It depends on the information you have.
The empirical formula tells us the simplest whole-number ratio of the different types of atoms in a compound.
FROM THE MOLECULAR FORMULA
If you know that the molecular formula of butane is C₄H₁₀, then you divide the subscripts by their highest common factor (2).
This gives you the empirical formula C₂H₅.
FROM PERCENTAGE COMPOSITION
You can calculate the empirical formula from percentage composition.
Example
Butane is 82.66 % C and 17.34 % by mass. What is its empirical formula?
Solution
Assume 100 g of butane. Then you have 82.66 g of C and 17.34 g of H.
Moles of C = 82.66 g C × ##(1″mol C”)/(12.01″g C”)## = 6.8826 mol C
Moles of H = 17.34 g H × ##(1″mol H”)/(1.008″g H”)## = 17.202 mol H
##”Moles of C”/”Moles of H” = (6.8826″mol”)/(17.202″mol”) = 1/2.4994 =2/4.9988 ≈ 2/5##
∴ The empirical formula of butane is C₂H₅.
FROM COMBUSTION ANALYSIS
You can calculate the empirical formula by doing a combustion analysis.
You burn a sample of butane and measure the masses of CO₂ and H₂O produced.
Example
The combustion of a sample of butane produces 1.6114 g of carbon dioxide and 0.8427 g of water. What is the empirical formula of butane?
Solution
Moles of C = 1.6114 g CO₂ × ##(1″mol CO₂”)/(44.01″g CO₂”) × (1″mol C”)/(1″mol CO₂”)## = 0.036 614 mol C
Moles of H = 0.8247 g H₂O × ##(1″mol H₂O”)/(18.02″g H₂O”) × (2″mol H”)/(1″mol H₂O”)## = 0.091 532 mol H
##”Moles of C”/”Moles of H” = (0.036 614″mol”)/(0.091 532″mol”) = 1/2.5000 =2/5.0000 ≈ 2/5##
∴ The empirical formula of butane is C₂H₅.
Very good explanation, large fond, video on this page: https://www.boundless.com/chemistry/textbooks/boundless-chemistry-textbook/atoms-molecules-and-ions-2/chemical-formulas-37/empirical-formulas-210-587/
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