Calculate the number of cations and anions in each of the following compounds: (a) \(0.764 \mathrm{~g}\) of CsI, (b) \(72.8 \mathrm{~g}\) of \(\mathrm{K}_{2} \mathrm{Cr}_{2}\mathrm{O}_{7}\) (c) \(6.54 \mathrm{~g}\) of\(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}\)

Understanding the molar mass of a compound is critical for stoichiometry calculations. The molar mass is the weight of one mole (Avogadro's number of atoms or molecules) of any chemical substance. To calculate it, one must sum up the atomic masses of each atom present in the molecular formula of the compound. For instance, in the exercise, the molar mass of CsI is computed by adding the atomic masses of Cesium (Cs) and Iodine (I), which are sourced from the periodic table.

The process is straightforward but requires accuracy and attention to detail:

• Identify each unique atom in the molecular formula.
• Find each atom's atomic mass typically expressed in grams per mole (g/mol).
• Multiply the atomic mass by the number of atoms of that element in the molecule.
• Add up these amounts to determine the total molar mass of the compound.

When done correctly, this step provides the basis for converting between moles and grams, which is essential in most stoichiometry problems.

The mole-to-mass conversion is a key stoichiometry skill, which allows us to quantify the actual amount of a substance involved in a chemical reaction. In our example exercise, we tackle this conversion by dividing the mass of the compound by its molar mass to determine the number of moles present.

This step is important because chemical equations are balanced based on moles rather than mass. By converting the mass of each compound into moles, we utilize a consistent and comparable unit that directly relates to the number of molecules or ions in the substance. The formula for this conversion is: \[ \text{Number of moles} = \frac{\text{Given mass (g)}}{\text{Molar mass (g/mol)}} \].
Understanding this relationship equips students with the ability to interpret and balance chemical equations and proceed with more intricate stoichiometry calculations.

Avogadro's number, approximately \(6.022 \times 10^{23}\), represents the number of particles found in one mole of a substance and is pivotal in transforming moles into actual numbers of atoms, molecules, or ions. In our textbook exercise, the final step uses Avogadro's number to calculate how many individual cations and anions are present in a given mass of a compound.

This is analogous to knowing how many eggs are in a dozen; similarly, with Avogadro's number, we can find out how many ions are in a 'mole's worth'. To complete this conversion, we multiply the number of moles of each ion by Avogadro's number: \[ \text{Number of ions} = \text{Number of moles} \times \text{Avogadro's number} \].
Practicing with this immense number helps students grasp the concept of moles and deal with quantities of substances at the molecular level, which is essential in chemistry.

Molecular formulas convey critical information about the composition of chemical compounds. They indicate the types and numbers of atoms present in a single molecule of the compound. For example, in our exercise, the molecular formula for potassium dichromate is \(\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7\), which indicates the presence of 2 potassium atoms, 2 chromium atoms, and 7 oxygen atoms within each molecule.

Working with molecular formulas allows students to understand the stoichiometry of a compound—the ratio of its components. This stoichiometry is crucial when calculating the number of moles of cations and anions in a compound, as shown in our exercise's steps. To elucidate further:

• Examine the subscript numbers in the formula to determine the ratio of the atoms present.
• Use this ratio to establish the mole relationship between the compound and its constituent ions or atoms.
• Ensure you understand the charge of ions to correctly account for the types of ions—cations or anions.

Mastery of molecular formulas is a stepping stone to predicting and balancing chemical reactions, which is a foundational aspect of chemical education.