Chapter 12.23 - Estimation of Halogens, Sulphur, Phosphorus and Oxygen

In the previous section, we saw estimation of carbon, hydrogen and nitrogen in organic compounds. In this section, we will see estimation of halogens, sulfur, phosphorus and oxygen.

Estimation of halogens (Cl, Br, I) is done using Carius method. It can be written in 4 steps:
1. A known mass (m) of an organic compound is heated with fuming nitric acid in the presence of silver nitrate.
• The heating is done in a hard glass tube known as Carius tube.
• The heating is done using a furnace.
2. Carbon in the organic compound gets oxidized to CO₂ and hydrogen gets oxidized to H₂O
• Both will be in gaseous form.
3. The halogen in the organic compound will be converted into the corresponding silver halide (AgX). It will be in solid form.
• It is washed and dried. Then it is weighed and mass m_AgX is noted.
4. Using this m_AgX, the mass of X and it’s percentage can be calculated. The following solved example shows the procedure.

Solved example 12.24
In Carius method of estimation of halogen, 0.15 grams of an organic compound gave 0.12 grams of AgBr. Find out the percentage of bromine in the compound.
Solution:
1. Molar mass of AgBr is (108 + 80) = 188 grams
• So 188 grams of AgBr will contain 80 grams of Br
⇒ 1 gram of AgBr will contain $\frac{80}{188}$ grams of Br
⇒ 0.12 grams of AgBr will contain $0.12 \times \frac{80}{188}$ grams of Br
2. Original mass of the organic compound was 0.15 grams.
• So the percentage of Br in this compound = $\frac{0.12 \times \frac{80}{188}}{0.15}\times 100$ = 34.04%

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Estimation of Sulfur is done using a Carius tube. It can be written in 4 steps:
1. A known mass (m) of an organic compound is heated with sodium peroxide or fuming nitric acid.
• The heating is done in a Carius tube.
• The heating is done using a furnace.
2. Sulfur in the organic compound gets oxidized to sulfuric acid
3. Excess 'barium chloride solution in water' is added.
• The sulfuric acid gets precipitated as barium sulfate.
• The precipitate is washed and dried. Then it is weighed and the mass m_B is noted.
4. Using this m_B, the mass of sulfur and it’s percentage can be calculated. The following solved example shows the procedure.

Solved example 12.25
In sulfur estimation, 0.157 grams of an organic compound gave 0.4813 grams of barium sulfate. Find out the percentage of sulfur in the compound.
Solution:
1. Molar mass of BaSO₄ is (137 + 32 + 64) = 233 grams
• So 233 grams of BaSO₄ will contain 32 grams of S
⇒ 1 gram of BaSO₄ will contain $\frac{32}{233}$ grams of S
⇒ 0.4813 grams of BaSO₄ will contain $0.4813 \times \frac{32}{233}$ grams of S
2. Original mass of the organic compound was 0.157 grams.
• So the percentage of S in this compound = $\frac{0.4813 \times \frac{32}{233}}{0.157}\times 100$ = 42.10%

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Estimation of phosphorus can be done by two methods.
• Method 1 can be written in 4 steps:
1. A known mass (m) of an organic compound is heated with fuming nitric acid.
2. Phosphorus in the organic compound gets oxidized to phosphoric acid
3. Ammonia and ammonium molybdate are added.
• The phosphoric acid gets precipitated as ammonium phosphomolybdate ((NH₄)₃PO₄.12MoO₃).
• The precipitate is weighed and the mass m_A is noted.
4. Using this m_A, the mass of phosphorus and it’s percentage can be calculated.
• It can be written in 2 steps:
(i) Molar mass of (NH₄)₃PO₄.12MoO₃ = 1877 grams
• So 1877 grams of (NH₄)₃PO₄.12MoO₃ will contain 31 grams of S
(Molar mass of P atom is 31 grams)
⇒ 1 gram of (NH₄)₃PO₄.12MoO₃ will contain $\frac{31}{1877}$ grams of P
⇒ m_A grams of (NH₄)₃PO₄.12MoO₃ will contain $m_A \times \frac{31}{1877}$ grams of P
(ii) Original mass of the organic compound was m grams.
• So the percentage of P in this compound = $\frac{m_A \times \frac{31}{1877}}{m}\times 100=\frac{31 \times m_A \times 100}{1877 \times m}$

• Method 2 can be written in 4 steps. First two steps are the same.
1. A known mass (m) of an organic compound is heated with fuming nitric acid.
2. Phosphorus in the organic compound gets oxidized to phosphoric acid
3. Magnesia mixture is added.
• The phosphoric acid gets precipitated as MgNH₄PO₄.
4. This precipitate is ignited. We get: Mg₂P₂O₇
• This is weighed and the mass m_A is noted.
5. Using this m_A, the mass of phosphorus and it’s percentage can be calculated.
• It can be written in 2 steps:
(i) Molar mass of Mg₂P₂O₇ = 222 grams
• So 222 grams of Mg₂P₂O₇ will contain 62 grams of S
(Molar mass of P atom is 31 grams)
⇒ 1 gram of Mg₂P₂O₇ will contain $\frac{62}{222}$ grams of P
⇒ m_A grams of Mg₂P₂O₇ will contain $m_A \times \frac{62}{222}$ grams of P
(ii) Original mass of the organic compound was m grams.
• So the percentage of P in this compound = $\frac{m_A \times \frac{62}{222}}{m}\times 100=\frac{62 \times m_A \times 100}{222 \times m}$

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Now we will see the estimation of oxygen.
• First we will see the indirect method. It can be written in 4 steps:
1. Consider the following percentages:
    ♦ Percentage of C = P_C %
    ♦ Percentage of H = P_H %
    ♦ Percentage of N = P_N %
    ♦ Percentage of X = P_X %
    ♦ Percentage of S = P_S %
    ♦ Percentage of P = P_P %
    ♦ Percentage of O = P_O %
2. Suppose that, the given organic compound contains C, H, N and O.
• Then we can write: (P_C + P_H + P_N + P_O) = 100
• From this we get: P_O = [100 - (P_C + P_H + P_N)]
3. Suppose that, the given organic compound contains C, H, S and O.
• Then we can write: (P_C + P_H + P_S + P_O) = 100
• From this we get: P_O = [100 - (P_C + P_H + P_S)]
4. Usually we follow this method to find the percentage of O.
• That is., we add the percentages of all other elements in the given compound.
• Then we subtract that sum from 100.

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Now we will see the direct method. It can be written in 11 steps:
1. A known mass (m) of an organic compound is heated in a stream of nitrogen gas.
2. The organic compound gets decomposed and we get a gaseous mixture. Oxygen is contained in this mixture.
3. This gaseous mixture is passed over red hot coke.
• All the oxygen in the mixture will be converted into CO
• The equation is: 2C + O₂ ⟶ 2CO
4. This mixture is then passed through warm iodine pentoxide (I₂O₅). All the CO will get oxidized to CO₂.
• The equation is: I₂O₅ + 5CO ⟶ I₂ + 5CO₂
5. We see that:
   ♦ In the equation in (3), CO is on the right side.
   ♦ In the equation in (4), CO is on the left side.
• Let us make the coefficients of CO equal.
6. We can multiply (3) by ‘5’, which is the coefficient of CO in (4)
We get: 10C + 5O₂ ⟶ 10CO
7. We can multiply (4) by ‘2’, which is the coefficient of CO in (3)
We get:  2I₂O₅ + 10CO ⟶ 2I₂ + 10CO₂
8. Adding (6) and (7), we get:
10C + 5O₂ + 2I₂O₅ + 10CO ⟶ 10CO +  2I₂ + 10CO₂
• Canceling 10 CO on either sides, we get:
10C + 5O₂ + 2I₂O₅ ⟶ 2I₂ + 10CO₂
9. In this equation, the I₂O₅ was externally added. It remains as such.
• So we can write:
Five moles of O₂ give ten moles of CO₂.
• This is same as:
One mole O₂ gives moles of CO₂.
10. Thus, by using the mass of CO₂ produced, we can find the mass of O₂ in the original organic compound.
• It can be explained in 5 steps:
(i) Let the mass of CO₂ produced be m_C grams
(ii) Molar mass of CO₂ = 44 grams.
• So number of moles of CO₂ produced = $\frac{m_C}{44}$
(iii) Availability of one mole of CO₂ means that, 0.5 moles of O₂ is present.
• So the availability of $\frac{m_C}{44}$ moles of CO₂ means that, $0.5 \times \frac{m_C}{44}$ moles of O₂ is present.
(iv) One mole of O₂ has a mass of 32 grams.
• So $0.5 \times \frac{m_C}{44}$ moles will have a mass of $0.5 \times \frac{m_C}{44} \times 32=\frac{32 \times m_C}{88}$ grams
(v) If m is the mass of the original organic compound, the percentage of O₂ will be given by: $\frac{\frac{32 \times m_C}{88}}{m} \times 100=\frac{32 \times m_C \times 100}{88 \times 100}$
11. Note that, there is I₂ in the final equation in (8).
• We see that, five moles of O₂ give two moles of I₂.
• So by using the mass of I₂ also, we can find the mass of O₂ in the original organic compound.

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• The link below gives the folder containing additional solved examples on this chapter.
• Parts 3 and 4 are related to purification, qualitative analysis and quantitative analysis

Additional solved examples

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We have completed the discussions in this chapter. In the next chapter we will see hydrocarbons.

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Chapter 1.3 - Empirical formula - Solved examples

In the previous section, we saw percentage composition. We also saw how to obtain the molecular formula from empirical formula. In this section, we will see some solved examples

Solved example 1.9

Calculate the mass percent of different elements present in sodium sulphate (Na₂SO₄)
Solution:
• We can use Eq.1.1:
Percentage mass of an element in a pure sample of any of it's compound 
= $\mathbf\small{\left(\frac{(nA)_{\rm{Element}} \times (GAM)_{\rm{Element}}}{(GMM)_{\rm{Compound}}}\right)\times 100}$
$\mathbf\small{(nA)_{\rm{Na}}}$ = 2
$\mathbf\small{(nA)_{\rm{S}}}$ = 1
$\mathbf\small{(nA)_{\rm{O}}}$ = 4
$\mathbf\small{(GAM)_{\rm{Na}}}$ = 22.99 g
$\mathbf\small{(GAM)_{\rm{S}}}$ = 32.06 g
$\mathbf\small{(GAM)_{\rm{O}}}$ = 16.00 g
$\mathbf\small{(GMM)_{\rm{Na_2SO_4}}}$ 
= (2 × 22.99) + (1 × 32.06) + (4 × 16.00) = 142.04 g 
(i) For sodium, we have:
Percentage mass of sodium = $\mathbf\small{\left(\frac{(nA)_{\rm{Na}} \times (GAM)_{\rm{Na}}}{(GMM)_{\rm{Na_2SO_4}}}\right)\times 100}$
= $\mathbf\small{\left(\frac{2 \times 22.99}{142.04}\right)\times 100=32.37\text{%}}$ 

(ii) For sulphur, we have:
Percentage mass of sulphur = $\mathbf\small{\left(\frac{(nA)_{\rm{S}} \times (GAM)_{\rm{S}}}{(GMM)_{\rm{Na_2SO_4}}}\right)\times 100}$
= $\mathbf\small{\left(\frac{1 \times 32.06}{142.04}\right)\times 100=22.57\text{%}}$

(iii) For oxygen, we have:
Percentage mass of oxygen = $\mathbf\small{\left(\frac{(nA)_{\rm{O}} \times (GAM)_{\rm{O}}}{(GMM)_{\rm{Na_2SO_4}}}\right)\times 100}$
= $\mathbf\small{\left(\frac{4 \times 16.00}{142.04}\right)\times 100=45.06\text{%}}$ 

Solved example 1.10
Determine the empirical formula and molecular formula of an oxide of iron, which has 69.9% iron and 30.1% oxygen by mass. Given that the molar mass of the oxide is 159.69 g mol^-1.
Solution:
1. First, we use Eq.1.2:
$\mathbf\small{\left(\frac{(nA)_{\rm{Element}}}{(GMM)_{\rm{Compound}}}\right)=\frac{\text{% of the element}}{\text{GAM of the element} \times 100}}$
• % of iron = 69.9%
• % of oxygen = 30.1%
• GAM of iron = 55.85 g
• GAM of oxygen = 16.00
(i) For iron, we have:
$\mathbf\small{\left(\frac{(nA)_{\rm{Fe}}}{(GMM)_{\rm{Compound}}}\right)=\frac{\text{69.9}}{\text{55.85} \times 100}=0.0125}$
(ii) For oxygen, we have:
$\mathbf\small{\left(\frac{(nA)_{\rm{O}}}{(GMM)_{\rm{Compound}}}\right)=\frac{\text{30.1}}{\text{16.00} \times 100}=0.0188}$
2. Now we take the ratio:
$\mathbf\small{\left(\frac{(nA)_{\rm{Fe}}}{(GMM)_{\rm{Compound}}}\right):\left(\frac{(nA)_{\rm{O}}}{(GMM)_{\rm{Compound}}}\right)}$
Since GMM_compound is common, the ratio is same as:
(nA)_Fe : (nA)_O
3. So we get:
(nA)_Fe : (nA)_O = 0.0125 : 0.0188
• A ratio must be in the form of whole numbers. To achieve that, we multiply/divide all values in a ratio by suitable numbers. The ratio will not change
• Dividing the right side by the smallest value '0.0125', we get:
(nA)_Fe : (nA)_O = 1 : 1.50
• Multiplying the right side by 2, we get:
(nA)_Fe : (nA)_O = 2 : 3
4. So empirical formula is: Fe₂O₃
• Empirical formula mass = (2 × 55.85) + (3 × 16.00) = 159.7 g
• Given that, molar mass = 159.69
• So we get: k = ^{Molecular formula mass}⁄_{Empirical formula mass} = ^159.69⁄_159.7 = 0.9999 = 1
5. So molecular formula is same as the empirical formula
• We can write:
The required molecular formula is Fe₂O₃.
6. Another method to find molecular formula without finding empirical formula:
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 iron, we have:
$\mathbf\small{\left(\frac{(nA)_{\rm{Fe}}}{(GMM)_{\rm{Compound}}}\right)=\frac{\text{% of Fe}}{\text{GAM of Fe} \times 100}}$
$\mathbf\small{\Rightarrow \left(\frac{(nA)_{\rm{Fe}}}{159.69}\right)=\frac{69.9}{55.85 \times 100}}$
$\mathbf\small{\Rightarrow (nA)_{\rm{Fe}}=\frac{69.9 \times 159.69}{55.85 \times 100}=1.998=2}$
(ii) For oxygen, we have:
$\mathbf\small{\left(\frac{(nA)_{\rm{O}}}{(GMM)_{\rm{Compound}}}\right)=\frac{\text{% of O}}{\text{GAM of O} \times 100}}$
$\mathbf\small{\Rightarrow \left(\frac{(nA)_{\rm{O}}}{159.69}\right)=\frac{30.1}{16.00 \times 100}}$
$\mathbf\small{\Rightarrow (nA)_{\rm{O}}=\frac{30.1\times 159.69}{16.00 \times 100}=3.004=3}$
• So the molecular formula is: Fe₂O₃

Solved example 1.11
The empirical formula of a compound is CH₂O and it's molecular mass is 150 g. Find it's molecular formula
Solution:
1. Empirical formula is CH₂O 
• So empirical formula mass = (1 × 12.01) + (2 × 1.008) + (1 × 16.00) = 30.026 g
2. Given that, molecular mass = 150 g
• So we get: k = ^{Molecular formula mass}⁄_{Empirical formula mass} = ¹⁵⁰⁄_30.026 = 4.995 = 5
3. So the molecular formula is C_1kH_2kO_1k = C₅H₁₀O₅.

Solved example 1.12
A compound on analysis was found to contain 18.5% carbon, 1.55% hydrogen, 55.04% chlorine and 24.81% oxygen. Find it's empirical formula
Solution:
1. First, we use Eq.1.2:
$\mathbf\small{\left(\frac{(nA)_{\rm{Element}}}{(GMM)_{\rm{Compound}}}\right)=\frac{\text{% of the element}}{\text{GAM of the element} \times 100}}$
• % of carbon = 18.5%
• % of hydrogen = 1.55%
• % of chlorine = 55.04%
• % of oxygen = 24.81%
• GAM of carbon= 12.01 g
• GAM of hydrogen = 1.008 g
• GAM of chlorine = 35.45 g
• GAM of oxygen = 16.00 g
(i) For carbon, we have:
$\mathbf\small{\left(\frac{(nA)_{\rm{C}}}{(GMM)_{\rm{Compound}}}\right)=\frac{\text{18.5}}{\text{12.01} \times 100}=0.0154}$
(ii) For hydrogen, we have:
$\mathbf\small{\left(\frac{(nA)_{\rm{H}}}{(GMM)_{\rm{Compound}}}\right)=\frac{\text{1.55}}{\text{1.008} \times 100}=0.0154}$
(iii) For chlorine, we have:
$\mathbf\small{\left(\frac{(nA)_{\rm{Cl}}}{(GMM)_{\rm{Compound}}}\right)=\frac{\text{55.04}}{\text{35.45} \times 100}=0.0155}$
(iv) For oxygen, we have:
$\mathbf\small{\left(\frac{(nA)_{\rm{O_2}}}{(GMM)_{\rm{Compound}}}\right)=\frac{\text{24.81}}{\text{16.00} \times 100}=0.0155}$
2. Now we take the ratio:
$\mathbf\small{\left(\frac{(nA)_{\rm{C}}}{(GMM)_{\rm{Compound}}}\right):\left(\frac{(nA)_{\rm{H}}}{(GMM)_{\rm{Compound}}}\right):\left(\frac{(nA)_{\rm{Cl}}}{(GMM)_{\rm{Compound}}}\right)}:\left(\frac{(nA)_{\rm{O}}}{(GMM)_{\rm{Compound}}}\right)$
Since GMM_compound is common, the ratio is same as:
(nA)_C : (nA)_H : (nA)_Cl : (nA)_O
3. So we get:
(nA)_C : (nA)_H : (nA)_Cl : (nA)_O = 0.0154 : 0.0154 : 0.0155 : 0.0155
• A ratio must be in the form of whole numbers. To achieve that, we multiply/divide all values in a ratio by suitable numbers. The ratio will not change
• Dividing the right side by the smallest value '0.0154', we get:
(nA)_C : (nA)_H : (nA)_Cl : (nA)_O = 1 : 1 : 1.006 : 1.006 = 1 : 1 : 1 : 1
4. So empirical formula is: CHClO

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• In the above problems, we see a peculiarity:
Whenever it is required to find the molecular formula, the molecular mass is given directly in the question. In a later section, we will see problems in which we have to find the molecular mass ourselves
We will see such problems after learning stoichiometry and molarity
Links to some difficult problems are given below:

Solved example 1.19

Solved example 1.29

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In the next section, we will see stoichiometry

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Chapter 1.2 - Molecular formula from Empirical formula

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

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 
• 2N_A hydrogen atoms
• 1N_A oxygen atoms
(iii) So in (⁵⁰⁄_18.02) moles, there will be:
(⁵⁰⁄_18.02) × 2N_A hydrogen atoms
(⁵⁰⁄_18.02) × 1N_A 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
⇒ N_A 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
⇒ N_A 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 
= $\mathbf\small{\left(\frac{(nA)_{\rm{Element}} \times (GAM)_{\rm{Element}}}{(GMM)_{\rm{Compound}}}\right)\times 100}$

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Let us apply this formula to the elements in ethanol:

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

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So now we can find the percentage mass of any element. Let us put it into practical use:

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:
$\mathbf\small{71.65=\left(\frac{(nA)_{\rm{Cl}} \times (GAM)_{\rm{Cl}}}{(GMM)_{\rm{Compound}}}\right)\times 100}$
$\mathbf\small{\Rightarrow 71.65=\left(\frac{(nA)_{\rm{Cl}} \times 35.45}{(GMM)_{\rm{Compound}}}\right)\times 100}$
$\mathbf\small{\Rightarrow \left(\frac{(nA)_{\rm{Cl}}}{(GMM)_{\rm{Compound}}}\right)=\frac{71.65}{35.45 \times 100}=0.0202}$

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The above 3 sets of calculations give us an idea to write a general equation
Eq.1.2:

$\mathbf\small{\left(\frac{(nA)_{\rm{Element}}}{(GMM)_{\rm{Compound}}}\right)=\frac{\text{% of the element}}{\text{GAM of the element} \times 100}}$

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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, 'GMM_Compound' 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

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In the next section, we will see some solved examples

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