What is the empirical formula of a compound that contains 60.0% C, 13.4% H, and 26.6% O?

Steps to Solve

1. Convert percentages to grams

Assume we have 100g of the compound. This makes the percentages directly equivalent to grams. Thus, we have 60.0g of C, 13.4g of H, and 26.6g of O.

2. Convert grams to moles

To convert grams to moles, divide the mass of each element by its molar mass (atomic weight):

• Carbon (C): molar mass = 12.01 g/mol
• Hydrogen (H): molar mass = 1.01 g/mol
• Oxygen (O): molar mass = 16.00 g/mol

Calculate the number of moles for each element:

• Moles of C = $\frac{60.0 \text{ g}}{12.01 \text{ g/mol}} \approx 5.00$ mol
• Moles of H = $\frac{13.4 \text{ g}}{1.01 \text{ g/mol}} \approx 13.27$ mol
• Moles of O = $\frac{26.6 \text{ g}}{16.00 \text{ g/mol}} \approx 1.66$ mol

3. Find the simplest whole number ratio

Divide each mole value by the smallest mole value (1.66 mol).

• Ratio of C = $\frac{5.00}{1.66} \approx 3.01 \approx 3$
• Ratio of H = $\frac{13.27}{1.66} \approx 8.00 \approx 8$
• Ratio of O = $\frac{1.66}{1.66} = 1$

4. Write the empirical formula The empirical formula follows the ratio calculated above. The empirical formula is $C_3H_8O$.