2O
Example 51
A pure sample of compound is found to contain 2.04 g of sodium, 2.65 × 1022 atoms of carbon and 0.132 mole of oxygen atoms. Determine the empirical formula of the compound.
Solution
Let us calculate the molar ratio of each type of atoms.
Mole of O atoms = 0.132
The atomic ratio Na : C : O is: 0.0887 : 0.044 : 0.132On dividing by the least, we get 2 : 1 : 3
Thus, the empirical formula is Na2CO3.
EXERCISE 7.7
1. A substance on analysis gave the following composition : C = 40%, H = 6.67%, O = 53.33%. Calculate its empirical formula. [Atomic masses: C = 12, H = 1 and O = 16]
2. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
3. Calculate empirical formula of a compound which has 48% carbon, 8% hydrogen, 28% nitrogen and 16% oxygen.
7.10 STOICHIOMETRIC CALCULATIONS
The word ‘stoichiometry’ is derived from two Greek words – stoicheion (meaning element) and metron (meaning measure). Stoichiometry, thus, deals with the calculation of masses (sometimes volumes also) of the reactants and the products involved in a chemical reaction. Before understanding how to calculate the amounts of reactants required or the products produced in a chemical reaction, let us study what information is available from the balanced chemical equation of a given reaction.
Balancing a Chemical Equation
According to the law of conservation of mass, a balanced chemical equation has the same number of atoms of each element on both sides of the equation. Many chemical equations can be balanced by trial and error. Let us take the reactions of a few metals and non-metals with oxygen to give oxides
4Fe(s) + 3O2(g) → 2Fe2O3(s)
(a) balanced equation 2Mg(s) + O2(g) → 2MgO(s)
(b) balanced equation P4(s) + O2(g) → P4O10(s)
c) unbalanced equation
Equations (a) and (b) are balanced since there are same number of metal and oxygen atoms on each side of equations.
However, equation (c) is not balanced. In this equation, phosphorus atoms are balanced but not the oxygen atoms. To balance it, we must place the coefficient 5 on the left of oxygen on the left side of the equation to balance the oxygen atoms appearing on the right side of the equation
. P4(s) + 5O2(g) → P4O10(s)
balanced equation
Now let us take combustion of propane, C3H8.
This equation can be balanced in steps.
Step 1:Write down the correct formulas of reactants and products. Here propane and oxygen are reactants, and carbon dioxide and water are products.
C3H8(g) + O2(g) → CO2(g) + H2O(l) unbalanced equation
Step 2: Balance the number of C atoms: Since 3 carbon atoms are in the reactant, therefore, three CO2 molecules are required on the right side.
C3H8(g) + O2(g) → 3CO2(g) + H2O(l)
Step 3:Balance the number of H atoms: On the left there are 8 hydrogen atoms in the reactants, however each molecule of water has two hydrogen atoms, so four molecules of water will be required for eight hydrogen atoms on the right side.
C3H8(g) + O2(g) → 3CO2(g) + 4H2O(l)
Step 4:Balance the number of O atoms: There are ten oxygen atoms on the right side(3 × 2 = 6 in CO2 and 4 × 1 = 4 in water).
Therefore, five O2 molecules are needed to supply the required ten oxygen atoms.
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
Step 5:Verify that the number of atoms of each element is balanced in the final equation. The equation shows three carbon atoms, eight hydrogen atoms, and ten oxygen atoms on each side.
All equations that have correct formulas for all reactants and products can be balanced. Always remember that subscripts in formulas of reactants and products cannot be changed to balance an equation.
Let us consider the combustion of methane. A balanced equation for this reaction is as given below:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Here, methane and oxygen are called reactants and carbon dioxide and water are called products. Note that all the reactants and the products are gases in the above reaction and this has been indicated by letter (g) in the brackets next to its formula. Similarly, in the case of solids and liquids, (s) and (l) are written respectively.
The coefficients 2 for O2 and H2O are called stoichiometric coefficients. Similarly, the coefficient for CH4 and CO2 is one in each case. They represent the number of molecules (and moles as well) taking part in the reaction or formed in the reaction.
Thus, according to the above chemical reaction,
•One mole of CH4(g) reacts with two moles of O2(g) to give one mole of CO2(g) and two moles of H2O(g)
•One molecule of CH4(g) reacts with 2 molecules of O2(g) to give one molecule of CO2(g) and 2 molecules of H2O(g)
•22.7 l of CH4(g) reacts with 45.4 l of O2(g) to give 22.7 l of CO2(g) and 45.4 l of H2O(g)
•16 g of CH4(g) reacts with 2 × 32 g of O2(g) to give 44 g of CO2 (g) and 2 × 18 g of H2O(g).
From these relationships, the given data can be inter converted as follows:
Example 52
How many grams of oxygen (O2) are required to completely react with 0.200 g of hydrogen (H2) to yield water (H2O)? Also calculate the amount of water formed. (At. mass H = 2; O = 32).
Solution
The balanced equation for the reaction is:
Example 53
What mass of zinc is required to produce hydrogen by reaction with HCl which is enough to produce 4 mol of ammonia according to the reactions
Zn + 2HCl —→ ZnCl2 + H2
3H2 + N2 —→ 2NH3
Solution
The given equations are:
Zn + 2HCl —→ ZnCl2 H2
3H2 + N2 —→ 2NH3
From the equations it is clear that
2 mol of NH3 require = 3 mol of H2,and 1 mol of H2 requires = 1 mol of Zn
or 3 mol of H2 require = 3 mol of Zn
Thus, 2 mol of NH3 require= 3 mol of Zn = 3 × 65 g of Zn∴ 4 mol of NH3 require
Example 54
Calculate the amount of water (g) produced by the combustion of 16 g of methane.
Solution
The balanced equation for combustion of methane is:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
(i) 16 g of CH4 corresponds to one mole.
(ii) From the above equation, 1 mol of CH4 (g) gives 2 mol of H2O(g).
2 mol of water (H2O) = 2 × (2 + 16)= 2 × 18 = 36 g1 mol H2O = 18 g H2O
Example 55
How many moles of methane are required to produce 22 g CO2(g) after combustion?
Solution
According to the chemical equation,CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)44 g CO2(g) is obtained from 16 g CH4(g).
[Q 1 mol CO2(g) is obtained from 1 mol of CH4(g)]mole of CO2(g)
Hence, 0.5 mol CO2(g) would be obtained from 0.5 mol CH4(g) or 0.5 mol of CH4(g) would be required to produce 22 g CO2(g).
EXERCISE 7.8
1. Balance the following chemical reactions:
(a)CO + O2→CO2
(b)KNO3→KNO2+ O2
(c)O3→O2
(d) NH4NO3→N2O +H2O
(e)CH3NH2+O2→CO2+H2O +N2
(f) Cr(OH)3+HClO4→Cr(ClO4)3+H2O
2. Write the balanced chemical equations of each reaction:
(a) Calcium carbide (CaC2) reacts with water to form calcium hydroxide (Ca(OH)2) and acetylene gas (C2H2).
(b) When potassium chlorate (KClO3) is heated, it decomposes to form KCl and oxygen gas (O2).
(c) C6H6combusts in air.
(d) C5H12O combusts in air.
7.11 LIMITING REACTANTS
ACTIVITY 7.4: Illustrating Limiting Reactants
Understanding Limiting Reagent
Imagine you are packing biscuits. In each biscuit packet, you put 10 biscuits.
For packing, you need 1 empty packet and 10 loose biscuits. This means1 empty packet + 10 loose biscuits →1 biscuit packet
Case 1: If you have 30 empty packets and 250 loose biscuits, how many packets of biscuits can you make? How many loose biscuits or empty packets will remain?
Case 2: If you have 15 empty packets and 200 loose biscuits, how many packets of biscuits can you make? How many loose biscuits or empty packet will remain?
In Case 1, you can make 25 packets of biscuits and 5 empty packets will be left over. Here, you do not have more biscuits to fill in the empty packets. So, biscuit is the limiting reagent.
In Case 2, you can make 15 packets of biscuits and 50 loose biscuits will be left over. Here, you do not have more empty packets to make biscuit packets. So, empty packet is the limiting reagent.
Many a time, the reactions are carried out when the reactants are not present in the amounts as required by a balanced chemical reaction. In such situations, one reactant is in excess over the other. The reactant which is present in the lesser amount gets consumed after sometime and after that no further reaction takes place whatever be the amount of the other reactant present. Hence, the reactant which gets consumed, limits the amount of product formed and is therefore called the limiting reagent.
In performing stoichiometric calculations, this aspect is also to be kept in mind
Example 56
How much magnesium sulphide can be obtained from 2.00 g of magnesium and 2.00 g of sulphur by the reaction Mg + S → MgS? Which is the limiting reagent? Calculate the amount of one of the reactants which remains un reacted.
Solution
First of all each of the masses are expressed in moles:
From the equation, Mg + S → MgS, it follows that one mole of Mg reacts with one mole of S. We are given more moles of Mg than of S therefore, Mg is in excess and some of it will remain un reacted when the reaction is over. S is the limiting reagent and will control the amount of product. From the equation we note that one mole of S gives one mole of MgS, so 0.0624 mole of S will react with 0.0624 mole of Mg to form 0.0624 mole of MgS
.Molar mass of MgS = 56.4 g
∴ Mass of MgS formed= 0.0624 × 56.4 g
= 3.52 g of MgS
Mole of Mg left un reacted= 0.0824 – 0.0624 moles of Mg
= 0.0200 moles of Mg
Mass of Mg left un reacted = moles of Mg × molar mass of Mg
= 0.0200 × 24.3 g of Mg = 0.486 g of Mg
Example 57
50.0 kg of N2(g) and 10.0 kg of H2(g) are mixed to produce NH3(g). Calculate the NH3(g) formed. Identify the limiting reagent in the production of NH3 in this situation.
Solution
A balanced equation for the above reaction is written as follows:
Calculation of moles:
According to the above equation, 1 mol N2(g) requires 3 mol H2(g), for the reaction. Hence, for 17.86 × 102 mol of N2, the moles of H2(g) required would be
But we have only 4.96 × 103 mol H2. Hence, hydrogen is the limiting reagent in this case. So NH3(g) would be formed only from that amount of available hydrogen, i.e.,4.96 × 103 mol
Since 3 mol H2(g) gives 2 mol NH3(g)
If they are to be converted to grams, it is done as follows:
Example 58
Copper reacts with silver nitrate solution according to the equation
Cu(s) + 2AgNO3(aq)→ Cu(NO3)2(aq) + 2Ag(s)
If 0.50 mol of copper is added to 1.5 mol of silver nitrate, which is the limiting reagent and how many moles of silver are formed?
Solution
Decide which is the limiting reagent.According to the equation, 1 mol Cu = 2 mol AgNO3, so 0.50 mol Cu = 2 × 0.50 = 1.0 mol AgNO3 but there are 0.50 mol Cu and 1.5 mol AgNO3. Therefore, AgNO3 is present in excess and Cu is the limiting reagent.Calculate how many moles of silver are formed.
Use the amount of the limiting reagent to find the amount of product. According to the equation, 1 mol Cu = 2 mol Ag. Therefore, 0.50 mol Cu ≡ 2 × 0.50 = 1.0 mol Ag.So, 1.0 mol Ag is formed.
EXERCISE 7.9
1. Identify the limiting reactant in the reaction of hydrogen and oxygen to form water if 61.0 g of O2 and 8.40 g of H2 are combined. Determine the amount (in grams) of excess reactant that remains after the reaction is complete.
2. Calculate the limiting reactant.Zinc metal reacts with hydrochloric acid by the following reaction:
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
If 0.30 mol Zn is added to hydrochloric acid containing 0.52 mol HCl, how many moles of H2 are produced?
7.12 THE GASEOUS STATE
This is the simplest state of matter. Throughout our life, we remain immersed in the ocean of air which is a mixture of gases. We spend our life in the lowermost layer of the atmosphere called troposphere, which is held to the surface of the earth by gravitational force. The thin layer of atmosphere is vital to our life. It shields us from harmful radiations and contains substances like oxygen, nitrogen, carbon dioxide, water vapor, etc.
Let us now focus our attention on the behavior of substances which exist in the gaseous state under normal conditions of temperature and pressure. A look at the periodic table shows that only eleven elements exist as gases under normal conditions (Figure 7.5).
The gaseous state is characterized by the following physical properties.
•Gases are highly compressible.
•Gases exert pressure equally in all directions.
•Gases have much lower density than the solids and liquids.
•The volume and the shape of gases are not fixed.
These assume volume and shape of the container.
•Gases mix evenly and completely in all proportions without any mechanical aid.
Simplicity of gases is due to the fact that the forces of interaction between their molecules are negligible. Their behavior is governed by same general laws, which were discovered as a result of their experimental studies. These laws are relationships between measurable properties of gases. Some of these properties like pressure, volume, temperature and mass are very important because relationships between these variables describe state of the gas. Interdependence of these variables leads to the formulation of gas laws.
7.13 THE GAS LAWS
The gas laws which we will study now are the result of research carried on for several centuries on the physical properties of gases. The first reliable measurement on properties of gases was made by Anglo-Irish scientist Robert Boyle in 1662. The law which he formulated is known as Boyle’s Law. Later on, attempts to fly in air with the help of hot air balloons motivated Jacques Charles and Joseph Lewis Gay Lussac to discover additional gas laws. Contribution from Avogadro and others provided lot of information about gaseous state.
STANDARD TEMPERATURE AND PRESSURE (S.T.P.)
Since volume of a given mass of a gas depends on temperature and pressure both, it is necessary to specify the values of p and T when the value of V is stated.In general, the comparison of the volumes of different gases is made with reference to standard temperature and pressure.The standard temperature is taken as 0°C (or 273.15 K).
The standard pressure according to the latest recommendations is taken as 1 bar (or 105 pascal). The molar volume of ideal gas at S.T.P. conditions is 22.71098 L mol–1 (≈ 22.7 L mol–1). These conditions are quite often abbreviated as S.T.P. meaning standard temperature and pressure.It is worth noting that previous S.T.P. conditions (0°C or 273.15 K temperature and 1 atm (= 1.01325 bar) pressure are still used in many books quite often. At these conditions, the molar volume of ideal gas is 22.413996 L mol–1 (≈ 22.4 L mol–1).
It may not be out of place to mention here, that Standard Ambient Temperature and Pressure (SATP) conditions are also used in some scientific data. The SATP conditions are 298.15 K (25°C)and 1 bar (105 Pa) pressure the molar volume of ideal gas at SATP conditions is24.789 L mol–1.
7.13.1 Gay Lussac Law
ACTIVITY 7.5: Illustrating Gay Lussac Law
•Take a balloon.
•Fill the balloon with air.
•Put the air filled balloon in a hot summer day outside your classroom.
•Observe the balloon.
In Activity 7.5, you will observe that after some time the inflated balloon will burst. The balloon bursts because the pressure inside the balloon increases due to high temperature.Gay Lussac’s Law states that when gases react, they do so in volumes which bear a simple ratio to one another provided temperature and pressure are kept constant.
For example,2H2(g) + O2(g) → 2H2O(l) Here, two volumes of hydrogen combine with 1 volume of oxygen, to form two volumes of water. H2 and O2 combine in simple ratio(2 : 1) to form 2 volumes of water.
7.13.2 Charles’ Law
ACTIVITY 7.6: Illustrating Charles’ Law
•Step 1: Fill a balloon with air.
•Step 2: Fit the balloon between two steady objects such as stacks of books so that it just barely touches the objects on each side. Keep the objects exactly where they are. Don’t move the objects.
•Step 3: Place the balloon in a freezer or ice box for thirty minutes to cool it.
•Step 4: Remove the balloon from the freezer or ice box.
•Step 5: Replace the balloon between the two objects of Step 2. Don’t move the objects.What did you observe?In this activity, you will observe that there should be extra space between the balloon and the two objects. The volume of the balloon has decreased because the air in the balloon is colder than it was initially.
Volume-temperature relationship
A French chemist J A C Charles made certain observations on the effect of temperature on the volumes of gases. He established the fact that a sample of any gas at 0°C would expand by
of its volume for each degree it was heated, and would contract by of its volume for each degree it was cooled. This was true for all gases.Based on Charles’ results Lord Kelvin, a Scottish physicist, reasoned that if gases lost of their volume at 0°C for every degree they were cooled, there must be a temperature of – 273°C where gases had no volume at all.He called – 273°C absolute zero at which molecules would have lost all their energy. As a result of this, a new scale of temperature, now known as Kelvin scale or absolute scale was developed. In this scale, 0 K corresponds to –273°C. Centigrade temperature is converted into Kelvin scale by adding 273 to the temperature in centigrade.
Thus, T = t + 273where T = Temperature in Kelvin scale or the absolute scale
t = Temperature in centigrade scale
The relationship of volume and temperature of a gas is then expressed in absolute scale. This is known as Charles’ law, which states:
Pressure remaining constant, the volume of a given mass of gas is directly proportional to its absolute temperature.All gases obey Charles’ law at very low pressures and high temperatures.Let the volume of a given mass of gas be V at the absolute temperature T,
then V ∝ T (pressure remaining constant)
The value of k depends on the mass and the nature of gas. Now, let the volumes of a given gas be V1 and V2 at the absolute temperatures T1 and T2 respectively, pressure remaining constant, then
Thus, using the absolute scale, we can now calculate the change in volume of a sample of a gas for any change in temperature. If the temperature is lowered, the volume will decrease; if the temperature is raised, the volume will increase.
Explanation of Charles’ Law by Kinetic Theory
On heating the gas, the kinetic energy of molecules increases. This means the molecules will move faster. Hence, the gas will expand provided pressure remains constant.
Example 59
What will be the volume of a gas at 0°C which occupies 200 mL at 27°C? (assume no change in pressure).
Solution
First, change the temperature to absolute scale.0°C = (0 + 273) K = 273 K
(i.e., T2)27°C = (27 + 273) K = 300 K (i.e., T1)V1 = 200 ml, V2 = ?By Charles’ law,
7.13.3 Boyle’s Law
ACTIVITY 7.7: Illustrating Boyle’s Law
Take a J shape tube as shown in Figure. Tube is partially filled with mercury. Pressure is now increased by putting more mercury into the open limb. The volume of air enclosed in the space above mercury in shorter limb is noted each time. It is found that as pressure increases, the volume of enclosed air gradually decreases from V1to V2.
Volume-pressure Relationship
In 1660, Robert Boyle formulated a quantitative relationship between pressure and volume of a given mass of air at constant temperature.
Boyle’s law states:The temperature remaining constant, the volume of a given mass of gas is inversely proportional to the pressure applied to it.This means, if the original pressure is increased five times, the original volume is reduced to its one-fifth.
The figure given above shows a sample of gas enclosed in a cylinder fitted with a moving piston. The pressure is increased by increasing the weight on the piston. With increasing pressure, the volume is reduced. When the weight is double, the new volume is one-half the original volume, and so on.
Mathematical Derivation of the Relationship
Let the given mass of gas occupies volume V at pressure P.
Then, according to Boyle’s law,
(temperature remaining constant)or, PV = k (a constant)The value of k depends on the nature and mass of gas.Now, let the volumes occupied by the same amount of gas be V1 and V at P1 and P2 respectively, then at constant temperature
P1V1 = k
P2V2 = k
or, P1V1 = P2V2 = k
or, P1V1 = P2V2
Boyle’s law is also manifested in the working of many devices that we use in daily life such as Tyre pressure gauze, aneroid barometer and cycle pump, etc.
Boyle’s law can also help to deduce relationship between density and pressure of the gas
which shows that pressure of fixed mass of a gas is directly proportional to density.
Explanation of Boyle’s Law of Kinetic Theory
•According to the kinetic theory of gases, the pressure exerted by a gas results from the combined bombardment of its molecules on the walls of the container.
•The number of bombardments depends on the concentration of molecules at a certain temperature.
•With the volume is reduced to one-fifth, the concentration of molecules is increased five times and hence the pressure is increased five times.
•Thus, the pressure of the gas becomes inversely proportional to the volume of a gas at constant temperature.Under ordinary conditions, Boyle’s law is entirely satisfactory. But at very low temperatures or under high pressures, deviations may occur. This is because the actual molecules are taking up a large part of the total volume of the gas at high pressures or at very low temperatures. Correction factors have been suggested to Boyle’s law. A gas which obeys Boyle’s law is said to show ideal behavior.
Example 60
A certain mass of a gas occupies 48 ml, at a pressure of 720 mm of Hg. What is the volume when the pressure is increased to 960 mm of Hg? (temperature remains constant).
Solution
V1 = 48 ml, V2 = ? P1 = 720 mm, P2 = 960 mm of Hg
By Boyle’s law, P1V1 = P2V2
Example 61
A gas occupies 1200 liters at 2 atm pressure. To what pressure must it be compressed to occupy 60 liters at the same temperature?
Solution
V1 = 1200 L, V2 = 60 L P1 = 2 atm, P2 = ?By Boyle’s law, P1V1 = P2V
The Gas Equation
Boyle’s law and Charles’ law can be combined to form one single relation. The equation is called combined gas law or general gas equation.It can be deduced as follows:
Then, according to mathematical law,
the volume, pressure and temperature of a given sample of gas at one set of conditions to those under any other set of conditions, we may simply write
where P1, V1 and T1; P2, V2 and T2; P3, V3and T3 respectively represent the pressure, volume and temperature values under given set of conditions.
Example 62
A gas measures 80 ml at 2.5 atm pressure and a temperature of 27°C. Calculate its volume at standard conditions of temperature and pressure.
Solution
P1 = 2.5 atm, P2 = 1 atm, V1 = 80 ml, V1 = ? T1 = (27 + 273) = 300 KT2 = 272 K
From the gas equation,
Example 63
The volume of a gas is 150 ml. at 17°C and 700 mm of Hg. What will be its volume at normal temperature pressure?
Solution
P1 = 700 mm of Hg, P2 = 760 mm of Hg V1 = 150 ml, initial, V2 = ?T1 = (17 + 273) = 290 K, T2 = 273KFrom the gas equation,
7.13.4 Avogadro’s Law—Volume Amount Relationship
This law describes the volume-amount relationship of gases at constant temperature and pressure. It was given by Amedeo Avogadro in 1811. It states that equal volumes of all the gases under similar conditions of temperature and pressure contain equal number of molecules.
For example, 1 mol of all the gases contains 6.023 × 1023 molecules. At the same time 1 mol of all the gases at 273.15 K (0°C) and 1 bar pressure occupy a volume of 22.7 l (22.7 × 10–3 m3). This means that as long as the temperature and pressure remain constant, the volume of the gas is directly proportional to the number of molecules. In other words, the amount of the gas molecules is constant. Mathematically, we can write V ∝ N (T and p are constant)
The number of molecules N is directly proportional to number of moles
∴ V ∝n(T and p are constant)or V = k4nk4 is constant of proportionality
Now, if m is the mass of the gas having molar mass equal to M, then the number of moles
are given as
where d is density of gas.
The above relationship implies that density of the gas at a given temperature and pressure is directly proportional to its molar mass.
7.13.5 Ideal Gas Equation
A gas that follows Boyle’s law, Charles’ law and Avogadro’s law strictly at all conditions, is called Ideal gas.It is assumed that inter-molecular forces are not present between the molecules of ideal gas.
The combination of various gas laws namely; Boyle’s law, Charles’ law and Avogadro’s law leads to the development of the mathematical relation which relates four variables pressure, volume, absolute temperature and number of moles of ideal gas. The equation so formulated is called ideal gas equation.Let us solve some numerical problems based on gas laws.
Note: In many of these problems, the conversion of temperature in Celsius scale to kelvin scale has been done by adding 273 instead of 273.15 in order to make the calculations simple.
Example 64
A sample of a gas is found to occupy a volume of 900 cm3 at 27°C. Calculate the temperature at which it will occupy a volume of 300 cm3, provided the pressure is kept constant.
Solution
Here, V1 = 900 cm3 V2 = 300 cm3T1 = (27 + 273) K = 300 K, T2 = ?
Example 65
It is desired to increase the volume of 80 cm3 of a gas by 20% without changing the pressure. To what temperature the gas be heated if its initial temperature is 25°C?
Solution
The desired increase in the volume of gas= 20% of 80 cm3
× 20 = 16 cm3Thus, the final volume of the gas = 80 + 16 = 96 cm3Now, V1 = 80 cm3 V2 = 96 cm3 T1 = 25°C = 298 K T2 = ?
Example 66
An iron tank contains helium at a pressure of 2.5 atmospheres at 25°C. The tank can withstand a maximum pressure of 10 atmospheres. The building in which tank has been placed catches fire. Predict whether, the tank will blow up first or melt. (The melting point of iron = 1535°C).
Solution
Let us proceed to calculate the pressure build up in the tank at the melting point of iron.Thus, p1 = 2.5 atm. p2 = ? T1 = 25°C = 298 K T2 = 1535°C = 1808 K
According to Charle’s law equation,
Since, pressure of the gas in the tank is much more than 10 atm at the melting point. Thus, the tank will blow up before reaching the melting point.
Example 67
When a ship is sailing in Pacific ocean where temperature is 23.4°C, a balloon is filled with 2.0 l of air. What will be the volume of the balloon when the ship reaches Indian ocean, where temperature is 26.1°C.
Solution
T1 = 23.4°C = 23.4 + 273.15 = 296.55 K
T2 = 26.1°C = 26.1 + 273.15 = 299.25 K
V1 = 2.0 l; V2 = ?Applying Charles’ law
Example 68
On a ship sailing in pacific ocean where temperature is 23.4°C, a balloon is filled with 2 l air. What will be the volume of the balloon when the ship reaches Indian ocean, where temperature is 26.1°C?
Solution
V1 = 2 l T2 = 26.1 + 273 T1 = (23.4 + 273) K = 299.1 K= 296.4 K
From Charles’ Law
Example 69
In a J tube partially filled with mercury the volume of air column is 4.2 ml and the mercury level in the two limbs is same. Some mercury is, now added to the tube so that the volume of air enclosed in shorter limb is now 2.8 ml. What is the difference in the levels of mercury in this situation. Atmospheric pressure is reported to be 1.0 bar.
Solution
Initial pressure p1 = 1 bar = 0.987 atm= 0.987 × 760 = 750.12 mm Hg
Final Pressure p2 = (p1 + h) V1 = 4.2 ml; V2 = 2.8 ml
Example 70
A balloon is filled with hydrogen at room temperature. It will burst if pressure exceeds 0.2 bar. If at 1 bar pressure the gas occupies 0.27 l Volume; up to what volume can the balloon be expanded by filling H2?
Solution
Here, p1 = 1 bar, p2 = 0.2 bar V1 = 0.27 l, V2 = ?
From Boyle’s law equation:
Since the balloon bursts at 0.2 bar pressure. Hence, the volume of balloon should remain less than 1.35 l
Example 71
A balloon is filled with hydrogen at room temperature. It will burst if pressure exceeds 0.2 bar. If at 1 bar pressure the gas occupies 2.27 l volume, up to what volume can the balloon be expanded?
Solution
According to Boyle’s law, p1V1 = p2V2 If p1 is 1 bar, V1 will be 2.27 l
Since balloon bursts at 0.2 bar pressure, the volume of balloon should be less than 11.35 l.
Type I. Solved Problems Based on Combined Gas Law:
Example 72
A sample of nitrogen gas occupies a volume of 320 cm3 at S.T.P. Calculate its volume at 66°C and 0.825 bar pressure.
Solution
Here, p1 = 1.00 bar
p2 = 0.825 bar
V1 = 320 cm3
V2 = ?
T1 = 273.15 K
T2 = 66°C = (66 + 273.15)
K = 339.15 K
Example 73
1.0 mol of pure dinitrogen gas at SATP conditions was put into a vessel of volume 0.025 m3, maintained at the temperature of 50°C. What is the pressure of the gas in the vessel?
Solution
Volume of 1.0 mol of a gas at SATP (V1) = 24.8 × 10–3 m3Initial condition (SATP) Final conditionV1 = 24.8 × 10–3m3V2 = 0.025 m3p1 = 1 barp2 = ?T1 = 298.15 KT2 = 50°C = 323.15 K
According to general gas equation,
Example 74
At 25°C and 760 mm of Hg pressure, a gas occupies 600 ml volume. What will be its pressure at a height where temperature is 10°C and volume of the gas is 640 ml.
Solution
p1 = 760 mm Hg,
V1 = 600 ml T1 = 25 + 273 = 298 K
V2 = 640 mland T2 = 10 + 273 = 283 K
According to Combined gas law,
Type II. Solved Problems Involving Molar Mass and Density
Example 75
The density of certain gaseous oxide at 1.5 bar pressure at 10°C is same as that of dioxygen at 20°C and 4.5 bar pressure. Calculate the molar mass of gaseous oxide.
Solution
Density of dioxygen (O2) at 4.5 bar pressure and 10°C
Type III. Solved Problems Involving Mass-Volume Relationship between Reactants and Products
Example 76
Isobutane (C4H10 ) undergoes combustion in oxygen according to the reaction :
2C4H10(g) + 13O2(g) →8CO2(g) + 10 H2O(l)When 10.00 L of isobutane is burnt at 27°C and 1 bar pressure, what volume of CO2 is produced at 80°C and 1.5 bar pressure.
Solution
Step I.Calculation of volume of CO2 at 27 ° C and 1 bar.From the chemical equation it is clear that2 l of C4H10 produce CO2 = 8 l(Similar condition of T and P)
Step II.Conversion of volume of CO2 at 80°C and 1.5 bar.Here, p1 = 1 bar p2 = 1.5 bar
V1 =40 l V2 = ? T1 = 300 K T2 = 80°C = 80 + 273 = 353 K
Using the gas equation,
7.13.6 Graham’s Law of Diffusion
ACTIVITY 7.8: To Study the Diffusion of Two Gases
1.Clamp a long glass tube about (1 m × 25 mm) dimensions in a horizontal position as shown in figure. Fix long strips of blue and red litmus papers with the help of a cello-tape. Insert cotton wool plugs at both ends.
2.With the help of a medicine dropper place 5 drops of concentrated hydrochloric acid on one cotton plug.
At the same time another student must place 5 drops of concentrated ammonia on the cotton plug at the other end. As quickly as possible start the stopwatch and cork at both ends of the tube.
3.When white ring of smoke is formed, stop the stopwatch and record the time. Mark the point where the ring appears first and measure the distance of this point from both ends.
Make the following record of observations
(a) Initial time (t1) when drops are introduced ......
(b) Final time (t2) when the white ring appears first ......
(c) Time taken for diffusion ‘t’ = (t2 – t1)(d) Distance traveled by HCl (x1) =
(e) Distance traveled by NH3 (x2) =
(f) Rate of diffusion of HCl (r1) = x1/t
(g) Rate of diffusion of NH3 (r2) = x2/t
Find the ratio r2/r1
(i) Compare this ratio to another ratio
(j) Verify Graham’s Law.
Thomas Graham put forward a generalization after studying the rates of diffusion of different gases, which is known after his name as Graham’s law of diffusion. The law states:
Under similar conditions of temperature and pressure, the rates of diffusion of gases are inversely proportional to the square roots of their densities.
The law can mathematically be put as
where r is the rate of diffusion and d is the density of the gas.
Now, if there are two gases A and B having r1 and r2 as their rates of diffusion and d1and d2, and their densities respectively. Then
We know that molecular mass is twice the vapor density. Therefore, the above expression may be written as
where M1 and M2 are the molecular masses of the gases having densities d1 and d2respectively. Thus, Graham’s law may also be stated as:
Under similar conditions of temperature and pressure, the rates of diffusion of gases are inversely proportional to the square root of their molecular masses.
Again, the rate of diffusion of the gas is equal to the volume of the gas which diffuses per unit time, i.e.,Rate of diffusion
If V1 and V2 are the volumes of the gases diffusing in the time t1 and t2 respectively under similar conditions, then
Thus, putting the values of r1 and r2 we arrive at the following formula:
Now, if V1 = V2 = V, we get
This implies that time taken for the diffusion of equal volumes of two gases under similar conditions of temperature and pressure is directly proportional to the square root of their densities or molecular masses.
Similarly, if t1 = t2 = t, then
It means that volumes of the two gases which diffuse in the same time under similar conditions are inversely proportional to the square roots of their densities or molecular masses.
Let us now apply Graham’s law to solve some numerical problems.
Example 77
Compare the rates of diffusion of 235UF6 and 238UF6.
Solution
Let the rate of diffusion of 235UF6 be r1 and rate of diffusion of 238UF6 be r2
Now, molecular mass of 235UF6(M1)= 235 + 6 × 19 = 349
Molecular mass of 238UF6(M2)= 238 + 6 × 19 = 35
Example 78
Relative densities of oxygen and carbon (IV) oxide are 16 and 22 respectively. If 25 cm3 of carbon (IV) oxide diffuses out in 75 seconds, what volume of oxygen will diffuse out in 96 seconds under similar conditions?
Solution
Here, volume of carbon (IV) oxide VCO2 = 25 cm3
Time taken tCO2 = 75 seconds
Let the volume of oxygen diffused VO2= V cm3
Time taken tO2= 96 seconds
Relative density of carbon (IV) oxide,dCO2= 22
Relative density of oxygen dO2 = 16 seconds
Thus, volume of oxygen diffused = 37.5 cm3
Example 79
A certain gaseous fluoride of phosphorus has a formula PFx. Under similar conditions, fluorine diffuses 1.82 times faster than the gaseous phosphorus fluoride. Find the value of x and formula of the phosphorus fluoride. (Atomic masses, P = 31.0; F = 19.0).
Solution
According to Graham’s law
Molecular mass of PFx from formula= 31 + 19x ...(ii)
Equate (i) and (ii)
31 + 19x = 125.8
∴ The formula of phosphorus fluoride =PF5
7.14 CALCULATION OF MOLAR GAS VOLUME UNDER STANDARD CONDITIONS
Example 80
Calculate number of moles in each of the following:
(i) 11 g of CO2
(ii) 3.01 × 1022 molecules of CO2
(iii) 1.12 litre of CO2 at S.T.P.
Solution
(i) Molecular mass of CO2= (12 + 2 × 16) = 44 u44 g of CO2 = 1 mole of CO2
(ii) 6.02 × 1023 molecules of CO2= 1 mole of CO2
3.01 × 1022 molecules of CO2
(iii) 22.4 liters of CO2 at S.T.P.= 1 mole of CO2
1.12 liters of CO2 at S.T.P.
Example 81
Find out volume of the following at S.T.P.:
(i) 14 g of nitrogen
(ii) 6.022 × 1022 molecules of ammonia (NH3)
(iii) 0.1 mole of sulphur dioxide (SO2).
Solution
(i) Molar mass of N2= 14 × 2 = 28 g mol–1
28 g of N2 occupy at S.T.P.= 22.4 liters
∴ 14 g of N2 occupy at S.T.P.
(ii) 6.022 × 1023 molecules of NH3 occupy at S.T.P.= 22.4 liters
∴ 6.022 × 1022 molecules of NH3occupy at S.T.P.
(iii) 1 mole of SO2 occupies at S.T.P. = 22.4 liters
∴ 0.1 mol of SO2 occupies at S.T.P.
EXERCISE 7.10
1. The relationship of volume and temperature of a gas is expressed in absolute scale. This is known as _________ .
2. Boyle’s law can help to deduce relationship between __________ and __________ of the gas.
3. What do you mean by ideal gas ?
4. The rates of diffusion of gases are inversely proportional to the square root of their densities; under similar conditions temperature and pressure.(True or False)
5. Under similar conditions of temperature and pressure, the rate of diffusion of gases are inversely proportional to the square root of their molecular masses. (True or False)
7.15 SUMMARY
•Avogadro’s Number is6.022×1023.
•The ratio of the average mass per atom of the naturally occurring form of an element to one-twelfth the mass of an atom of carbon-12 is known as relative atomic mass.
•The ratio of the average mass of one molecule of an element or compound to one twelfth of the mass of an atom of carbon-12 is known as relative molecular mass.
•Formula mass is the sum of the atomic masses of atoms in a molecule.
•Molar mass is the mass of one mole of a substance, usually expressed in grams. The molar mass of a substance in grams is numerically equal to the atomic, molecular, or formula mass of the substance expressed in amu. Molar mass and Avogadro’s number can be used to inter convert among mass, moles, and number of particles (atoms, molecules, ions, formula units, etc.).
•The amount of substance present in a given volume of a solution is expressed in mol per volume.
•The empirical formula can be used to calculate per cent composition. The empirical formula and molar mass can be used to determine the molecular formula.
•A chemical formula that gives the total number of atoms of each element in each molecule of a substance is called molecular formula.
•The quantitative study of the reactants required or the products formed is called stoichiometry. Using stoichiometric calculations, the amounts of one or more reactant(s) required to produce a particular amount of product can be determined and vice versa.
•The limiting reactant is the reactant that is consumed completely in a chemical reaction. An excess reactant is the reactant that is not consumed completely. The maximum amount of product that can form depends on the amount of limiting reactant.
•Gay Lussac Law states that when gases react, they do so in volumes which bear a simple ratio to one another provided temperature and pressure are kept constant.
•Charles’ law is a relationship between volume and absolute temperature under isobaric condition. It states that volume of a fixed amount of gas is directly proportional to its absolute temperature.
•Boyle’s law states that under isothermal condition, pressure of a fixed amount of a gas is inversely proportional to its volume.
•Avogadro law states that equal volumes of all gases under same conditions of temperature and pressure contain equal number of molecules.
•Graham’s law states that the rate of diffusion of a gas is inversely proportional to the square root of its molecular mass.
7.16 GLOSSARY
•Diffusion:the spreading of something more widely.
•Limiting reactant: the reactant that is consumed completely in a chemical reaction.
•Mass: the amount of matter in a particle or object.
•Reactant: a substance that takes part in and undergoes change during a reaction.
•Stoichiometry: the calculation of mass.
7.17 UNIT ASSESSMENT
I. Multiple Choice Questions
1. Mole is the link between the number of ................... and ................... of atoms.
(a) Particles, mass (b) Electrons, mass (c) Protons, mass (d) Particle, protons
2. The number of particles present in 1 mole of any substance is
(a) 6.022 × 1023 (b) 60.22 × 1022 (c) 60.22 × 1022 (d) 6.022 × 1020
3. The molecular mass of nitric acid is
(a) 36 u (b) 48 u (c) 63 u (d) 84 u
4. The formula unit mass of NaCl is
(a) 55.8 u (b) 85.5 u (c) 58.5 u (d) 55.5 u
5. Empirical formula of H2O2 is
(a) H2O2 (b) H2O (c) HO (d) None of these
6. The mathematical relationship between pressure and temperature was given by
(a) Robert Boyle (b) Gay Laussac (c) Amedeo Avogadro (d) A.C. Charles
7. The universal gas constant is denoted by
(a) R (b) V (c) N (d) T
8. Charles’s law states that
(a) V ∝ T (b) V ∝ P (c) V ∝ 1/p (d) V ∝ 1/t
9. Graham’s law is the relation between
(a) Rate of diffusion and density (b) Rate of diffusion and volume
(c) Rate of diffusion and mass (d) Both (a) and (c)
II. Open Ended Questions
1. What does the term mole mean?
2. Describe relative atomic mass and relative molecular mass.
3. Give the relationship between number of moles, mass and molar mass.
4. Define empirical formula.
5. Give an example in which molecular formula is same as empirical formula.
6. Explain limiting reagent.
7. With the help of an activity; explain Gay Lussac law.
8. Illustrate Charles’ law and Boyle’s law.
9. Explain ideal gas equation.
10. State Graham’s law of diffusion.
III. Numerical
1. A vessel of 120 ml capacity contains a certain amount of gas at 35ºC and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 ml at 35ºC. What would be its pressure?
2. A sample of helium has volume of 520 ml at 100°C. Calculate the temperature at which the volume will become 260 ml. Assume that pressure is constant.
3. Calculate the density of ammonia (NH3) at 30°C and 5 bar pressure.
4. Calculate the volume occupied by 8.8 g of CO2 at 31.1°C and 1 bar pressure.(R = 0.083 bar dm3k–1 mol–1)
IV. Practical-based Questions
1. Complete the following representation.
2. The following figure depicts .................... law.
3. The following figure illustrates Charles’ law. (True or False)
4. How many burgers can be made? Which part is limiting reagent?
PROJECT
Carry out an experiment of burning magnesium ribbon in air in order to determine the per cent composition of magnesium in magnesium oxide, and write a report on the findings.