Question

Samples of several compounds are decomposed, and the masses of their constituent elements are measured. Calculate the empirical formula for each compound. (a) \(2.677 \mathrm{~g} \mathrm{Ba}, 3.115 \mathrm{~g} \mathrm{Br}\) (b) \(1.651 \mathrm{~g} \mathrm{Ag}, 0.1224 \mathrm{~g} \mathrm{O}\) (c) \(0.672 \mathrm{~g} \mathrm{Co}, 0.569 \mathrm{~g}\) As, \(0.486 \mathrm{~g} \mathrm{O}\)

Short answer

The empirical formula for (a) is BaBr, for (b) is Ag2O, and for (c) is CoAsO2.

Step-by-step solution

Convert Mass to Moles for Compound (a)

To find the empirical formula, we first convert the mass of each element to moles using the molar mass (atomic weight) of each element. For Ba, the molar mass is 137.33 g/mol, and for Br, it is 79.904 g/mol. The calculations are as follows:\[\text{moles of Ba} = \frac{2.677\text{ g}}{137.33 \text{ g/mol}}\] and \[\text{moles of Br} = \frac{3.115\text{ g}}{79.904 \text{ g/mol}}\].

Determine the Simplest Whole Number Ratio for Compound (a)

Divide the number of moles of each element by the smallest number of moles to find the simplest whole number ratio. We round to the nearest whole number if the result is close to a whole number.

Write the Empirical Formula for Compound (a)

Use the simplest whole number ratio to write the empirical formula by assigning subscripts to each element based on the ratio.

Convert Mass to Moles for Compound (b)

The mass of Ag is converted to moles using its molar mass, 107.87 g/mol, and the mass of O is converted using 16.00 g/mol molar mass. Perform the conversion:\[\text{moles of Ag} = \frac{1.651\text{ g}}{107.87 \text{ g/mol}}\] and \[\text{moles of O} = \frac{0.1224\text{ g}}{16.00 \text{ g/mol}}\].

Determine the Simplest Whole Number Ratio for Compound (b)

Calculate the ratio of moles of Ag to moles of O by dividing each by the smaller of the two values.

Write the Empirical Formula for Compound (b)

Assign the ratio as subscripts to Ag and O in the empirical formula.

Convert Mass to Moles for Compound (c)

Convert the mass of Co, As, and O to moles using their respective molar masses: 58.93 g/mol for Co, 74.92 g/mol for As, and 16.00 g/mol for O.

Determine the Simplest Whole Number Ratio for Compound (c)

Calculate the ratio of moles of each element by dividing by the smallest number of moles among the three.

Write the Empirical Formula for Compound (c)

Write the empirical formula using the whole number ratios as subscripts for each of Co, As, and O.

Key concepts

Molar Mass

Molar mass is a fundamental concept in chemistry, often referred to as molecular weight. It represents the mass of one mole of a substance, usually expressed in grams per mole (g/mol). The molar mass of each element is found on the periodic table as the atomic weight and reflects the average mass of all isotopes of that element.

For instance, in the exercise, we calculate the molar mass of barium (Ba) which is 137.33 g/mol, bromine (Br) which is 79.904 g/mol, and so on for other elements. Knowing the molar mass allows us to convert between the mass of a sample and the number of moles, forming the first step in finding the empirical formula of a compound.

Moles

The concept of moles is central to the understanding of chemical reactions and composition. A mole is defined as the amount of substance that contains the same number of entities (such as atoms or molecules) as there are atoms in 12 grams of carbon-12. This number, Avogadro's number, is approximately 6.022 x 10²³.

In the provided exercise, we've converted the mass of each element into moles using their molar masses. This is a crucial step because it normalizes measurements, allowing for comparison and combination of different elements in a chemically meaningful way. Understanding moles is key to making sense of the stoichiometry of reactions and the composition of compounds.

Whole Number Ratio

Entities in compounds always combine in fixed whole number ratios, not fractions. This is a reflection of the discrete nature of atoms and molecules. The whole number ratio is the essence of an empirical formula, which indicates the simplest ratio of atoms in a compound.

Once the number of moles for each element in the compound is known, we divide each value by the smallest number of moles among them. The resulting values are rounded to the nearest whole numbers to determine the empirical formula, as demonstrated in the steps for each compound in the original exercise. These ratios are vital for illustrating the proportional relationship of each element within a compound.

Chemical Composition

Chemical composition refers to the identity and the relative number of the atoms that constitute a particular compound. It can be described using an empirical formula, which represents the simplest whole number ratio of elements in the compound. The chemical composition is fundamental for understanding the properties of a substance and how it behaves in chemical reactions.

Using the exercise as an example, once the number of moles of each element is determined and the whole number ratio is calculated, the empirical formula that represents the chemical composition is written. This simple and high-level view of a compound's makeup is often used in stoichiometrical calculations in chemistry.

Stoichiometry

Stoichiometry is the section of chemistry that relates to the quantitative relationships between reactants and products in a chemical reaction. It uses the conservation of mass and the concept of moles to help predict the amounts of substances consumed and produced in reactions.

In the context of our exercise, stoichiometry would come into play after establishing the empirical formula, where we could use it to calculate how much of each element would react or be produced in a given chemical reaction. It serves as the backbone for further calculations in chemistry, such as determining the theoretical yield or limiting reactants in reactions.