Question

Direct reaction of iodine \(\left(I_{2}\right)\) and chlorine \(\left(C l_{2}\right)\) produces an iodine chloride, \(I_{x} C l_{y},\) a bright yellow solid. If you completely consume \(0.678 \mathrm{g}\) 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?

Short answer

The empirical formula is \( ICl_6 \), and the molecular formula is \( I_3Cl_9 \).

Step-by-step solution

Calculate Moles of Iodine

First, we need to calculate the moles of iodine consumed in the reaction. Given that the molar mass of iodine, \( I_2 \), is approximately 253.8 g/mol (2 * 126.9 g/mol for each I atom):\[ \text{Moles of } I_2 = \frac{0.678 \, g}{253.8 \, g/mol} \approx 0.00267 \, mol \]

Determine Mass of Chlorine in Product

Next, we find the mass of chlorine in the iodine chloride product. We know the total mass of product, \( I_xCl_y \), is 1.246 g. Subtracting the mass of iodine, we find:\[ \text{Mass of } Cl = 1.246 \, g - 0.678 \, g = 0.568 \, g \]

Calculate Moles of Chlorine

Using the molar mass of chlorine, \( Cl_2 \) (approximately 70.9 g/mol for \( Cl_2 \), but since we're looking for \( Cl \) in \( I_xCl_y \), use 35.45 g/mol for a single chlorine atom):\[ \text{Moles of } Cl = \frac{0.568 \, g}{35.45 \, g/mol} \approx 0.01602 \, mol \]

Find the Empirical Formula

To find the empirical formula, we compare the mole ratio of iodine to chlorine. Divide each by the smallest value:\[ \text{Ratio of } I : Cl = \frac{0.00267}{0.00267} : \frac{0.01602}{0.00267} = 1 : 6 \]This results in \( ICl_6 \) as the empirical formula.

Calculate Molar Mass from Empirical Formula

The molar mass for one formula unit of \( ICl_6 \) is calculated as follows:\[ 1(126.9) + 6(35.45) = 126.9 + 212.7 = 339.6 \, g/mol \]

Determine Molecular Formula

Given a molar mass of 467 g/mol, compare it with the empirical formula molar mass:\[ \text{Molecular formula factor} = \frac{467}{339.6} \approx 1.376 \]Since the factor is about 1.376, round to 1.375 as closest integer: 1.5, which indicates we multiply the empirical formula by 1.5 to find \( (ICl_6) \times 1.5 = I_3Cl_9 \).

Key concepts

Molecular Formula

The molecular formula of a compound indicates the actual number of atoms of each element in a molecule. It is crucial differences from the empirical formula, which provides the simplest whole-number ratio of atoms in a compound. In this case, the empirical formula of the compound was found to be \( ICl_6 \). However, the actual molecular formula was determined to be \( I_3Cl_9 \) based on the compound's molar mass. This determination involves comparing the empirical formula mass with the actual molar mass, which allows chemists to identify the multiplicative factor needed to scale the empirical formula to reflect the true nature of the molecule. Thus, understanding the relationship between empirical and molecular formulas is key in analyzing substances, ensuring accuracy in representing compounds at a molecular level.

Mole Calculations

Mole calculations are fundamental in chemistry. They are used to convert between mass, moles, and numbers of atoms or molecules. In the problem, determining the moles of iodine and chlorine was the first step.

• We calculated the moles of iodine by dividing its mass by its molar mass (253.8 g/mol).
• Similarly, knowing the mass of chlorine after reaction, we calculated the moles of chlorine using its atomic molar mass (35.45 g/mol).

This conversion is essential as it provides a quantitative way of understanding chemical processes. Remember, the mole is the bridge between the atomic scale and the real world, allowing us to handle substances in manageable amounts.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. In this scenario, iodine \((I_2)\) reacted with chlorine \((Cl_2)\) to form an iodine chloride \(I_xCl_y\). This reaction is direct, implying a straightforward combination of reactants.

• Iodine was completely consumed in the reaction.
• The excess of chlorine ensured that all iodine reacted.

Such reactions not only require balancing but also understanding the stoichiometry to predict the amounts of products formed. Comprehending the formation of iodine chloride in this reaction helps elucidate chemical compound synthesis and analysis, vital in practical chemistry applications.

Stoichiometry

Stoichiometry serves as the quantitative backbone of chemistry, guiding the calculations needed to predict the outcome of chemical reactions. It involves using balanced chemical equations to relate moles of reactants to moles of products. In our example:

• The ratio of moles of iodine to chlorine helped determine the empirical formula \(ICl_6\).
• Stoichiometric calculations further aided in adjusting this empirical formula to find the molecular formula.

Effective stoichiometry ensures accurate and efficient use of reactants and helps in scaling reactions from laboratory to industrial scales. Understanding stoichiometry is pivotal for anyone aiming to master the calculations in chemical reactions, be it in academic settings or applied research.