Question

What are the empirical formulas of the following compounds? (a) \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{OH})_{2}\) (d) \(\mathrm{B}_{2} \mathrm{H}_{6}\) (b) \(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{8}\) (e) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (c) \(\mathrm{C}_{4} \mathrm{H}_{10}\)

Short answer

(a) \(\mathrm{C}_{2} \mathrm{H}_{4}\mathrm{O}_{2}\), (d) \(\mathrm{BH}_{3}\), (b) \(\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{8}\), (e) \(\mathrm{C}_{2} \mathrm{H}_{6}\mathrm{O}\), (c) \(\mathrm{C}_{2} \mathrm{H}_{5}\)

Step-by-step solution

Identify and Simplify for Compound (a)

To determine the empirical formula, we need to find the simplest whole number ratio of atoms in the compound. For \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{OH})_{2}\), the existing numbers are already in the lowest whole number ratio. Therefore, the empirical formula is the same as the molecular formula: \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{OH})_{2}\).

Identify and Simplify for Compound (d)

For \(\mathrm{B}_{2} \mathrm{H}_{6}\), we can see that the subscripts are both divisible by 2. Dividing by 2 gives \(\mathrm{B}_{2/2} \mathrm{H}_{6/2}\), which simplifies to \(\mathrm{B}\mathrm{H}_{3}\). Thus, the empirical formula is \(\mathrm{B}\mathrm{H}_{3}\).

Identify and Simplify for Compound (b)

For \(\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{8}\), all subscripts are already at their lowest ratio with no common divisors other than 1. Therefore, the empirical formula is identical to its molecular formula, which is \(\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{8}\).

Identify and Simplify for Compound (e)

The compound \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) consists of a hydroxyl group (-OH) bonded to an ethyl group (\mathrm{C}_{2} \mathrm{H}_{5}). The empirical formula represents the simplest whole number ratio, but here it is not simple because the -OH group cannot be simplified along with the ethyl group. So the empirical formula remains as \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) or \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}\).

Identify and Simplify for Compound (c)

In the case of \(\mathrm{C}_{4} \mathrm{H}_{10}\), we look for the greatest common divisor of the subscripts, which is 2. Dividing the subscripts by 2 gives us \(\mathrm{C}_{4/2} \mathrm{H}_{10/2}\), simplifying to \(\mathrm{C}_{2} \mathrm{H}_{5}\). Therefore, the empirical formula for this compound is \(\mathrm{C}_{2} \mathrm{H}_{5}\).

Key concepts

Chemical Compound Composition

Understanding the makeup of chemical compounds is essential when diving into the world of chemistry. A chemical compound is composed of two or more elements that are chemically bonded together. These elements combine in fixed ratios which are represented by chemical formulas.

The chemical formula provides information about the types and numbers of atoms in the smallest representative unit of the compound. For example, water (H₂O) is composed of two hydrogen atoms bonded with one oxygen atom. In the provided exercise, the aim was to find the empirical formulas for a set of compounds by examining these ratios.

An empirical formula is the simplest whole number ratio of the elements within a compound. It may not necessarily represent the actual numbers of atoms within a molecule but provides a basic estimation of the composition. For instance, the molecular formula of glucose is C₆H₁₂O₆, but its empirical formula is CH₂O, depicting that for every carbon atom, there are two hydrogen atoms and one oxygen atom.

Molecular Formulas

While the empirical formula gives us the simplest representation, the molecular formula of a compound indicates the exact number of atoms of each element in one molecule of the compound. It could be the same as the empirical formula or a multiple of it. For instance, ethene has a molecular formula of C₂H₄ which happens to be the same as its empirical formula because it is already in the lowest whole number ratio.

In the exercises provided, analyzing the molecular formulas was a necessary step in determining the empirical formulas. It is important to note that while the empirical formula is helpful for calculation purposes in stoichiometry, the molecular formula provides a clear understanding of the exact structure of the molecule. Molecular formulas are particularly useful when determining the molecular weights and when you need to understand the interactions between atoms within a molecule.

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative aspect of chemical reactions. It is essentially about the calculation of reactants and products involved in chemical processes. Stoichiometry uses balanced chemical equations, empirical formulas, and molecular formulas to perform these calculations.

For example, when balancing a chemical equation, it's vital to ensure that the number of atoms for each element is equal on both sides of the equation, reflecting the law of conservation of mass. This is where your empirical formula knowledge comes into play – it helps predict how much of each reactant will be needed and the amount of product that will be produced.

The earlier exercise establishes the empirical formulas, which are often used in stoichiometric calculations to determine the amount of a substance needed or produced. A firm grasp of stoichiometry is essential in the practical application of chemistry, be it in a lab synthesizing new materials or in an industrial process where efficiency and accuracy are critical.