Problem 84 - General Chemistry

Question

A soluble iodide was dissolved in water. Then an excess of silver nitrate, \(\mathrm{AgNO}_{3}\), was added to precipitate all of the iodide ion as silver iodide, AgI. If \(1.545 \mathrm{~g}\) of the soluble iodide gave \(2.185 \mathrm{~g}\) of silver iodide, how many grams of iodine are in the sample of soluble iodide? What is the mass percentage of iodine, \(\mathrm{I}\), in the compound?

Step-by-Step Solution

Answer

The sample contains 1.181 g of iodine; mass percentage of iodine is 76.43%.

Step 1: Write the chemical reaction equation

The chemical reaction for the precipitation of iodide ion as silver iodide is: \( \text{I}^- + \text{Ag}^+ \rightarrow \text{AgI} \). This means each iodide ion forms one molecule of \(\text{AgI}\) with silver ions.

Step 2: Calculate moles of silver iodide (AgI)

First, calculate the molar mass of \(\text{AgI}\): \(\text{Ag} = 107.87\ g/mol\) and \(\text{I} = 126.90\ g/mol\). \[ \text{Molar mass of AgI} = 107.87 + 126.90 = 234.77\ g/mol \]Next, use the mass of \(\text{AgI}\) from the problem:\[ \text{Moles of AgI} = \frac{2.185 \text{ g}}{234.77 \text{ g/mol}} = 0.00931 \text{ moles} \]

Step 3: Determine moles of iodine

From the equation \( \text{I}^- + \text{Ag}^+ \rightarrow \text{AgI} \), each mole of \(\text{AgI}\) contains one mole of iodine. Therefore, the moles of iodine in the sample are the same as the moles of \(\text{AgI}\): \[ 0.00931 \text{ moles of I} \]

Step 4: Calculate the mass of iodine

Find the mass of iodine using its molar mass (126.90 g/mol):\[ \text{Mass of Iodine} = 0.00931 \text{ moles} \times 126.90 \text{ g/mol} = 1.181 \text{ g} \]

Step 5: Calculate mass percentage of iodine in the compound

Use the initial mass of the soluble iodide sample:\[ \text{Mass percentage of iodine} = \left(\frac{1.181 \text{ g}}{1.545 \text{ g}}\right) \times 100\% = 76.43\% \]

Key Concepts

Molar Mass Calculation

Molar mass is an essential concept in chemistry, often acting as a bridge between the mass of a substance and the amount in moles. To calculate the molar mass of a compound, simply sum the atomic masses of all atoms present in its formula. For example, in the case of silver iodide (AgI), first find the atomic masses:

• Silver (Ag) has an atomic mass of 107.87 g/mol.
• Iodine (I) has an atomic mass of 126.90 g/mol.

Add these values together to arrive at the molar mass of AgI:\[\text{Molar mass of AgI} = 107.87 + 126.90 = 234.77 \text{ g/mol}\]Molar mass allows us to convert grams to moles, making it possible to relate the mass of a substance to its quantity at the molecular level. This is fundamental when calculating quantities in chemical reactions.

Chemical Reaction Equation

Chemical reaction equations are a concise way of representing chemical reactions. They illustrate the transformation of reactants into products. In our problem, the equation is:\[I^- + Ag^+ \rightarrow AgI\]This equation shows that one iodide ion combines with one silver ion to form one molecule of silver iodide (AgI). Knowing this equation helps us understand the stoichiometry of the reaction, which is crucial in determining how one substance can be transformed into another.By balancing the coefficients, we know exactly how many moles of reactants combine to form a specific amount of product. For every mole of silver iodide formed, one mole of iodide ions is consumed. This stoichiometric relationship is central to problem-solving in chemistry.

Mass Percentage Calculation

The mass percentage of an element within a compound tells us how much of the compound's mass is due to that specific element. It's a handy way to express composition in chemical formulas. To calculate the mass percentage, use the formula:\[\text{Mass percentage of element} = \left(\frac{\text{mass of element in compound}}{\text{total mass of compound}}\right) \times 100\%\]In our case:

• The mass of iodine in the compound was found to be 1.181 g.
• The total mass of the soluble iodide used was 1.545 g.

Substitute these values to find the mass percentage of iodine:\[\text{Mass percentage of iodine} = \left(\frac{1.181 \text{ g}}{1.545 \text{ g}}\right) \times 100\% = 76.43\%\]This means that 76.43% of the total mass of the initial iodide sample consisted of iodine. Mass percentage is a critical parameter for understanding purities and compositions in mixtures and compounds.

Precipitation Reaction

Precipitation reactions are chemical reactions in which an insoluble solid, called a precipitate, forms in solution. Our problem involves a classic precipitation reaction:\[I^- + Ag^+ \rightarrow AgI\]When aqueous solutions of silver nitrate and a soluble iodide mix, silver iodide (AgI) precipitates out of the solution as a solid. These reactions are important for separating components or identifying ion presence.Knowing the complete chemical equation and the solubility rules helps predict whether a precipitate will form. In this scenario, silver iodide is not soluble in water, making it perfect for isolating iodide ions. Because the stoichiometry of this reaction is one-to-one, every mole of silver ion will capture one mole of iodide ion, resulting in the formation of a solid AgI precipitate.