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

---

• 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

---

In the next section, we will see stoichiometry

PREVIOUS           CONTENTS          NEXT


Copyright©2019 Higher Secondary Chemistry. blogspot.in - All Rights Reserved