How to Determine the Empirical Formula of a Compound: Step-by-Step Guide

The empirical formula of a compound represents the simplest whole-number ratio of the atoms of each element in the substance. To determine it, you start with the mass or percent composition of each element, convert those values to moles, then find the simplest mole ratio. This method is widely used in chemistry to identify unknown substances and verify purity. The process is straightforward and relies on basic stoichiometry.

Gather Experimental Data

You need the mass or percentage by mass of each element in the compound. This data often comes from combustion analysis or elemental analysis. For example, when a hydrocarbon burns completely, it produces carbon dioxide and water. From the masses of these products, you can calculate the masses of carbon and hydrogen in the original sample.

Suppose a 1.000 g sample of a compound containing carbon, hydrogen, and oxygen is combusted, yielding 2.444 g of CO₂ and 1.000 g of H₂O. First, calculate the mass of carbon from CO₂. Since CO₂ is 12.01/44.01 carbon by mass, the mass of carbon is:

(12.01 g/mol ÷ 44.01 g/mol) × 2.444 g ≈ 0.667 g C

For hydrogen, H₂O is 2.016/18.02 hydrogen by mass, so:

(2.016 g/mol ÷ 18.02 g/mol) × 1.000 g ≈ 0.112 g H

The mass of oxygen is found by subtracting the masses of carbon and hydrogen from the total sample mass:

1.000 g − (0.667 g + 0.112 g) = 0.221 g O

Now you have the elemental masses: 0.667 g C, 0.112 g H, and 0.221 g O.

Convert Mass to Moles

Next, convert each mass to moles using molar masses: C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00 g/mol.

• Moles of C: 0.667 g ÷ 12.01 g/mol ≈ 0.0555 mol
• Moles of H: 0.112 g ÷ 1.008 g/mol ≈ 0.111 mol
• Moles of O: 0.221 g ÷ 16.00 g/mol ≈ 0.0138 mol

Find the Simplest Mole Ratio

Divide each mole value by the smallest number of moles, which is 0.0138 (oxygen):

• C: 0.0555 ÷ 0.0138 ≈ 4.02
• H: 0.111 ÷ 0.0138 ≈ 8.04
• O: 0.0138 ÷ 0.0138 = 1.00

The values are very close to whole numbers: 4, 8, and 1. So the empirical formula is C₄H₈O.

If ratios are not close to whole numbers, multiply to get integers. For example, a ratio of 1:1.5 would become 2:3 when multiplied by 2. But in this case, rounding 4.02 and 8.04 to 4 and 8 is justified because the values are within acceptable error.

Verify the Empirical Formula

Check your result by calculating the percent composition of C₄H₈O and comparing it to the original data.

Molar mass of C₄H₈O = (4 × 12.01) + (8 × 1.008) + 16.00 = 72.10 g/mol

• % C = (48.04 ÷ 72.10) × 100 ≈ 66.6%
• % H = (8.064 ÷ 72.10) × 100 ≈ 11.2%
• % O = (16.00 ÷ 72.10) × 100 ≈ 22.2%

From the original sample:

• % C = (0.667 ÷ 1.000) × 100 = 66.7%
• % H = (0.112 ÷ 1.000) × 100 = 11.2%
• % O = (0.221 ÷ 1.000) × 100 = 22.1%

The values match closely, confirming the empirical formula is correct.

Key Takeaways

• The empirical formula shows the simplest whole-number ratio of elements in a compound.
• Start with mass or percent composition, then convert each element’s mass to moles.
• Divide all mole values by the smallest to find the simplest ratio; multiply if needed to get whole numbers.
• Always verify your result by checking if the calculated composition matches the original data.