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Question: How can I calculate the empirical formula of butane?...

How can I calculate the empirical formula of butane?

Explanation

Solution

Hint Empirical formula is the simplest formula which provides the lowest whole number ratio of atoms which exist in the compound. The relative number of atoms of every element in the compound is given by this formula.

Complete Step by step solution: Butane is an alkane represented with the formula C4H10{{C}_{4}}{{H}_{10}}. Empirical formula of butane can be calculated by two methods:
1. From molecular formula: If we know the molecular formula of butane i.e. C4H10{{C}_{4}}{{H}_{10}} then by dividing the subscripts by their HCF i.e. highest common factor which is 2 in this case. By this we can consider that the empirical formula of butane is C2H5{{C}_{2}}{{H}_{5}}.
2. From percentage composition: Butane consists of 82.66% carbon and 17.34% hydrogen. To find out composition we have to consider the moles of carbon and hydrogen.
Let us assume 100g of butane which consists of 82.66 g of carbon and 17.34 g of hydrogen. Then moles of carbon and hydrogen can be calculated as:
Moles of carbon = 82.66×1 mol of C12.01g C=6.8826 mol C82.66\times \dfrac{1\ \text{mol of C}}{12.01g\ \text{C}}=6.8826\ \text{mol C}
Moles of hydrogen = 17.34×1 mol of H1.008g H=17.202 mol H17.34\times \dfrac{1\ \text{mol of H}}{1.008g\ H}=17.202\ \text{mol H}
Empirical formula can be calculated by dividing the moles of carbon and hydrogen which can be shown as:
Moles of CMoles of H=6.882617.202=12.4994=24.99825\dfrac{Moles\ \text{of C}}{Moles\ \text{of H}}=\dfrac{6.8826}{17.202}=\dfrac{1}{2.4994}=\dfrac{2}{4.998}\simeq \dfrac{2}{5}
Hence from both the methods we get that empirical formula of butane is C2H5{{C}_{2}}{{H}_{5}}.

Note: At room temperature and atmospheric pressure butane exist as a gas which is highly flammable, colorless and easily liquefied gas which vaporize at room temperature. Butane is generally used as a fuel gas in gasoline and by mixing with propane it can be used in LPG too.