Question

The compound As \(_{2} \mathrm{L}_{4}\) is synthesized by reaction of arsenic metal with arsenic triodide. If a solid cubic block of arsenic \(\left(d=5.72 \mathrm{g} / \mathrm{cm}^{3}\right)\) that is 3.00 \(\mathrm{cm}\) on edge is allowed to react with \(1.01 \times 10^{24}\) molecules of arsenic triodide, what mass of \(\mathrm{As}_{2} \mathrm{L}_{4}\) can be prepared? If the percent yield of \(\mathrm{As}_{2} \mathrm{L}_{4}\) was 75.6\(\%\) what mass of \(\mathrm{As}_{2} \mathrm{I}_{4}\) was actually isolated?

Short answer

The theoretical mass of As\(_{2}\)I\(_{4}\) that can be prepared is approximately 1103.47 g, and the actual mass of As\(_{2}\)I\(_{4}\) isolated, given a 75.6% yield, is approximately 833.62 g.

Step-by-step solution

Calculate the mass of solid arsenic

Since the density, d = mass/volume, we can calculate the mass of the solid arsenic cube by using the given dimensions: mass = density × volume mass = 5.72 g/cm³ × (3.00 cm)³ mass = 5.72 g/cm³ × 27.00 cm³ mass ≈ 154.44 g The mass of the solid arsenic cube is 154.44 g.

Calculate moles of arsenic and arsenic triiodide

To calculate the moles of arsenic from the mass, we'll use the formula: moles = mass/molar_mass, where the molar mass of As = 74.92 g/mol. moles of As = 154.44 g / 74.92 g/mol ≈ 2.06 mol We are given the number of molecules of arsenic triiodide, 1.01 × 10²⁴. To convert this to moles, we'll use Avogadro's number (6.022 × 10²³ molecules/mol): moles of AsI₃ = (1.01 × 10²⁴) / (6.022 × 10²³) moles of AsI₃ ≈ 1.68 mol

Calculate the theoretical mass of As\(_{2}\)I\(_{4}\)

First, we must determine the stoichiometry between As and AsI₃ based on the chemical reaction. Since 1 mole of As reacts with 1 mole of AsI₃ to produce 1 mole of As\(_{2}\)I\(_{4}\), we have: As + AsI₃ → As\(_{2}\)I\(_{4}\) Now since As is in excess, we will base our calculations on the moles of AsI₃: moles of AsI₃ = 1.68 mol Now, we can find the theoretical mass of As\(_{2}\)I\(_{4}\) produced, using its molar mass (2 moles of As = 2 × 74.92 g/mol, and 4 moles of I = 4 × 126.90 g/mol): molar_mass of As\(_{2}\)I\(_{4}\) = 2 × 74.92 g/mol + 4 × 126.90 g/mol = 656.84 g/mol mass of As\(_{2}\)I\(_{4}\) = moles of As\(_{2}\)I\(_{4}\) × molar_mass of As\(_{2}\)I\(_{4}\) mass of As\(_{2}\)I\(_{4}\) = 1.68 mol × 656.84 g/mol ≈ 1103.47 g The theoretical mass of As\(_{2}\)I\(_{4}\) that can be prepared is approximately 1103.47 g.

Calculate the actual mass of As\(_{2}\)I\(_{4}\) isolated

The percent yield of As\(_{2}\)I\(_{4}\) is given as 75.6%. We can use this and the theoretical mass to calculate the actual mass of As\(_{2}\)I\(_{4}\) isolated: actual_mass = (percent_yield/100) × theoretical_mass actual_mass = 0.756 × 1103.47 g actual_mass ≈ 833.62 g The actual mass of As\(_{2}\)I\(_{4}\) isolated is approximately 833.62 g.

Key concepts

Chemical reactions

Understanding chemical reactions is key to grasping stoichiometry. In any chemical reaction, reactants transform into products through the rearrangement of atoms. This process is represented by a chemical equation, which illustrates the substances involved and their respective quantities. For a reaction to proceed, the reactants must be present in the correct proportions, or stoichiometric ratios, to ensure complete conversion into the desired products.

In the given exercise, the reaction involves arsenic (As) and arsenic triiodide (AsI₃) forming arsenic tetraiodide (As₂I₄). The balanced chemical equation is:

- As + AsI₃ → As₂I₄

This equation tells us that one mole of arsenic reacts with one mole of arsenic triiodide to produce one mole of arsenic tetraiodide. Understanding these proportions is essential for predicting how much of each compound is needed or produced.

Percent yield

Percent yield is a measure of the efficiency of a chemical reaction. It compares the actual amount of product obtained (actual yield) to the amount that was theoretically possible (theoretical yield), expressed as a percentage.

The formula to calculate percent yield is:

- \( \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \% \)

In our example, the theoretical mass of arsenic tetraiodide produced was 1103.47 g, but only 833.62 g was isolated. The percent yield is given as 75.6%. This indicates that only 75.6% of the predicted amount was actually recovered, which can be due to various factors like side reactions, incomplete reactions, or loss during product recovery.

Recognizing the percent yield helps chemists assess reaction conditions and optimize processes to enhance efficiency.

Mole calculations

Mole calculations are fundamental in stoichiometry and involve using the mole concept to relate masses, particles, and volumes in chemical reactions. The mole is a standard scientific unit for measuring large quantities, particularly atoms and molecules, allowing for precise stoichiometric calculations.

In this exercise, mole calculations were used to determine the number of moles of arsenic and arsenic triiodide. The moles of arsenic were calculated based on its mass and molar mass:\[\text{moles of As} = \frac{154.44\, \text{g}}{74.92\, \text{g/mol}} \approx 2.06\, \text{mol} \]

The moles of arsenic triiodide were converted from molecules using Avogadro's number:\[\text{moles of AsI}_3 = \frac{1.01 \times 10^{24}}{6.022 \times 10^{23}} \approx 1.68\, \text{mol} \]

These calculations provided the basis for determining the limiting reactant and the theoretical yield of the product. Mastery of mole calculations allows students to predict the outcomes of chemical reactions accurately.