In the previous section, we completed a discussion on empirical and molecular formulas. In this section, we will see stoichiometry
• On many occasions, we will want to know the 'quantity of the reactants and products' taking part in a reaction. For example:
♦ How many grams of chlorine is required?
♦ How many litres of hydrogen is required?
♦ How many grams of oxygen can be obtained?
♦ How many litres of ammonia can be obtained?
■ Stoichiometry deals with the calculation of masses and volumes of the reactants and products in a chemical reaction
• The word stoichiometry is derived from two Greek word:
♦ 'Stoichion' which means element
♦ 'Metron' which means measurement
• Recall how we balance a chemical equation. Detailed notes can be seen here.
• At the end of those notes, the steps for balancing the following equation are given:
C3H8 + O2 ⟶ CO2 + H2O
• This equation is for the combustion of propane. It is not balanced
• The balanced equation is: C3H8 + 5O2 ⟶ 3CO2 + 4H2O
■ Once we obtain a balanced equation, we can get many important information from it
Let us see an example:
• The balanced chemical equation for the combustion of methane is:
CH4 (g) + 2O2 (g) ⟶ CO2 (g) + 2H2O (g).
The 6 information that we get are:
1. All the reactants and products are in the gaseous state in the above reaction
• This is clear from the '(g)' written next to each of the reactants and products
• In the same way, solid state is indicated by (s) and liquid state is indicated by (l)
2. The coefficient for the reactants and products are:
A coefficient of '1' for CH4
A coefficient of '2' for O2
A coefficient of '1' for CO2
A coefficient of '2' for H2O.
■ These coefficients are called stoichiometric coefficients
3. one molecule of methane (g) reacts with two molecules of oxygen (g)
As a result of the reaction, one molecule of carbon dioxide (g) and two molecules of water (g) are formed
4. one mole of methane (g) reacts with two moles of oxygen (g)
As a result of the reaction, one mole of carbon dioxide (g) and two moles of water (g) are formed
5. 22.7 L of methane (g) reacts with 45.4 L of oxygen (g)
As a result of the reaction, 22.7 L of carbon dioxide (g) and 45.4 L of water (g) are formed
6. 16 g of methane (g) reacts with 64 g of oxygen (g)
As a result of the reaction, 44 g of carbon dioxide (g) and 36 g of water (g) are formed
• Further more, we can use the relation Density = Mass⁄Volume if required
Solved example 1.13
Calculate the amount of water produced by the combustion of 16 g of methane
Solution:
1. Combustion is the reaction of a hydrocarbon with oxygen (Details here)
• The products obtained are carbon dioxide and water in gaseous form
2. So we can write:
CH4 (g) + O2 (g) ⟶ CO2 (g) + H2O (g)
• But this is not a balanced equation. The balanced equation is:
CH4 (g) + 2O2 (g) ⟶ CO2 (g) + 2H2O (g)
3. We see that, one mol of methane reacts with two mols of oxygen
• So first we find the number of moles of methane available
• One mole of methane = (1 × 12.01) + (4 × 1.008) = 16.042 g
4. We have 16 grams of methane
• So number of mols of methane available = 16⁄16.042 = 0.997 ≈ 1
5. The ratio between CH4 and CO2 in the balanced equation is 1 : 1
• The ratio between CH4 and H2O in the balanced equation is 1 : 2
• So we will get 1 mol of CO2 (g) and 2 moles of H2O (g)
6. One mole of H2O = (2 × 1.008) + (1 × 16.00) = 18.016 g
• So two moles = 2 × 18.016 = 36.032 g
Solved example 1.13
How many moles of methane are required to produce 22 g of CO2 (g) after combustion?
Solution:
1. The balanced equation for the combustion of methane is:
CH4 (g) + 2O2 (g) ⟶ CO2 (g) + 2H2O (g)
2. We see that, the ratio between CH4 and CO2 in the balanced equation is 1 : 1
• So one mole of CH4 (g) gives one mole of CO2 (g)
3. One mole of CO2 = (1 × 12.01) + (2 × 16.00) = 44.01 g
• So 22 g of CO2 is 0.5 moles of CO2.
• To get half mole of CO2, we will need half mole of CH4
• We have seen that, from a balanced equation, we can obtain the quantities required for the reaction to take place
• We can also obtain the quantities of the products that will become available when the reaction becomes complete
• But some times, the 'quantity of one of the reactants available' may be less than 'it's required quantity'
• This 'reactant which is has fallen short', will be consumed first
• Once it is completely consumed, the reaction will come to a stop
• The reaction will not proceed even if the other reactant is available in plenty
■ The reactant, which gets consumed first, limits the amount of product formed and is, therefore,
called the limiting reagent.
• In performing stoichiometric calculations, the limiting reagent plays a crucial role
Solved example 1.14
50.0 kg of N2 (g) and 10.0 kg of H2 (g) are mixed to produce NH3 (g). Calculate the amount of NH3 (g) formed. Identify the limiting reagent in the production of NH3 in this situation.
Solution:
1. The balanced equation is: N2 (g) + 3H2 (g) ⟶ 2NH3 (g)
• One mole of N2 reacts with 3 moles of H2 to give 2 moles of NH3 (g)
2. Number of moles of N2 in 50.0 kg of N2 = 50000⁄28 = 1785.71
• Number of moles of H2 in 10.0 kg of H2 = 10000⁄2.016 = 4960.32
3. One mole of N2 reacts with 3 moles of H2
• So 1785.71 moles of N2 will need (3 × 1785.71) = 5357.13 moles of H2
4. But only 4960.32 moles of H2 are available
• That is., H2 has fallen short from it's required quantity
■ H2 is the limiting reagent
5. Three moles H2 of will give 2 moles of NH3
• So one mole of H2 will give 2⁄3 mole of NH3
• So 4960.32 moles of H2 will give (2⁄3 × 4960.32) moles of NH3
6. One mole of NH3 = (1 × 14) + (3 × 1) = 17.00 g
• So (2⁄3 × 4960.32) moles of NH3 = (2⁄3 × 4960.32) × 17 = 56216.96 g = 56.22 kg
Solved example 1.15
Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen
(iii) 2 moles of carbon are burnt in 16 g of dioxygen
Solution:
• The balanced chemical equation is: C (s) + O2 (g) ⟶ CO2 (g)
Part (i):
1. Carbon is burnt in air. So oxygen is available in plenty
• Carbon is the limiting reagent. All the carbon will be used up
2. One mole of C will give one mole of CO2 (g)
• One mole of CO2 = (1 × 12.01) + (2 × 16.00) = 44.01 g
Part (ii):
1. One mole of carbon is burnt in 16 g of dioxygen.
• From the balanced equation, we see that, 1 mole of C requires 1 mole of O2
2. One mole of O2 is 32 grams of O2
• But we have only 16 grams of O2
• That means we have only 0.5 mole of O2.
• That is., O2 has fallen short from it's required quantity
• O2 is the limiting reagent
3. One mole of O2 will give one mole of CO2
• So half mole of O2 will give half mole of CO2
• Half mole of CO2 is 22.005 g
Part (iii):
• Two moles of carbon are burnt in 16 g of dioxygen.
• In part (ii) we saw that 16 g of dioxygen is not sufficient even for one mole of C
• So in this case also, O2 is the limiting reagent
• We will get 22.005 g of CO2.
Solved example 1.16
In a reaction A + B2 ⟶ AB2
Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
Solution:
Part (i):
300 atoms of A + 200 molecules of B
1. From the given equation, it is clear that:
One atom of A reacts with two atoms of B
2. So if there are n atoms of A, there must be 2n atoms of B
• Given that 300 atoms of A are available. So there must be 600 atoms of B
3. The quantity of B is given in terms of molecules. There are 200 molecules of B available
• 200 molecules of B will give 400 atoms of B (∵ B2 indicates a diatomic molecule)
• This 400 falls short of the required 600
• So B is the limiting reagent
Part (ii):
2 mol A + 3 mol B
1. Two mols of A will require 2 mols of B
• But 3 mols of B are available
2. For 3 mols of B, 3 mols of A is required
• But only 2 mols of A are available
• This falls short of the required 3 mols
• So A is the limiting reagent
Part (iii):
100 atoms of A + 100 molecules of B
1. One atom of A reacts with two atoms of B
• So if there are n atoms of A, there must be 2n atoms of B
2. Given that 100 atoms of A are available. So there must be 200 atoms of B
• The quantity of B is given in terms of molecules. There are 100 molecules of B available
• 100 molecules of B will give 200 atoms of B (∵ B2 indicates a diatomic molecule)
3. This 200 is the exact number required
• So there is no limiting reagent. Both A and B will be completely used up
Part (iv):
5 mol A + 2.5 mol B
1. five mols of A will require five mols of B
2. But only 2.5 mols of B are available
• This falls short of the required 5 mols
• So B is the limiting reagent
Part (v):
2.5 mol A + 5 mol B
1. 2.5 mols of A will require 2.5 mols of B
2. But 5 mols of B are available
• For 5 mols of B, 5 mols of A is required
3. But only 2.5 mols of A are available
• This falls short of the required 5 mols
• So A is the limiting reagent
Solved example 1.17
Dinitrogen and dihydrogen react with each other to produce ammonia according
to the following chemical equation:
N2 (g) + 3H2 (g) ⟶ 2NH3 (g)
(i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts
with 1.00 ×103 g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?
Solution:
Part (i):
1. One mole of N2 reacts with 3 mols of H2
• One mol of N2 is 28 g
2. So in 2000 g of N2, there will be (2000⁄28) mols of N2.
• So we will need [3×(2000⁄28)] = 214.286 mols of H2.
3. Number of mols available in 1000 g of H2 = (1000⁄2.016) = 496.03
• 496.03 is in excess of the required 214.286 mols
• So H2 is not the limiting reagent. It is N2
4. (2000⁄28) mols of N2 will give (4000⁄28) mols of NH3
• One mole of NH3 is 17 grams
• So (4000⁄28) mols of NH3 is [(4000⁄28)×17] = 2428.57 g
Part (ii):
In this reaction, N2 is the limiting reagent. Some portion of H2 will remain unreacted
Part (iii):
1. From part (i), we have:
• 214.286 mols of H2 is required
• 496.03 mols of H2 is available
2. So excess quantity = (496.03-214.286) = 281.744 mols of H2
• 1 mol of H2 = 2.016 g
• So 281.744 mols = (281.744 × 2.016) = 567.995904 g ≈ 568 g
Solved example 1.18
36 g of carbon and 128 g of O2 are mixed to produce carbon monoxide. Calculate the amount of CO produced and identify the limiting reagent
Solution:
1. The balanced equation is: 2C (s) + O2 (g) ⟶ 2CO (g)
• So 2 mols of C reacts with 1 mol of O2
2. 36 g of carbon = (36⁄12) = 3 mols
• 128 g of O2 = (128⁄32) = 4 mols
3. Four mols of O2 will require 8 mols of C
• But we have only 3 mols of C. So C is the limiting reagent
• The ratio between C and CO in the equation is 1 : 1
• So 3 mols of C will give 3 mols of CO
4. one mol of C = (1 × 12) + (1 × 16.00) = 28 g
• So 3 mols = (3 × 28) = 84 g
Next we will see a problem involving stoichiometry and 'finding molecular formula'
Solved example 1.19
An organic compound contains C, H and O only. When 0.3 g of that compound undergoes combustion, 0.44 g of CO2 and 0.18 g of H2O is produced. If the molar mass of that compound is 60 g, what is it's molecular formula?
Solution:
1. Given that, 0.44 g of CO2 is produced. This is greater than the original 0.3 g
• This increase occurs because, 'extra oxygen from the atmosphere' also combines with the carbon in the compound
2. Consider any sample of CO2
• Using Eq.1.1, we can find the % mass of carbon in it
We have:
Percentage mass of an element in a pure sample of any of it's compound
= $\mathbf\small{\left(\frac{(nA)_{\rm{Element}} \times (GAM)_{\rm{Element}}}{(GMM)_{\rm{Compound}}}\right)\times 100}$
3. Substituting the values, we get:
Percentage mass of carbon in a pure sample of CO2
= $\mathbf\small{\left(\frac{1 \times 12}{44}\right)\times 100=\frac{300}{11}\text{%}}$
4. So (300⁄11) % of 0.44 g is C
⇒ Mass of carbon in the 0.44 g of CO2 = (0.44 × 3⁄11) = 0.12 g
5. The only source of this C is the original organic compound
So we can write:
Mass of C in the original organic compound = 0.12 g
6. Consider any sample of H2O
• Using Eq.1.1 above, we can find the % mass of H in it
7. Substituting the values, we get:
Percentage mass of H in a pure sample of H2O
= $\mathbf\small{\left(\frac{2 \times 1}{18}\right)\times 100=\frac{100}{9}\text{%}}$
8. So (100⁄9) % of 0.18 g is H
⇒ Mass of H in the 0.18 g of H2O = (0.18 × 1⁄9) = 0.02 g
9. The only source of this H is the original organic compound
So we can write:
Mass of H in the original organic compound = 0.02 g
10. Total mass of C and H in the original organic compound = (0.12+0.02) = 0.14 g
• Given that, mass of the original organic compound = 0.3 g
• So mass of O in the original organic compound = (0.3-0.14) = 0.16 g
11. Now we can find the percentages:
% mass of C in the original organic compound = (0.12⁄0.3) × 100 = 40%
% mass of H in the original organic compound = (0.02⁄0.3) × 100 = (20⁄3)%
% mass of C in the original organic compound = (0.16⁄0.3) × 100 = (160⁄3)%
12. Now we use Eq.1.2:
• On many occasions, we will want to know the 'quantity of the reactants and products' taking part in a reaction. For example:
♦ How many grams of chlorine is required?
♦ How many litres of hydrogen is required?
♦ How many grams of oxygen can be obtained?
♦ How many litres of ammonia can be obtained?
■ Stoichiometry deals with the calculation of masses and volumes of the reactants and products in a chemical reaction
• The word stoichiometry is derived from two Greek word:
♦ 'Stoichion' which means element
♦ 'Metron' which means measurement
• Before beginning the discussion on stoichiometry, we must first understand all the important features of balanced chemical equations
• At the end of those notes, the steps for balancing the following equation are given:
C3H8 + O2 ⟶ CO2 + H2O
• This equation is for the combustion of propane. It is not balanced
• The balanced equation is: C3H8 + 5O2 ⟶ 3CO2 + 4H2O
■ Once we obtain a balanced equation, we can get many important information from it
Let us see an example:
• The balanced chemical equation for the combustion of methane is:
CH4 (g) + 2O2 (g) ⟶ CO2 (g) + 2H2O (g).
The 6 information that we get are:
1. All the reactants and products are in the gaseous state in the above reaction
• This is clear from the '(g)' written next to each of the reactants and products
• In the same way, solid state is indicated by (s) and liquid state is indicated by (l)
2. The coefficient for the reactants and products are:
A coefficient of '1' for CH4
A coefficient of '2' for O2
A coefficient of '1' for CO2
A coefficient of '2' for H2O.
■ These coefficients are called stoichiometric coefficients
3. one molecule of methane (g) reacts with two molecules of oxygen (g)
As a result of the reaction, one molecule of carbon dioxide (g) and two molecules of water (g) are formed
4. one mole of methane (g) reacts with two moles of oxygen (g)
As a result of the reaction, one mole of carbon dioxide (g) and two moles of water (g) are formed
5. 22.7 L of methane (g) reacts with 45.4 L of oxygen (g)
As a result of the reaction, 22.7 L of carbon dioxide (g) and 45.4 L of water (g) are formed
6. 16 g of methane (g) reacts with 64 g of oxygen (g)
As a result of the reaction, 44 g of carbon dioxide (g) and 36 g of water (g) are formed
• Thus we see that, the balanced chemical equation gives us valuable information in terms of moles, mass and volume
Now we will see some solved examples
Calculate the amount of water produced by the combustion of 16 g of methane
Solution:
1. Combustion is the reaction of a hydrocarbon with oxygen (Details here)
• The products obtained are carbon dioxide and water in gaseous form
2. So we can write:
CH4 (g) + O2 (g) ⟶ CO2 (g) + H2O (g)
• But this is not a balanced equation. The balanced equation is:
CH4 (g) + 2O2 (g) ⟶ CO2 (g) + 2H2O (g)
3. We see that, one mol of methane reacts with two mols of oxygen
• So first we find the number of moles of methane available
• One mole of methane = (1 × 12.01) + (4 × 1.008) = 16.042 g
4. We have 16 grams of methane
• So number of mols of methane available = 16⁄16.042 = 0.997 ≈ 1
5. The ratio between CH4 and CO2 in the balanced equation is 1 : 1
• The ratio between CH4 and H2O in the balanced equation is 1 : 2
• So we will get 1 mol of CO2 (g) and 2 moles of H2O (g)
6. One mole of H2O = (2 × 1.008) + (1 × 16.00) = 18.016 g
• So two moles = 2 × 18.016 = 36.032 g
Solved example 1.13
How many moles of methane are required to produce 22 g of CO2 (g) after combustion?
Solution:
1. The balanced equation for the combustion of methane is:
CH4 (g) + 2O2 (g) ⟶ CO2 (g) + 2H2O (g)
2. We see that, the ratio between CH4 and CO2 in the balanced equation is 1 : 1
• So one mole of CH4 (g) gives one mole of CO2 (g)
3. One mole of CO2 = (1 × 12.01) + (2 × 16.00) = 44.01 g
• So 22 g of CO2 is 0.5 moles of CO2.
• To get half mole of CO2, we will need half mole of CH4
Limiting reagent
• We can also obtain the quantities of the products that will become available when the reaction becomes complete
• But some times, the 'quantity of one of the reactants available' may be less than 'it's required quantity'
• This 'reactant which is has fallen short', will be consumed first
• Once it is completely consumed, the reaction will come to a stop
• The reaction will not proceed even if the other reactant is available in plenty
■ The reactant, which gets consumed first, limits the amount of product formed and is, therefore,
called the limiting reagent.
• In performing stoichiometric calculations, the limiting reagent plays a crucial role
Solved example 1.14
50.0 kg of N2 (g) and 10.0 kg of H2 (g) are mixed to produce NH3 (g). Calculate the amount of NH3 (g) formed. Identify the limiting reagent in the production of NH3 in this situation.
Solution:
1. The balanced equation is: N2 (g) + 3H2 (g) ⟶ 2NH3 (g)
• One mole of N2 reacts with 3 moles of H2 to give 2 moles of NH3 (g)
2. Number of moles of N2 in 50.0 kg of N2 = 50000⁄28 = 1785.71
• Number of moles of H2 in 10.0 kg of H2 = 10000⁄2.016 = 4960.32
3. One mole of N2 reacts with 3 moles of H2
• So 1785.71 moles of N2 will need (3 × 1785.71) = 5357.13 moles of H2
4. But only 4960.32 moles of H2 are available
• That is., H2 has fallen short from it's required quantity
■ H2 is the limiting reagent
5. Three moles H2 of will give 2 moles of NH3
• So one mole of H2 will give 2⁄3 mole of NH3
• So 4960.32 moles of H2 will give (2⁄3 × 4960.32) moles of NH3
6. One mole of NH3 = (1 × 14) + (3 × 1) = 17.00 g
• So (2⁄3 × 4960.32) moles of NH3 = (2⁄3 × 4960.32) × 17 = 56216.96 g = 56.22 kg
Solved example 1.15
Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen
(iii) 2 moles of carbon are burnt in 16 g of dioxygen
Solution:
• The balanced chemical equation is: C (s) + O2 (g) ⟶ CO2 (g)
Part (i):
1. Carbon is burnt in air. So oxygen is available in plenty
• Carbon is the limiting reagent. All the carbon will be used up
2. One mole of C will give one mole of CO2 (g)
• One mole of CO2 = (1 × 12.01) + (2 × 16.00) = 44.01 g
Part (ii):
1. One mole of carbon is burnt in 16 g of dioxygen.
• From the balanced equation, we see that, 1 mole of C requires 1 mole of O2
2. One mole of O2 is 32 grams of O2
• But we have only 16 grams of O2
• That means we have only 0.5 mole of O2.
• That is., O2 has fallen short from it's required quantity
• O2 is the limiting reagent
3. One mole of O2 will give one mole of CO2
• So half mole of O2 will give half mole of CO2
• Half mole of CO2 is 22.005 g
Part (iii):
• Two moles of carbon are burnt in 16 g of dioxygen.
• In part (ii) we saw that 16 g of dioxygen is not sufficient even for one mole of C
• So in this case also, O2 is the limiting reagent
• We will get 22.005 g of CO2.
Solved example 1.16
In a reaction A + B2 ⟶ AB2
Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
Solution:
Part (i):
300 atoms of A + 200 molecules of B
1. From the given equation, it is clear that:
One atom of A reacts with two atoms of B
2. So if there are n atoms of A, there must be 2n atoms of B
• Given that 300 atoms of A are available. So there must be 600 atoms of B
3. The quantity of B is given in terms of molecules. There are 200 molecules of B available
• 200 molecules of B will give 400 atoms of B (∵ B2 indicates a diatomic molecule)
• This 400 falls short of the required 600
• So B is the limiting reagent
Part (ii):
2 mol A + 3 mol B
1. Two mols of A will require 2 mols of B
• But 3 mols of B are available
2. For 3 mols of B, 3 mols of A is required
• But only 2 mols of A are available
• This falls short of the required 3 mols
• So A is the limiting reagent
Part (iii):
100 atoms of A + 100 molecules of B
1. One atom of A reacts with two atoms of B
• So if there are n atoms of A, there must be 2n atoms of B
2. Given that 100 atoms of A are available. So there must be 200 atoms of B
• The quantity of B is given in terms of molecules. There are 100 molecules of B available
• 100 molecules of B will give 200 atoms of B (∵ B2 indicates a diatomic molecule)
3. This 200 is the exact number required
• So there is no limiting reagent. Both A and B will be completely used up
Part (iv):
5 mol A + 2.5 mol B
1. five mols of A will require five mols of B
2. But only 2.5 mols of B are available
• This falls short of the required 5 mols
• So B is the limiting reagent
Part (v):
2.5 mol A + 5 mol B
1. 2.5 mols of A will require 2.5 mols of B
2. But 5 mols of B are available
• For 5 mols of B, 5 mols of A is required
3. But only 2.5 mols of A are available
• This falls short of the required 5 mols
• So A is the limiting reagent
Solved example 1.17
Dinitrogen and dihydrogen react with each other to produce ammonia according
to the following chemical equation:
N2 (g) + 3H2 (g) ⟶ 2NH3 (g)
(i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts
with 1.00 ×103 g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?
Solution:
Part (i):
1. One mole of N2 reacts with 3 mols of H2
• One mol of N2 is 28 g
2. So in 2000 g of N2, there will be (2000⁄28) mols of N2.
• So we will need [3×(2000⁄28)] = 214.286 mols of H2.
3. Number of mols available in 1000 g of H2 = (1000⁄2.016) = 496.03
• 496.03 is in excess of the required 214.286 mols
• So H2 is not the limiting reagent. It is N2
4. (2000⁄28) mols of N2 will give (4000⁄28) mols of NH3
• One mole of NH3 is 17 grams
• So (4000⁄28) mols of NH3 is [(4000⁄28)×17] = 2428.57 g
Part (ii):
In this reaction, N2 is the limiting reagent. Some portion of H2 will remain unreacted
Part (iii):
1. From part (i), we have:
• 214.286 mols of H2 is required
• 496.03 mols of H2 is available
2. So excess quantity = (496.03-214.286) = 281.744 mols of H2
• 1 mol of H2 = 2.016 g
• So 281.744 mols = (281.744 × 2.016) = 567.995904 g ≈ 568 g
Solved example 1.18
36 g of carbon and 128 g of O2 are mixed to produce carbon monoxide. Calculate the amount of CO produced and identify the limiting reagent
Solution:
1. The balanced equation is: 2C (s) + O2 (g) ⟶ 2CO (g)
• So 2 mols of C reacts with 1 mol of O2
2. 36 g of carbon = (36⁄12) = 3 mols
• 128 g of O2 = (128⁄32) = 4 mols
3. Four mols of O2 will require 8 mols of C
• But we have only 3 mols of C. So C is the limiting reagent
• The ratio between C and CO in the equation is 1 : 1
• So 3 mols of C will give 3 mols of CO
4. one mol of C = (1 × 12) + (1 × 16.00) = 28 g
• So 3 mols = (3 × 28) = 84 g
Next we will see a problem involving stoichiometry and 'finding molecular formula'
Solved example 1.19
An organic compound contains C, H and O only. When 0.3 g of that compound undergoes combustion, 0.44 g of CO2 and 0.18 g of H2O is produced. If the molar mass of that compound is 60 g, what is it's molecular formula?
Solution:
1. Given that, 0.44 g of CO2 is produced. This is greater than the original 0.3 g
• This increase occurs because, 'extra oxygen from the atmosphere' also combines with the carbon in the compound
2. Consider any sample of CO2
• Using Eq.1.1, we can find the % mass of carbon in it
We have:
Percentage mass of an element in a pure sample of any of it's compound
= $\mathbf\small{\left(\frac{(nA)_{\rm{Element}} \times (GAM)_{\rm{Element}}}{(GMM)_{\rm{Compound}}}\right)\times 100}$
3. Substituting the values, we get:
Percentage mass of carbon in a pure sample of CO2
= $\mathbf\small{\left(\frac{1 \times 12}{44}\right)\times 100=\frac{300}{11}\text{%}}$
4. So (300⁄11) % of 0.44 g is C
⇒ Mass of carbon in the 0.44 g of CO2 = (0.44 × 3⁄11) = 0.12 g
5. The only source of this C is the original organic compound
So we can write:
Mass of C in the original organic compound = 0.12 g
6. Consider any sample of H2O
• Using Eq.1.1 above, we can find the % mass of H in it
7. Substituting the values, we get:
Percentage mass of H in a pure sample of H2O
= $\mathbf\small{\left(\frac{2 \times 1}{18}\right)\times 100=\frac{100}{9}\text{%}}$
8. So (100⁄9) % of 0.18 g is H
⇒ Mass of H in the 0.18 g of H2O = (0.18 × 1⁄9) = 0.02 g
9. The only source of this H is the original organic compound
So we can write:
Mass of H in the original organic compound = 0.02 g
10. Total mass of C and H in the original organic compound = (0.12+0.02) = 0.14 g
• Given that, mass of the original organic compound = 0.3 g
• So mass of O in the original organic compound = (0.3-0.14) = 0.16 g
11. Now we can find the percentages:
% mass of C in the original organic compound = (0.12⁄0.3) × 100 = 40%
% mass of H in the original organic compound = (0.02⁄0.3) × 100 = (20⁄3)%
% mass of C in the original organic compound = (0.16⁄0.3) × 100 = (160⁄3)%
12. Now we use Eq.1.2:
$\mathbf\small{\left(\frac{(nA)_{\rm{Element}}}{(GMM)_{\rm{Compound}}}\right)=\frac{\text{% of the element}}{\text{GAM of the element} \times 100}}$
• Substituting the values for C, we get:
$\mathbf\small{\left(\frac{(nA)_{\rm{C}}}{60}\right)=\frac{40}{12 \times 100}}$
⇒ (nA)C = 2
• Substituting the values for H, we get:
$\mathbf\small{\left(\frac{(nA)_{\rm{H}}}{60}\right)=\frac{\frac{20}{3}}{1 \times 100}}$
⇒ (nA)H = 4
• Substituting the values for O, we get:
$\mathbf\small{\left(\frac{(nA)_{\rm{H}}}{60}\right)=\frac{\frac{20}{3}}{1 \times 100}}$
⇒ (nA)H = 2
13. So the molecular formula is:
C2H4O2
In the next section, we will see reactions in solutions
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