Chemistry » Quantitative Aspects of Chemical Change » Stoichiometric Calculations Continued

Molecular and Empirical Formulae

Molecular and Empirical Formulae

Molecular and empirical formulae were introduced in another lesson. The empirical formula is the simplest formula of a compound (and represents the ratio of atoms of each element in a compound). The molecular formula is the full formula of the compound (and represents the total number of atoms of each element in a compound). You should also recall from grade 10 the percent composition of a substance. This is the percentage by molecular mass that each element contributes to the overall formula. For example water (\(\text{H}_{2}\text{O}\)) has the following percentage composition: \(\text{89}\%\) oxygen and \(\text{11}\%\) hydrogen.

Example: Empirical and Molecular Formula

Question

Vinegar, which is used in our homes, is a dilute form of acetic acid. A sample of acetic acid has the following percentage composition: \(\text{39.9}\%\) carbon, \(\text{6.7}\%\) hydrogen and \(\text{53.4}\%\) oxygen.

  1. Determine the empirical formula of acetic acid.

  2. Determine the molecular formula of acetic acid if the molar mass of acetic acid is \(\text{60.06}\) \(\text{g·mol$^{-1}$}\).

Step 1: Find the mass

In \(\text{100}\) \(\text{g}\) of acetic acid, there is: \(\text{39.9}\) \(\text{g}\) \(\text{C}\), \(\text{6.7}\) \(\text{g}\) \(\text{H}\) and \(\text{53.4}\) \(\text{g}\) \(\text{O}\).

Step 2: Find the moles

\(n = \frac{m}{M}\)

\begin{align*} n_{\text{C}} & = \frac{\text{39.9}}{\text{12}} = \text{3.325}\text{ mol}\\ n_{\text{H}} & = \frac{\text{6.7}}{\text{1.01}} = \text{6.6337}\text{ mol}\\ n_{\text{O}} & = \frac{\text{53.4}}{\text{16}} = \text{3.3375}\text{ mol} \end{align*}

Step 3: Find the empirical formula

To find the empirical formula we first note how many moles of each element we have. Then we divide the moles of each element by the smallest of these numbers, to get the ratios of the elements. This ratio is rounded off to the nearest whole number.

\(\text{C}\)

\(\text{H}\)

\(\text{O}\)

\(\text{3.325}\)

\(\text{6.6337}\)

\(\text{3.3375}\)

\(\dfrac{\text{3.325}}{\text{3.325}}\)

\(\dfrac{\text{6.6337}}{\text{3.325}}\)

\(\dfrac{\text{3.3375}}{\text{3.325}}\)

1

2

1

The empirical formula is \(\text{CH}_{2}\text{O}\).

Step 4: Find the molecular formula

The molar mass of acetic acid using the empirical formula (\(\text{CH}_{2}\text{O}\)) is \(\text{30.02}\) \(\text{g·mol$^{-1}$}\). However the question gives the molar mass as \(\text{60.06}\) \(\text{g·mol$^{-1}$}\). Therefore the actual number of moles of each element must be double what it is in the empirical formula (\(\dfrac{\text{60.06}}{\text{30.02}} = \text{2}\)).The molecular formula is therefore \(\text{C}_{2}\text{H}_{4}\text{O}_{2}\) or \(\text{CH}_{3}\text{COOH}\)

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