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 1.1 - Mole concept and Molar masses

In the previous section, we saw some basics about nature of matter. In this section, we will see properties of matter and their measurements. Also later in this section we will see problems on molecular and atomic masses

Properties of substances can be classified into two categories:
(i) Physical properties
Color, odour, melting point, boiling point etc., are examples of physical properties  
(ii) Chemical properties
Composition, combustibility, reactivity with acids and bases etc., are examples of chemical properties

■ Both physical properties and chemical properties of a substance can be measured or observed 
But there is a difference:
• For measuring or observing physical properties of a substance, we do not need to change the identity or composition of that substance
• For measuring or observing chemical properties, a chemical change must occur. After the chemical change, the substance will be be transformed to another substance

This difference will be clear from the two examples given below:
Example 1:
• To find the boiling point of a liquid substance, we heat it
• The heat is supplied until the liquid starts to boil
• The 'temperature at which the boiling begins' is noted down as the boiling point
• Even after supplying the heat, the liquid remains the same. It is not transformed into another substance
Example 2:
• To find the combustibility of carbon, we must burn it
• Burning or combustion is a reaction with oxygen
• The carbon reacts with oxygen and becomes carbon dioxide
• So carbon is changed into a new substance

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Chemists study the behaviour of substances based on the physical and chemical properties. So measurements of these properties are very essential.

Let us see how the measurements are done:
• Properties like area, volume, density, etc., are quantitative in nature
• Any quantitative measurement or observation is represented by a number followed by a unit in which it is measured
• For example, height of a wall is 2 m
    ♦ ‘2’ is the number
    ♦ ‘m’ denotes meter, the unit in which length is measured

■ We use the SI system of units for measurements. A detailed description about the importance of the units and the system of units can be seen here. The SI system is also described in detail

■ Notes on measurements of mass and time can be seen here

■ Notes on measurements of heat can be seen here

■ Notes on accuracy and precision can be seen here

■ Notes on significant figures precision can be seen here

■ Notes on errors in measurements can be seen here

■ Notes on Dimensional analysis can be seen here

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Molecular mass is the the sum of atomic masses of the elements present in a molecule. We have learnt about them in our previous classes. Detailed notes about atomic mass, molecular mass, mole concept and molarity can be seen here.

Let us see some solved examples:
Solved example 1.1
Calculate the molar mass of the following:
(i) H₂O (ii) CO₂ (iii) CH₄
Solution:
Molar mass is the mass of one mole
(i) H₂O
Mass of one mole of H₂O
= (2 × 1.008) + 16.00 = 18.02 g
(ii) CO₂
Mass of one mole of CO₂
= 12.01 + (2 × 16.00) = 44.01 g
(ii) CH₄
Mass of one mole of CH₄
= 12.01 + (4 × 1.008) = 16.04 g

Solved example 1.2
Calculate the atomic mass (average) of chlorine using the following data:
In a sample of chlorine:
(i) 75.77% is the isotope $\mathbf\small{{^{35}C}l}$ 
(17 protons and 18 neutrons)
    ♦ It has a molar mass of 34.9689
(ii) 24.23% is the isotope $\mathbf\small{{^{37}C}l}$ 
(17 protons and 20 neutrons)
    ♦ It has a molar mass of 36.9659
Solution:
From the above data, we get:
Average atomic mass of chlorine = (34.9689 × 0.7577) + (36.9659 × 0.2423) = 35.4528 g mole^-1

Solved example 1.3
Calculate the atomic mass (average) of argon using the following data:
In a sample of argon:
(i) 0.337% is the isotope $\mathbf\small{{^{36}Ar}}$ 
(18 protons and 18 neutrons)
    ♦ It has a molar mass of 35.96755
(ii) 0.063% is the isotope $\mathbf\small{{^{38}Ar}}$ 
(18 protons and 20 neutrons)
    ♦ It has a molar mass of 37.96272
(iii) 99.600% is the isotope $\mathbf\small{{^{40}Ar}}$ 
(18 protons and 22 neutrons)
    ♦ It has a molar mass of 39.9624
Solution:
From the above data, we get:
Average atomic mass of argon 
= (35.96755 × 0.00337) + (37.96272 × 0.00063) + (39.9624 × 0.9960) = 39.9477 g mole^-1

Solved example 1.4
How much copper can be obtained from 100 g of copper sulphate (CuSO₄)?
Solution:
1. One mole of CuSO₄ = 63.54 + 32.06 + (16.00 × 4) = 159.60 g
2. So no. of moles in 100 g = (¹⁰⁰⁄_159.60) = 0.6266 moles
3. So no. of molecules of CuSO₄ in 100 g = (0.6266 × N_A)
(Where N_A is the Avogadro number)
4. So no. of atoms of Cu = (0.6266 × N_A)
5. Mass of one mole of Cu = 63.54 g
• That is., N_A atoms of Cu has a mass of 63.54 g
• So 1 atom of Cu has a mass of $\mathbf\small{\frac{63.54}{N_A}}$ g
6. So from (4), we get:
Total mass of the available Cu atoms = $\mathbf\small{0.6266 \times N_A \times \frac{63.54}{N_A}=0.6266 \times 63.54 = 39.81\,g}$

Solved example 1.5
In three moles of ethane (C₂H₆), calculate the following:
(i) Number of moles of carbon atoms
(ii) Number of moles of hydrogen atoms
(iii) Number of molecules of ethane
Solution:
Part (i):
1. One mole of C₂H₆ contains N_A molecules of C₂H₆
• So 3 moles of C₂H₆ will contain (3N_A) molecules of C₂H₆
2. In (3N_A) molecules of C₂H₆, there will be (2 × 3N_A) atoms of carbon
• That is., (6N_A) atoms of carbon
3. N_A atoms of carbon is one mole of carbon
• So 6N_A atoms of carbon is 6 moles of carbon
Part (ii):
1. One mole of C₂H₆ contains N_A molecules of C₂H₆
• So 3 moles of C₂H₆ will contain (3N_A) molecules of C₂H₆
2. In (3N_A) molecules of C₂H₆, there will be (6 × 3N_A) atoms of hydrogen
• That is., (18N_A) atoms of hydrogen
3. N_A atoms of hydrogen is one mole of hydrogen
• So 18N_A atoms of hydrogen is 18 moles of hydrogen
Part (iii):
• One mole of C₂H₆ contains N_A molecules of C₂H₆
• So 3 moles of C₂H₆ will contain (3N_A) molecules of C₂H₆
• That is., (3 × 6.02 × 10²³) molecules of C₂H₆.

Solved example 1.6
Which one of the following will have the largest number of atoms?
(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g of Cl2(g)
Solution:
Part (i):
1 mole of Au has a mass of 196.97 g
⇒ 196.97 g of gold will contain N_A atoms of gold
⇒ 1 g of gold will contain $\mathbf\small{\left(\frac{N_A}{196.97}\right)}$ atoms of gold
Part (ii):
1 mole of Na has a mass of 22.99 g
⇒ 22.99 g of sodium will contain N_A atoms of sodium
⇒ 1 g of sodium will contain $\mathbf\small{\left(\frac{N_A}{22.99}\right)}$ atoms of sodium
Part (iii):
1 mole of Li has a mass of 6.94 g
⇒ 6.94 g of lithium will contain N_A atoms of lithium
⇒ 1 g of lithium will contain $\mathbf\small{\left(\frac{N_A}{6.94}\right)}$ atoms of lithium
Part (iv):
• Cl₂ is a molecule of chlorine
1 mole of chlorine molecules has a mass of (35.45 × 2) = 70.9 g
⇒ 70.9 g of chlorine will contain N_A molecules of chlorine
⇒ 1 g of chlorine will contain $\mathbf\small{\left(\frac{N_A}{70.9}\right)}$ molecules of chlorine
⇒ 1 g of chlorine will contain $\mathbf\small{\left(\frac{2N_A}{70.9}\right)}$ atoms of chlorine
⇒ 1 g of chlorine will contain $\mathbf\small{\left(\frac{N_A}{35.45}\right)}$ atoms of chlorine
• All the four results have the same numerator
• So the result with the smallest denominator will be the largest
■ Thus we can write:
Among the given four samples, the lithium sample will have the largest number of atoms 

Solved example 1.7
What will be the mass of one $\mathbf\small{{^{12}C}}$ atom in g?
Solution:
• Molar mass of $\mathbf\small{{^{12}C}}$ = 12 g
⇒ 1 mole of $\mathbf\small{{^{12}C}}$ has a mass of 12 g
⇒ N_A atoms of $\mathbf\small{{^{12}C}}$ have a mass of 12 g
⇒ 1 atom of $\mathbf\small{{^{12}C}}$ has a mass of $\mathbf\small{\left(\frac{12}{N_A}\right)}$ g
• $\mathbf\small{\frac{12}{N_A}=\frac{12}{6.022 \times 10^{23}}=1.9927 \times 10^{-23}}$ g

Solved example 1.8
Calculate the number of atoms in each of the following:
(i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He
Solution:
Part (i):
• 1 mole of Ar atoms contains N_A atoms of Ar
⇒ 52 moles of Ar will contain (52N_A) = (52 × 6.022 × 10²³) = 313.144 × 10²³ = 3.13 × 10²⁵ atoms
Part (ii):
1. 1 u = 1.6605 × 10^-25 g
52 u = 52 × 1.6605 × 10^-25 g
2. Molar mass of Ar is 39.95 g
⇒ 39.95 g of Ar will contain N_A atoms
⇒1 g of Ar will contain $\mathbf\small{\left(\frac{N_A}{39.95}\right)}$ atoms
3. So (52 × 1.6605 × 10^-25) g will contain $\mathbf\small{52 \times 1.6605 \times 10^{-25} \times \left(\frac{N_A}{39.95}\right)}$ atoms
⇒ 52 u of Ar will contain $\mathbf\small{52 \times 1.6605 \times 10^{-25} \times \left(\frac{6.022 \times 10^{23}}{39.95}\right)=13.02}$ atoms
• So 52 u of Ar will contain approximately 13 Ar atoms
Part (iii):
1. Molar mass of He is 4.00 g
⇒ 4.00 g of He contains N_A atoms of He
2. 1 g of He contains $\mathbf\small{\left(\frac{N_A}{4}\right)}$ atoms of He
⇒ 52 g of He will contain:
$\mathbf\small{52 \times\left(\frac{N_A}{4}\right)=52 \times\left(\frac{6.022 \times 10^{23}}{4}\right)=7.8286 \times 10^{24}}$ atoms of He

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We will see more problems related to mole concept in the coming sections. In the next section, we will see percentage composition

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