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Chapter 3: Problem 140

A compound X contains 63.3 percent manganese (Mn) and 36.7 percent O by mass. When \(X\) is heated, oxygen gas is evolved and a new compound Y containing 72.0 percent \(\mathrm{Mn}\) and 28.0 percent \(\mathrm{O}\) is formed. (a) Determine the empirical formulas of X and Y. (b) Write a balanced equation for the conversion of \(\mathrm{X}\) to \(\mathrm{Y}\)

Short Answer

Expert verified
The empirical formulas of X and Y are MnO\( _2 \) and Mn\( _3 \)O\( _4 \), respectively. The balanced equation for the conversion of X to Y is 2 MnO\( _2 \) -> Mn\( _3 \)O\( _4 \)+ O\( _2 \)

Step by step solution

01

Determine the Empirical Formula for compound X

Assume a 100 g sample of X. This will contain 63.3 g of Mn and 36.7 g of O.\nMoles of Mn = 63.3 g Mn / (54.94 g/mol Mn) = 1.15 mol Mn\nMoles of O = 36.7 g O / (16.00 g/mol O) = 2.29 mol O\nThe ratio of Mn to O is 1:2 so the empirical formula is MnO\( _2 \).
02

Determine the Empirical Formula for compound Y

Assume a 100 g sample of Y. This will contain 72.0 g of Mn and 28.0 g of O. \nMoles of Mn = 72.0 g Mn / (54.94 g/mol Mn) = 1.31 mol Mn\nMoles of O = 28.0 g O / (16.00 g/mol O) = 1.75 mol O\nThe ratio of Mn to O is close to 3:4. The smallest whole number ratio of Mn to O is 3:4 so the empirical formula is Mn\( _3 \)O\( _4 \).
03

Write a balanced equation for the conversion of X to Y

Starting with the empirical formulas, propose a chemical equation.\nMnO\( _2 \) -> Mn\( _3 \)O\( _4 \)+ O\( _2 \)\nThis equation is not yet balanced. Balancing this equation gives:\n2 MnO\( _2 \) -> Mn\( _3 \)O\( _4 \)+ O\( _2 \)

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Key Concepts

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

Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It is founded on the law of conservation of mass, which states that matter cannot be created or destroyed in an isolated system. When applying stoichiometry to determine the empirical formula of a compound, the first step involves converting mass percentages into moles, which are the fundamental unit for measuring quantity in chemistry.

For example, to find the empirical formula of compound X in our exercise, we start by assuming a 100 g sample to make the math straightforward, as the mass percentages then directly translate to grams. After converting the masses of manganese and oxygen into moles, we find their simplest whole-number ratio by dividing by the smallest number of moles obtained. Here, the ratio of 1.15 mol Mn to 2.29 mol O simplifies to approximately 1:2, leading to the empirical formula MnO2.

Understanding stoichiometry is crucial because it allows chemists to predict the quantities of substances consumed and produced in a given reaction, which is essential for scaling up reactions for industrial applications and for experiments in the lab.
Chemical Composition
Chemical composition refers to the identity and relative number of the elements that make up a chemical compound. The composition is usually represented using empirical formulas, which show the simplest whole-number ratio of the atoms within a substance. Deducing the chemical composition is a step-by-step process that involves first determining the percentage of each element in the compound and then translating these percentages into the mole ratio of the elements.

In the case of compound Y, we again take a 100 g sample for ease of calculation. The sample consists of 72.0% manganese and 28.0% oxygen by mass. Converting these percentages to moles, we find that the ratio of moles of manganese to moles of oxygen is 1.31 to 1.75. Simplifying this ratio to the smallest whole numbers gives us 3:4, indicating that the empirical formula for compound Y is Mn3O4.

Knowing the chemical composition is fundamental, as it not only determines the properties of the compound but also provides information necessary for preparing compounds through chemical reactions, analyzing reaction yields, and understanding the molecular structure.
Balancing Chemical Equations
Balancing chemical equations is essential for describing the changes that occur during a chemical reaction. An equation must be balanced to reflect the conservation of mass, ensuring that the number of atoms of each element is the same on both sides of the equation. To balance an equation, coefficients are placed before the formulas of reactants and products to adjust the number of moles of each substance accordingly.

In our exercise, the conversion of compound X (MnO2) to compound Y (Mn3O4) requires a balanced chemical equation. Initially, the proposed conversion is MnO2 -> Mn3O4 + O2, which is unbalanced. Through balancing, we add coefficients to make sure there are equal amounts of Mn and O on both sides, resulting in the equation 2 MnO2 -> Mn3O4 + O2.

Mastering the skill of balancing equations is invaluable in chemistry as it is required for correct stoichiometric calculations, determination of reaction stoichiometry, and ensuring that chemical reactions represented on paper or in laboratory experiments proceed according to the expected ratios of reactants and products.

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Most popular questions from this chapter

Because of its detrimental effect on the environment, the lead compound described in Problem 3.148 has been replaced by methyl tert-butyl ether (a compound of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\) ) to enhance the performance of gasoline. (This compound is also being phased out because of its contamination of drinking water.) When \(12.1 \mathrm{~g}\) of the compound are burned in an apparatus like the one shown in Figure \(3.6,30.2 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(14.8 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are formed. What is the empirical formula of the compound?

The following reaction is stoichiometric as written$$\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}+\mathrm{NaOC}_{2}\mathrm{H}_{5} \longrightarrow \mathrm{C}_{4}\mathrm{H}_{8}+\mathrm{C}_{2} \mathrm{H}_{5}\mathrm{OH}+\mathrm{NaCl}$$ but it is often carried out with an excess of \(\mathrm{NaOC}_{2} \mathrm{H}_{5}\) to react with any water present in the reaction mixture that might reduce the yield. If the reaction shown was carried out with \(6.83 \mathrm{~g}\) of \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}\), how many grams of \(\mathrm{NaOC}_{2} \mathrm{H}_{5}\) would be needed to have a 50 percent molar excess of that reactant?

A sample of iron weighing \(15.0 \mathrm{~g}\) was heated with potassium chlorate \(\left(\mathrm{KClO}_{3}\right)\) in an evacuated container. The oxygen generated from the decomposition of \(\mathrm{KClO}_{3}\) converted some of the Fe to \(\mathrm{Fe}_{2} \mathrm{O}_{3}\). If the combined mass of Fe and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) was \(17.9 \mathrm{~g}\), calculate the mass of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) formed and the mass of \(\mathrm{KClO}_{3}\) decomposed.

Monosodium glutamate (MSG), a food-flavor enhancer, has been blamed for "Chinese restaurant syndrome," the symptoms of which are headaches and chest pains. MSG has the following composition by mass: 35.51 percent \(\mathrm{C}, 4.77\) percent \(\mathrm{H}\), 37.85 percent \(\mathrm{O}, 8.29\) percent \(\mathrm{N},\) and 13.60 percent Na. What is its molecular formula if its molar mass is about \(169 \mathrm{~g}\) ?

Balance the following equations using the method outlined in Section 3.7 : (a) \(\mathrm{C}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}\) (b) \(\mathrm{CO}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}\) (c) \(\mathrm{H}_{2}+\mathrm{Br}_{2} \longrightarrow \mathrm{HBr}\) (d) \(\mathrm{K}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{KOH}+\mathrm{H}_{2}\) (e) \(\mathrm{Mg}+\mathrm{O}_{2} \longrightarrow \mathrm{MgO}\) (f) \(\mathrm{O}_{3} \longrightarrow \mathrm{O}_{2}\) (g) \(\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\) (h) \(\mathrm{N}_{2}+\mathrm{H}_{2} \longrightarrow \mathrm{NH}_{3}\) (i) \(\mathrm{Zn}+\mathrm{AgCl} \longrightarrow \mathrm{ZnCl}_{2}+\mathrm{Ag}\) (j) \(\mathrm{S}_{8}+\mathrm{O}_{2} \longrightarrow \mathrm{SO}_{2}\) (k) \(\mathrm{NaOH}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}\) (l) \(\mathrm{Cl}_{2}+\mathrm{NaI} \longrightarrow \mathrm{NaCl}+\mathrm{I}_{2}\) \((\mathrm{m}) \mathrm{KOH}+\mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow \mathrm{K}_{3} \mathrm{PO}_{4}+\mathrm{H}_{2} \mathrm{O}\) (n) \(\mathrm{CH}_{4}+\mathrm{Br}_{2} \longrightarrow \mathrm{CBr}_{4}+\mathrm{HBr}\)
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