Question

The amount of iron and manganese in an alloy is determined by precipitating the metals with 8 -hydroxyquinoline, \(\mathrm{C}_{9} \mathrm{H}_{7} \mathrm{NO}\). After weighing the mixed precipitate, the precipitate is dissolved and the amount of 8-hydroxyquinoline determined by another method. In a typical analysis a 127.3 -mg sample of an alloy containing iron, manganese, and other metals is dissolved in acid and treated with appropriate masking agents to prevent an interference from other metals. The iron and manganese are precipitated and isolated as \(\mathrm{Fe}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}\) and \(\mathrm{Mn}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2},\) yielding a total mass of \(867.8 \mathrm{mg}\). The amount of 8 -hydroxyquinolate in the mixed precipitate is determined to be \(5.276 \mathrm{mmol}\). Calculate the \(\% \mathrm{w} / \mathrm{w} \mathrm{Fe}\) and \(\% \mathrm{w} / \mathrm{w} \mathrm{Mn}\) in the alloy.

Short answer

Calculate molar masses, use mole ratios, set up and solve the equations to find Fe and Mn percentages.

Step-by-step solution

Understanding the Chemical Formulas

Identify the chemical formulas involved. We have two precipitates: \( \mathrm{Fe}\left(\mathrm{C}_{9}\mathrm{H}_{6}\mathrm{NO}\right)_{3} \) and \( \mathrm{Mn}\left(\mathrm{C}_{9}\mathrm{H}_{6}\mathrm{NO}\right)_{2} \). Each precipitate contains 8-hydroxyquinoline, and we need to find their molar masses to use in calculations.

Molar Mass Calculation

Calculate the molar mass of 8-hydroxyquinoline \( \mathrm{C}_{9}\mathrm{H}_{7}\mathrm{NO} \) as 145.16 g/mol. Now, calculate the molar mass of \( \mathrm{Fe}\left(\mathrm{C}_{9}\mathrm{H}_{6}\mathrm{NO}\right)_{3} \), adding iron’s molar mass (55.85 g/mol) to three times that of 8-hydroxyquinoline minus three hydrogens. Similarly, calculate for \( \mathrm{Mn}\left(\mathrm{C}_{9}\mathrm{H}_{6}\mathrm{NO}\right)_{2} \).

Mole Ratio from Total Precipitate

Determine the mol ratio from the mass of 8-hydroxyquinoline in the total precipitation. The total moles of 8-hydroxyquinoline are given as 5.276 mmol, distributed among the precipitates.

Set Up Equations for Masses

Set up the equations for the mass balance based on the moles of the ligands: \( 3x + 2y = 5.276 \) (where \( x \) and \( y \) are moles of \( \mathrm{Fe}\left(\mathrm{C}_{9}\mathrm{H}_{6}\mathrm{NO}\right)_{3} \) and \( \mathrm{Mn}\left(\mathrm{C}_{9}\mathrm{H}_{6}\mathrm{NO}\right)_{2} \) respectively), and \( M_{Fe} x + M_{Mn} y = 867.8 \, \text{mg} \).

Solving the System of Equations

Solve the system of equations to find the values of \( x \) and \( y \). Substitute the known values into these equations and solve for the mole numbers of each metal complex.

Calculate Mass of Fe and Mn

Using the molar amounts from the solutions, convert the moles into grams using their respective molar masses. This step gives us the mass of iron and manganese in the precipitate.

Determine Percentage Composition

Calculate the percentage weight by dividing the mass of each metal by the initial 127.3 mg sample mass, then multiply by 100 to obtain the percentage composition. Thus, calculate \( \% \mathrm{w} / \mathrm{w} \mathrm{Fe} \) and \( \% \mathrm{w} / \mathrm{w} \mathrm{Mn} \).

Key concepts

Precipitation Method

The precipitation method is a pivotal technique in analytical chemistry used to separate and identify components in a mixture. This approach involves converting a substance into an insoluble solid, called a precipitate, by reacting it with a precipitating agent. In the context of the given exercise, iron and manganese are isolated from a metal alloy using 8-hydroxyquinoline as a precipitating agent. This compound reacts with the metals to form specific precipitates - \( \mathrm{Fe}\left(\mathrm{C}_{9}\mathrm{H}_{6}\mathrm{NO}\right)_{3} \) and \( \mathrm{Mn}\left(\mathrm{C}_{9}\mathrm{H}_{6}\mathrm{NO}\right)_{2} \). Key steps in this method:

• Addition of a precise quantity of 8-hydroxyquinoline to the dissolved metal sample.
• Formation and isolation of solid precipitates by the metals bonding with the reagent.
• Separation of these precipitates from the solution for further analysis.

The success of this technique hinges on correct chemical conditions, such as pH and temperature, to ensure complete precipitation of the target metals.

Molar Mass Calculation

Molar mass calculation is crucial for translating the chemical formula of a compound into a practical value used in quantitative analysis. Determining the molar mass lets chemists convert moles into grams, a necessary step in analyzing compound compositions.Let's consider the compound 8-hydroxyquinoline, with its chemical formula \( \mathrm{C}_{9}\mathrm{H}_{7}\mathrm{NO} \). The molar mass is calculated by summing the atomic masses of all atoms in the formula:

• Carbon: \( 9 \times 12.01 \text{ g/mol} \)
• Hydrogen: \( 7 \times 1.008 \text{ g/mol} \)
• Nitrogen: \( 1 \times 14.01 \text{ g/mol} \)
• Oxygen: \( 1 \times 16.00 \text{ g/mol} \)

The total comes out to 145.16 g/mol.For the compounds \( \mathrm{Fe}\left(\mathrm{C}_{9}\mathrm{H}_{6}\mathrm{NO}\right)_{3} \) and \( \mathrm{Mn}\left(\mathrm{C}_{9}\mathrm{H}_{6}\mathrm{NO}\right)_{2} \), similar calculations incorporate the metal's molar mass:

• Add the metal molar mass to three or two times the hydroxyquinoline molar mass (minus hydrogens).
• For example, for iron, add 55.85 g/mol to three times the 8-hydroxyquinoline mass.

This makes further mass calculations possible and allows determination of metal content after isolation.

Percentage Composition

Percentage composition helps determine the proportion of each component in a mixture, providing specific insights into the contents of a sample. This is especially useful when assessing the purity or formulation of chemical products.To calculate the percentage composition, follow these steps:

• Determine the exact mass of the metal in the precipitate using molar mass and mole conversions from the chemical reactions.
• The mass is then compared to the original sample mass (127.3 mg in the exercise).
• This ratio is multiplied by 100 to convert it to a percentage.

In this exercise, after obtaining the masses of iron and manganese, their individual percentage weights in the alloy can be calculated. For instance, if the mass of iron obtained is 50 mg, its percentage composition is \( \frac{50}{127.3} \times 100 \), resulting in the percentage weight of iron in the alloy.By calculating these percentages, you learn the concentration of each metal in the sample, tying back to the overall composition and quality of the alloy.