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

Understand the Problem

The question is asking to determine the empirical formula of a compound given its percent composition by mass. To solve this, we need to convert the percentages to grams, then to moles, and finally find the simplest whole number ratio of the moles of each element.

Answer

$C_3H_8O$

Steps to Solve

Convert percentages to grams Assume we have 100 g of the compound. This makes the percentages directly equivalent to grams. So, we have 60.0 g of C, 13.4 g of H, and 26.6 g of O.
Convert grams to moles Divide the mass of each element by its molar mass to find the number of moles. Molar mass of C = 12.01 g/mol Molar mass of H = 1.008 g/mol Molar mass of O = 16.00 g/mol

Moles of C = $\frac{60.0 \text{ g}}{12.01 \text{ g/mol}} = 4.99 \text{ mol}$ Moles of H = $\frac{13.4 \text{ g}}{1.008 \text{ g/mol}} = 13.3 \text{ mol}$ Moles of O = $\frac{26.6 \text{ g}}{16.00 \text{ g/mol}} = 1.66 \text{ mol}$

Find the simplest whole number ratio Divide each mole value by the smallest mole value to find the simplest ratio. The smallest mole value is 1.66.

Ratio of C = $\frac{4.99}{1.66} = 3.01 \approx 3$ Ratio of H = $\frac{13.3}{1.66} = 8.01 \approx 8$ Ratio of O = $\frac{1.66}{1.66} = 1$

Write the empirical formula The empirical formula is $C_3H_8O_1$, which can be written as $C_3H_8O$.

$C_3H_8O$

More Information

The empirical formula represents the simplest whole-number ratio of atoms in a compound. It doesn't necessarily represent the actual number of atoms in a molecule of the compound, which is given by the molecular formula.

Tips

A potential mistake is not dividing by the smallest mole number. Also, rounding too early can cause errors in the final ratio.

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