CHEM - Finding Empirical and Molecular Formulas

Empirical formula: gives the simplest whole number ratio of atoms of each element in a compound

Molecular formula: gives the actual number of atoms of each element in a compound

Examples:

• The molecular formula of glucose is C₆H₁₂O₆. The empirical formula of glucose is CH₂O.
• The molecular formula of ethane is C₂H₆. The empirical formula of glucose is CH₃.
• The molecular formula of hydrogen peroxide is H₂O₂. The empirical formula of hydrogen peroxide is HO.
• The molecular formula of methane is CH₄. The empirical formula of methane is CH₄.

Let's take a step-by-step look at how to determine the empirical and molecular formula of an unknown compound 😁

Example 1

Find the molecular formula of a compound that has a molar mass of 92 g/mol and is composed of 30.4% nitrogen and 69.6% oxygen.

Step 1. Calculate the mass of each element of the compound. Assume you have 100 g of the compound.

In 100 g of the compound,

• mass of nitrogen: 30.4 g
• mass of oxygen: 69.6 g

Step 2. Divide the mass of each element by its relative atomic mass to find the moles of each element.

• relative atomic mass of nitrogen: 14.0 g/mol
• relative atomic mass of oxygen: 16.0 g/mol
• moles of nitrogen: 30.4 ÷ 14.0 = 2.17
• moles of oxygen: 69.6 ÷ 16.0 = 4.35

• nitrogen : oxygen = 1 : 2

• The empirical formula is NO₂.

• mass from empirical formula = 14.0 x 1 + 16.0 x 2 = 46 g/mol

• 92 ÷ 46 = 2

• The molecular formula is N₂O₄.

In summary:

Example 2

A compound with the following composition has a molar mass of 60 g/mol: 39.97% carbon; 13.41% hydrogen; 46.62% nitrogen. Find the molecular formula.

Step 1. Calculate the mass of each element of the compound. Assume you have 100 g of the compound.

In 100 g of the compound,

• mass of carbon: 39.97 g
• mass of hydrogen: 13.41 g
• mass of nitrogen: 46.62 g

Step 2. Divide the mass of each element by its relative atomic mass to find the moles of each element.

• relative atomic mass of carbon: 12.0 g/mol
• relative atomic mass of hydrogen: 1.0 g/mol
• relative atomic mass of nitrogen: 14.0 g/mol
• moles of carbon: 39.97 ÷ 12.0 = 3.33
• moles of hydrogen: 13.41 ÷ 1.0 = 13.41
• moles of nitrogen: 46.62 ÷ 14.0 = 3.33

• carbon : hydrogen : nitrogen = 1 : 4: 1

• The empirical formula is CH₄N.

• mass from empirical formula = 12.0 x 1 + 1.0 x 4 + 14.0 x 1 = 30 g/mol

• 60 ÷ 30 = 2

• The molecular formula is C₂H₈N₂.

In summary:

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