# How can I calculate the empirical formula of butane?

Jul 29, 2014

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 × $\left(1 \text{mol C")/(12.01"g C}\right)$ = 6.8826 mol C

Moles of H = 17.34 g H × $\left(1 \text{mol H")/(1.008"g H}\right)$ = 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 × $\left(1 \text{mol CO₂")/(44.01"g CO₂") × (1"mol C")/(1"mol CO₂}\right)$ = 0.036 614 mol C

Moles of H = 0.8247 g H₂O × $\left(1 \text{mol H₂O")/(18.02"g H₂O") × (2"mol H")/(1"mol H₂O}\right)$ = 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.