Chemical formulae such as a molecular formula and an empirical formula is the language of chemistry. These give us a wise insight into the chemical composition of a compound. A molecular formula represents the total number of atoms of each element present in a molecule. An empirical formula on the other hand simplifies this information provided in the molecular formula. Keep reading the article to know how it is done.
What is empirical formula
An empirical formula is generally used to show what elements are present in a molecule in the simplest notation. It does not provide any structural information regarding the position of atoms present in a molecule and/or the chemical bonds binding the atoms together.
Empirical formula definition
The empirical formula of a compound is defined as the simplest whole number ratio of all the elements present in one molecule or a formula unit of that compound.
Some interesting facts about empirical formula
- The formula given for an ionic compound is always its empirical formula. For example, sodium chloride is an ionic compound represented by NaCl. NaCl is also the empirical formula for sodium chloride.
- The empirical and molecular formulae for simple inorganic molecules are often the same such as CO2 for carbon dioxide, CH4 for methane and NH3 for ammonia.
- Complex organic molecules generally have distinct empirical and molecular formulae. For example, the molecular formula for octane is C8H18 while its empirical formula is C4H9.If the molecular formula for an organic compound, dodecane is C12H26, can you determine its empirical formula? It is C6H13. But how is it determined? Let’s discuss.
How to calculate empirical formula
An empirical formula can be calculated through chemical stoichiometry. If we know which elements are present in a molecule and in what ratio, we can calculate the molecule’s empirical formula. The calculation depends on the information provided.
Case 1: Molecular formula of a compound is given
This is the simplest example of determining an empirical formula. If the molecular formula of the compound is already known, we can find its empirical formula just by dividing the number of atoms of each element with a whole number such that we obtain the simplest integer values.
Case 2: Mass of each element present in the sample of a compound is given
Considering the complete combustion of 1.55 grams of phosphorus produces 3.55 grams of an oxide of phosphorus. We can use this information to deduce the empirical formula of this oxide of phosphorus, performing the calculations given below.
Step I: Carefully study the information already provided. The mass of phosphorus (P) is given i.e., 1.55 grams. All the phosphorus burns to produce its oxide which means 1.55 grams P will be present in the oxide.
Step II: Subtract this mass of phosphorus from the total mass of phosphorus oxide to find the mass of oxygen (O) present in the oxide.
Mass of phosphorus oxide = 3.55 grams
Mass of phosphorus present in it = 1.55 grams
Mass of oxygen present in it = 3.55-1.55 = 2.0 grams
Step III: Find the number of moles of each element in the compound by dividing the mass (in grams) of each with its atomic mass.
Step IV: Divide each value by the lowest figure.
Step V: Now multiply each value with the smallest integer that can convert 2.5 into a whole number i.e., 2 in this case.
Step VI: Construct the empirical formula by using the resulting numbers as subscripts for each element.
Result: The empirical formula for the oxide of phosphorus obtained in this chemical reaction is P2O5.
Case 3: Determining empirical formula from analysis of percentage composition
If the percentage composition of all the elements present in a compound is given. This data can be sufficiently used to determine the empirical formula of this compound. For instance, a compound PABA based on carbon, hydrogen, nitrogen and oxygen consists of C (61.31%), H (5.14%), N (10.21%) and O (23.33%). The step-by-step guide below gives a procedure to calculate its empirical formula.
Step I: Assuming 100 grams of PABA, the mole ratio for each element can be calculated by dividing their % age composition with the respective atomic masses.
Step II: Divide all the values with the smallest number of moles, rounding off answers to the nearest whole number values.
Result: The numerical values obtained gives the simplest whole number mole ratio thus the empirical formula for the compound under investigation is C7H7NO2.
Test Yourself by calculating the empirical formulae from the data provided in the following set of questions.
Q1) Write the empirical formula for
a) Hydrazine (N2H4)
b) Magnesium chloride (MgCl2)
c) Acetamide (CH3CONH2)
(Answers: a) NH2 , b) MgCl2 , c) C2H5NO)
Q2) A student was performing some gravimetric analysis in his chemistry lab. He recorded the following data on his notebook.
Mass of crucible (empty) = 28.288 grams
Mass of crucible + potassium = 28.709 grams
Mass of crucible + potassium oxide formed after reaction = 28.793 grams
How can you use this data to help the student in calculating the empirical formula of potassium oxide formed during the chemical reaction?
(Answer: empirical formula for potassium oxide= K2O)
Practice more on calculating empirical formulae using the following sources: