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To find the formula of a compound composed of iron and carbon monoxide, \(\mathrm{Fe}_{x}(\mathrm{CO})_{y}\), the compound is burned in pure oxygen to give \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) and \(\mathrm{CO}_{2} .\) If you burn \(1.959 \mathrm{g}\) of \(\mathrm{Fe}_{x}(\mathrm{CO})_{y}\) and obtain \(0.799 \mathrm{g}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) and \(2.200 \mathrm{g}\) of \(\mathrm{CO}_{2}\) what is the empirical formula of \(\mathrm{Fe}_{x}(\mathrm{CO})_{y^{2}}\)

Short Answer

Expert verified
The empirical formula is \(\mathrm{Fe(CO)_5}\).

Step by step solution

01

Calculate moles of products

First, we need to calculate the number of moles of each product obtained from burning the compound. **For FeFe_2O_3:**The molar mass of Fe_2O_3 is approximately 159.7 g/mol. The moles of Fe_2O_3 are given by:\[moles\,of\,Fe_2O_3 = \frac{0.799 g}{159.7 g/mol} \approx 0.00500\, mol\]**For CO_2:**The molar mass of CO_2 is approximately 44.0 g/mol. The moles of CO_2 are given by:\[moles\,of\,CO_2 = \frac{2.200 g}{44.0 g/mol} = 0.0500\, mol\]
02

Determine moles of Fe and C in products

Each molecule of Fe_2O_3 contains 2 Fe atoms, so we multiply the moles of Fe_2O_3 by 2 to find moles of Fe:\[moles\,of\,Fe = 2 \times 0.00500\, mol = 0.0100\, mol\]Each molecule of CO_2 contains 1 C atom, so moles of C are equal to moles of CO_2:\[moles\,of\,C = 0.0500\, mol\]
03

Calculate masses of Fe and C in original compound

Using the moles calculated, we now find the mass of Fe and C in the original compound.**For Iron (Fe):**The molar mass of Fe is approximately 55.8 g/mol. Thus, the mass of Fe is:\[mass\,of\,Fe = 0.0100\, mol \times 55.8 g/mol = 0.558 g\]**For Carbon (C):**The molar mass of C is approximately 12.0 g/mol. Thus, the mass of C is:\[mass\,of\,C = 0.0500\, mol \times 12.0 g/mol = 0.600 g\]
04

Calculate mass of oxygen in the original compound

Now, we find the mass of oxygen in the original compound by subtracting the masses of Fe and C from the total mass of the compound.\[mass\,of\,O = 1.959\, g - (0.558 + 0.600)\, g = 0.801\, g\]
05

Determine moles of O in original compound

Now calculate the moles of oxygen in the original compound. The molar mass of O is approximately 16.0 g/mol.\[moles\,of\,O = \frac{0.801\, g}{16.0\, g/mol} = 0.0501\, mol\]
06

Determine empirical formula

We have moles of Fe, O, and C as 0.0100, 0.0501, and 0.0500 respectively. Divide each by the smallest number of moles to obtain the simplest ratio:**Fe:**\[\frac{0.0100}{0.0100} = 1.00\]**C:**\[\frac{0.0500}{0.0100} = 5.00\]**O:**\[\frac{0.0501}{0.0100} = 5.01 \approx 5.00\]Thus, the empirical formula is \(\mathrm{Fe(CO)_5}\).

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Key Concepts

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

Chemical Reactions
Chemical reactions are processes in which substances, known as reactants, are transformed into different substances, called products. Understanding these reactions is essential for determining the empirical formulas of compounds. In the given exercise, the compound \( \mathrm{Fe}_x(\mathrm{CO})_y \) reacts with oxygen during combustion. This chemical reaction leads to the generation of \( \mathrm{Fe}_2\mathrm{O}_3 \) and \( \mathrm{CO}_2 \).To understand chemical reactions better, it's crucial to recognize:- The identity of the reactants and the products. The reactants in the exercise are the iron compound and oxygen.- The law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction. This means the total mass of the reactants is equal to the total mass of the products.- The role of balanced chemical equations that depict the reactants turning into products.In this exercise, identifying the correct products formed by burning the compound helps determine its empirical formula using further calculations.
Mole Calculation
The concept of mole calculation is pivotal for estimating the amounts of substances involved in a chemical reaction. A mole is a unit in chemistry that represents a specific number of entities (usually atoms, ions, or molecules). One mole is approximately \(6.022 \times 10^{23}\) entities, Avogadro's number.Steps to calculate moles include:- Determining the mass of a substance and knowing its molar mass, which is the mass of one mole of that substance.- Using the formula:\[moles = \frac{\text{mass in grams}}{\text{molar mass in g/mol}}\]Through these calculations, from the original solution, students learn to calculate moles of products like \( \mathrm{Fe}_2\mathrm{O}_3 \) and \( \mathrm{CO}_2 \).The exercise gives the masses of these products, letting us calculate moles by dividing each mass by their respective molar masses. Accurately calculating the moles of each element, such as iron (Fe) and carbon (C), is fundamental to determining the empirical formula.
Combustion Analysis
Combustion analysis is a laboratory method used to determine the composition of organic compounds. It involves burning the compound in an oxygen-rich environment to produce gases like carbon dioxide (\(\mathrm{CO}_2\)) and water (\(\mathrm{H}_2\mathrm{O}\)). These products are then analyzed to infer the original compound's formula. In the exercise, the combustion of \( \mathrm{Fe}_x(\mathrm{CO})_y \) results in \( \mathrm{Fe}_2\mathrm{O}_3 \) and \( \mathrm{CO}_2 \).Some key points about combustion analysis:- It is useful for compounds that contain carbon and hydrogen, and potentially other elements.- By measuring the masses of gases produced, like \(\mathrm{CO}_2\), you can calculate the amount of carbon that was present in the original compound.- This type of analysis often involves complex calculations to backtrack from the gaseous products to the empirical formula of the starting compound.In this step-by-step solution, the combustion analysis of the unidentified compound allows students to compute the amount of carbon, based on the mass of \(\mathrm{CO}_2\). Subsequently, combining this information with the mole calculations provides an empirical formula, illustrating how comprehensive combustion analysis can be.

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Most popular questions from this chapter

If \(10.0 \mathrm{g}\) of carbon is combined with an exact, stoichiometric amount of oxygen \((26.6 \mathrm{g})\) to produce carbon dioxide, what is the theoretical yield of \(\mathrm{CO}_{2}\), in grams?

Menthol, from oil of mint, has a characteristic odor. The compound contains only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O} .\) If \(95.6 \mathrm{mg}\) of menthol burns completely in \(\mathrm{O}_{2},\) and gives \(269 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and \(110 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O},\) what is the empirical formula of menthol?

Ammonia gas can be prepared by the following reaction: \(\mathrm{CaO}(\mathrm{s})+2 \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CaCl}_{2}(\mathrm{s})\) If \(112 \mathrm{g}\) of \(\mathrm{CaO}\) and \(224 \mathrm{g}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) are mixed, the theoretical yield of \(\mathrm{NH}_{3}\) is \(68.0 \mathrm{g}\) (Study Question 20 ). If only \(16.3 \mathrm{g}\) of \(\mathrm{NH}_{3}\) is actually obtained, what is its percent yield?

Sulfuric acid can be produced from a sulfide ore such as iron pyrite by the following sequence of reactions: $$\begin{aligned} 4 \mathrm{FeS}_{2}(\mathrm{s})+11 \mathrm{O}_{2}(\mathrm{g}) & \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+8 \mathrm{SO}_{2}(\mathrm{g}) \\ 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) & \longrightarrow 2 \mathrm{SO}_{3}(\mathrm{g}) \\ \mathrm{SO}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell) & \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(\ell) \end{aligned}$$ Starting with \(525 \mathrm{kg}\) of \(\mathrm{FeS}_{2}\) (and an excess of other reactants), what mass of pure \(\mathrm{H}_{2} \mathrm{SO}_{4}\) can be prepared?

Balance the following equations: (a) \(\operatorname{Cr}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s})\) (b) \(\mathrm{Cu}_{2} \mathrm{S}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{SO}_{2}(\mathrm{g})\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}(\ell)+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g})\)

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