Empirical and Molecular Formula: Definition, Questions and Examples

Edited By Shivani Poonia | Updated on Sep 19, 2024 07:04 PM IST

The chemical formula may be of two types- empirical formula and molecular formula. The formula which gives the simplest whole number ratio of the atoms of various elements present in one molecule of the compound is called the empirical formula whereas the formula which gives the actual number of atoms of various elements present in one molecule of the compound is called molecular formula. This formula is critical for understanding the precise structure and characteristics of a substance, essential for tasks ranging from chemical synthesis to pharmacological applications.

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This Story also Contains

What is a Chemical Formula?
Empirical Formula
Molecular Formula
Some Solved Example
Summary

Empirical and Molecular Formula: Definition, Questions and Examples

What is a Chemical Formula?

A chemical formula represents the combination of atoms of all the elements that make up a compound. It represents the relative ratio of atoms of its constituent elements. In the case of a compound, it represents one molecule, one mole, and one gram molecular weight of the compound.
For example, CuSO₄.5H₂O represents one molecule, one mole, and one gram molecular weight of hydrated copper sulfate.

Empirical Formula

It is the simplest ratio of the number of atoms of different elements present in one molecule of a compound. It does not represent the actual number of atoms of different elements present in one molecule of the compound.

How to find out the empirical formula and the molecular formula in case the % composition of the compound is given to us

Step 1. Conversion of a mass percent to grams.

Step 2. Convert grams into the number of moles of each element.

Step 3. Divide the mole value obtained above by the smallest number.

Step 4. Write the empirical formula by mentioning the numbers obtained above after writing the symbols of respective elements.

Step 5. Write molecular formula (Molecular formula) with the help of the information given.

Molecular formula is a whole number multiple of the empirical formula

Molecular formula $=(\text { Empirical formula })_{\mathrm{n}}$

where n is the whole number.

Molecular Formula

It shows the actual number of atoms of different elements present in one molecule of a compound.

n = (Molecular weight) / (Empirical formula weight)
Molecular weight can be directly given or some other information like Vapour density can be given which will enable us to calculate the molecular weight.
Molecular weight = 2 x Vapour Density
For some compounds, the molecular formula and empirical formula may be the same also.

Some Solved Example

Qu 1. A 5.325 g sample of Methyl Benzoate, a compound used in the manufacturing of perfumes, is found to contain 3.758g of Carbon, 0.316g of Hydrogen and 1.251g of Oxygen. What is the empirical formula of the compound?

1) C₄H₄O

2) C₂H₄O

3) C₂H₂O

4) C₄H₃O

Solution

For glucose, the empirical formula is CH₂O

Element	%	Mole ratio	Simplest mole ratio
C	$\frac{3.758 \times 100}{5.325}=70.57$	$\frac{70.57}{12}=5.88$	$\frac{5.88}{1.47}=4$
H	$\frac{0.316 \times 100}{5.325}=5.93$	$\frac{5.93}{1}=5.93$	$\frac{5.93}{1.47}=4$
O	$\frac{1.251 \times 100}{5.325}=23.50$	$\frac{23.50}{16}=1.47$	$\frac{1.47}{1.47}=1$

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So, the Empirical formula is C₄H₄O

Hence, the answer is an option (1).

Qu 2: An organic compound containing C & H has 92.3% C. Its empirical formula is:

1) CH₂

2) CH₃

3) CH₄

4) CH

Solution

Element	%	Atomic Mass	Relative no. of atoms	Simplest ratio
C	92.30	12	7.69	1
H	7.70	1	7.70	1

So, its empirical formula is CH

Hence the answer is an option (4).

Qu 3. The ratio of mass percent of C and H of an organic compound (C_XH_YO_Z) is 6 : 1. If one molecule of the above compound (C_XH_YO_Z) contains half as much oxygen as required to burn one molecule of compound C_XH_Y completely to CO₂ and H₂O. The empirical formula of compound C_XH_YO_Z is :

1) C₂H₄O₃
2) C₃H₆O₃
3) C₂H₄O
4) C₃H₄O₂

Solution

The mass ratio of C & H is 6 : 1

So mole ratio of C & H is 1 : 2

$\therefore$ X : Y = 1 : 2

To burn one molecule of $C_X H_Y$

$\mathrm{CH}_2+\frac{3}{2} \mathrm{O}_2 \rightarrow \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}$

$\frac{3}{2}$ molecule of O₂ is required i.e. 3 atoms of O are required so $X: Y: Z=1: 2: \frac{3}{2} \Rightarrow X: Y: Z=2: 4: 3$

So empirical formula is $\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_3$

Hence, the answer is an option (1).

Qu 3. A compound has an empirical formula C₂H₄O. An independent analysis gave a value of 132 for its molecular mass. What is the correct molecular formula?

1) C₄H₄O₅

2) C₁₀H₁₂

3) C₆H₁₂O₃

4) C₄H₈O₅

Solution

$\mathrm{n}=\frac{\text { Molecular Mass }}{\text { Empirical Formula mass }}=\frac{132}{44}=3$

Therefore, Molecular formula is (C₂H₄O)₃ = (C₆H₁₂O₃)

Hence, the answer is an option (3).

Qu 4. An organic compound contains 49.3% C, 6.84% H and the remaining is oxygen and its vapor density is 73. The molecular formula of the compound is

1) C₃H₈O₂

2) C₆H₉O

3) C₄H₁₀O₂

4) C₆H₁₀O₄

Solution

Molecular mass $=2 \times 73=146$

% of oxygen = 100 - (49.3 + 6.84)
= 43.86%

Thus we have,

$C=\frac{\%}{100} \times \frac{\text { Molecular mass }}{\text { Atomic mass }}=\frac{49.3}{100} \times \frac{146}{12}=6$

$H=\frac{\%}{100} \times \frac{\text { Molecular mass }}{\text { Atomic mass }}=\frac{6.84}{100} \times \frac{146}{1}=10$

$O=\frac{\%}{100} \times \frac{\text { Molecular mass }}{\text { Atomic mass }}=\frac{43.86}{100} \times \frac{146}{16}=4$

Thus, Molecular Formula = C₆H₁₀O₄

Hence, the answer is an option (4).

Qu 5. A compound on analysis was found to have the following composition: (i) Sodium 14.31%, (ii) Sulphur 9.97%, (iii) Oxygen 69.5%, (iv) Hydrogen 6.22%. Calculate the molecular formula of the compound assuming that the whole of Hydrogen in the compound is present as water of Crystallization. The molecular mass of the compound is 322.

1) Na₂SH₂₀O₁₀

2) Na₂SH₂₀O₁₄

3) Na₂SH₂₀O₇

4) Na₂SH₃₀O₁₆

Solution

Element	%age	Atomic mass	Relative no of atoms	Simplest Ratio
Na	14.31	23	0.622	$\frac{0.622}{0.311}=2$
S	9.97	32	0.311	$\frac{0.311}{0.311}=1$
H	6.22	1	6.22	$\frac{6.22}{0.311}=20$
O	69.50	16	4.34	$\frac{4.34}{0.311}=14$

Empirical formula = Na₂SH₂₀O₁₄

Empirical formula mass $=2 \times 23+32+20 \times 1+14 \times 16=322$

Molecular mass = 322

Molecular Formula = Na₂SH₂₀O₁₄

Hence, the answer is an option (2).

Summary

In a chemical formula, the actual number of atoms are represented in the compound. chemical formula shows the exact number of atoms present in it and their composition and the ratio of each element in a molecule. and the empirical formula represents the simple whole number of the ratio of atoms inside the compound. it show the number of atoms but not in exact quantity in the relative proportion. Empirical formula and molecular formula are both related to each other by molecular mass. empirical formula can be used to find the chemical formula with the help of molecular mass.