Question

For the following compounds whose molecular formulas are given, indicate the empirical formula: (a) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \quad\) glucose (b) \(\mathrm{C}_{8} \mathrm{H}_{18} \quad\) octane (c) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3} \quad\) lactic acid (d) \(\mathrm{C}_{25} \mathrm{H}_{52} \quad\) paraffin (e) \(\mathrm{C}_{12} \mathrm{H}_{4} \mathrm{Cl}_{4} \mathrm{O}_{2} \quad\) dioxin (a powerful poison)

Short answer

(a) \(\mathrm{CH}_{2}\mathrm{O}\) (b) \(\mathrm{C}_{4}\mathrm{H}_{9}\) (c) \(\mathrm{CH}_{2}\mathrm{O}\) (d) \(\mathrm{C}_{25}\mathrm{H}_{52}\) (e) \(\mathrm{C}_{6}\mathrm{H}_{2} \mathrm{Cl}_{2} \mathrm{O}\)

Step-by-step solution

Understanding Empirical Formulas

The empirical formula of a compound is the simplest ratio of the elements present in it. To determine the empirical formula, simplify the molecular formula so that the subscript numbers are the smallest possible integers.

Simplify Glucose \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)

Divide each subscript in the molecular formula by the greatest common divisor (GCD). For glucose, the GCD of 6, 12, and 6 is 6. Therefore, the empirical formula is: \[ \mathrm{C}_{6 / 6} \mathrm{H}_{12 / 6} \mathrm{O}_{6 / 6} = \mathrm{C}_{1} \mathrm{H}_{2} \mathrm{O}_{1} \Rightarrow \mathrm{CH}_{2}\mathrm{O} \]

Simplify Octane \(\mathrm{C}_{8} \mathrm{H}_{18}\)

Identify the GCD of 8 and 18, which is 2. Simplify the molecular formula by dividing each subscript by 2: \[ \mathrm{C}_{8 / 2} \mathrm{H}_{18 / 2} = \mathrm{C}_{4} \mathrm{H}_{9} \]

Simplify Lactic Acid \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}\)

For lactic acid, the GCD of 3, 6, and 3 is 3. Simplify by dividing each subscript by 3: \[ \mathrm{C}_{3 / 3} \mathrm{H}_{6 / 3} \mathrm{O}_{3 / 3} = \mathrm{C}_{1} \mathrm{H}_{2} \mathrm{O}_{1} \Rightarrow \mathrm{CH}_{2}\mathrm{O} \]

Simplify Paraffin \(\mathrm{C}_{25} \mathrm{H}_{52}\)

The GCD of 25 and 52 is 1, so the empirical formula is the same as the molecular formula: \[ \mathrm{C}_{25} \mathrm{H}_{52} \]

Simplify Dioxin \(\mathrm{C}_{12} \mathrm{H}_{4} \mathrm{Cl}_{4} \mathrm{O}_{2}\)

The GCD of 12, 4, 4, and 2 is 2. Simplify by dividing each subscript by 2: \[ \mathrm{C}_{12 / 2} \mathrm{H}_{4 / 2} \mathrm{Cl}_{4 / 2} \mathrm{O}_{2 / 2} = \mathrm{C}_{6} \mathrm{H}_{2} \mathrm{Cl}_{2} \mathrm{O}_{1} \Rightarrow \mathrm{C}_{6} \mathrm{H}_{2} \mathrm{Cl}_{2} \mathrm{O} \]

Key concepts

simplifying molecular formulas

Simplifying molecular formulas is essential in chemistry to better understand the composition of basic compounds. Molecular formulas represent the actual number of atoms of each element in a molecule. However, these can often be complex. To simplify them, we convert the molecular formula to an empirical formula, which shows the simplest whole-number ratio of the constituent elements. For instance, if we look at glucose with the molecular formula \(\text{C}_6 \text{H}_{12} \text{O}_6\), the empirical formula is \(\text{CH}_2\text{O}\). This is achieved by dividing each subscript by the greatest common divisor (GCD). In this case, the GCD for 6, 12, and 6 is 6. This process helps in identifying the basic building blocks of the compound more easily.

greatest common divisor in chemistry

The greatest common divisor (GCD) is a useful mathematical tool in chemistry for simplifying molecular formulas to empirical formulas. The GCD helps in finding the smallest whole-number ratio of elements by dividing the subscripts of each element in the molecular formula by the largest integer that can evenly divide them. For instance, consider octane with the molecular formula \(\text{C}_8 \text{H}_{18}\). The GCD of 8 and 18 is 2. Dividing the subscripts by 2 simplifies the formula to \(\text{C}_4 \text{H}_9\), which is the empirical formula. Similarly, for lactic acid \(\text{C}_3 \text{H}_6 \text{O}_3\), the GCD is 3, simplifying it to \(\text{CH}_2\text{O}\). Recognizing and applying the GCD allows chemists to better understand the underlying structure of complex molecules.

basic chemical compounds

Understanding and identifying basic chemical compounds require knowing their molecular and empirical formulas. Basic compounds consist of combinations of elements represented in a specific ratio. For example, paraffin's molecular formula is \(\text{C}_{25} \text{H}_{52}\). Since the GCD of 25 and 52 is 1, its empirical formula remains the same as its molecular formula. Another example includes dioxin, a complex and hazardous compound, with a molecular formula \(\text{C}_{12} \text{H}_4 \text{Cl}_4 \text{O}_2\). Dividing each subscript by the GCD of 2 gives an empirical formula of \(\text{C}_6 \text{H}_2 \text{Cl}_2 \text{O}\). By simplifying molecular formulas, we can better understand the properties and behaviors of these compounds in various reactions and their effects in different environments.