Empirical Formula: In the study of a chemical system, we need to represent elements and compounds very frequently. Therefore, in chemistry, the elements and compounds are represented in abbreviated forms. The abbreviated representation of an element or a compound is called chemical formula. Still, there is another way of representing compounds by their simple whole-number ratio of different types of atoms present in one compound molecule. 

The easiest definition of empirical formula is that it is the simplest ratio of the number of atoms involved in the compound’s formation. Any compound’s chemical formula can be defined using one of two types of formulas: molecular formulas and empirical formulas. Both the empirical formula and the molecular formula represent the atom’s number and identity. In this article, we will study in detail the empirical formula and how to calculate it.

Empirical Formula Definition

• The empirical formula is the simplest formula for a compound, defined as the ratio of subscripts of the formula’s smallest conceivable an entire number of parts. It is sometimes referred to as the simplest formula.
• It is the formula of a compound expressed with the smallest integer subscript.
• It provides details about the atom ratio in the compound. The simplest formula of a compound is directly related to its per cent composition.

How to Calculate Empirical Formula?

To calculate the empirical formula, first determine the relative masses of the elements present. You have the option of using mass data in grams or percent composition.
For percent composition, we assume that the total percentage of a compound is equal to \(100\) percent and that the percent composition is the same in grams.
The compound, for example, has a total mass of \(100\) grams. If a compound contains \(68\) percent carbon, \(9\) percent hydrogen, and \(23\) percent oxygen, we would assume \(68\) grams of carbon, \(9\) grams of hydrogen, and \(23\) grams of oxygen.

The steps for determining a compound’s empirical formula are as follows:

1st Step: Calculate the mass of each element in grams.
Element percentage \( = \) mass in grams \( = {\text{m}}\)
2nd Step: Count the number of moles of each type of atom that is present.
\({\rm{m/atomic mass}}\,{\rm{ = }}\,{\rm{molar quantity }}\left( {\rm{M}} \right)\)
3rd Step: Divide the number of moles of each element from the smallest number of moles found in the previous step.
\({\rm{Atomic Ratio}}\,{\rm{ = }}\,{\rm{M/least M value }}\left( {\rm{R}} \right)\)
4th Step: Converting numbers to whole numbers is as simple as multiplying one by the smallest number, which yields only whole numbers. The simplest formula utilises these whole numbers as subscripts.
Empirical Formula \( = {{\text{R}}^*}\) whole number

Relationship between Empirical Formula and Molecular Formula

A compound’s molecular formula is a whole number multiple of its empirical formula. Thus, Molecular Formula \( = {\text{n}} \times \) Empirical Formula
where \({\text{n}}\) is an integer having values \(1,2,3,4,\) etc.
The value of n for a particular compound is given by:
\({\text{n=}}\frac{{{\text{Molecular}}\,{\text{mass}}}}{{{\text{Empirical}}\,{\text{formula}}\,{\text{mass}}}}\)

Empirical Formula Mass

The empirical formula mass is the sum of atomic masses of all atoms present in the empirical formula of the compound.
For example, for glucose, the value of \({\text{n}}\) can be obtained as follows: The empirical formula of glucose is \({\text{C}}{{\text{H}}_2}{\text{O}}\) while its molecular formula is \({{\text{C}}_6}{{\text{H}}_{12}}{{\text{O}}_6}.\)
Therefore, the empirical formula mass \( = 12.010\left({2 \times 1.008} \right) + 16 = 30.026\)
And the molecular mass \( = \left({6 \times 12.01}\right) + \left({12 \times 1.008} \right) + \left({6 \times 16} \right) = 180\)
Here,
\({\text{n=}}\frac{{{\text{Molecular}}\,{\text{mass}}}}{{{\text{Empirical}}\,{\text{formula}}\,{\text{mass}}}} = \frac{{180}}{{30}} = 6\)

Difference between an Empirical Formula and Molecular Formula

The empirical formula is distinct from the molecular formula in that it represents the simplest ratio of atoms involved in the compound. In contrast, the molecular formula represents the total number of atoms of an element present in the compound. The simplest formula represents the percentage of elements in a compound.

Solved Examples of Empirical Formula

Q.1. The empirical formula of aluminium oxide, which has \(1.08\,{\text{g}}\) of aluminium, combines chemically with \(0.96\,{\text{g}}\) of oxygen.
Ans: Mass of aluminium \( = 1.08\,{\text{g}}\)
Mass of oxygen \(0.96\,{\text{g}}\)
Number of moles \( = {\text{mass}}/{\text{atomic}}\,{\text{mass}}\)
No. of moles of aluminum \( = 1.08/27 = 0.04\)
Number of moles of oxygen \( = 0.96/16 = 0.06\)
Ratio of Al moles \( = 0.04/0.04 = 1\)
Ratio of oxygen moles \( = 0.06/0.04 = 1.5\)
Since the ratio must contain the simplest whole number, the ratio is \(2:3.\) Thus, the simplest formula is \({\text{A}}{{\text{l}}_2}{{\text{O}}_3}.\)

Calculation of Empirical Formula from the Percentage Composition

Q.2. A compound was discovered to contain \(32.65\% \) sulphur, \(65.32\% \) oxygen, and \(2.04\% \) hydrogen. What is the compound’s simplest formula?
Ans: Step 1) Convert the percentage to grams.
\(32.65\) percent \( = 32.65\,{\text{g}}\) of \({\text{S}}\)
\(65.3\) percent \( = 65.3\,{\text{g}}\,{\text{O}}\)
\(2.04\) percent \( = 2.04\,{\text{g}}\) of \({\text{H}}\)
Step 2) Next, divide each given mass by its molar mass.
\(32.65{\mkern 1mu} {\rm{g}}/32\,{\mkern 1mu} {\rm{g}}\,{\mkern 1mu} {\rm{mo}}{{\rm{l}}^{ – 1}} = 1.0203{\mkern 1mu} \,{\rm{moles}}{\mkern 1mu} \,{\rm{S}}\)
\(65.3{\mkern 1mu} {\rm{g}}/16{\mkern 1mu} \,{\rm{g}}{\mkern 1mu} \,{\rm{mo}}{{\rm{l}}^{ – 1}} = 4.08{\mkern 1mu} \,{\rm{moles}}\,{\mkern 1mu} {\rm{O}}\)
\(2.04{\mkern 1mu} {\rm{g}}/1.008{\mkern 1mu} \,{\rm{g}}{\mkern 1mu} \,{\rm{mo}}{{\rm{l}}^{ – 1}} = 2.024{\mkern 1mu} \,{\rm{moles}}{\mkern 1mu} \,{\rm{H}}\)
Step 3) Next, take the smallest Answer in moles from the previous step and divide all of the others by it. Remember to round off to the nearest whole number when calculating \( \times 0.9\) numbers:
\(1.0203\) moles of \({\text{S}}/1.2 = 0203 = 1\)
\(4.08\) moles of \({\text{O}}/1.0203 = 3.998 \simeq 4\)
\(2024\) moles of \({\text{H}}/1.0203 \simeq 2\)
Step 4) Finally, the coefficients calculated in the previous step will become the chemical formula’s subscripts.
\({\text{S=1}}\)
\({\text{O=4}}\)
\({\text{H=2}}\)
Therefore, the empirical formula will become \({{\text{H}}_2}{\text{S}}{{\text{O}}_4}.\)

Calculation of Molecular Formulas from the Simplest Formula

Q.3. A compound contains \(4.07\% \) hydrogen, \(24.27\% \) carbon and \(71.65\% \) chlorine. Its molar mass is \(98.96\,{\text{g}}.\) What are its empirical formula and molecular formula?
Ans: Step 1) Convert the percentage to grams.
\(4.07\% \) hydrogen \( = 4.07\,{\text{g}}\) of \({\text{H}}\)
\(24.27\% \) carbon \( = 24.27\,{\text{g}}\) of \({\text{C}}\)
\(71.65\% \) chlorine \( = 71.65\,{\text{g}}\) of \({\text{Cl}}\)
Step 2) Next, divide each given mass by its molar mass.
\(4.07\,{\text{g}}\) of \({\rm{H}}/1{\mkern 1mu} \,{\rm{g}}{\mkern 1mu} \,{\rm{mo}}{{\rm{l}}^{ – 1}} = 4.07{\mkern 1mu} \,{\rm{moles}}\)
\(24.27\,{\text{g}}\) of \({\rm{C}}/1{\mkern 1mu} 2{\mkern 1mu} \,{\rm{g}}{\mkern 1mu} \,{\rm{mo}}{{\rm{l}}^{ – 1}} = 2.02{\mkern 1mu} \,{\rm{moles}}\)
\(71.65\,{\text{g}}\) of \({\rm{Cl}}/35.5{\mkern 1mu} \,{\rm{g}}{\mkern 1mu} \,{\rm{mo}}{{\rm{l}}^{ – 1}} = 2.02{\mkern 1mu} \,{\rm{moles}}\)
Step 3) Next, take the smallest answer in moles from the previous step and divide all of the others by it,
\(4.07\) moles of \({\text{H}}/2.02 = 2\)
\(2.02\) moles of \({\text{C}}/2.02 = 1\)
\(2.02\) moles of \({\text{Cl}}/2.02 = 1\)
Step 4) Finally, the coefficients calculated in the previous step will become the chemical formula’s subscripts.
\({\text{H}} = 2\)
\({\text{C}} = 2\)
\({\text{Cl}} = 1\)
Therefore, the empirical formula of the compound will be \({\text{C}}{{\text{H}}_2}{\text{Cl}}{\text{.}}\)
Empirical mass of \({\text{C}}{{\text{H}}_2}{\text{Cl=12 + 2}} \times {\text{1 + 35}}{\text{.5=49}}{\text{.5}}\)
\({\text{n}} = 2\)
Molecular Formula \({\text{=n}} = \times {\text{E}}.{\text{F=2}} \times {\text{C}}{{\text{H}}_2}{\text{Cl}} = {{\text{C}}_2}{{\text{H}}_4}{\text{C}}{{\text{l}}_2}.\)

Summary

The empirical formula is a chemical formula that represents the simplest ratio of atoms in the chemical formula of the compound. Molecular Formula \( = {\text{n}} \times \) Empirical Formula is the general relationship between the empirical and molecular formulas. This can be used to determine the compound’s empirical formula as well as its molecular formula. In the relationship, n represents the compound’s molecular mass to empirical mass ratio. The empirical formula’s main role in experimental chemistry is to allow statisticians to predict the likely outcome when complete data is not available.

Also, Check –

• Molecular Formula

FAQs on Empirical Formula

Frequently asked questions related to the simplest formula are listed as follows:

Q.1: Define the molecular formula.
A: The molecular formula represents the total number of different atoms present in one molecule of the given compound. Benzene, for example, has the molecular formula \({{\text{C}}_6}{{\text{H}}_6}.\) This means that one molecule of benzene is made up of six carbon atoms and six hydrogen atoms.

Q.2: State the steps of finding the empirical formula.
A: The steps for determining a compound’s empirical formula are mentioned in the above article.

Q.3: What is the empirical mass?
A: The empirical mass is the sum of atomic masses of all atoms present in the compound’s empirical formula.

Q.4: Why do we use the empirical formula?
A: Empirical formulas are the most basic notational form. They have the smallest whole-number ratio between the compound elements. In contrast to molecular formulae, they will not know the total number of atoms in a single molecule.

Q.5: Why is the empirical rule useful?
A: In most cases, the empirical rule is used to help determine outcomes when not all of the data is available. It allows statisticians – or those studying the data – to predict where the data will fall once all of the information is available. The empirical rule can also determine how standard a set of data is.

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