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# Empirical Formula

We require some way to express the compound or molecule like an atom is expressed with the help of atomic symbols. We know a compound is composed of different atoms which can be expressed in terms of atomic symbols. Hence a molecule or compound can also be expressed with the help of atomic symbols only. But we should know the number of constituent atoms or percentage of them to write the molecular formula.

A chemical formula or molecular formula helps in the identification of each constituent element by using chemical symbols. It also indicates the number of atoms present in each molecule of that compound. The numbers of constituent atoms can be indicated with the help of subscript after the chemical symbol of atoms such as one methane molecule contains one carbon atom and 4 hydrogen atoms, therefore the chemical formula can be written as $CH_4$. Similarly one glucose molecule contains 6 C-atoms, 12 H-atoms and 6 O-atoms. Hence the molecular formula will be $C_{6}H_{12}O_{6}$.

Chemical formulas can be different types such as molecular formula, empirical formula and structural formula. A molecular formula only shows the number of constituent elements whereas structural formula indicates the structure of the molecule. Let’s discuss about empirical formulas.

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## Empirical Formula Definition

Empirical formula is the simplest ratio of all the elements present in a compound. Empirical formula only gives the ratio in which the elements are present in a compound.

The empirical formula of a compound is the “Simplest formula which expresses its percentage composition”.

So, any number of compounds can have the same empirical formula, since many compounds can be made with the same ratio of elements in them.

## Determining Empirical Formula

Determining empirical formula is explained in detail below :
1. Some chemists carry out chemical reactions called synthesis to produce new compounds for many different uses.
2. Chemists must be able to determine the identity of the substance produced by these synthesis reactions.
3. One procedure that the chemists use to identify an unknown substance is the elemental analysis.
4. This technique tells the chemists about the elements present in the compound that is synthesized and the percent by mass of these elements.
5. The information gained from such an analysis is the “percent composition” of the substance synthesized.
6. For example, a compound that is 88.7% Oxygen and 11.3% Hydrogen has a percentage composition that matches that of water, H2O. A compound that is 60.7% chlorine and 39.3% sodium has a percentage composition that matches Sodium chloride, NaCl.
7. These data along with the further tests are used to identify an unknown compound.

## Multiples of Empirical Formula

For a compound, the empirical formula is the same as the molecular formula. Such is the case of Water, H2O and Nitric acid, HNO3. But, for some compounds, molecular formula is the whole number multiple of empirical formula.

For example,

 Compound Compound Empirical formula Molecular formula Acetic acid CH2O (CH2O) x 2 = C2H4O2 Glucose CH2O (CH2O) X 6 =C6H12O6

So, molecular formulas are calculated using empirical formula and molar mass data. The empirical formula is a multiple of the molecular formula, as shown in the relationship:

n(empirical formula) = Molecular formula

where, ‘n’ will always be a whole number and it is occasionally a very large number. With acetic acid, it is 2, but with glucose, it is 6. Likewise, true molar mass should be a multiple of the empirical formula mass.

## Empirical Formula Examples

Example: 1

A hydrocarbon was found to contain the following percentage composition.

C = 92.3%, H = 7.69%. Calculate its empirical formula (C=12, H=1)

Solution:

The given problem is solved in the following table

 Element Symbol Relative atomic mass % Composition Relative number of atoms Simple ratio Simple whole number ratio Carbon C 12 93.3 $\frac{92.3}{12}$ = 7.69 $\frac{7.69}{7.69}$ 1 Hydrogen H 1 7.69 $\frac{7.69}{1}$ = 7.69 $\frac{7.69}{7.69}$ 1

Empirical formula = C1H1 = CH

Example: 2

A substance was found to have the following percentage composition;
Ca = 40% C =12%, O = 48%. Calculate its empirical formula. (Relative atomic mass of Ca=40, C=12, O=16)

Solution:
The given problem is solved in the following table.

 Element Symbol Relative atomic mass % composition Relative number of atoms Simple ratio Simple whole number ratio Calcium Ca 40 40 $\frac{40}{40}$ = 1 1 1 Carbon C 12 12 $\frac{12}{12}$ = 1 1 1 Oxygen O 16 48 $\frac{48}{16}$ = 3 3 3

Empirical formula = Ca1C1O3 = CaCO3

Example: 3

An organic compound was found to contain the following percentage composition:

H = 2.22%, C = 26.67%, O = 71.11%. Find its empirical formula. (Relative atomic mass of H=1, C=12, O=16)

solution:

The given problem is solved in the following table.

 Element Symbol Relative atomic mass % composition Relative number of atoms Simple ratio Simple whole number ratio Hydrogen H 1 2.22 $\frac{2.22}{1}$ = 2.22 $\frac{2.22}{2.22}$ = 1 1 Carbon C 12 26.67 $\frac{26.67}{12}$ = 2.22 $\frac{2.22}{2.22}$ = 1 1 Oxygen O 16 71.11 $\frac{71.11}{16}$ = 4.44 $\frac{4.44}{2.22}$ = 2 2

Empirical formula = H1C1O2 = HCO2.

Example: 4

A compound was found to have the following percentage composition:

Nitrogen = 21.21%, Hydrogen= 6.06%, Sulfur = 24.24% and Oxygen = 48.48%

(N=14, H=1, S=32, O=16). Find its empirical formula.

Solution:

The given problem is solved in the following table.

 Element Symbol Relative atomic mass % composition Relative number of atoms Simple ratio Simple whole number ratio Nitrogen N 14 21.21 $\frac{21.21}{14}$ = 1.15 $\frac{1.5}{0.75}$ = 2 2 Hydrogen H 1 6.06 $\frac{6.06}{1}$ = 6.06 $\frac{6.06}{0.75}$ = 8.08 8 Sulfur S 32 24.24 $\frac{24.24}{32}$ = 0.75 $\frac{0.75}{0.75}$ = 1 1 Oxygen O 16 48.48 $\frac{48.48}{16}$ = 3.03 $\frac{3.03}{0.75}$ = 4.04 4

Empirical formula = N2H8S1O4 = N2H8SO4.

## Empirical Formula Problems

Percentage composition data allows us to calculate the simplest ratio among the elements in a compound, or the empirical formula.

For example

Ammonium nitrite has a true formula of NH4NO2 and consists of ammonium, NH4+ and nitrite, NO2- , ions in a 1:1 ratio. Using the data from an elemental analysis, a chemist would calculate the empirical formula to be NH2O because this formula represents the simplest ratio of all the elements present.

For many compounds, the empirical formula and the true formula are the same. To calculate an empirical formula from the percent composition, we need to convert the mass percent of each element to grams, and then to moles. Then, from the moles, we can arrive at the simplest whole number ratio among the elements in a compound. The method is to divide each amount by the smallest of all the amounts..

This process will give a coefficient of 1 for the atoms present in smallest amount. After this step, additional multiplication may be necessary to convert all coefficients to simple whole numbers. These numbers will form subscripts in the empirical formula.

### Solved Examples

Question 1: One of the substances in a new alkaline battery is composed of 63.0% Manganese and 37.0% oxygen by mass. Determine the empirical formula of the compound in the battery.
Solution:

Step: 1   List the things given in the question:
Percentage of Mn = 63%
Percentage of O = 37.0%

Step: 2   Set up the problem:
Converting the given percentages to grams by assuming that we have exactly 100 grams of unknown substance.
63.0% Mn x 100 grams = 63.0 grams of Mn
37.0 % of O x 100 grams = 37.0 grams of Oxygen.
We need the molar masses of Manganese to convert the grams of manganese and oxygen into moles.

Molar mass of Mn = 54.94g/mol
Molar mass of O = 16.00 g/mol

Step: 3 calculating the number of moles:
Moles of Mn         = 63.0 g of Mn x$\frac{1\ mole\ of\ Mn}{54.94 g of Mn}$
= 1.14670 moles of Mn = 1.15 moles of Mn.
Moles of Oxygen   = 37.0 g of O x $\frac{1 mol of O}{15g of O}$
= 2.31 moles of Oxygen.

To determine the simplest ratio of moles in the compound, we need to divide the amount of each element by the smallest amount. The smallest amount here is of Mn – 1.15.
$\frac{1.15\ mol\ of\ Mn}{1.15}$ = 1.00 mol Mn
$\frac{1.15\ mol\ of\ O}{1.15}$ = 2.01 moles of O
So, according to the calculations above, the elements, Mn and O are in the ratio of 1: 2.
The empirical formula, therefore, would be: MnO2.

Question 2: Chemical analysis of clear liquid shows that it is 60.0% C, 13.4% H and 26.6% O. Calculate the empirical formula of this substance.
Solution:

Step: 1    Let us list what we know
Percentage of C = 60.0%
Percentage of O = 26.6%
Percentage of H = 13.4%
Empirical formula = C? H? O?
Step: 2 set up the problem
Assuming that we have 100 grams of the liquid, let us convert the percentages of these elements into grams.

60.0% of C x 100g = 60.0 grams of Carbon
13.4 % of H x 100g = 13.4 grams of Hydrogen
26.6% of O x 100g = 26.6 grams of Oxygen
To convert the mass of each element into amount in moles, we need to have the molar masses of these elements from the periodic table.

Molar mass of C = $\frac{12.01g}{mol}$.
Molar mass of H = $\frac{1.01g}{mol}$.
Molar mass of O = $\frac{16.00 g}{mol}$.

Step: 3 calculating the number of moles
Let us calculate the number of moles of each of these elements from the molar masses given:
60.0 g of C x $\frac{1 mol\ of C}{12.00g\ of\ C}$
= 5.00 moles of C.
13.4 g of H x $\frac{1 mol\ of\ H}{1.01g\ of\ H}$ = 13.3 moles of H
26.6 g of O x$\frac{1 mol\ of\ O}{16.00g\ of\ O}$  = 1.66 moles of O
The moles of the elements present are all fractions. We need to divide the smallest number with all the number of moles given since the empirical formula is the simplest whole number ratio of all the elements present.
The smallest number is 1.66. So,

$\frac{1.66\ mol\ of O}{1.66}$ = 1.00 moles of O
$\frac{5.00\ mol\ of\ C}{1.66}$ = 3.01 moles of C
$\frac{13.3\ mol\ of H}{1.66}$  = 8.01 moles of H.

The empirical formula of this liquid would be:
C3H8O

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