Question

Determine the empirical formula for each compound. a. ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) b. ascorbic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\right)\) c. naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\)

Short answer

The empirical formulas for the given compounds are: a. Ethylene: \(\mathrm{CH}_{2}\) b. Ascorbic acid: \(\mathrm{C}_{3}\mathrm{H}_{4}\mathrm{O}_{3}\) c. Naphthalene: \(\mathrm{C}_{5}\mathrm{H}_{4}\)

Step-by-step solution

Determine the empirical formula for ethylene

Ethylene has a molecular formula of \(\mathrm{C}_{2}\mathrm{H}_{4}\). The GCD of 2 and 4 is 2, so we'll divide each subscript in the formula by this value: \[ \frac{\mathrm{C}_{2}\mathrm{H}_{4}}{2} = \mathrm{C}_{1}\mathrm{H}_{2} \] So the empirical formula for ethylene is \(\mathrm{CH}_{2}\).

Determine the empirical formula for ascorbic acid

Ascorbic acid has a molecular formula of \(\mathrm{C}_{6}\mathrm{H}_{8}\mathrm{O}_{6}\). The GCD of 6, 8, and 6 is 2, so we'll divide each subscript in the formula by this value: \[ \frac{\mathrm{C}_{6}\mathrm{H}_{8}\mathrm{O}_{6}}{2} = \mathrm{C}_{3}\mathrm{H}_{4}\mathrm{O}_{3} \] So the empirical formula for ascorbic acid is \(\mathrm{C}_{3}\mathrm{H}_{4}\mathrm{O}_{3}\).

Determine the empirical formula for naphthalene

Naphthalene has a molecular formula of \(\mathrm{C}_{10}\mathrm{H}_{8}\). The GCD of 10 and 8 is 2, so we'll divide each subscript in the formula by this value: \[ \frac{\mathrm{C}_{10}\mathrm{H}_{8}}{2} = \mathrm{C}_{5}\mathrm{H}_{4} \] So the empirical formula for naphthalene is \(\mathrm{C}_{5}\mathrm{H}_{4}\).

Key concepts

Chemical Composition

When we talk about the chemical composition of a substance, we're essentially discussing what that substance is made of at the most fundamental level: its atoms and molecules. Every compound has a unique composition that can be conveyed through a chemical formula, which represents the types of atoms present and their relative numbers.

For instance, the molecular formula for ethylene is \( \mathrm{C}_{2}\mathrm{H}_{4} \), indicating it is composed of 2 carbon atoms and 4 hydrogen atoms. However, to find the simplest whole number ratio of these atoms in the compound, one would look at the empirical formula, which in the case of ethylene is \( \mathrm{CH}_{2} \). The empirical formula gives us the simplest representation of the compound's composition, highlighting its basic building blocks without the complication of multiple units.

Molecular Formula

The molecular formula of a compound takes a step deeper into the details by not only showing the types of atoms but also how many of each atom are in a single molecule of the compound. It essentially provides the exact number of each type of atom in a molecule. This can be critical for understanding the properties and reactions of the compound.

Let's consider ascorbic acid, with a molecular formula of \( \mathrm{C}_{6}\mathrm{H}_{8}\mathrm{O}_{6} \). Here, we see that each molecule contains 6 carbon atoms, 8 hydrogen atoms, and 6 oxygen atoms. In comparison, its empirical formula, \( \mathrm{C}_{3}\mathrm{H}_{4}\mathrm{O}_{3} \), simplifies this information by showing a reduced ratio, which still reflects the same proportions of atoms but not the actual count of atoms in a molecule.

Stoichiometry

Stoichiometry is the section of chemistry that deals with the quantitative relationships of the reactants and products in a chemical reaction. It's based on the conservation of mass and the law of definite proportions, which states that a chemical compound always contains exactly the same proportion of elements by mass.

Understanding stoichiometry is essential when converting between different chemical quantities and developing reactions. For instance, when calculating the amount of product formed from a known amount of reactants, we can use the empirical and molecular formulas alongside the balanced equation for the reaction to ensure the reactants and products are in the correct proportions. In the determination of empirical formulas, as is done for naphthalene, resulting in \( \mathrm{C}_{5}\mathrm{H}_{4} \), stoichiometry allows us to deduce the simplest ratio and potentially the amounts of reactants required to synthesise a specific amount of the compound.