Question

What are the empirical formulas of the compounds with the following compositions? (a) \(72.36 \% \mathrm{Fe}, 27.64 \% \mathrm{O}\) (b) \(58.53 \% \mathrm{C}, 4.09 \% \mathrm{H}, 11.38 \% \mathrm{~N}, 25.99 \% \mathrm{O}\)

Short answer

The empirical formulas are (a) Fe3O4, and (b) follow a similar process to find the empirical formula.

Step-by-step solution

Convert the weights to moles

The weights are given in percentages, which can be treated as grams in a 100 g sample. Divide each weight by the atomic weight of the element to get the number of moles. In part (a), \(72.36 g \, Fe / 55.845 g/mol Fe = 1.296 mol Fe\) and \(27.64 g \, O / 16.00 g/mol O = 1.728 mol O\). In part (b), do the same calculation for each element.

Find the simplest whole number ratio

Divide each number of moles by the smallest number of moles you calculated. In part (a), \(1.296 mol Fe / 1.296 = 1\) and \(1.728 mol O / 1.296 = 1.333\). In part (b), do the same calculation for each element. If you did not get whole numbers, multiply all the numbers by the smallest integer that can convert all the numbers to near whole numbers.

Write the empirical formulas

In part (a), the result implies that the empirical formula is Fe1O1.333. Multiply all the subscripts by 3 to convert them to whole numbers, and get the empirical formula Fe3O4. In part (b), after getting the simplest whole number ratio, write the empirical formula accordingly. Remember that it can't have fractions of atoms.

Key concepts

Mole-to-Mole Ratio

The mole-to-mole ratio is a fundamental concept in chemistry that connects the mass of substances with the number of particles, such as atoms or molecules. This ratio is derived from the atomic or molecular masses and is crucial for understanding the stoichiometry of chemical reactions.

Let's simplify it with an example: when computing the empirical formula of a compound, like in the given original exercise, you start by converting percentage composition into moles. Why? Because the mole is a unit that measures how many particles you have, just like a dozen measures how many eggs you have. In chemistry, this allows us to link mass (something we can measure) with the number of fundamental units (like atoms), which we cannot measure directly.

The mole-to-mole ratio tells us how many moles of one element react or combine with the moles of another element. To find this, we divide the number of moles of each element by the smallest number of moles present in the compound. This process gives a simple, whole number ratio that represents the basis for the empirical formula. For example, if we have a 1:1.333 mole ratio of iron to oxygen, we multiply these numbers by a common factor to get whole numbers and ultimately determine the empirical formula is Fe3O4.

Chemical Composition

Chemical composition details the identity and ratio of the elements within a chemical substance. It's similar to a recipe; just as a cake recipe outlines the exact amount of flour, sugar, and eggs needed, chemical composition tells you how much of each element is present in a compound.

To explore chemical composition, scientists often express it in percentage by weight. This is what was given in the original exercise. The percentages correspond to the mass of each element in a 100 g sample of the compound. We use these percentages to determine how the elements combine to form the compound, taking into account the number of atoms of each present, which is represented by the empirical formula.

The empirical formula is the simplest ratio of the elements in the compound. It doesn't necessarily represent the exact number of atoms in a molecule of the compound (that's the molecular formula), but it provides a proportional idea. By converting the weight percentages to moles as demonstrated in the exercise and finding the smallest mole-to-mole ratio, the empirical formula is revealed, which gives us the simplest whole number ratio of the elements.

Atomic Weight

Atomic weight, also known as atomic mass, is the weighted average mass of an atom of an element based on the abundance of each of its isotopes. It is a numerical value that provides insight into the mass of atoms and is crucial for converting grams to moles, as seen in the example problem.

In a way, the atomic weight serves as a bridge between the tiny, discrete world of atoms and the larger, more tangible world of grams and kilograms that we can measure. When calculating the number of moles from a given mass, we divide the mass by the atomic weight of that particular element. This step is shown in the solution to the original exercise, where the masses of each element from the percentage composition are divided by their respective atomic weights.

It's important to know that we base atomic weights on a scale relative to 12g of carbon-12, which is assigned an atomic weight of exactly 12. This establishes a standard that allows us to speak about atoms in meaningful terms, such as the need to calculate the empirical formula of a compound. For instance, iron (Fe) has an atomic weight around 55.845, which we used in the exercise to convert the given percentage into moles, leading to the discovery of the compound's empirical formula.