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Empirical Formula

25th Aug 2019 @ 22 min read

Analytical Chemistry

In chemistry, particularly analytical chemistry, the empirical formula is very useful to establish the relative relationship among constituent elements of a compound. Earlier, it was a challenge to determine the elemental composition of a compound. French chemist Antoine Lavoisier was well known for quantitative elemental analysis.

Portrait of Antoine Lavoisier and His Wife
Figure 1: Portrait of Antoine Lavoisier and His Wife
[Image Source: Public Domain]

Elemental analysis can be qualitative and quantitative. A qualitative analysis tells us which elements are present in a sample while a quantitative analysis determines the mass composition of elements in a sample. Analysis methods include mass spectroscopy, gravimetry, neutral activation analysis etc.

By knowing the mass composition and the element composition in a given sample, we can generate the empirical formula.

Definition of Empirical Formula

It is the formula formed by atoms of a compound (or a molecule) such that atoms of an element are in the simplest relative ratio.

According to the golden book of the IUPAC, it is “formed by the juxtaposition of the atomic symbols with their appropriate subscripts to give the simplest possible formula expressing the composition of a compound.”

The empirical formula of a compound tells which elements are present in a compound and the relative mass composition of the elements. It does not talk about the number of atoms of an element present in a molecule.

Empirical Formula vs Molecular Formula

The empirical formula is not the molecular formula. The difference between both is presented in the table below.

Table 1: Difference between Molecular Formula vs Empirical Formula
Empirical FormulaMolecular Formula
It presents the simplest positive integer ratio of elements present in a compound.The molecular formula presents the actual number of atoms of an element in a compound.
It only tells the relative number of elements in a compound.It tells the actual number of atoms of an element in a compound.
Examples are glucose, CH2O; ethene CH2; nitrogen monoxide NO.Examples are glucose, C6H12O6; ethene, C2H4; dinitrogen dioxide, N2O2.

Note:

  • Different compounds can have the same empirical formula. For example, ethylene C2H4 and propylene C3H6 have the same empirical formula, CH2. In the same manner, compounds can have the same molecular formula. For example, ethanol and dimethyl ether have the same molecular formula, C2H6.
  • None of them talks about the structure of a compound. The structure of a compound is understood by the structural formula.
  • The empirical formula and the molecular formula can be the same for many compounds. For example, ethanol has the same empirical and molecular formula; it is C2H6.

List of Compounds

The table list some compounds with their empirical and molecular formula.

Table 2: List of Compounds with their Empirical and Molecular Formula
CompoundsEmpirical FormulaMolecular Formula
Calcium SulphateCaSO4CaSO4
Sulphuric AcidH2SO4H2SO4
Sodium ChlorideNaClNaCl
Dinitrogen Dioxide1NON2O2
Nitrogen Monoxide1NONO
MethaneCH4CH4
Ethanol2C2H6OC2H6O
Dimethyl Ether2C2H6OC2H6O
Acetylene3CHC2H2
Benzene3CHC6H6
Glucose4CH2OC6H12O6
Glyceraldehyde4CH2OC3H6O3
Propanal5C3H6OC3H6O
Acetone5C3H6OC3H6O
Formaldehyde6CH2OCH2O
Acetic Acid6CH2OC2H4O2
NitrobenzeneC6H5NO2C6H5NO2
The compounds superscripted by the same number (1, 2, 3, 4, 5, 6) have the same empirical and/or molecular formula.

To Identify Unknown Compound

The word “empirical” suggests experimental observations. We may identify unknown compounds using elemental analysis (experiments) and predict the respective empirical formula (or molecular formula if the molar mass of a compound is known). Through elemental analysis, we can determine which elements are in a sample and their relative mass composition. By knowing these, we can follow the steps below to generate the empirical or molecular formula.

Mass Percentages to Empirical Formula

Step 1: Convert mass percentages to the gram by considering 100 g of a sample. For example, 74.83 % of carbon becomes 74.83 g of carbon and 25.17 % of hydrogen becomes 25.17 g of hydrogen.

Step 2: Convert the gram of elements into the mole by dividing the gram to the molar mass of an element. For example, 74.83 g of carbon is 74.83 ÷ 12.011 g mol−1 = 6.230 mol of carbon and 25.17 g of hydrogen is 25.17 ÷ 1.008 g mol−1 = 24.97 mol of hydrogen.

Step 3: Take ratios of each value of moles to the smallest value of moles and then approximate the ratios to the nearest whole number. For example, 6.230 mol is the smallest. So, the ratio is 24.97 ÷ 6.230 = 4.008. The ratio is approximated to the closest whole number, 4.035 ≈ 4.

Note: If a ratio can not be approximated, try to multiply it with the smallest number such that the product is a whole number. And multiply the remaining ratios with the same smallest number. For example, if a ratio is 1.333, multiply it with 3, which is the smallest number that will result in a whole number. Thus, 1.333 × 3 ≈ 4.

Step 4: Now, the empirical formula is made by placing each of the whole numbers as the subscript to respective elements.

Empirical Formula to Molecular Formula

Once the empirical formula is estimated, we can also find the molecular formula if the molar mass is known.

Step 5: Determine the ratio of the molar mass to the empirical formula mass. Approximate the ratio to the closest whole number and multiply the whole number to the empirical formula to get the molecular formula.

Let take a proper example to make the above steps clearer.

Example 1: To Determine the Molecular Formula

Given Data: Elemental analysis shows a compound has carbon and hydrogen. And the mass percentages are 82.66 % of carbon and 17.34 % of hydrogen. Also, the molar mass of the compound is 58.12 g mol−1.

Now, follow the above steps.

Step 1: Consider a 100 g of the compound. So, it contains 82.66 g of carbon and 17.34 g of hydrogen.

Step 2: The molar mass of carbon and hydrogen is 12.011 g mol−1 and 1.008 g mol−1. The moles of carbon and hydrogen are calculated as follows:

n_\text{C} =\frac{82.66 \,\text{g}}{12.011 \,\text{g}\,\text{mol}^{-1}} =6.882\,0 \,\text{mol} n_\text{H} =\frac{17.34 \,\text{g}}{1.008 \,\text{g}\,\text{mol}^{-1}} =17.202 \,\text{mol}

Step 3: nC = 6.882 0 mol is the smallest number. So, The ratios is \frac{n_\text{H}}{n_\text{C}}.

\frac{n_\text{H}}{n_\text{C}} =\frac{17.202}{6.882\,0} =2.499\,6 \approx 2.5

Now, 2.5 is not a whole number. So, to make it a whole number we multiply it by 2.

\frac{n_\text{H}}{n_\text{C}} = 2.5 = 2.5 \times \frac{2}{2} =\frac{5}{2}

Thus, the mole of carbon to the mole of hydrogen ratio is 5 : 2.

Step 4: We can write the empirical formula by placing the numbers as the subscript to the element’s symbols.

Therefore, the empirical formula is C2H5.

Step 5: The molar mass of the compound is known to us, M = 58.12 g mol−1. The empirical mass of the compound is obtained by adding the molar mass of individual elements.

M_\text{empirical} &=2 \,M_\text{C} + 5 \,M_\text{H} \\ &=2 \times 12.011+5\times 1.008 \\ &=29.062 \,\text{g}\,\text{mol}^{-1}

Let the ratio of the molar mass to empirical mass be r.

r =\frac{M}{M_\text{empirical}} =\frac{58.12}{29.062} =1.999 \approx 2

Thus, multiplying 2 to the empirical formula, 2 × C2H5 = C4H10.

Finally, the molecular formula is C4H10.

The unknown compound is butane. But we cannot determine which butane is it; it can be n-butane or isobutane.

Isomers of Butane
Figure 2: Isomers of Butane

Note: From the above example it is clear the empirical or molecular formula is not helpful to identify isomers of a compound.

Let consider another example.

Example 2: To Determine the Molecular Formula

Given Data: The mass composition of carbon, hydrogen, and oxygen is 66.63 %, 11.18 %, and 22.19 % respectively. The molar mass of the compound is 144.214 g mol−1.

Step 1: Consider a 100 g of the compound. So, it contains 66.63 g of carbon, 11.18 g of hydrogen, and 22.19 g of oxygen.

Step 2: The molar mass of carbon, hydrogen, and oxygen is 12.011 g mol−1, 1.008 g mol−1, and 15.999 g mol−1. The moles of carbon, hydrogen, and oxygen are calculated as follows:

n_\text{C} =\frac{66.63 \,\text{g}}{12.011 \,\text{g}\,\text{mol}^{-1}} =5.547 \,\text{mol} n_\text{H} =\frac{11.18 \,\text{g}}{1.008 \,\text{g}\,\text{mol}^{-1}} =11.091\,\text{mol} n_\text{O} =\frac{22.19 \,\text{g}}{15.999 \,\text{g}\,\text{mol}^{-1}} =1.387\,0\,\text{mol}

Step 3: nO = 1.387 0 mol is the smallest number. So, The ratios are \frac{n_\text{C}}{n_\text{O}} and \frac{n_\text{H}}{n_\text{O}}.

\frac{n_\text{C}}{n_\text{O}} =\frac{5.547}{1.387\,0} =3.999 \approx 4 \frac{n_\text{H}}{n_\text{O}} =\frac{11.091}{1.387\,0} =7.996\,4 \approx 8

Thus, the mole ratio of carbon to oxygen and hydrogen to oxygen is 4 : 1 and 8 : 1.

Step 4: We can write the empirical formula by placing the numbers as the subscript to the element’s symbols.

Therefore, the empirical formula is C4H8O.

Step 5: The molar mass of the compound is known to us, M = 144.214 g mol−1. The empirical mass of the compound is obtained by adding the molar mass of individual elements.

M_\text{empirical} &=4 \,M_\text{C} + 8 \,M_\text{H} + M_\text{O} \\ &=4 \times 12.011+8 \times 1.008 + 15.999 \\ &=72.107 \,\text{g}\,\text{mol}^{-1}

Let the ratio of the molar mass to empirical mass be r.

r =\frac{M}{M_\text{empirical}} =\frac{144.214}{72.107} =2

Thus, multiplying 2 to the empirical formula, 2 × C4H8O = C8H16O2.

Finally, the molecular formula is C8H16O2.

There are many compounds with the molecular formula C8H16O2. So, the identity of the compound is still unknown, but some of them are mentioned below.

Some Compounds with the molecular formula C8H16O2
Figure 3: Some Compounds with the molecular formula C8H16O2

Solved Problems

Problem 1: To Determine the Molecular Formula

Given Data: An experiment was conducted and it is known that the sample contains carbon, hydrogen, nitrogen, and oxygen. The mass composition of carbon, hydrogen, nitrogen, and oxygen is 42.87 %, 2.40 %, 16.66 %, and 38.07 % respectively. The molar mass of the compound is 168.096 g mol−1.

Step 1: Consider a 100 g of the compound. So, it contains 42.87 g of carbon, 2.40 g of hydrogen, 16.66 g of nitrogen, and 38.07 g of oxygen.

Step 2: The molar mass of carbon, hydrogen, nitrogen, and oxygen is 12.011 g mol−1, 1.008 g mol−1, 14.007 g mol−1, and 15.999 g mol−1. The moles of carbon, hydrogen, nitrogen, and oxygen are calculated as follows:

n_\text{C} =\frac{42.87 \,\text{g}}{12.011 \,\text{g}\,\text{mol}^{-1}} =3.569\,2 \,\text{mol} n_\text{H} =\frac{2.40 \,\text{g}}{1.008 \,\text{g}\,\text{mol}^{-1}} =2.381\,0\,\text{mol} n_\text{N} =\frac{16.66 \,\text{g}}{14.007 \,\text{g}\,\text{mol}^{-1}} =1.189\,4\,\text{mol} n_\text{O} =\frac{38.07 \,\text{g}}{15.999 \,\text{g}\,\text{mol}^{-1}} =2.379\,5\,\text{mol}

Step 3: nN = 1.189 4 mol is the smallest number. So, The ratios are \frac{n_\text{C}}{n_\text{N}}, \frac{n_\text{H}}{n_\text{N}}, and \frac{n_\text{O}}{n_\text{N}}.

\frac{n_\text{C}}{n_\text{N}} =\frac{3.569\,2}{1.189\,4} =3.002 \approx 3 \frac{n_\text{H}}{n_\text{N}} =\frac{2.381\,0}{1.189\,4} =2.003 \approx 2 \frac{n_\text{O}}{n_\text{N}} =\frac{2.379\,5}{1.189\,4} =2.001 \approx 2

Thus, the mole ratio of carbon to nitrogen, hydrogen to nitrogen, and oxygen to nitrogen is 3 : 1, 2 : 1, and 2 : 1.

Step 4: We can write the empirical formula by placing the numbers as the subscript to the element’s symbols.

Therefore, the empirical formula is C3H2NO2.

Step 5: The molar mass of the compound is known to us, M = 168.096 g mol−1. The empirical mass of the compound is obtained by adding the molar mass of individual elements.

M_\text{empirical} &=3 \,M_\text{C} + 2 \,M_\text{H} + M_\text{N} +2 \,M_\text{O} \\ &=3 \times 12.011+2 \times 1.008 + 14.007 + 2 \times 15.999 \\ &=84.054 \,\text{g}\,\text{mol}^{-1}

Let the ratio of the molar mass to empirical mass be r.

r =\frac{M}{M_\text{empirical}} =\frac{168.096}{84.054} =1.999\, 9 \approx 2

Thus, multiplying 2 to the empirical formula, 2 × C3H2NO2 = C6H4N2O4.

Finally, the molecular formula is C6H4N2O4.

There are many compounds with the molecular formula C6H4N2O4. So, the identity of the compound is still unknown, but some of them having the same molecular formula are mentioned below.

Some Compounds with the molecular formula C6H4N2O4
Figure 4: Some Compounds with the molecular formula C6H4N2O4

Problem 2: To Determine the Empirical Formula

Given Data: A compound has the mass composition of 27.9 % of iron, 24.1 % of sulphur, and 48.0 % of oxygen. The molar mass of the compound is unknown.

Step 1: Consider a 100 g of the compound. So, it contains 27.9 g of iron, 24.1 g of sulphur, and 48.0 g of oxygen.

Step 2: The molar mass of iron, sulphur, and oxygen is 55.845 g mol−1, 32.065 g mol−1, and 15.999 g mol−1. The moles of iron, sulphur, and oxygen are calculated as follows:

n_\text{Fe} =\frac{27.9 \,\text{g}}{55.845 \,\text{g}\,\text{mol}^{-1}} =0.499\,6 \,\text{mol} n_\text{S} =\frac{24.1 \,\text{g}}{32.065 \,\text{g}\,\text{mol}^{-1}} =0.751\,6\,\text{mol} n_\text{O} =\frac{48.0 \,\text{g}}{15.999 \,\text{g}\,\text{mol}^{-1}} =3.000\,\text{mol}

Step 3: nFe = 0.499 6 mol is the smallest number. So, The ratios are \frac{n_\text{S}}{n_\text{Fe}} and \frac{n_\text{O}}{n_\text{Fe}}.

\frac{n_\text{S}}{n_\text{Fe}} =\frac{0.751\,6}{0.499\,6} =1.5044 \approx 1.5 \frac{n_\text{O}}{n_\text{Fe}} =\frac{3.000}{0.499\,6} =6.005 \approx 6

1.5 is not a whole number. So, we need to multiply by 2 to get a whole number.

\frac{n_\text{S}}{n_\text{Fe}} = 1.5 =1.5 \times \frac{2}{2} =\frac{3}{2}

We need to multiply the same number to \frac{n_\text{O}}{n_\text{Fe}}.

\frac{n_\text{O}}{n_\text{Fe}} =6 =6 \times\frac{2}{2} =\frac{12}{2}

Thus, the mole ratio of sulphur to iron and oxygen to iron is 3 : 2 and 12 : 2.

Step 4: We can write the empirical formula by placing the numbers as the subscript to the element’s symbols.

Therefore, the empirical formula is Fe2S3O12.

Problem 3: To Determine the Empirical Formula

Given Data: An ionic compound has the mass composition of 60.30 % of magnesium and 39.70 % of oxygen.

Step 1: Consider 100 g of the compound. So, it contains 60.30 g of magnesium and 39.70 g of oxygen.

Step 2: The molar mass of magnesium and oxygen is 24.305 g mol−1 and 15.999 g mol−1. The moles of magnesium and oxygen are calculated as follows:

n_\text{Mg} =\frac{60.30 \,\text{g}}{24.305 \,\text{g}\,\text{mol}^{-1}} =2.481\,0 \,\text{mol} n_\text{O} =\frac{39.70 \,\text{g}}{15.999 \,\text{g}\,\text{mol}^{-1}} =2.481\,4\,\text{mol}

Step 3: nMg = 2.481 0 mol is the smallest number. So, The ratios are \frac{n_\text{O}}{n_\text{Mg}}

\frac{n_\text{O}}{n_\text{Mg}} =\frac{2.481\,4}{2.481\,0} =1.000\,1 \approx 1

Thus, the mole ratio of oxygen to magnesium is 1 : 1.

Step 4: We can write the empirical formula by placing the numbers as the subscript to the element’s symbols.

Therefore, the empirical formula is MgO.

If the molar mass of the compound is 40.304 g mol−1, the compound is magnesium oxide.

Practice Problems

Problem 1: To Determine the Empirical Formula

Given Data: The mass composition of a sample is 52.67 % of carbon, 9.33 % of hydrogen, 6.82 % of nitrogen, and 31.18 % of oxygen.

Problem 2: To Determine the Empirical Formula

Given Data: This compound is a cobalt complex. It has the mass composition of 6.78 % of hydrogen, 31.42 % of nitrogen, 39.76 % of chlorine, and 22.04 % of cobalt. The molar mass of the compound is unknown.

Problem 3: To Determine the Molecular Formula

Given Data: The molar mass of a compound is 119.38 g mol−1. It has the mass composition of 10.06 % of carbon, 0.85 % of hydrogen, and 89.09 % of chlorine.

Problem 4: To Determine the Molecular Formula

Given Data: The compound is an acid having the molar mass of 98.08 g mol−1. It has the mass composition of 2.06 % of hydrogen, 32.69 % of sulphur, and 65.25 % of oxygen.

  1. C9H19NO4
  2. H18N6Cl3Co
  3. CHC3 (chloroform)
  4. H2SO4 (sulphuric acid)

Summary

  • The empirical formula is the simplest formula of the relative ratio of elements in a compound.
  • It informs which elements are present in a compound and their relative percentages. But the number of atoms of an element is always unknown. Also, it does not tell anything about the structure, isomers, or properties of a compound.
  • It is determined from elemental analysis.
  • It is different from the molecular formula. The molecular formula presents the actual number of atoms of an element in a compound. For example, the molecular formula of hydrogen peroxide is H2O2, but its empirical formula is HO.
  • The empirical formula is determined from the mass percentage composition, which is obtained from elemental analysis. From the empirical formula, the molecular formula is calculated using the molar mass.
From Elemental Analysis to Molecular Formula
Figure 5: From Elemental Analysis to Molecular Formula

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