---

Example 3.11: Determining a Compound’s Empirical Formula from the Masses of Its Elements

A sample of the black mineral hematite ( Figure 12), an oxide of iron found in many iron ores, contains 34.97 g of iron and 15.03 g of oxygen. What is the empirical formula of hematite?
Figure 12: Hematite is an iron oxide that is used in jewelry. (credit: Mauro Cateb)

Solution

For this problem, we are given the mass in grams of each element.

\(m_{\mathrm{\ce{Fe}}}\) \(= 34.97\ \mathrm{g}\)


\(m_{\mathrm{\ce{O}}}\) \(= 15.03\ \mathrm{g}\)


\(\mathrm{ratio}_{\mathrm{molar}}\) = ?


Begin by finding the chemical amount of each:

\(n_{\mathrm{\ce{Fe}}}\) \(= \dfrac{m_{\mathrm{\ce{Fe}}}}{M_{\mathrm{\ce{Fe}}}}\)

\(\ \ \ =\dfrac{34.97\ \mathrm{g}}{55.845\ \frac{\mathrm{g}}{\mathrm{mol}}}\)

\(\ \ \ =0.6262\ \mathrm{mol}\)


\(n_{\mathrm{\ce{O}}}\) \(= \dfrac{m_{\mathrm{\ce{O}}}}{M_{\mathrm{\ce{O}}}}\)

\(\ \ \ =\dfrac{15.03\ \mathrm{g}}{15.999\ \frac{\mathrm{g}}{\mathrm{mol}}}\)

\(\ \ \ =0.9394\ \mathrm{mol}\)


There is less iron than oxygen, so calculate the oxygen-to-iron molar ratio:

\(\mathrm{ratio}_{\mathrm{\ce{O:Fe}}}\) \(= \dfrac{n_{\mathrm{\ce{O}}}}{n_{\mathrm{\ce{Fe}}}}\)

\(\ \ \ =\dfrac{0.9394\ \mathrm{mol}}{0.6262\ \mathrm{mol}}\)

\(\ \ \ =1.500\)


The ratio is of iron to oxygen is 1 : 1.500, so the formula is


Finally, multiply all coefficients by two to get the smallest possible whole number subscripts while still maintaining the correct iron-to-oxygen ratio:


The empirical formula is \(\ce{Fe2O3}\).