Question

On analysis, a certain compound was found to contain \(254 \mathrm{~g}\) of iodine (at. mass 127 ) and \(80 \mathrm{~g}\) oxygen (at. mass 16 ). What is the formula of the compound ? (a) IO (b) \(\mathrm{I}_{2} \mathrm{O}\) (c) \(\mathrm{I}_{5} \mathrm{O}_{3}\) (d) \(\mathrm{I}_{2} \mathrm{O}_{\mathrm{5}}\)

Short answer

The formula of the compound is \(\mathrm{I}_{2} \mathrm{O}_{5}\).

Step-by-step solution

Calculate moles of iodine

To find the moles of iodine in the compound, use the formula: moles = mass / molar mass. For iodine, the molar mass is 127 g/mol, so the moles are calculated as follows: moles of iodine = 254 g / 127 g/mol.

Calculate moles of oxygen

Repeat the process for oxygen, where the molar mass is 16 g/mol: moles of oxygen = 80 g / 16 g/mol.

Determine the simplest mole ratio

Determine the simplest whole number ratio between the moles of iodine and oxygen by dividing each by the smallest number of moles calculated in the previous steps.

Write the empirical formula

Using the simplest mole ratio, write the empirical formula of the compound. If necessary, multiply the ratios by the smallest number to get whole numbers and match the options given.

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationship between reactants and products in a chemical reaction. In the context of empirical formula calculation, stoichiometry involves determining the simplest ratio of elements in a compound, based on their molar amounts.

For instance, in the given exercise, stoichiometry is used to find the most basic ratio of iodine to oxygen in the compound. We achieve this by calculating the moles of each element, and then expressing these quantities in the smallest whole number ratio that reflects the proportions in which these elements combine to form the compound.

In the real-world application of stoichiometry, it's essential for chemists to predict how much product a reaction will yield, or how much of each reactant is needed, which requires a solid understanding of this concept.

Molar Mass

Molar mass is the mass of one mole of a substance, typically measured in grams per mole (g/mol). The molar mass of an element is numerically equal to its atomic mass and is essential in converting between the mass of a substance and the number of moles.

In applying the concept of molar mass to the exercise at hand, we use the atomic masses of iodine (127 g/mol) and oxygen (16 g/mol) to calculate how many moles of each element are present in the sample. The molar mass acts as a bridge between the mass of a compound found in a laboratory setting and the mole concept, which is crucial for stoichiometric calculations.

Accurate determination of molar masses is critical in chemical analysis and helps ensure precise stoichiometric calculations in various chemical processes, including empirical formula determination.

Mole Concept

The mole concept is a fundamental principle in chemistry that provides a method for counting particles at the atomic and molecular level. One mole is defined as exactly 6.022 x 10^23 particles, known as Avogadro's number. This concept allows chemists to count atoms, molecules, or ions in a given mass of a substance.

In the exercise scenario, the mole concept is applied to measure the number of moles of iodine and oxygen using their respective masses and molar masses. This step is vital to determine the simple mole ratio between these two elements, which directly leads to establishing the empirical formula of the compound.

Understanding the mole concept is indispensable for students as it lays the groundwork for interpreting chemical reactions and performing calculations related to the amounts of reactants and products.

Chemical Compound Analysis

Chemical compound analysis involves identifying and quantifying the elements or compounds in a chemical mixture. The empirical formula obtained from such analysis reflects the simplest whole number ratio of atoms of each element in a compound, but not necessarily the actual numbers of atoms in the molecules.

In the context of the exercise, chemical compound analysis is what provides the mass of iodine and oxygen, which are crucial for calculating the empirical formula. This type of analysis is essential in various fields, including pharmacology, materials science, and environmental science, where understanding the composition of a substance can have significant practical implications.

Students must appreciate the importance of careful and accurate measurement in chemical analysis, as well as the calculations that follow, to determine the characteristics of unknown substances, ultimately revealing the empirical formula of compounds.