In the previous section, we saw how to convert mass into 'number of moles'. We also saw how to convert a given number of atoms/molecules into 'number of moles'. In this section we will see the reverse. That is., conversion of 'number of moles' into mass.
■ First we will see the calculation of mass of atoms from no. of moles:
• We know that one mole of 'atoms of a particular element' will have a mass of GAM of that atom
• So we can write:
Eq.10.2:
Total mass of a sample of atoms = No. of moles in that sample × GAM
■ Now we will see the calculation of mass of molecules from no. of moles:
• We know that one mole of 'molecules of a particular compound' will have a mass of GMM of that molecule
• So we can write:
Eq.10.3:
Total mass of a sample of molecules = No. of moles in that sample × GMM
Solved example 10.15
Solved example 10.16
Solved example 10.18
Solved example 10.20
Calculate the following:
(a) Number of moles in 1 kg of water
(b) Number of moles in 500 g of CaCO3
(c) Number of moles and total number of atoms in 88 g of CO2
(GMM/GFM: H2O = 18 g, CaCO3 = 100 g, CO2 = 44 g, )
Solution:
(a) One GMM of H2O = 18 g
• So one mole of H2O has a mass of 18 g.
• In other words, if we take 18 grams of H2O molecules, it will contain 1 mole
• So, if we take 1 g of H2O molecules, it will contain 1⁄18 mole
• Thus, no. of moles in 1000 g = 1000 × 1⁄18 = 1000⁄18 moles
(b) One GMM of CaCO3 = 100 g
• So one mole of CaCO3 has a mass of 100 g.
• In other words, if we take 100 grams of CaCO3 molecules, it will contain 1 mole
• So, if we take 1 g of CaCO3 molecules, it will contain 1⁄100 mole
• Thus, no. of moles in 500 g = 500 × 1⁄100 = 500⁄100 = 5 moles
(c) One GMM of CO2 = 44 g
• So one mole of CO2 has a mass of 44 g.
• In other words, if we take 44 grams of CO2 molecules, it will contain 1 mole
• So, if we take 1 g of CO2 molecules, it will contain 1⁄44 mole
• Thus, no. of moles in 88 g = 88 × 1⁄44 = 88⁄44 = 2 moles
• So no. of molecules = 2 × 6.022×1023
• Each of these molecules have 3 atoms. (1 atom of carbon and 2 atoms of oxygen)
• So no. of atoms = 3 × 2 × 6.022×1023
In the next section we will learn about STP
■ First we will see the calculation of mass of atoms from no. of moles:
• We know that one mole of 'atoms of a particular element' will have a mass of GAM of that atom
• So we can write:
Eq.10.2:
Total mass of a sample of atoms = No. of moles in that sample × GAM
■ Now we will see the calculation of mass of molecules from no. of moles:
• We know that one mole of 'molecules of a particular compound' will have a mass of GMM of that molecule
• So we can write:
Eq.10.3:
Total mass of a sample of molecules = No. of moles in that sample × GMM
Now we will see some solved examples:
Solved example 10.9
Given that GMM of water is 18 g. Find the number of molecules in 180 grams of water. Also find the total number of atoms in this 180 g of water
Solution:
1. One mole of 'molecules of a particular compound' will consist of NA molecules of that compound
2. And it will have a mass of GMM of that molecule
3. So one GMM of water will have NA molecules
4. No. of GMMs in 180 g = 180⁄18 =10
5. Thus total no. of molecules = 10 × NA = 10 × 6.022×1023.
6. Each of these molecules will have two hydrogen atoms and one oxygen atom
7. So total number of hydrogen atoms = 2 × 10 × 6.022×1023
8. Total number of oxygen atoms = 1 × 10 × 6.022×1023.
Solved example 10.10
What will be the mass (in grams) of 10,00,000 molecules of water
Solution:
Solution:
1. One mole of 'molecules of a particular compound' will consist of NA molecules of that compound
2. And it will have a mass of GMM of that molecule
3. So one GMM of water will have NA molecules
4. One GMM of water = 18 grams.
5. So mass of one mole of water = 18 grams.
6. That is., mass of NA molecules = 18 grams
That is., mass of 6.022×1023 molecules = 18 grams
7. So mass of one molecule = 18⁄(6.022 × 1023) grams
8. Thus mass of 10,00,000 molecules
= 18⁄(6.022 × 1023) × 1000000
= 180000000⁄(6.022 × 1023) grams
4. One GMM of water = 18 grams.
5. So mass of one mole of water = 18 grams.
6. That is., mass of NA molecules = 18 grams
That is., mass of 6.022×1023 molecules = 18 grams
7. So mass of one molecule = 18⁄(6.022 × 1023) grams
8. Thus mass of 10,00,000 molecules
= 18⁄(6.022 × 1023) × 1000000
= 180000000⁄(6.022 × 1023) grams
Solved example 10.11
How many molecules are present in 710 g of chlorine gas (Cl2)? What will be the total number of atoms in it?
Solution:
1. One mole of Cl2 will contain NA molecules.
2. That is., one GMM of Cl2 (= 2×35.5 = 71 grams) will contain NA molecules
3. No. of GMMs in 710 grams = 710⁄71 = 10
4. So no. of molecules = 10×NA = 10×6.022×1023
5. Each Cl2 molecule will contain two atoms. So no. of atoms = 2×10×6.022×1023
1. One mole of Cl2 will contain NA molecules.
2. That is., one GMM of Cl2 (= 2×35.5 = 71 grams) will contain NA molecules
3. No. of GMMs in 710 grams = 710⁄71 = 10
4. So no. of molecules = 10×NA = 10×6.022×1023
5. Each Cl2 molecule will contain two atoms. So no. of atoms = 2×10×6.022×1023
Solved example 10.12
How many molecules are present in 90 g of glucose? What will be the total number of atoms in it?
Solution:
1. One mole of glucose (C6H12O6) will contain NA molecules.
2. That is., one GMM of C6H12O6 [=(6×12)+(12×1)+(6×16) =180 grams] will contain NA molecules
3. No. of GMMs in 90 grams = 90⁄180 = 0.5
4. So no. of molecules = 0.5×NA = 0.5×6.022×1023
5. Each C6 H12O6 molecule will contain (6+12+6 = 24) atoms.
So no. of atoms = 24×0.5×6.022×1023
1. One mole of glucose (C6H12O6) will contain NA molecules.
2. That is., one GMM of C6H12O6 [=(6×12)+(12×1)+(6×16) =180 grams] will contain NA molecules
3. No. of GMMs in 90 grams = 90⁄180 = 0.5
4. So no. of molecules = 0.5×NA = 0.5×6.022×1023
5. Each C6 H12O6 molecule will contain (6+12+6 = 24) atoms.
So no. of atoms = 24×0.5×6.022×1023
Solved example 10.13
Calculate the mass of each of the following in grams:
(i) One nitrogen atom (ii) One nitrogen molecule (iii) One sulphuric acid molecule
Given GAMs: H = 1 g, C = 12 g, S = 32 g, O = 16 g, Cl = 35.5 g, N = 14 g
Solution of part (i):
1. One mole of nitrogen atoms will contain NA atoms.
2. That is., one GAM (= 14 g) of N will contain NA atoms
3. So, mass of NA atoms of nitrogen is 14 grams
4. Thus, mass of one atom of nitrogen = 14⁄NA = 14⁄(6.022 × 1023) grams
Solution of part (ii):
1. One mole of nitrogen molecules will contain NA molecules.
2. That is., one GMM (= 28 g) of N will contain NA molecules
3. So, mass of NA molecules of nitrogen is 28 grams
4. Thus, mass of one molecule of nitrogen = 28⁄NA = 28⁄(6.022 × 1023) grams
Solution of part (iii):
1. One mole of sulphuric acid (H2SO4) molecules will contain NA molecules.
2. That is., one GMM [=(2×1)+(1×32)+(4×16) =98 grams] of H2SO4 will contain NA molecules
3. So, mass of NA molecules of H2SO4 is 98 grams
4. Thus, mass of one molecule of H2SO4 = 98⁄NA = 98⁄(6.022 × 1023) grams
Now we will see some solved examples based on what we have learned so far in this chapter
Solved example 10.14
Suppose 20 molecules of hydrogen are allowed to react with 20 molecules of oxygen to produce water.
(a) Which of the reactant molecules get consumed first?
(b) The molecules of which reactant will remain unreacted? How many?
Solution:
The balanced chemical equation for the reaction is:
2H2 + O2 → 2H2O.
1. For every one oxygen molecule, there must be two hydrogen molecules. So if there are 20 oxygen molecules, there must be 40 hydrogen molecules.
2. But we have only 20 hydrogen molecules. All of them will be used up by 10 oxygen molecules.
3. The remaining 10 oxygen molecules do not have any hydrogen molecules to react with. So we can write:
• The hydrogen molecules get consumed first
• Some molecules of oxygen will remain unreacted. Number of oxygen molecules which will remain unreacted is 10
Solved example 10.15
Which among the following is used as the basis of expressing atomic mass?
(Hydrogen, carbon-12, carbon-14, Oxygen-16)
Solved example 10.16
The atomic mass of helium is 4 and that of oxygen is 16. How many grams of helium is required to get as many number of atoms as are present in 40 grams of oxygen?
Solution:
■ Consider 'x' grams of helium. We want to satisfy the following condition:
The number of helium atoms in x grams of helium
= The number of oxygen atoms in 40 grams of oxygen
1. GAM of oxygen = 16 grams.
2. So 16 g of oxygen will contain NA atoms of oxygen
So number of atoms in 1 g of oxygen = NA⁄16
3. Thus number of atoms in 40 g of oxygen = NA⁄16×40 = 2.5NA
4. GAM of helium = 4 grams.
5. So 4 g of helium will contain NA atoms of helium
So number of atoms in 1 g of helium = NA⁄4
6. Thus number of atoms in x g of oxygen = NA⁄4 × x = 0.25xNA
7. (3) and (6) must be equal. So we can write: 2.5NA = 0.25xNA.
⇒ 2.5 = 0.25x ⇒ x = 2.5⁄0.25 ⇒ x = 10 g
Solved example 10.17
Find the gram atomic masses in each of the following
(a) 100 g of helium (b) 200 g of oxygen (c) 70 g of nitrogen (d) 1 g of calcium
(Atomic masses: He = 4, O = 16, N = 14, Ca = 40)
Solution:
(a) Atomic mass of helium = 4 u
So GAM of helium = 4 g
No. of GAMs in 100 g of helium = 100⁄4 = 25
(b) Atomic mass of oxygen = 16 u
So GAM of oxygen = 16 g
No. of GAMs in 100 g of oxygen = 200⁄16 = 12.5
(c) Atomic mass of nitrogen = 14 u
So GAM of nitrogen = 14 g
No. of GAMs in 100 g of nitrogen = 70⁄14 = 5
(d) Atomic mass of calcium = 40 u
So GAM of calcium = 40 g
No. of GAMs in 1 g of calcium = 1⁄40 = 0.025
Solved example 10.18
Calculate the gram molecular mass / gram formula mass of the following:
(a) HNO3 (b) CaCl2 (c) Na2SO4 (d) NH4NO3
(Gram atomic masses: H = 1 g, N = 14 g, O = 16 g, Na = 23 g, S = 32 g, Cl = 35.5 g, Ca = 40 g)
Solution:
(a) GAM of HNO3 = (1×1) + (1×14) + (3×16) = 1 + 14 + 48 = 63 g
(b) GAM of CaCl2 = (1×40) + (2×35.5) = 40 + 71 = 111 g
(c) GAM of Na2SO4 = (2×23) + (1×32) + (4×16) = 46 + 32 + 64 = 142 g
(d) GAM of NH4NO3 = (1×14) + (4×1) + (1×14) + (3×16) = 14 + 4 + 14 + 48 = 80 g
Solved example 10.19
Given below are few samples:
(a) 400 g of water (H2O) (b) 400 g of carbon (C) (c) 400 g og helium (He) (d) 400 g of hydrogen (H2) (e) 400 g of glucose (C6 H12O6)
(i) Find the number of moles in each
(ii) Arrange the samples in the increasing order of their number of moles
(Gram molecular masses: He = 4 g, C = 12 g, H2 = 2 g, H2O = 18 g, C6 H12O6 = 180 g)
Solution:
(a) One GMM of H2O = 18 g
• So one mole of H2O has a mass of 18 g.
• In other words, if we take 18 grams of H2O molecules, it will contain 1 mole
• So, if we take 1 g of H2O molecules, it will contain 1⁄18 mole
• Thus, no. of moles in 400 g = 400 × 1⁄18 = 400⁄18 moles
(b) One GMM of C = 12 g
• So one mole of C has a mass of 12 g.
• In other words, if we take 12 grams of C molecules, it will contain 1 mole
• So, if we take 1 g of C molecules, it will contain 1⁄12 mole
• Thus, no. of moles in 400 g = 400 × 1⁄12 = 400⁄12 moles
(c) One GMM of He = 4 g
• So one mole of He has a mass of 4 g.
• In other words, if we take 4 grams of He molecules, it will contain 1 mole
• So, if we take 1 g of He molecules, it will contain 1⁄4 mole
• Thus, no. of moles in 400 g = 400 × 1⁄4 = 400⁄4 moles
(d) One GMM of H2 = 2 g
• So one mole of H2 has a mass of 2 g.
• In other words, if we take 2 grams of H2 molecules, it will contain 1 mole
• So, if we take 1 g of H2 molecules, it will contain 1⁄2 mole
• Thus, no. of moles in 400 g = 400 × 1⁄2 = 400⁄2 moles
(e) One GMM of C6H12O6 = 180 g
• So one mole of C6H12O6 has a mass of 180 g.
• In other words, if we take 180 grams of C6H12O6 molecules, it will contain 1 mole
• So, if we take 1 g of C6H12O6 molecules, it will contain 1⁄180 mole
• Thus, no. of moles in 400 g = 400 × 1⁄180 = 400⁄180 moles
(ii) Let us write the number of moles in order:
• Water - 400⁄18, • Carbon - 400⁄12, • Helium - 400⁄4, • Hydrogen - 400⁄2, • Glucose - 400⁄180
• We find that, the all are fractions with the same numerator.
• So, if we want an increasing order of number of moles, we must have a decreasing order of denominators. Thus we can write:
• Glucose - 400⁄180, • Water - 400⁄18, • Carbon - 400⁄12, • Helium - 400⁄4, • Hydrogen - 400⁄2,
Solved example 10.20
(a) Number of moles in 1 kg of water
(b) Number of moles in 500 g of CaCO3
(c) Number of moles and total number of atoms in 88 g of CO2
(GMM/GFM: H2O = 18 g, CaCO3 = 100 g, CO2 = 44 g, )
Solution:
(a) One GMM of H2O = 18 g
• So one mole of H2O has a mass of 18 g.
• In other words, if we take 18 grams of H2O molecules, it will contain 1 mole
• So, if we take 1 g of H2O molecules, it will contain 1⁄18 mole
• Thus, no. of moles in 1000 g = 1000 × 1⁄18 = 1000⁄18 moles
(b) One GMM of CaCO3 = 100 g
• So one mole of CaCO3 has a mass of 100 g.
• In other words, if we take 100 grams of CaCO3 molecules, it will contain 1 mole
• So, if we take 1 g of CaCO3 molecules, it will contain 1⁄100 mole
• Thus, no. of moles in 500 g = 500 × 1⁄100 = 500⁄100 = 5 moles
(c) One GMM of CO2 = 44 g
• So one mole of CO2 has a mass of 44 g.
• In other words, if we take 44 grams of CO2 molecules, it will contain 1 mole
• So, if we take 1 g of CO2 molecules, it will contain 1⁄44 mole
• Thus, no. of moles in 88 g = 88 × 1⁄44 = 88⁄44 = 2 moles
• So no. of molecules = 2 × 6.022×1023
• Each of these molecules have 3 atoms. (1 atom of carbon and 2 atoms of oxygen)
• So no. of atoms = 3 × 2 × 6.022×1023
In the next section we will learn about STP
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