Problem 56 - Chemistry: An Atoms-Focused Approach

Question

A sample of an iron-containing compound is \(22.0 \%\) iron, \(50.2 \%\) oxygen, and \(27.8 \%\) chlorine by mass. What is the empirical formula of this compound?

Step-by-Step Solution

Answer

The empirical formula of the iron-containing compound is FeO8Cl2.

Step 1: Convert mass percentages to grams

Consider a 100g sample of the compound, since this will make working with the mass percentages easier. The masses of the elements are: - Iron (Fe): \(22.0 \%\) of 100g \(\implies\) 22.0g - Oxygen (O): \(50.2 \%\) of 100g \(\implies\) 50.2g - Chlorine (Cl):\(27.8 \%\) of 100g \(\implies\) 27.8g

Step 2: Convert grams to moles for each element

Using the molar mass of each element, convert the masses obtained in step 1 to moles: - Moles of Fe: \(\frac{22.0}{55.85 \text{ g/mol}} = 0.3938 \text{ mol}\) - Moles of O: \(\frac{50.2}{16.00 \text{ g/mol}} = 3.1375 \text{ mol}\) - Moles of Cl: \(\frac{27.8}{35.45 \text{ g/mol}} = 0.7844 \text{ mol}\)

Step 3: Find the simplest whole number ratio of moles

Determine the smallest value of moles found in step 2: \(\text{min}(0.3938, 3.1375, 0.7844) = 0.3938\) Now divide the moles of each element by this smallest value to find the ratio: - Ratio of Fe: \(\frac{0.3938}{0.3938} = 1\) - Ratio of O: \(\frac{3.1375}{0.3938} \approx 7.97 \approx 8\) - Ratio of Cl: \(\frac{0.7844}{0.3938} \approx 1.99 \approx 2\)

Step 4: Write the empirical formula

Based on the whole number ratios found in step 3, the empirical formula of the iron-containing compound is FeO\(_8\)Cl\(_2\).