Empirical Formula

Learning Objectives

When you finish reading this text, you will be able to:

• define what empirical formula is


• determine a compound's empirical formula


• determine a compound's chemical formula using its empirical formula and its formula mass.

Definition

The empirical formula of a compound indicates the smallest integer ratio of the compound's constituent atoms.

It simply tells us the mole-to-mole ratio of the atoms present in a compound.

Calculation of Empirical Formula

To be able to calculate the empirical formula of a compound, you need to:

Step 1	Determine the individual masses of the compound's constituent atoms.
Step 2	Convert these masses into moles.
Step 3	Divide each of these amounts in moles by the one with the smallest value.
Step 4	If the results are not whole numbers, then multiply these values by an integer that will yield the smallest whole numbers.

Example 1 and Example 2 show the step-by-step calculation of empirical formula from the known masses of atoms, and from the percent composition, of a compound, respectively.

Example 3 shows how to calculate the empirical formula of a hydrate.

Example 1 Determining the Empirical Formula of an Unknown Compound From the Masses of Atoms

Example 1  

A sample of a certain compound contains the following: 0.747 g of Na, 0.195 g of C, and 0.780 g of O.

Using the step by step guide outlined above, we will determine the empirical formula of the given compound.

Step 1 Masses of the compound's constituent atoms are given in the problem.

\begin{align}Na\ \ \ 0.747\ g\\C\ \ \ 0.195\ g\\O\ \ \ 0.780\ g\end{align}

Step 2 Convert these masses into moles by dividing each mass by the molar mass of its corresponding atom.

`mo\l\e Na:`

`(0.747 g)/(22.98977 g//mo\l\e) =  0.0325 mo\l\e`

`mo\l\e C:`

`(0.195 g)/(12.011 g//mo\l\e)=  0.0162 mo\l\e`

`mo\l\e O:`

`(0.780 g)/(15.9994 g//mo\l\e)=  0.0488 mo\l\e`

Step 3 Divide these amounts in mole by the one with the smallest value.

`Na:`

`(0.0325 mo\l\e)/(0.0162 mo\l\e) = 2.0`

`C:`

`(0.0162 mo\l\e)/(0.0162 mo\l\e) = 1.0`

`O:`

`(0.0488 mo\l\e)/(0.0162 mo\l\e) = 3.0`

Step 4 Since the results are whole numbers, our calculation is done.

The empirical formula of the compound is

Na₂CO₃.

Example 2 Determining the Empirical Formula of an Unknown Compound From Percent Composition

When the individual masses of the compound's atoms are not given explicitly, they can still be obtained using percent composition.

Example 2  

An organic halide has a percent composition of 12.82% C, 56.76% Cl and 30.42% F.

Let us find its empirical formula.

Step 1 We can determine the masses of the constituent atoms by multiplying their weight percentages by any amount we desire.

For convenience, we will use 100-g sample. In this way, the weights in grams are numerically equal to the weight percentages:

\begin{align}0.1282\ x\ 100\ g\ =\ 12.82\ g\ \ C\ \\0.5676\ x\ 100\ g\ =\ 56.76\ g\ \ Cl\\0.3042\ x\ 100\ g\ =\ 30.42\ g\ \ F\ \end{align}

Step 2 Convert these masses into moles by dividing each mass by the molar mass of its corresponding atom.

`mo\l\e C:`

`(12.82 g)/(12.011 g//mo\l\e) =  1.0674 mo\l\e`

`mo\l\e Cl:`

`(56.76 g)/(35.453 g//mo\l\e) =  1.6010 mo\l\e`

`mo\l\e F:`

`(30.42 g)/(18.9984 g//mo\l\e) =  1.6012 mo\l\e`

Step 3 Divide these amounts in mole by the one with the smallest value.

`C:`

`(1.0674 mo\l\e)/(1.0674 mo\l\e) = 1.0`

`Cl:`

`(1.6010 mo\l\e)/(1.0674 mo\l\e) = 1.5`

`F:`

`(1.6012 mo\l\e)/(1.0674 mo\l\e) = 1.5`

Step 4 Here, we obtain non-integer results:

C₁Cl_1.5F_1.5

So we multiply them with an integer that will give the smallest whole number. That number is 2.

C_{1 x 2}Cl_{1.5 x 2}F_{1.5 x 2}

Hence, the empirical formula is

C₂Cl₃F₃.

Example 3 Determining the Empirical Formula of a Hydrate

A hydrate is a compound with water molecules adsorbed on its surface. The problem is about finding the number of water molecules bound to the compound.

Example 3  

A 5.00-g sample of the hydrated CoCl₂ was analyzed and found to contain 2.27 g of H₂O.

We are going to determine the compound's empirical formula.

Step 1 Instead of using the masses of the atoms, we use the masses of CoCl₂ and H₂O.

`H_2O\ \ \ \ \ \ 2.27\ g`

`CoCl_2\ \ \ 5.00\ g - 2.27\ g = 2.73\ g`

Step 2 To convert these masses into moles, we divide them by their respective formula masses instead.

`mo\l\e\ \ H_2O:`

`(2.27\ g)/(18.0152\ g//mo\l\e)=0.1260\ mo\l\e`

`mo\l\e\ \ CoCl_2:`

`(2.73\ g)/(129.8392\ g//mo\l\e)=0.0210\ mo\l\e`

Step 3 Divide these amounts in mole by the one with the smaller value.

`H_2O:`

`(0.1260\ mo\l\e )/(0.0210\ mo\l\e) = 6.0`

`CoCl_2:`

`(0.0210\ mo\l\e )/(0.0210\ mo\l\e) = 1.0`

Step 4 Since we obtain whole numbers, the empirical formula is

(CoCl₂)₁(H₂O)₆

which can be rewritten as

CoCl₂·6H₂O

The compound is cobalt(II) chloride hexahydrate.

Use of Empirical Formula

Empirical formula is used in determining the chemical formula of a compound.

A compound's chemical formula is a multiple of its empirical formula.

Depending on the compound, the chemical formula may consist of one or more empirical formula units. (See table below.)

Table of Some Compounds With Their Chemical and Empirical Formulas

Show Table  ∨

Compound	Formula	Empirical Formula
benzene	C6H6	CH
acetylene	C2H2	CH
mercury(I) nitrate	Hg2(NO3)2	HgNO3
ammonium oxalate	(NH4)2C2O4	NH4CO2

From the table, the first two examples are molecular compounds and the last two examples are ionic compounds.

Looking at their formulas, we can see that:

• benzene has 6 empirical formula units, (CH)₆


• acetylene has 2 empirical formula units, (CH)₂


• mercury(I) nitrate has 2 empirical formula units, (HgNO₃)₂


• and ammonium oxalate has 2 empirical formula units, (NH₄CO₂)₂

Facts you should be aware of regarding the empirical formula of compounds:

Compounds having different chemical formulas can have the same empirical formulas as shown by the first two compounds in the table.

For the majority of ionic compounds, their empirical formulas are their chemical formulas also. The last two compounds given in the table are just some of the very few exceptions to this.

For a compound whose chemical formula consists of more than one empirical formula unit, deriving its correct chemical formula requires a knowledge of its empirical formula and its formula mass.

The formula mass is divided by the empirical formula mass to obtain the number of empirical formula units (efu):

`efu=(f\o\r\m\u\la\ mass)/(em\p\irical\ \f\o\r\m\u\la\ mass)`

where:

• formula mass, usually determined by experiment, is the sum of the atomic masses of the compound's constituent atoms


• empirical formula mass is the sum of the atomic masses of the atoms in the empirical formula

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NOTE: Formula mass is a generic term for the molar mass of molecular and ionic compounds.

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The subscripts of the empirical formula are multiplied by this calculated number of efu in order to obtain the correct chemical formula of the compound.

Example 4 Determining the Molecular Formula of an Unknown Molecular Compound

Show Example 4  ∨

An organic compound was found by analysis to contain the following composition: 54.53% C, 9.15% H, 36.32% O. Its molecular mass was determined to be 88.106 amu.

We'll determine its molecular formula.

Step 1 Following the method outlined in Step 1 of Example 2, we obtain the following masses:

\begin{align}C\ \ \ 54.53\ g\\H\ \ \ \ \ 9.15\ g\\O\ \ \ 36.32\ g\end{align}

Step 2 To convert these masses into moles, we divide them by their respective molar masses.

`mo\l\e\ \ C:`

`(54.53\ g)/(12.011\ g//mo\l\e)=4.5400\ mo\l\e`

`mo\l\e\ \ H:`

`(9.15\ g)/(1.0079\ g//mo\l\e)=9.0783\ mo\l\e`

`mo\l\e\ \ O:`

`(36.32\ g)/(15.9994\ g//mo\l\e)=2.2701\ mo\l\e`

Step 3 Divide these amounts in mole by the one with the smallest value.

`C:`

`(4.5400\ mo\l\e)/(2.2701\ mo\l\e) = 2.0`

`H:`

`(9.0783\ mo\l\e)/(2.2701\ mo\l\e) = 4.0`

`O:`

`(2.2701\ mo\l\e)/(2.2701\ mo\l\e) = 1.0`

Step 4 Since we obtained whole numbers, the empirical formula is

C₂H₄O.

Step 5 Now, we calculate the number of empirical formula units by dividing the given molecular mass, 88.106 amu, by the empirical formula mass which is 2(12.011) + 4(1.0079) + 15.9994 = 44.053 amu.

`efu=(f\o\r\m\u\la\ mass)/(em\p\irical\ \f\o\r\m\u\la\ mass)`

`\ \ \ \ \ \ =88.106/44.053`

`\ \ \ \ \ \ =2`

Hence, the empirical formula is to be multiplied by 2,

(C₂H₄O)₂.

The molecular formula, therefore, is

C₄H₈O₂.