In the previous section, we saw some basics about nature of matter. In this section, we will see properties of matter and their measurements. Also later in this section we will see problems on molecular and atomic masses
Properties of substances can be classified into two categories:
(i) Physical properties
Color, odour, melting point, boiling point etc., are examples of physical properties
(ii) Chemical properties
Composition, combustibility, reactivity with acids and bases etc., are examples of chemical properties
■ Both physical properties and chemical properties of a substance can be measured or observed
But there is a difference:
• For measuring or observing physical properties of a substance, we do not need to change the identity or composition of that substance
• For measuring or observing chemical properties, a chemical change must occur. After the chemical change, the substance will be be transformed to another substance
This difference will be clear from the two examples given below:
Example 1:
• To find the boiling point of a liquid substance, we heat it
• The heat is supplied until the liquid starts to boil
• The 'temperature at which the boiling begins' is noted down as the boiling point
• Even after supplying the heat, the liquid remains the same. It is not transformed into another substance
Example 2:
• To find the combustibility of carbon, we must burn it
• Burning or combustion is a reaction with oxygen
• The carbon reacts with oxygen and becomes carbon dioxide
• So carbon is changed into a new substance
Let us see how the measurements are done:
• Properties like area, volume, density, etc., are quantitative in nature
• Any quantitative measurement or observation is represented by a number followed by a unit in which it is measured
• For example, height of a wall is 2 m
♦ ‘2’ is the number
♦ ‘m’ denotes meter, the unit in which length is measured
■ We use the SI system of units for measurements. A detailed description about the importance of the units and the system of units can be seen here. The SI system is also described in detail
■ Notes on measurements of mass and time can be seen here
■ Notes on measurements of heat can be seen here
■ Notes on accuracy and precision can be seen here
■ Notes on significant figures precision can be seen here
■ Notes on errors in measurements can be seen here
■ Notes on Dimensional analysis can be seen here
Properties of substances can be classified into two categories:
(i) Physical properties
Color, odour, melting point, boiling point etc., are examples of physical properties
(ii) Chemical properties
Composition, combustibility, reactivity with acids and bases etc., are examples of chemical properties
■ Both physical properties and chemical properties of a substance can be measured or observed
But there is a difference:
• For measuring or observing physical properties of a substance, we do not need to change the identity or composition of that substance
• For measuring or observing chemical properties, a chemical change must occur. After the chemical change, the substance will be be transformed to another substance
Example 1:
• To find the boiling point of a liquid substance, we heat it
• The heat is supplied until the liquid starts to boil
• The 'temperature at which the boiling begins' is noted down as the boiling point
• Even after supplying the heat, the liquid remains the same. It is not transformed into another substance
Example 2:
• To find the combustibility of carbon, we must burn it
• Burning or combustion is a reaction with oxygen
• The carbon reacts with oxygen and becomes carbon dioxide
• So carbon is changed into a new substance
Chemists study the behaviour of substances based on the physical and chemical properties. So measurements of these properties are very essential.
Let us see how the measurements are done:
• Properties like area, volume, density, etc., are quantitative in nature
• Any quantitative measurement or observation is represented by a number followed by a unit in which it is measured
• For example, height of a wall is 2 m
♦ ‘2’ is the number
♦ ‘m’ denotes meter, the unit in which length is measured
■ We use the SI system of units for measurements. A detailed description about the importance of the units and the system of units can be seen here. The SI system is also described in detail
■ Notes on measurements of mass and time can be seen here
■ Notes on measurements of heat can be seen here
■ Notes on accuracy and precision can be seen here
■ Notes on significant figures precision can be seen here
■ Notes on errors in measurements can be seen here
■ Notes on Dimensional analysis can be seen here
Molecular mass is the the sum of atomic masses of the elements present in a molecule. We have learnt about them in our previous classes. Detailed notes about atomic mass, molecular mass, mole concept and molarity can be seen here.
Let us see some solved examples:
Solved example 1.1
Calculate the molar mass of the following:
(i) H2O (ii) CO2 (iii) CH4
Solution:
Molar mass is the mass of one mole
(i) H2O
Mass of one mole of H2O
= (2 × 1.008) + 16.00 = 18.02 g
(ii) CO2
Mass of one mole of CO2
= 12.01 + (2 × 16.00) = 44.01 g
(ii) CH4
Mass of one mole of CH4
= 12.01 + (4 × 1.008) = 16.04 g
Solved example 1.2
Calculate the atomic mass (average) of chlorine using the following data:
In a sample of chlorine:
(i) 75.77% is the isotope $\mathbf\small{{^{35}C}l}$
(17 protons and 18 neutrons)
♦ It has a molar mass of 34.9689
(ii) 24.23% is the isotope $\mathbf\small{{^{37}C}l}$
(17 protons and 20 neutrons)
♦ It has a molar mass of 36.9659
Solution:
From the above data, we get:
Average atomic mass of chlorine = (34.9689 × 0.7577) + (36.9659 × 0.2423) = 35.4528 g mole-1
Solved example 1.3
Calculate the atomic mass (average) of argon using the following data:
In a sample of argon:
(i) 0.337% is the isotope $\mathbf\small{{^{36}Ar}}$
(18 protons and 18 neutrons)
♦ It has a molar mass of 35.96755
(ii) 0.063% is the isotope $\mathbf\small{{^{38}Ar}}$
(18 protons and 20 neutrons)
♦ It has a molar mass of 37.96272
(iii) 99.600% is the isotope $\mathbf\small{{^{40}Ar}}$
(18 protons and 22 neutrons)
♦ It has a molar mass of 39.9624
Solution:
From the above data, we get:
Average atomic mass of argon
= (35.96755 × 0.00337) + (37.96272 × 0.00063) + (39.9624 × 0.9960) = 39.9477 g mole-1
Solved example 1.4
How much copper can be obtained from 100 g of copper sulphate (CuSO4)?
Solution:
1. One mole of CuSO4 = 63.54 + 32.06 + (16.00 × 4) = 159.60 g
2. So no. of moles in 100 g = (100⁄159.60) = 0.6266 moles
3. So no. of molecules of CuSO4 in 100 g = (0.6266 × NA)
(Where NA is the Avogadro number)
4. So no. of atoms of Cu = (0.6266 × NA)
5. Mass of one mole of Cu = 63.54 g
• That is., NA atoms of Cu has a mass of 63.54 g
• So 1 atom of Cu has a mass of $\mathbf\small{\frac{63.54}{N_A}}$ g
6. So from (4), we get:
Total mass of the available Cu atoms = $\mathbf\small{0.6266 \times N_A \times \frac{63.54}{N_A}=0.6266 \times 63.54 = 39.81\,g}$
Solved example 1.5
In three moles of ethane (C2H6), calculate the following:
(i) Number of moles of carbon atoms
(ii) Number of moles of hydrogen atoms
(iii) Number of molecules of ethane
Solution:
Part (i):
1. One mole of C2H6 contains NA molecules of C2H6
• So 3 moles of C2H6 will contain (3NA) molecules of C2H6
2. In (3NA) molecules of C2H6, there will be (2 × 3NA) atoms of carbon
• That is., (6NA) atoms of carbon
3. NA atoms of carbon is one mole of carbon
• So 6NA atoms of carbon is 6 moles of carbon
Part (ii):
1. One mole of C2H6 contains NA molecules of C2H6
• So 3 moles of C2H6 will contain (3NA) molecules of C2H6
2. In (3NA) molecules of C2H6, there will be (6 × 3NA) atoms of hydrogen
• That is., (18NA) atoms of hydrogen
3. NA atoms of hydrogen is one mole of hydrogen
• So 18NA atoms of hydrogen is 18 moles of hydrogen
Part (iii):
• One mole of C2H6 contains NA molecules of C2H6
• So 3 moles of C2H6 will contain (3NA) molecules of C2H6
• That is., (3 × 6.02 × 1023) molecules of C2H6.
Solved example 1.6
Which one of the following will have the largest number of atoms?
(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g of Cl2(g)
Solution:
Part (i):
1 mole of Au has a mass of 196.97 g
⇒ 196.97 g of gold will contain NA atoms of gold
⇒ 1 g of gold will contain $\mathbf\small{\left(\frac{N_A}{196.97}\right)}$ atoms of gold
Part (ii):
1 mole of Na has a mass of 22.99 g
⇒ 22.99 g of sodium will contain NA atoms of sodium
⇒ 1 g of sodium will contain $\mathbf\small{\left(\frac{N_A}{22.99}\right)}$ atoms of sodium
Part (iii):
1 mole of Li has a mass of 6.94 g
⇒ 6.94 g of lithium will contain NA atoms of lithium
⇒ 1 g of lithium will contain $\mathbf\small{\left(\frac{N_A}{6.94}\right)}$ atoms of lithium
Part (iv):
• Cl2 is a molecule of chlorine
1 mole of chlorine molecules has a mass of (35.45 × 2) = 70.9 g
⇒ 70.9 g of chlorine will contain NA molecules of chlorine
⇒ 1 g of chlorine will contain $\mathbf\small{\left(\frac{N_A}{70.9}\right)}$ molecules of chlorine
⇒ 1 g of chlorine will contain $\mathbf\small{\left(\frac{2N_A}{70.9}\right)}$ atoms of chlorine
⇒ 1 g of chlorine will contain $\mathbf\small{\left(\frac{N_A}{35.45}\right)}$ atoms of chlorine
• All the four results have the same numerator
• So the result with the smallest denominator will be the largest
■ Thus we can write:
Among the given four samples, the lithium sample will have the largest number of atoms
Solved example 1.7
What will be the mass of one $\mathbf\small{{^{12}C}}$ atom in g?
Solution:
• Molar mass of $\mathbf\small{{^{12}C}}$ = 12 g
⇒ 1 mole of $\mathbf\small{{^{12}C}}$ has a mass of 12 g
⇒ NA atoms of $\mathbf\small{{^{12}C}}$ have a mass of 12 g
⇒ 1 atom of $\mathbf\small{{^{12}C}}$ has a mass of $\mathbf\small{\left(\frac{12}{N_A}\right)}$ g
• $\mathbf\small{\frac{12}{N_A}=\frac{12}{6.022 \times 10^{23}}=1.9927 \times 10^{-23}}$ g
Solved example 1.8
Calculate the number of atoms in each of the following:
(i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He
Solution:
Part (i):
• 1 mole of Ar atoms contains NA atoms of Ar
⇒ 52 moles of Ar will contain (52NA) = (52 × 6.022 × 1023) = 313.144 × 1023 = 3.13 × 1025 atoms
Part (ii):
1. 1 u = 1.6605 × 10-25 g
52 u = 52 × 1.6605 × 10-25 g
2. Molar mass of Ar is 39.95 g
⇒ 39.95 g of Ar will contain NA atoms
⇒1 g of Ar will contain $\mathbf\small{\left(\frac{N_A}{39.95}\right)}$ atoms
3. So (52 × 1.6605 × 10-25) g will contain $\mathbf\small{52 \times 1.6605 \times 10^{-25} \times \left(\frac{N_A}{39.95}\right)}$ atoms
⇒ 52 u of Ar will contain $\mathbf\small{52 \times 1.6605 \times 10^{-25} \times \left(\frac{6.022 \times 10^{23}}{39.95}\right)=13.02}$ atoms
• So 52 u of Ar will contain approximately 13 Ar atoms
Part (iii):
1. Molar mass of He is 4.00 g
⇒ 4.00 g of He contains NA atoms of He
2. 1 g of He contains $\mathbf\small{\left(\frac{N_A}{4}\right)}$ atoms of He
⇒ 52 g of He will contain:
$\mathbf\small{52 \times\left(\frac{N_A}{4}\right)=52 \times\left(\frac{6.022 \times 10^{23}}{4}\right)=7.8286 \times 10^{24}}$ atoms of He
Let us see some solved examples:
Solved example 1.1
Calculate the molar mass of the following:
(i) H2O (ii) CO2 (iii) CH4
Solution:
Molar mass is the mass of one mole
(i) H2O
Mass of one mole of H2O
= (2 × 1.008) + 16.00 = 18.02 g
(ii) CO2
Mass of one mole of CO2
= 12.01 + (2 × 16.00) = 44.01 g
(ii) CH4
Mass of one mole of CH4
= 12.01 + (4 × 1.008) = 16.04 g
Solved example 1.2
Calculate the atomic mass (average) of chlorine using the following data:
In a sample of chlorine:
(i) 75.77% is the isotope $\mathbf\small{{^{35}C}l}$
(17 protons and 18 neutrons)
♦ It has a molar mass of 34.9689
(ii) 24.23% is the isotope $\mathbf\small{{^{37}C}l}$
(17 protons and 20 neutrons)
♦ It has a molar mass of 36.9659
Solution:
From the above data, we get:
Average atomic mass of chlorine = (34.9689 × 0.7577) + (36.9659 × 0.2423) = 35.4528 g mole-1
Solved example 1.3
Calculate the atomic mass (average) of argon using the following data:
In a sample of argon:
(i) 0.337% is the isotope $\mathbf\small{{^{36}Ar}}$
(18 protons and 18 neutrons)
♦ It has a molar mass of 35.96755
(ii) 0.063% is the isotope $\mathbf\small{{^{38}Ar}}$
(18 protons and 20 neutrons)
♦ It has a molar mass of 37.96272
(iii) 99.600% is the isotope $\mathbf\small{{^{40}Ar}}$
(18 protons and 22 neutrons)
♦ It has a molar mass of 39.9624
Solution:
From the above data, we get:
Average atomic mass of argon
= (35.96755 × 0.00337) + (37.96272 × 0.00063) + (39.9624 × 0.9960) = 39.9477 g mole-1
Solved example 1.4
How much copper can be obtained from 100 g of copper sulphate (CuSO4)?
Solution:
1. One mole of CuSO4 = 63.54 + 32.06 + (16.00 × 4) = 159.60 g
2. So no. of moles in 100 g = (100⁄159.60) = 0.6266 moles
3. So no. of molecules of CuSO4 in 100 g = (0.6266 × NA)
(Where NA is the Avogadro number)
4. So no. of atoms of Cu = (0.6266 × NA)
5. Mass of one mole of Cu = 63.54 g
• That is., NA atoms of Cu has a mass of 63.54 g
• So 1 atom of Cu has a mass of $\mathbf\small{\frac{63.54}{N_A}}$ g
6. So from (4), we get:
Total mass of the available Cu atoms = $\mathbf\small{0.6266 \times N_A \times \frac{63.54}{N_A}=0.6266 \times 63.54 = 39.81\,g}$
Solved example 1.5
In three moles of ethane (C2H6), calculate the following:
(i) Number of moles of carbon atoms
(ii) Number of moles of hydrogen atoms
(iii) Number of molecules of ethane
Solution:
Part (i):
1. One mole of C2H6 contains NA molecules of C2H6
• So 3 moles of C2H6 will contain (3NA) molecules of C2H6
2. In (3NA) molecules of C2H6, there will be (2 × 3NA) atoms of carbon
• That is., (6NA) atoms of carbon
3. NA atoms of carbon is one mole of carbon
• So 6NA atoms of carbon is 6 moles of carbon
Part (ii):
1. One mole of C2H6 contains NA molecules of C2H6
• So 3 moles of C2H6 will contain (3NA) molecules of C2H6
2. In (3NA) molecules of C2H6, there will be (6 × 3NA) atoms of hydrogen
• That is., (18NA) atoms of hydrogen
3. NA atoms of hydrogen is one mole of hydrogen
• So 18NA atoms of hydrogen is 18 moles of hydrogen
Part (iii):
• One mole of C2H6 contains NA molecules of C2H6
• So 3 moles of C2H6 will contain (3NA) molecules of C2H6
• That is., (3 × 6.02 × 1023) molecules of C2H6.
Solved example 1.6
Which one of the following will have the largest number of atoms?
(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g of Cl2(g)
Solution:
Part (i):
1 mole of Au has a mass of 196.97 g
⇒ 196.97 g of gold will contain NA atoms of gold
⇒ 1 g of gold will contain $\mathbf\small{\left(\frac{N_A}{196.97}\right)}$ atoms of gold
Part (ii):
1 mole of Na has a mass of 22.99 g
⇒ 22.99 g of sodium will contain NA atoms of sodium
⇒ 1 g of sodium will contain $\mathbf\small{\left(\frac{N_A}{22.99}\right)}$ atoms of sodium
Part (iii):
1 mole of Li has a mass of 6.94 g
⇒ 6.94 g of lithium will contain NA atoms of lithium
⇒ 1 g of lithium will contain $\mathbf\small{\left(\frac{N_A}{6.94}\right)}$ atoms of lithium
Part (iv):
• Cl2 is a molecule of chlorine
1 mole of chlorine molecules has a mass of (35.45 × 2) = 70.9 g
⇒ 70.9 g of chlorine will contain NA molecules of chlorine
⇒ 1 g of chlorine will contain $\mathbf\small{\left(\frac{N_A}{70.9}\right)}$ molecules of chlorine
⇒ 1 g of chlorine will contain $\mathbf\small{\left(\frac{2N_A}{70.9}\right)}$ atoms of chlorine
⇒ 1 g of chlorine will contain $\mathbf\small{\left(\frac{N_A}{35.45}\right)}$ atoms of chlorine
• All the four results have the same numerator
• So the result with the smallest denominator will be the largest
■ Thus we can write:
Among the given four samples, the lithium sample will have the largest number of atoms
Solved example 1.7
What will be the mass of one $\mathbf\small{{^{12}C}}$ atom in g?
Solution:
• Molar mass of $\mathbf\small{{^{12}C}}$ = 12 g
⇒ 1 mole of $\mathbf\small{{^{12}C}}$ has a mass of 12 g
⇒ NA atoms of $\mathbf\small{{^{12}C}}$ have a mass of 12 g
⇒ 1 atom of $\mathbf\small{{^{12}C}}$ has a mass of $\mathbf\small{\left(\frac{12}{N_A}\right)}$ g
• $\mathbf\small{\frac{12}{N_A}=\frac{12}{6.022 \times 10^{23}}=1.9927 \times 10^{-23}}$ g
Solved example 1.8
Calculate the number of atoms in each of the following:
(i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He
Solution:
Part (i):
• 1 mole of Ar atoms contains NA atoms of Ar
⇒ 52 moles of Ar will contain (52NA) = (52 × 6.022 × 1023) = 313.144 × 1023 = 3.13 × 1025 atoms
Part (ii):
1. 1 u = 1.6605 × 10-25 g
52 u = 52 × 1.6605 × 10-25 g
2. Molar mass of Ar is 39.95 g
⇒ 39.95 g of Ar will contain NA atoms
⇒1 g of Ar will contain $\mathbf\small{\left(\frac{N_A}{39.95}\right)}$ atoms
3. So (52 × 1.6605 × 10-25) g will contain $\mathbf\small{52 \times 1.6605 \times 10^{-25} \times \left(\frac{N_A}{39.95}\right)}$ atoms
⇒ 52 u of Ar will contain $\mathbf\small{52 \times 1.6605 \times 10^{-25} \times \left(\frac{6.022 \times 10^{23}}{39.95}\right)=13.02}$ atoms
• So 52 u of Ar will contain approximately 13 Ar atoms
Part (iii):
1. Molar mass of He is 4.00 g
⇒ 4.00 g of He contains NA atoms of He
2. 1 g of He contains $\mathbf\small{\left(\frac{N_A}{4}\right)}$ atoms of He
⇒ 52 g of He will contain:
$\mathbf\small{52 \times\left(\frac{N_A}{4}\right)=52 \times\left(\frac{6.022 \times 10^{23}}{4}\right)=7.8286 \times 10^{24}}$ atoms of He
We will see more problems related to mole concept in the coming sections. In the next section, we will see percentage composition
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