The empirical formula of a chemical compound is a simple expression of the relative number of each type of atom in it. In contrast, the molecular formula of a chemical compound gives the actual number of atoms of each element found in a molecule of that compound.
Definition: Empirical formula
The empirical formula of a chemical compound gives the relative number of each type of atom in it.
Definition: Molecular formula
The molecular formula of a chemical compound gives the exact number of atoms of each element in one molecule of that compound.
The compound ethanoic acid for example, has the molecular formula CH3COOH or simply C2H4O2. In one molecule of this acid, there are two carbon atoms, four hydrogen atoms and two oxygen atoms. The ratio of atoms in the compound is 2:4:2, which can be simplified to 1:2:1. Therefore, the empirical formula for this compound is CH2O. The empirical formula contains the smallest whole number ratio of the elements that make up a compound.
Knowing either the empirical or molecular formula of a compound, can help to determine its composition in more detail. The opposite is also true. Knowing the composition of a substance can help you to determine its formula. There are three different types of composition problems that you might come across:
Problems where you will be given the formula of the substance and asked to calculate the percentage by mass of each element in the substance.
Problems where you will be given the percentage composition and asked to calculate the formula.
Problems where you will be given the products of a chemical reaction and asked to calculate the formula of one of the reactants. These are usually referred to as combustion analysis problems.
Worked Example 63: Calculating the percentage by mass of elements in a compound
Question: Calculate the percentage that each element contributes to the overall mass of sulfuric acid (H2SO4).
Step 1 : Write down the relative atomic mass of each element in the compound.
Hydrogen = 1.008 × 2 = 2.016 u
Sulfur = 32.07 u
Oxygen = 4 × 16 = 64 u
Step 2 : Calculate the molecular mass of sulfuric acid.
Use the calculations in the previous step to calculate the molecular mass of sulfuric acid.
Mass = 2.016 + 32.07 + 64 = 98.09u
Step 3 : Convert the mass of each element to a percentage of the total mass of the compound
Use the equation:
Percentage by mass = atomic mass / molecular mass of H2SO4 × 100%
2.016/98.09 x 100% = 2.06%
32.07/98.09 x 100% = 32.69%
64/98.09 x 100% = 65.25%
(You should check at the end that these percentages add up to 100%!)
In other words, in one molecule of sulfuric acid, hydrogen makes up 2.06% of the mass of the compound, sulfur makes up 32.69% and oxygen makes up 65.25%.
Worked Example 64: Determining the empirical formula of a compound Question: A compound contains 52.2% carbon (C), 13.0% hydrogen (H) and 34.8% oxygen (O). Determine its empirical formula.
Step 1 : If we assume that we have 100 g of this substance, then we can convert each element percentage into a mass in grams.
Carbon = 52.2 g, hydrogen = 13 g and oxygen = 34.8 g
Step 2 : Convert the mass of each element into number of moles
n = m/M
Therefore, see Step 2 attachment.
Step 3 : Convert these numbers to the simplest mole ratio by dividing by the smallest number of moles
In this case, the smallest number of moles is 2.18. Therefore... see Step 3 attachment.
Therefore the empirical formula of this substance is: C2H6O. Do you recognise this compound?
Worked Example 65: Determining the formula of a compound
Question: 207 g of lead combines with oxygen to form 239 g of a lead oxide. Use this information to work out the formula of the lead oxide (Relative atomic masses: Pb = 207 u and O = 16 u).
Step 1 : Calculate the mass of oxygen in the reactants
239 − 207 = 32gStep 2 : Calculate the number of moles of lead and oxygen in the reactants. n = m/M Lead 207/207 = 1mol Oxygen 32/16 = 2mol
Step 3 : Deduce the formula of the compound
The mole ratio of Pb:O in the product is 1:2, which means that for every atom of lead, there will be two atoms of oxygen. The formula of the compound is PbO2.
Worked Example 66: Empirical and molecular formula
Question: Vinegar, which is used in our homes, is a dilute form of acetic acid. A sample of acetic acid has the following percentage composition: 39.9% carbon, 6.7% hyrogen and 53.4% oxygen.
1. Determine the empirical formula of acetic acid.
2. Determine the molecular formula of acetic acid if the molar mass of acetic acid is 60g/mol.
Step 1 : Calculate the mass of each element in 100 g of acetic acid. In 100g of acetic acid, there is 39.9 g C, 6.7 g H and 53.4 g O
Step 2 : Calculate the number of moles of each element in 100 g of acetic acid.
n = m/M
See example 66, step 2.
Step 3 : Divide the number of moles of each element by the lowest number to get the simplest mole ratio of the elements (i.e. the empirical formula) in acetic acid.
Empirical formula is CH2O
Step 4 : Calculate the molecular formula, using the molar mass of acetic acid.
The molar mass of acetic acid using the empirical formula is 30 g/mol. Therefore the actual number of moles of each element must be double what it is in the emprical formula.
The molecular formula is therefore C2H4O2 or CH3COOH
Complete the following exercise for homework:
Exercise: Moles and empirical formulae
1. Calcium chloride is produced as the product of a chemical reaction.
(a) What is the formula of calcium chloride?
(b) What percentage does each of the elements contribute to the mass of a
molecule of calcium chloride?
(c) If the sample contains 5 g of calcium chloride, what is the mass of calcium
in the sample?
(d) How many moles of calcium chloride are in the sample?
2. 13g sulfide? of zinc combines with 6.4g of sulfur.What is the empirical formula of zinc sulfide?
(a) What mass of zinc sulfide will be produced?
(b) What percentage does each of the elements in zinc sulfide contribute to
(c) Determine the formula of zinc sulfide.
3. A calcium mineral consisted of 29.4% calcium, 23.5% sulphur and 47.1% oxygen by mass. Calculate the empirical formula of the mineral.
4. A chlorinated hydrocarbon compound when analysed, consisted of 24.24% car- bon, 4.04% hydrogen, 71.72% chlorine. The molecular mass was found to be 99 from another experiment. Deduce the empirical and molecular formula.