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Example 2.6: Empirical and Molecular Formulas

Molecules of glucose (blood sugar) contain 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. What are the molecular and empirical formulas of glucose?

Solution

The molecular formula is:


(because one molecule actually contains 6 C, 12 H, and 6 O atoms)
The empirical formula shows the simplest whole-number ratio of the elements in a compound. We can take the numbers in the molecular formula and divide them all by the smallest number to get a tentative empirical formula. If all the numbers are whole numbers, we are done. If there are fractions such as halves or thirds, we have to multiply everything again (by 2 or 3, respectively) to arrive at the simplest whole-number ratio.

\(N_{\mathrm{\ce{C}}}\) \(= 6\)


\(N_{\mathrm{\ce{H}}}\) \(= 12\)


\(N_{\mathrm{\ce{O}}}\) \(= 6\)


\(\mathrm{smallest}\) \(= N_{\mathrm{\ce{C}}}\)    or \(N_{\mathrm{\ce{O}}}\)

\(\ \ \ =6\)

\(\ \ \ =6\)


\(N_{\mathrm{\ce{C},tentative}}\) \(= \dfrac{N_{\mathrm{\ce{C}}}}{\mathrm{smallest}}\)

\(\ \ \ =\dfrac{6}{6}\)

\(\ \ \ =1\)


\(N_{\mathrm{\ce{H},tentative}}\) \(= \dfrac{N_{\mathrm{\ce{H}}}}{\mathrm{smallest}}\)

\(\ \ \ =\dfrac{12}{6}\)

\(\ \ \ =2\)


\(N_{\mathrm{\ce{O},tentative}}\) \(= \dfrac{N_{\mathrm{\ce{O}}}}{\mathrm{smallest}}\)

\(\ \ \ =\dfrac{6}{6}\)

\(\ \ \ =1\)


In this case, all numbers are whole, so the simplest whole-number ratio C to H to O atoms in glucose is 1:2:1. (You are welcome to figure this out in your head if it is obvious to you from just looking at the formula.)
The empirical formula is: