Problem 141 - Chemistry and Chemical Reactivity

Question

Direct reaction of iodine \(\left(\mathrm{I}_{2}\right)\) and chlorine \(\left(\mathrm{Cl}_{2}\right)\) produces an iodine chloride, \(\mathrm{I}_{x} \mathrm{Cl}_{y},\) a bright yellow solid. If you completely consume 0.678 g of \(\mathrm{I}_{2}\) in a reaction with excess \(\mathrm{Cl}_{2}\) and produce \(1.246 \mathrm{g}\) of \(\mathrm{I}_{x} \mathrm{Cl}_{y},\) what is the empirical formula of the compound? A later experiment showed that the molar mass of \(\mathrm{I}_{x} \mathrm{Cl}_{y}\) was \(467 \mathrm{g} / \mathrm{mol} .\) What is the molecular formula of the compound?

Step-by-Step Solution

Answer

The empirical and molecular formula is \(\mathrm{ICl}_6\).

Step 1: Calculate Moles of Iodine

First, find the number of moles of iodine (\(\mathrm{I}_2\)) used in the reaction. The molar mass of iodine \(\mathrm{I}_2\) is approximately 254 g/mol (since each iodine atom is about 127 g/mol, and \(\mathrm{I}_2\) consists of two iodine atoms). Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). \[ \text{Moles of } \mathrm{I}_2 = \frac{0.678 \text{ g}}{254 \text{ g/mol}} = 0.00267 \text{ mol} \]

Step 2: Determine Mass of Chlorine Used

Next, calculate the mass of chlorine (\(\mathrm{Cl}_2\)) that reacted by subtracting the mass of iodine used from the total mass of the iodine chloride product. \[ \text{Mass of } \mathrm{Cl}_2 = 1.246 \text{ g} - 0.678 \text{ g} = 0.568 \text{ g} \]

Step 3: Calculate Moles of Chlorine

Find the number of moles of chlorine atoms needed. Chlorine gas is \(\mathrm{Cl}_2\), with a molar mass of 70.9 g/mol, but since the mass references chloride atoms, use the atomic mass of chlorine which is approximately 35.45 g/mol. \[ \text{Moles of } \mathrm{Cl} = \frac{0.568 \text{ g}}{35.45 \text{ g/mol}} = 0.0160 \text{ mol} \]

Step 4: Determine Empirical Formula

Determine the ratio of moles of iodine to chlorine. Divide both by the smallest number of moles calculated, which is 0.00267. \[ \text{Ratio of } \mathrm{I} = \frac{0.00267}{0.00267} = 1 \] \[ \text{Ratio of } \mathrm{Cl} = \frac{0.0160}{0.00267} \approx 6 \] Thus, the empirical formula is \(\mathrm{ICl}_6\).

Step 5: Calculate Empirical Formula Molar Mass

Calculate the molar mass of the empirical formula \(\mathrm{ICl}_6\). \[ \text{Molar mass of } \mathrm{ICl}_6 = 127 \cdot 1 + 35.45 \cdot 6 = 127 + 212.7 = 339.7 \text{ g/mol} \]

Step 6: Determine Molecular Formula

Compare the empirical formula molar mass with the given molecular molar mass (467 g/mol). Divide the molecular mass by the empirical formula molar mass to find the multiplier. \[ \text{Multiplier} = \frac{467}{339.7} \approx 1.375 \] Since it should be a whole number, round to the nearest integer which is 1, signifying the empirical formula is the same as the molecular formula. Therefore, \(\mathrm{ICl}_6\) is both the empirical and molecular formula.

Key Concepts

Chemical Reactions

Chemical reactions are transformations where substances, known as reactants, are converted into different substances, called products. In the exercise, iodine \((\mathrm{I}_2)\) and chlorine \((\mathrm{Cl}_2)\) react to form iodine chloride \((\mathrm{I}_x\mathrm{Cl}_y)\). Such reactions can involve the breaking and forming of chemical bonds, often leading to the formation of one or more new compounds.

Understanding a chemical reaction requires recognizing the reactants involved, the products formed, and the type of reaction. Here, the reaction between iodine and chlorine is a synthesis reaction, where simpler molecules combine to form a more complex compound. This type of reaction is common and can often be identified through the combination of elements or simpler molecules into one larger compound.

In practice, knowing the specific details of chemical reactions helps chemists synthesize desired compounds, understand reaction mechanisms, and even optimize industrial processes.

Stoichiometry

Stoichiometry is a fundamental concept linking the quantities of reactants and products in chemical reactions. It is based on the principle that matter is conserved in reactions, meaning the total number of atoms on each side of a chemical equation must be equal.

In this exercise, stoichiometry helps us determine how much chlorine reacted by calculating the difference in mass before and after the reaction. By understanding stoichiometry, students can predict the amounts of products formed from given reactants, balance chemical equations, and quantify yield efficiency.

Key steps include:

• Using molar masses to convert between mass and moles.
• Determining the absolute amount of each element consumed or produced.
• Using ratios derived from balanced equations to relate different substances involved.

Mastering stoichiometry provides a foundational skill for anyone studying chemistry, enabling precise calculations in both lab and theoretical settings.

Molar Mass Calculation

Molar mass is crucial when working with chemical equations since it allows the conversion between the mass of a substance and the amount in moles. The molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol).

For iodine \((\mathrm{I}_2)\), the molar mass is calculated by adding the atomic masses of two iodine atoms, resulting in approximately 254 g/mol. Similarly, for chlorine being part of iodine chloride \((\mathrm{Cl})\), we use the atomic mass \(35.45 \text{ g/mol}\) for calculations.

Accurate molar mass calculations rely on understanding periodic table values, as these provide the necessary atomic masses. With these values, chemists calculate reactant or product amounts, analyze yields, and ensure reactions are properly balanced, which is essential in research and industrial applications.

Whether calculating the molar mass of molecules like \((\mathrm{ICl}_6)\) or determining unknown compounds, these techniques are vital in grasping and applying chemical principles.

Empirical Formula Determination

Empirical formulas are the simplest whole number ratios of elements within a compound. Determining an empirical formula involves analyzing the molar amounts of each atom and finding their simplest ratio.

In the problem, we calculate moles of iodine and chlorine, and use them to establish that the ratio is 1 iodine atom to 6 chlorine atoms, giving us \((\mathrm{ICl}_6)\) as the empirical formula.

Key steps in this process include:

• Calculating the moles of each element initially present.
• Dividing these quantities by the smallest mole number.
• Producing a set of integers reflecting element ratios.

Once established, empirical formulas provide insight into chemical compositions, helping predict compound properties and behaviors.

Finally, comparing the empirical formula mass with the actual molar mass provides validation and sometimes guides to the molecular formula, as seen in more complex analysis.