/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 147 Iron forms a sulfide with the ap... [FREE SOLUTION] | 91Ó°ÊÓ

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Iron forms a sulfide with the approximate formula \(\mathrm{Fe}_{7} \mathrm{~S}_{8} .\) Assume that the oxidation state of sulfur is -2 and that iron atoms exist in both +2 and +3 oxidation states. What is the ratio of Fe(II) atoms to Fe(III) atoms in this compound?

Short Answer

Expert verified
The ratio of Fe(II) to Fe(III) atoms is 5:2.

Step by step solution

01

Understanding the Problem

We have a compound with the formula \(\mathrm{Fe}_{7}\mathrm{S}_{8}\). We need to find the ratio of iron atoms in +2 (Fe(II)) to those in +3 (Fe(III)) oxidation states, given that sulfur has an oxidation state of -2.
02

Calculation of Total Charge from Sulfur

There are 8 sulfur atoms, each with an oxidation state of -2. Therefore, the total negative charge contributed by sulfur is \(8 \times (-2) = -16\).
03

Setting Up the Equation for Iron Charges

The total charge from iron atoms must balance the -16 charge from sulfur. Let \(x\) be the number of Fe(II) atoms and \(y\) be the number of Fe(III) atoms. We know that \(x + y = 7\) (total iron atoms) and \(2x + 3y = 16\) (to balance sulfur's charge).
04

Solve the System of Equations

Use the equations \(x + y = 7\) and \(2x + 3y = 16\). Substituting \(y = 7 - x\) into the second equation gives: \[2x + 3(7 - x) = 16\] \[2x + 21 - 3x = 16\] \[-x + 21 = 16\] \[x = 5\]. So, there are 5 Fe(II) atoms. Using \(y = 7 - x\), \(y = 2\), so there are 2 Fe(III) atoms.
05

Calculate the Ratio

The ratio of Fe(II) atoms to Fe(III) atoms is given by \(x : y = 5 : 2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fe(II) and Fe(III) ratio
In iron sulfide compounds, it's crucial to determine the balance between the different iron oxidation states. For \(\text{Fe}_7\text{S}_8\), the problem is to find the ratio between Fe(II) with an oxidation state of +2 and Fe(III) with an oxidation state of +3. This is important because the ratio affects the compound's chemical properties and reactivity.
To determine this, we start by establishing that there are 7 iron atoms in the compound. The distinction between Fe(II) and Fe(III) depends on how many electrons the iron atoms have lost. Fe(II) has lost 2 electrons, while Fe(III) has lost 3.
By setting up a system of equations based on charge balancing and atom count, we find that 5 of these iron atoms are in the Fe(II) state, and 2 are in the Fe(III) state. Hence, the ratio of Fe(II) to Fe(III) is 5:2. This ratio is intrinsic to maintaining the consistent structural and electrochemical integrity of the compound.
oxidation states
Oxidation states are vital in understanding how atoms interact in chemical compounds. They indicate the degree of oxidation, meaning the loss of electrons, an atom undergoes in forming a compound. In our compound, \(\text{Fe}_7\text{S}_8\), sulfur is in a -2 oxidation state. This means sulfur atoms have gained two electrons each, resulting in a total negative charge of -16 for the 8 sulfur atoms.
For iron, it exists in two oxidation states within this compound: +2 and +3. This combination of oxidation states includes iron atoms with different electron counts, which impacts their chemical reactions and how they bind within the compound. Understanding these oxidation states helps us predict the behavior of iron in different environments and its capability to form different types of compounds.
charge balance calculation
Charge balance calculation is essential for determining the stoichiometry of ionic compounds. In \(\text{Fe}_7\text{S}_8\), we ensured that the total positive charge from the iron atoms balances the total negative charge from the sulfur atoms.
The total negative charge from the 8 sulfur atoms is \(-16\). To balance this, we use the concept of total positive charge from iron atoms. Assume \(x\) represents the number of Fe(II) atoms and \(y\) represents the number of Fe(III) atoms. Then, \(2x + 3y = 16\) is established to balance the -16 from sulfur.
This equation alongside \(x + y = 7\) (since there are 7 total iron atoms), allows us to solve for \(x\) and \(y\). Solving these simultaneous equations helps us identify the contributions of each type of iron atom to total charge balance.
iron sulfide compound composition
Understanding the composition of an iron sulfide compound like \(\text{Fe}_7\text{S}_8\) sheds light on its structural and chemical properties. This compound comprises both sulfur and iron, with the latter existing in two different oxidation states.
The presence of two different forms of iron demonstrates a mixed-valence characteristic in such compounds, which can lead to unique magnetic and electrical properties. Knowing the exact numbers of each type, as determined from our calculations—5 Fe(II) and 2 Fe(III)—helps predict how this compound behaves under various chemical conditions.
Furthermore, the chemical formula \(\text{Fe}_7\text{S}_8\) reflects a specific arrangement and balance essential for the compound's stability, affecting its usefulness and application across various industries, such as metallurgy and electronics.

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Most popular questions from this chapter

A 3.75-g sample of iron ore is transformed to a solution of iron(II) sulfate, \(\mathrm{FeSO}_{4}\), and this solution is titrated with \(0.150 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (potassium dichromate). If it requires \(43.7 \mathrm{~mL}\) of potassium dichromate solution to titrate the iron(II) sulfate solution, what is the percentage of iron in the ore? The reaction is $$ \begin{array}{r} 6 \mathrm{FeSO}_{4}(a q)+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(a q)+7 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \\ 3 \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+7 \mathrm{H}_{2} \mathrm{O}(l)+ \\ \mathrm{K}_{2} \mathrm{SO}_{4}(a q) \end{array} $$

Classify each of the following reactions as a combination reaction, decomposition reaction, displacement reaction, or combustion reaction. (a) When they are heated, ammonium dichromate crystals, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7},\) decompose to give nitrogen, water vapor, and solid chromium(III) oxide, \(\mathrm{Cr}_{2} \mathrm{O}_{3}\). (b) When aqueous ammonium nitrite, \(\mathrm{NH}_{4} \mathrm{NO}_{2}\), is heated, it gives nitrogen and water vapor. (c) When gaseous ammonia, \(\mathrm{NH}_{3}\), reacts with hydrogen chloride gas, HCl, fine crystals of ammonium chloride, \(\mathrm{NH}_{4} \mathrm{Cl}\), are formed (d) Aluminum added to an aqueous solution of sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4},\) forms a solution of aluminum sulfate, \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} .\) Hydrogen gas is released.

Determine the volume of sulfuric acid solution needed to prepare \(37.4 \mathrm{~g}\) of aluminum sulfate, \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\), by the reaction $$ 2 \mathrm{Al}(s)+3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+3 \mathrm{H}_{2}(g) $$ The sulfuric acid solution, whose density is \(1.104 \mathrm{~g} / \mathrm{mL}\) contains \(15.0 \% \mathrm{H}_{2} \mathrm{SO}_{4}\) by mass. 4.160 Determine the volume of sodium hydroxide solution needed to prepare \(26.2 \mathrm{~g}\) sodium phosphate, \(\mathrm{Na}_{3} \mathrm{PO}_{4}\), by the reaction $$ 3 \mathrm{NaOH}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{Na}_{3} \mathrm{PO}_{4}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ The sodium hydroxide solution, whose density is \(1.133 \mathrm{~g} /\) \(\mathrm{mL},\) contains \(12.0 \% \mathrm{NaOH}\) by mass.

How many milliliters of \(0.126 \mathrm{M} \mathrm{HClO}_{4}\) (perchloric acid) are required to give \(0.150 \mathrm{~mol} \mathrm{HClO}_{4} ?\)

Nickel(II) sulfate solution reacts with sodium hydroxide solution to produce a precipitate of nickel(II) hydroxide and a solution of sodium sulfate. Write the molecular equation for this reaction. Then write the corresponding net ionic equation.

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