In the previous section, we saw some solved examples on mole concept and molar masses. In this section, we will see percentage composition. Also we will see how to obtain molecular formula from empirical formula
We will write the details in steps:
1. We know how to find answers to the kind of questions given below:
• How many atoms are present in 50 grams of oxygen?
• How many molecules are present in 85 grams of water?
• How many carbon atoms are present in 20 grams of glucose?
■ Those are questions related to 'number' of atoms or molecules
• Now we will try to answer the questions related to 'percentages'
2. Consider an example
(i) Take 50 g of water
• We know that water is a combination of oxygen and hydrogen
• The 50 g pure water will consist of only oxygen and hydrogen
(ii) Consider the following questions:
• What is the contribution of oxygen towards making up the 50 g?
• What is the contribution of hydrogen towards making up the 50 g?
In other words:
• What percentage of the 50 g, is 'the mass of oxygen'?
• What percentage of the 50 g, is 'the mass of hydrogen'?
3. Let us try to answer the two questions:
(i) The molar mass of water is 18.02 g
• So one mole of water molecules will have a mass of 18.02 g
• So number of moles of water molecules in 50 grams = 50⁄18.02
(ii) In one mole, there will be
• 2NA hydrogen atoms
• 1NA oxygen atoms
(iii) So in (50⁄18.02) moles, there will be:
(50⁄18.02) × 2NA hydrogen atoms
(50⁄18.02) × 1NA oxygen atoms atoms
(iv) So we can write:
• Total mass of hydrogen atoms in the 50 g sample
= [(50⁄18.02) × 2NA] × mass of one hydrogen atom
• Total mass of oxygen atoms in the 50 g sample
= [(50⁄18.02) × 1NA] × mass of one oxygen atom
(v) So now we have to find the two masses:
♦ The mass of one hydrogen atom
♦ The mass of one oxygen atom
• Molar mass of hydrogen = 1.008 g
⇒ NA atoms of hydrogen has a mass of 1.008 g
⇒ 1 atom of hydrogen has a mass of $\mathbf\small{ \left(\frac{1.008}{N_A}\right)}$ g
• Molar mass of oxygen = 16.00 g
⇒ NA atoms of oxygen has a mass of 16.00 g
⇒ 1 atom of oxygen has a mass of $\mathbf\small{ \left(\frac{16.00}{N_A}\right)}$ g
(vi) Substituting these in (iv), we get:
• Total mass of hydrogen atoms in the 50 g sample
= $\mathbf\small{\left(\frac{50}{18.02}\right)\times 2\times N_A \times \left(\frac{1.008}{N_A}\right)}$
=$\mathbf\small{\left(\frac{50}{18.02}\right)\times 2 \times 1.008}$
• Total mass of oxygen atoms in the 50 g sample
= $\mathbf\small{\left(\frac{50}{18.02}\right)\times 1\times N_A \times \left(\frac{16.00}{N_A}\right)}$
=$\mathbf\small{\left(\frac{50}{18.02}\right)\times 1 \times 16.00}$
(vii) So percentage mass of hydrogen atoms in the 50 g sample
= $\mathbf\small{\left(\frac{\left(\frac{50}{18.02}\right)\times 2 \times 1.008}{50}\right)\times 100}$
= $\mathbf\small{\left(\frac{2 \times 1.008}{18.02}\right)\times 100}$
• Similarly, percentage mass of oxygen atoms in the 50 g sample
= $\mathbf\small{\left(\frac{\left(\frac{50}{18.02}\right)\times 1 \times 16.00}{50}\right)\times 100}$
= $\mathbf\small{\left(\frac{16.00}{18.02}\right)\times 100}$
(viii) We see that the quantity '50 g' which is the original mass of the sample, has vanished
• Only general terms are remaining
• The reader may write the steps with different original masses. It will become clear that, what ever be the original mass, it will indeed vanish
4. Since there are only general terms remaining, we can write general equations:
■ Percentage mass of hydrogen in any sample of pure water
= $\mathbf\small{\left(\frac{(nA)_H \times (GAM)_H}{(GMM)_{H_2O}}\right)\times 100}$
• Where:
♦ $\mathbf\small{(nA)_H}$ is the number of hydrogen atoms in a water molecule
♦ $\mathbf\small{{(GAM)_H}}$ is the gram atomic mass of hydrogen
♦ $\mathbf\small{(GMM)_{H_2O}}$ is the gram molecular mass of hydrogen
■ Percentage mass of oxygen in any sample of pure water
= $\mathbf\small{\left(\frac{(nA)_O \times (GAM)_O}{(GMM)_{H_2O}}\right)\times 100}$
• Where:
♦ $\mathbf\small{(nA)_O}$ is the number of hydrogen atoms in a water molecule
♦ $\mathbf\small{{(GAM)_O}}$ is the gram atomic mass of hydrogen
♦ $\mathbf\small{(GMM)_{H_2O}}$ is the gram molecular mass of hydrogen
5. This gives us an idea to write a general formula which is applicable to any element in any compound
It can be written as:
Eq.1.1:
Percentage mass of an element in a pure sample of any of it's compound
= $\mathbf\small{\left(\frac{(nA)_{\rm{Element}} \times (GAM)_{\rm{Element}}}{(GMM)_{\rm{Compound}}}\right)\times 100}$
What is the percentage of carbon, hydrogen and oxygen in ethanol?
Solution:
1. We will first write the required values:
• GAM of carbon is 12.01 g
• GAM of hydrogen is 1.008 g
• GAM of oxygen is 16.00 g
• Molecular formula of ethanol is: C2H5OH
So the GMM of ethanol
= (2 × 12.01) + (6 × 1.008) + (1 × 16.00) = 46.068 g
2. We have:
Percentage mass of an element in a pure sample of any of it's compound
= $\mathbf\small{\left(\frac{(nA)_{\rm{Element}} \times (GAM)_{\rm{Element}}}{(GMM)_{\rm{Compound}}}\right)\times 100}$
(i) Applying the formula to carbon, we get:
Percentage mass of carbon in any pure sample of ethanol
= $\mathbf\small{\left(\frac{(nA)_{\rm{Carbon}} \times (GAM)_{\rm{carbon}}}{(GMM)_{\rm{Ethanol}}}\right)\times 100}$
= $\mathbf\small{\left(\frac{2 \times 12.01}{46.068}\right)\times 100}$ = 52.14%
(ii) Applying the formula to hydrogen, we get:
Percentage mass of hydrogen in any pure sample of ethanol
= $\mathbf\small{\left(\frac{6 \times 1.008}{46.068}\right)\times 100}$ = 13.13%
(iii) Applying the formula to oxygen, we get:
Percentage mass of oxygen in any pure sample of ethanol
= $\mathbf\small{\left(\frac{1 \times 12.00}{46.068}\right)\times 100}$ = 34.73%
3. Let us see an application of the above results:
• If we have 75 grams of ethanol:
♦ (75 × 0.5214) = 39.10 g will be carbon
♦ (75 × 0.1313) = 9.85 g will be hydrogen
♦ (75 × 0.3473) = 26.05 g will be oxygen
Check: (39.10 + 9.85 + 26.05) = 75 g
Percentage composition
We will write the details in steps:
1. We know how to find answers to the kind of questions given below:
• How many atoms are present in 50 grams of oxygen?
• How many molecules are present in 85 grams of water?
• How many carbon atoms are present in 20 grams of glucose?
■ Those are questions related to 'number' of atoms or molecules
• Now we will try to answer the questions related to 'percentages'
2. Consider an example
(i) Take 50 g of water
• We know that water is a combination of oxygen and hydrogen
• The 50 g pure water will consist of only oxygen and hydrogen
(ii) Consider the following questions:
• What is the contribution of oxygen towards making up the 50 g?
• What is the contribution of hydrogen towards making up the 50 g?
In other words:
• What percentage of the 50 g, is 'the mass of oxygen'?
• What percentage of the 50 g, is 'the mass of hydrogen'?
3. Let us try to answer the two questions:
(i) The molar mass of water is 18.02 g
• So one mole of water molecules will have a mass of 18.02 g
• So number of moles of water molecules in 50 grams = 50⁄18.02
(ii) In one mole, there will be
• 2NA hydrogen atoms
• 1NA oxygen atoms
(iii) So in (50⁄18.02) moles, there will be:
(50⁄18.02) × 2NA hydrogen atoms
(50⁄18.02) × 1NA oxygen atoms atoms
(iv) So we can write:
• Total mass of hydrogen atoms in the 50 g sample
= [(50⁄18.02) × 2NA] × mass of one hydrogen atom
• Total mass of oxygen atoms in the 50 g sample
= [(50⁄18.02) × 1NA] × mass of one oxygen atom
(v) So now we have to find the two masses:
♦ The mass of one hydrogen atom
♦ The mass of one oxygen atom
• Molar mass of hydrogen = 1.008 g
⇒ NA atoms of hydrogen has a mass of 1.008 g
⇒ 1 atom of hydrogen has a mass of $\mathbf\small{ \left(\frac{1.008}{N_A}\right)}$ g
• Molar mass of oxygen = 16.00 g
⇒ NA atoms of oxygen has a mass of 16.00 g
⇒ 1 atom of oxygen has a mass of $\mathbf\small{ \left(\frac{16.00}{N_A}\right)}$ g
(vi) Substituting these in (iv), we get:
• Total mass of hydrogen atoms in the 50 g sample
= $\mathbf\small{\left(\frac{50}{18.02}\right)\times 2\times N_A \times \left(\frac{1.008}{N_A}\right)}$
=$\mathbf\small{\left(\frac{50}{18.02}\right)\times 2 \times 1.008}$
• Total mass of oxygen atoms in the 50 g sample
= $\mathbf\small{\left(\frac{50}{18.02}\right)\times 1\times N_A \times \left(\frac{16.00}{N_A}\right)}$
=$\mathbf\small{\left(\frac{50}{18.02}\right)\times 1 \times 16.00}$
(vii) So percentage mass of hydrogen atoms in the 50 g sample
= $\mathbf\small{\left(\frac{\left(\frac{50}{18.02}\right)\times 2 \times 1.008}{50}\right)\times 100}$
= $\mathbf\small{\left(\frac{2 \times 1.008}{18.02}\right)\times 100}$
• Similarly, percentage mass of oxygen atoms in the 50 g sample
= $\mathbf\small{\left(\frac{\left(\frac{50}{18.02}\right)\times 1 \times 16.00}{50}\right)\times 100}$
= $\mathbf\small{\left(\frac{16.00}{18.02}\right)\times 100}$
(viii) We see that the quantity '50 g' which is the original mass of the sample, has vanished
• Only general terms are remaining
• The reader may write the steps with different original masses. It will become clear that, what ever be the original mass, it will indeed vanish
4. Since there are only general terms remaining, we can write general equations:
■ Percentage mass of hydrogen in any sample of pure water
= $\mathbf\small{\left(\frac{(nA)_H \times (GAM)_H}{(GMM)_{H_2O}}\right)\times 100}$
• Where:
♦ $\mathbf\small{(nA)_H}$ is the number of hydrogen atoms in a water molecule
♦ $\mathbf\small{{(GAM)_H}}$ is the gram atomic mass of hydrogen
♦ $\mathbf\small{(GMM)_{H_2O}}$ is the gram molecular mass of hydrogen
■ Percentage mass of oxygen in any sample of pure water
= $\mathbf\small{\left(\frac{(nA)_O \times (GAM)_O}{(GMM)_{H_2O}}\right)\times 100}$
• Where:
♦ $\mathbf\small{(nA)_O}$ is the number of hydrogen atoms in a water molecule
♦ $\mathbf\small{{(GAM)_O}}$ is the gram atomic mass of hydrogen
♦ $\mathbf\small{(GMM)_{H_2O}}$ is the gram molecular mass of hydrogen
5. This gives us an idea to write a general formula which is applicable to any element in any compound
It can be written as:
Eq.1.1:
Percentage mass of an element in a pure sample of any of it's compound
= $\mathbf\small{\left(\frac{(nA)_{\rm{Element}} \times (GAM)_{\rm{Element}}}{(GMM)_{\rm{Compound}}}\right)\times 100}$
Let us apply this formula to the elements in ethanol:
Solution:
1. We will first write the required values:
• GAM of carbon is 12.01 g
• GAM of hydrogen is 1.008 g
• GAM of oxygen is 16.00 g
• Molecular formula of ethanol is: C2H5OH
So the GMM of ethanol
= (2 × 12.01) + (6 × 1.008) + (1 × 16.00) = 46.068 g
2. We have:
Percentage mass of an element in a pure sample of any of it's compound
= $\mathbf\small{\left(\frac{(nA)_{\rm{Element}} \times (GAM)_{\rm{Element}}}{(GMM)_{\rm{Compound}}}\right)\times 100}$
(i) Applying the formula to carbon, we get:
Percentage mass of carbon in any pure sample of ethanol
= $\mathbf\small{\left(\frac{(nA)_{\rm{Carbon}} \times (GAM)_{\rm{carbon}}}{(GMM)_{\rm{Ethanol}}}\right)\times 100}$
= $\mathbf\small{\left(\frac{2 \times 12.01}{46.068}\right)\times 100}$ = 52.14%
(ii) Applying the formula to hydrogen, we get:
Percentage mass of hydrogen in any pure sample of ethanol
= $\mathbf\small{\left(\frac{6 \times 1.008}{46.068}\right)\times 100}$ = 13.13%
(iii) Applying the formula to oxygen, we get:
Percentage mass of oxygen in any pure sample of ethanol
= $\mathbf\small{\left(\frac{1 \times 12.00}{46.068}\right)\times 100}$ = 34.73%
3. Let us see an application of the above results:
• If we have 75 grams of ethanol:
♦ (75 × 0.5214) = 39.10 g will be carbon
♦ (75 × 0.1313) = 9.85 g will be hydrogen
♦ (75 × 0.3473) = 26.05 g will be oxygen
Check: (39.10 + 9.85 + 26.05) = 75 g
So now we can find the percentage mass of any element. Let us put it into practical use:
Suppose that, a sample of a substance is brought before a chemist. The identity of the substance is not known. So the task of the chemist is to identify it. That is., he has to find the answers to the following questions:
(i) Is the substance an element?
♦ If yes, which element?
(ii) If the substance is a compound,
♦ Which are the elements present in it?
♦ How many atoms of each element are present in it?
Answers to (ii) will enable the chemist to write the molecular formula. Thus the true identity of the substance will be revealed
• Let us write the procedure to find the molecular formula. We will write it in steps:
1. Conduct the 'required chemical tests' to find:
■ The mass percentage of each element
• Let it be as follows:
♦ 4.07% of the mass is hydrogen
(In other words, if we take 100 g from the sample, there will be 4.07 g of hydrogen in it)
♦ 24.27% of the mass is carbon
(In other words, if we take 100 g from the sample, there will be 24.27 g of carbon in it)
♦ 71.65% of the mass is chlorine
(In other words, if we take 100 g from the sample, there will be 71.65 g of chlorine in it)
• We will learn about the 'required chemical tests' to find the above percentages, in later chapters
2. Now we apply the Eq.1.1 in reverse. We get:
(i) For hydrogen:
$\mathbf\small{4.07=\left(\frac{(nA)_{\rm{H}} \times (GAM)_{\rm{H}}}{(GMM)_{\rm{Compound}}}\right)\times 100}$
$\mathbf\small{\Rightarrow 4.07=\left(\frac{(nA)_{\rm{H}} \times 1.008}{(GMM)_{\rm{Compound}}}\right)\times 100}$
$\mathbf\small{\Rightarrow \left(\frac{(nA)_{\rm{H}}}{(GMM)_{\rm{Compound}}}\right)=\frac{4.07}{1.008 \times 100}=0.0404}$
(ii) For carbon:
$\mathbf\small{24.27=\left(\frac{(nA)_{\rm{C}} \times (GAM)_{\rm{C}}}{(GMM)_{\rm{Compound}}}\right)\times 100}$
$\mathbf\small{\Rightarrow 24.27=\left(\frac{(nA)_{\rm{C}} \times 12.01}{(GMM)_{\rm{Compound}}}\right)\times 100}$
$\mathbf\small{\Rightarrow \left(\frac{(nA)_{\rm{C}}}{(GMM)_{\rm{Compound}}}\right)=\frac{24.27}{12.01 \times 100}=0.0202}$
(iii) For chlorine:
$\mathbf\small{71.65=\left(\frac{(nA)_{\rm{Cl}} \times (GAM)_{\rm{Cl}}}{(GMM)_{\rm{Compound}}}\right)\times 100}$
$\mathbf\small{\Rightarrow 71.65=\left(\frac{(nA)_{\rm{Cl}} \times 35.45}{(GMM)_{\rm{Compound}}}\right)\times 100}$
$\mathbf\small{\Rightarrow \left(\frac{(nA)_{\rm{Cl}}}{(GMM)_{\rm{Compound}}}\right)=\frac{71.65}{35.45 \times 100}=0.0202}$
Empirical formula for Molecular formula
Suppose that, a sample of a substance is brought before a chemist. The identity of the substance is not known. So the task of the chemist is to identify it. That is., he has to find the answers to the following questions:
(i) Is the substance an element?
♦ If yes, which element?
(ii) If the substance is a compound,
♦ Which are the elements present in it?
♦ How many atoms of each element are present in it?
Answers to (ii) will enable the chemist to write the molecular formula. Thus the true identity of the substance will be revealed
• Let us write the procedure to find the molecular formula. We will write it in steps:
1. Conduct the 'required chemical tests' to find:
■ The mass percentage of each element
• Let it be as follows:
♦ 4.07% of the mass is hydrogen
(In other words, if we take 100 g from the sample, there will be 4.07 g of hydrogen in it)
♦ 24.27% of the mass is carbon
(In other words, if we take 100 g from the sample, there will be 24.27 g of carbon in it)
♦ 71.65% of the mass is chlorine
(In other words, if we take 100 g from the sample, there will be 71.65 g of chlorine in it)
• We will learn about the 'required chemical tests' to find the above percentages, in later chapters
2. Now we apply the Eq.1.1 in reverse. We get:
(i) For hydrogen:
$\mathbf\small{4.07=\left(\frac{(nA)_{\rm{H}} \times (GAM)_{\rm{H}}}{(GMM)_{\rm{Compound}}}\right)\times 100}$
$\mathbf\small{\Rightarrow 4.07=\left(\frac{(nA)_{\rm{H}} \times 1.008}{(GMM)_{\rm{Compound}}}\right)\times 100}$
$\mathbf\small{\Rightarrow \left(\frac{(nA)_{\rm{H}}}{(GMM)_{\rm{Compound}}}\right)=\frac{4.07}{1.008 \times 100}=0.0404}$
(ii) For carbon:
$\mathbf\small{24.27=\left(\frac{(nA)_{\rm{C}} \times (GAM)_{\rm{C}}}{(GMM)_{\rm{Compound}}}\right)\times 100}$
$\mathbf\small{\Rightarrow 24.27=\left(\frac{(nA)_{\rm{C}} \times 12.01}{(GMM)_{\rm{Compound}}}\right)\times 100}$
$\mathbf\small{\Rightarrow \left(\frac{(nA)_{\rm{C}}}{(GMM)_{\rm{Compound}}}\right)=\frac{24.27}{12.01 \times 100}=0.0202}$
(iii) For chlorine:
$\mathbf\small{71.65=\left(\frac{(nA)_{\rm{Cl}} \times (GAM)_{\rm{Cl}}}{(GMM)_{\rm{Compound}}}\right)\times 100}$
$\mathbf\small{\Rightarrow 71.65=\left(\frac{(nA)_{\rm{Cl}} \times 35.45}{(GMM)_{\rm{Compound}}}\right)\times 100}$
$\mathbf\small{\Rightarrow \left(\frac{(nA)_{\rm{Cl}}}{(GMM)_{\rm{Compound}}}\right)=\frac{71.65}{35.45 \times 100}=0.0202}$
3. In the above steps, we obtained 3 quantities
♦ The numerator as well as denominator on the left sides are not known
♦ The numerator as well as denominator on the right sides are known
• We now take the ratio of their left sides:
$\mathbf\small{\left(\frac{(nA)_{\rm{H}}}{(GMM)_{\rm{Compound}}}\right):\left(\frac{(nA)_{\rm{C}}}{(GMM)_{\rm{Compound}}}\right):}\left(\frac{(nA)_{\rm{Cl}}}{(GMM)_{\rm{Compound}}}\right)$
• The ratio of the right sides is:
0.0404 : 0.0202 : 0.0202
4. But in the left side, 'GMMCompound' is common. So we get:
(nA)H : (nA)C : (nA)Cl = 0.0404 : 0.0202 : 0.0202
5. The right side in (4) can be reduced to the lowest form
For that, we divide all values by the lowest value '0.0202'
We get:
(nA)H : (nA)C : (nA)Cl = 0.0404⁄0.0202 : 0.0202⁄0.0202 : 0.0202⁄0.0202 = 2 : 1 : 1
6. So we can write:
• In the given sample, hydrogen, carbon and chlorine are present in the ratio 2 : 1 : 1
• Thus we can write the empirical formula of the given compound as: C1H2Cl1
■ Empirical formula is a formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.
7. So our next aim is to find the actual number of each atom
• For that, we can make use of the empirical formula
• The 'subscripts in the empirical formula' are in a ratio
• Recall the properties of ratios:
We can multiply or divide all values in a ratio by any factor 'k'. The ratio will not change
• In our present case, a 'suitable value for k' will give the actual number of various atoms
• That means., the actual molecular formula of the compound in our present case is: C(1k)H(2k)Cl(1k)
8. We have to find this 'suitable k'
We have:
(i) Empirical formula = C1H2Cl1
• So empirical formula mass
= (1 × 12.01) + (2 × 1.008) + (1 × 35.45) = 49.48 g
(ii) Molecular formula = C(1k)H(2k)Cl(1k)
• So molecular formula mass
= (1k × 12.01) + (2k × 1.008) + (1k × 35.45) = 49.48k g
• Dividing (ii) by (i), we get:
Molecular formula mass⁄Empirical formula mass = 49.48k⁄49.48= k
9. So we can easily find 'k'
• But there is a problem:
The 'molecular formula mass' on the left side is not known
• In this situation, there are two options:
(i) The 'molecular formula mass' will be given in the question
OR
(ii) Enough data will be given so that we can calculate the 'molecular formula mass' ourselves
■ Note that, the 'molecular formula mass' is same as the GMM
10. In our present case, the molecular formula mass (same as molar mass or GMM) is given as 98.96 g
• So we get:
Molecular formula mass⁄Empirical formula mass = 98.96⁄49.48 = k = 2
11. Once we find 'k', we can easily write the molecular formula using the result in (7)
The actual molecular formula is:
C(1k)H(2k)Cl(1k) = C2H4Cl2
12. Alternate method:
We have Eq.1.2:
$\mathbf\small{\left(\frac{(nA)_{\rm{Element}}}{(GMM)_{\rm{Compound}}}\right)=\frac{\text{% of the element}}{\text{GAM of the element} \times 100}}$
Using this equation, we can directly obtain $\mathbf\small{(nA)_{\rm{Element}}}$
(i) For carbon, we have:
$\mathbf\small{\left(\frac{(nA)_{\rm{C}}}{(GMM)_{\rm{Compound}}}\right)=\frac{\text{% of C}}{\text{GAM of C} \times 100}}$
$\mathbf\small{\Rightarrow \left(\frac{(nA)_{\rm{C}}}{98.96}\right)=\frac{24.27}{12.01 \times 100}}$
$\mathbf\small{\Rightarrow (nA)_{\rm{C}}=\frac{24.27 \times 98.96}{12.01 \times 100}=1.999=2}$
(ii) For hydrogen, we have:
$\mathbf\small{\left(\frac{(nA)_{\rm{H}}}{(GMM)_{\rm{Compound}}}\right)=\frac{\text{% of H}}{\text{GAM of H} \times 100}}$
$\mathbf\small{\Rightarrow \left(\frac{(nA)_{\rm{H}}}{98.96}\right)=\frac{4.07}{1.008 \times 100}}$
$\mathbf\small{\Rightarrow (nA)_{\rm{H}}=\frac{4.07 \times 98.96}{1.008 \times 100}=3.995=4}$
(iii) For chlorine, we have:
$\mathbf\small{\left(\frac{(nA)_{\rm{Cl}}}{(GMM)_{\rm{Compound}}}\right)=\frac{\text{% of Cl}}{\text{GAM of Cl} \times 100}}$
$\mathbf\small{\Rightarrow \left(\frac{(nA)_{\rm{Cl}}}{98.96}\right)=\frac{71.65}{35.45 \times 100}}$
$\mathbf\small{\Rightarrow (nA)_{\rm{Cl}}=\frac{71.65 \times 98.96}{35.35 \times 100}=2.005=2}$
• So the molecular formula is: C2H4Cl2
• In this method, there is no need to find the empirical formula
♦ The numerator as well as denominator on the left sides are not known
♦ The numerator as well as denominator on the right sides are known
• We now take the ratio of their left sides:
$\mathbf\small{\left(\frac{(nA)_{\rm{H}}}{(GMM)_{\rm{Compound}}}\right):\left(\frac{(nA)_{\rm{C}}}{(GMM)_{\rm{Compound}}}\right):}\left(\frac{(nA)_{\rm{Cl}}}{(GMM)_{\rm{Compound}}}\right)$
• The ratio of the right sides is:
0.0404 : 0.0202 : 0.0202
4. But in the left side, 'GMMCompound' is common. So we get:
(nA)H : (nA)C : (nA)Cl = 0.0404 : 0.0202 : 0.0202
5. The right side in (4) can be reduced to the lowest form
For that, we divide all values by the lowest value '0.0202'
We get:
(nA)H : (nA)C : (nA)Cl = 0.0404⁄0.0202 : 0.0202⁄0.0202 : 0.0202⁄0.0202 = 2 : 1 : 1
6. So we can write:
• In the given sample, hydrogen, carbon and chlorine are present in the ratio 2 : 1 : 1
• Thus we can write the empirical formula of the given compound as: C1H2Cl1
■ Empirical formula is a formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.
7. So our next aim is to find the actual number of each atom
• For that, we can make use of the empirical formula
• The 'subscripts in the empirical formula' are in a ratio
• Recall the properties of ratios:
We can multiply or divide all values in a ratio by any factor 'k'. The ratio will not change
• In our present case, a 'suitable value for k' will give the actual number of various atoms
• That means., the actual molecular formula of the compound in our present case is: C(1k)H(2k)Cl(1k)
8. We have to find this 'suitable k'
We have:
(i) Empirical formula = C1H2Cl1
• So empirical formula mass
= (1 × 12.01) + (2 × 1.008) + (1 × 35.45) = 49.48 g
(ii) Molecular formula = C(1k)H(2k)Cl(1k)
• So molecular formula mass
= (1k × 12.01) + (2k × 1.008) + (1k × 35.45) = 49.48k g
• Dividing (ii) by (i), we get:
Molecular formula mass⁄Empirical formula mass = 49.48k⁄49.48= k
9. So we can easily find 'k'
• But there is a problem:
The 'molecular formula mass' on the left side is not known
• In this situation, there are two options:
(i) The 'molecular formula mass' will be given in the question
OR
(ii) Enough data will be given so that we can calculate the 'molecular formula mass' ourselves
■ Note that, the 'molecular formula mass' is same as the GMM
10. In our present case, the molecular formula mass (same as molar mass or GMM) is given as 98.96 g
• So we get:
Molecular formula mass⁄Empirical formula mass = 98.96⁄49.48 = k = 2
11. Once we find 'k', we can easily write the molecular formula using the result in (7)
The actual molecular formula is:
C(1k)H(2k)Cl(1k) = C2H4Cl2
12. Alternate method:
We have Eq.1.2:
$\mathbf\small{\left(\frac{(nA)_{\rm{Element}}}{(GMM)_{\rm{Compound}}}\right)=\frac{\text{% of the element}}{\text{GAM of the element} \times 100}}$
Using this equation, we can directly obtain $\mathbf\small{(nA)_{\rm{Element}}}$
(i) For carbon, we have:
$\mathbf\small{\left(\frac{(nA)_{\rm{C}}}{(GMM)_{\rm{Compound}}}\right)=\frac{\text{% of C}}{\text{GAM of C} \times 100}}$
$\mathbf\small{\Rightarrow \left(\frac{(nA)_{\rm{C}}}{98.96}\right)=\frac{24.27}{12.01 \times 100}}$
$\mathbf\small{\Rightarrow (nA)_{\rm{C}}=\frac{24.27 \times 98.96}{12.01 \times 100}=1.999=2}$
(ii) For hydrogen, we have:
$\mathbf\small{\left(\frac{(nA)_{\rm{H}}}{(GMM)_{\rm{Compound}}}\right)=\frac{\text{% of H}}{\text{GAM of H} \times 100}}$
$\mathbf\small{\Rightarrow \left(\frac{(nA)_{\rm{H}}}{98.96}\right)=\frac{4.07}{1.008 \times 100}}$
$\mathbf\small{\Rightarrow (nA)_{\rm{H}}=\frac{4.07 \times 98.96}{1.008 \times 100}=3.995=4}$
(iii) For chlorine, we have:
$\mathbf\small{\left(\frac{(nA)_{\rm{Cl}}}{(GMM)_{\rm{Compound}}}\right)=\frac{\text{% of Cl}}{\text{GAM of Cl} \times 100}}$
$\mathbf\small{\Rightarrow \left(\frac{(nA)_{\rm{Cl}}}{98.96}\right)=\frac{71.65}{35.45 \times 100}}$
$\mathbf\small{\Rightarrow (nA)_{\rm{Cl}}=\frac{71.65 \times 98.96}{35.35 \times 100}=2.005=2}$
• So the molecular formula is: C2H4Cl2
• In this method, there is no need to find the empirical formula
In the next section, we will see some solved examples
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