Problem 97 - Chemistry: An Atoms-Focused Approach

Question

A number of chemical reactions have been proposed for the formation of organic compounds from inorganic precursors, including the following: $$\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{FeS}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{FeS}_{2}(s)+\mathrm{HCO}_{2} \mathrm{H}(\ell)$$ a. Identify the ions in \(\mathrm{FeS}\) and \(\mathrm{FeS}_{2}\) b. What are the names of \(\mathrm{FeS}\) and \(\mathrm{FeS}_{2} ?\) c. How much \(\mathrm{HCO}_{2} \mathrm{H}\) is obtained by reacting \(1.00 \mathrm{g}\) FeS, \(0.50 \mathrm{g} \mathrm{H}_{2} \mathrm{S},\) and \(0.50 \mathrm{g} \mathrm{CO}_{2}\) if the reaction results in a \(50.0 \%\) yield?

Step-by-Step Solution

Answer

a. In FeS, the ions present are Fe^2+ and S^2-. In FeS2, the ions are Fe^2+ and 2 S^2-. b. FeS is called Iron(II) sulfide, and FeS₂ is called Iron(II) disulfide. c. Given the amounts of reactants and considering a 50.0% yield, 0.313 g of HCO2H is produced.

Step 1: Identify ions in FeS and FeS2

To identify the ions in these compounds, examine the elements in each compound. Iron (Fe) is a transition metal that can form cations. The most common oxidation states for iron are +2 and +3. The sulfur (S) and its corresponding ion, the sulfide ion (S^2-), have a -2 charge. For FeS: Iron (Fe) will have a +2 charge to balance with the -2 charge of sulfide (S^2-). For FeS2: In this case, iron (Fe) will still have a +2 charge. However, to balance with the two sulfide ions (each with a -2 charge), we need to have two S^2- ions. So the ions in FeS are Fe^2+ and S^2-, and in FeS2, they are Fe^2+ and 2 S^2-.

Step 2: Name FeS and FeS2

Now that we've identified the ions in the compounds, we can name them. FeS: We have Fe^2+ and S^2-, so the name of the compound is Iron(II) sulfide FeS2: We still have Fe^2+ but now there are 2 S^2- ions. The name of the compound is Iron(II) disulfide.

Step 3: Calculate the amount of HCO2H produced

We are given the following amounts of reactants: FeS = 1.00 g H2S = 0.50 g CO2 = 0.50 g First, we need to determine the limiting reactant. To do this, we'll convert each given mass to moles using their respective molar masses. FeS = 1.00 g / (55.85 g/mol + 32.07 g/mol) = 0.01204 mol H2S = 0.50 g / (2×1.01 g/mol + 32.07 g/mol) = 0.01372 mol CO2 = 0.50 g / (12.01 g/mol + 2×16.00 g/mol) = 0.01120 mol Comparing the molar ratios: 0.01204 mol FeS / 1 = 0.01204 0.01372 mol H2S / 1 = 0.01372 0.01120 mol CO2 / 1 = 0.01120 CO2 is the limiting reactant as it has the lowest ratio. Now, we'll use the stoichiometry of the balanced chemical equation to find moles of HCO2H produced. 1 mol CO2 : 1 mol HCO2H 0.01120 mol CO2 × (1 mol HCO2H/1 mol CO2) = 0.01120 mol HCO2H produced However, we're told that the reaction has a 50.0% yield. So, we'll multiply the amount of HCO2H obtained by 0.50. 0.01120 mol HCO2H × 0.50 = 0.00560 mol HCO2H Finally, we'll convert the moles of HCO2H back to grams using the molar mass. 0.00560 mol HCO2H × (1.01 g/mol + 12.01 g/mol + 2(16.00 g/mol) + 1.01 g/mol) = 0.313 g HCO2H So, 0.313 g of HCO2H is obtained with the given amounts of reactants and a 50.0% yield.

Key Concepts

Inorganic Precursors

Inorganic precursors play an essential role in the synthesis of various compounds, especially in the formation of organic compounds. They are substances that do not contain carbon-hydrogen bonds, typical of inorganic chemistry. In the given chemical reaction: \[\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{FeS}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{FeS}_{2}(s)+\mathrm{HCO}_{2} \mathrm{H}(\ell)\] the inorganic precursors involved are \(\mathrm{H}_{2} \mathrm{S}\), \(\mathrm{FeS}\), and \(\mathrm{CO}_{2}\).

• \(\mathrm{H}_{2} \mathrm{S}\) (hydrogen sulfide) is a gas with distinct odor, often associated with the smell of rotten eggs.
• \(\mathrm{FeS}\) (iron(II) sulfide) is a solid commonly found in the earth's crust as a mineral known as pyrite or "fool's gold."
• \(\mathrm{CO}_{2}\) (carbon dioxide) is a well-known gas prevalent in the earth's atmosphere, crucial for life on earth.

Inorganic precursors serve as the initial reactants in many chemical processes to help form organic products and structures. Understanding their properties allows us to predict how they interact in reactions.

Organic Compounds

Organic compounds are central to chemistry, characterized by containing carbon atoms. The product \(\mathrm{HCO}_{2} \mathrm{H}\), also known as formic acid, is an organic compound produced in the reaction from inorganic precursors. Formic acid is the simplest carboxylic acid, with a chemical formula \(\mathrm{HCO}_{2} \mathrm{H}\). It is known for its usefulness in various industrial applications such as leather processing and as a preservative. Its formation from inorganic starting materials showcases how inorganic chemistry integrates with organic compounds in producing diverse materials. The transition from inorganic to organic compounds in reactions like these highlights the chemical versatility and complexity inherent in the synthesis processes that drive biochemical and industrial reactions.

Limiting Reactant

The concept of a limiting reactant is crucial in understanding how much product can be formed in a chemical reaction. It represents the reactant that will be entirely used up first, thus limiting the amount of product that can be formed. In the equation:- \(\mathrm{H}_{2} \mathrm{S}\)- \(\mathrm{FeS}\)- \(\mathrm{CO}_{2}\)To identify the limiting reactant, we needed to compare the number of moles of each reactant. The moles are calculated by dividing the given mass by the molar mass of each compound:

• \(\mathrm{FeS}: \frac{1.00 \text{ g}}{55.85 + 32.07 \text{ g/mol}} = 0.01204 \text{ mol}\)
• \(\mathrm{H}_{2} \mathrm{S}: \frac{0.50 \text{ g}}{2 \, (1.01) + 32.07 \text{ g/mol}} = 0.01372 \text{ mol}\)
• \(\mathrm{CO}_{2}: \frac{0.50 \text{ g}}{12.01 + 2 \, (16) \text{ g/mol}} = 0.01120 \text{ mol}\)

The compound with the smallest mole value, \(\mathrm{CO}_{2}\), is identified as the limiting reactant. It determines the maximum amount of \(\mathrm{HCO}_{2} \mathrm{H}\) that can be formed, dictating the efficiency of the reaction. Recognizing and calculating the limiting reactant in a reaction helps anticipate potential inefficiencies and adjust reactant quantities for desired product yields.

Stoichiometry

Stoichiometry is essential for quantifying relationships in chemical reactions. It utilizes balanced chemical equations to determine the amounts of reactants and products. By using stoichiometry, we can predict how much product forms from given reactants. In the provided reaction, stoichiometry helps us understand the relation:\[1 \text{ mol } \mathrm{CO}_{2} \rightarrow 1 \text{ mol } \mathrm{HCO}_{2} \mathrm{H}\]This simple 1:1 ratio means one mole of \(\mathrm{CO}_{2}\) will produce one mole of \(\mathrm{HCO}_{2} \mathrm{H}\) if the reaction goes to completion. However, real-world reactions often don't yield 100%. Given a 50% yield, only half of the calculated \(\mathrm{HCO}_{2} \mathrm{H}\) amount will actually be realized: \[0.01120 \text{ mol } \times 0.5 = 0.00560 \text{ mol}\]The mass of this \(\mathrm{HCO}_{2} \mathrm{H}\) can then be calculated using its molar mass, providing an anticipated mass which aids industries in maximizing efficiencies and minimizing resource costs.