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Chapter 3: Problem 106

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 \(\mathrm{Y}\) 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 \(X\) to \(Y\).

Short Answer

Expert verified
The empirical formula of compound X is \( MnO_2 \), the empirical formula of compound Y is \( Mn_3O_4 \), the balanced equation for the conversion of X to Y is \( 3MnO_2 \rightarrow 2Mn_3O_4 + O_2 \).

Step by step solution

01

Determine the empirical formula for compound X

First, convert the percentage composition to grams (assuming we have 100g of the compound so the percentage is equivalent to grams). There are 63.3g of Mn and 36.7g of Oxygen. Then, convert the mass of each element to mole by dividing by its molar mass. Molar mass of Mn is approximately 54.94g/mol and Oxygen is approximately 16g/mol. \( \frac{63.3g}{54.94g/mol} \approx 1.15mol \) for Mn and \( \frac{36.7g}{16g/mol} \approx 2.29mol \) for Oxygen. Write the ratio of moles, and divide by the smallest number of moles to get the simplest whole number ratio, which gives \( \frac{1.15}{1.15} : \frac{2.29}{1.15} \approx 1:2 \), so the formula for compound X is \( MnO_2 \)
02

Determine the empirical formula for compound Y

Use a similar method as in step 1 with percentage compositions of 72.0% Mn and 28.0% Oxygen. This gives \( \frac{72.0g}{54.94g/mol} \approx 1.31mol \) for Mn and \( \frac{28.0g}{16g/mol} \approx 1.75mol \) for Oxygen. Get the simplest whole number ratio, which gives \( \frac{1.31}{1.31} : \frac{1.75}{1.31} \approx 1:1.33 \) Since empirical formula must contains whole numbers, 1:1.33 is approximately equal to 3:4, so the formula for compound Y is \( Mn_3O_4 \)
03

Write a balanced chemical reaction for the conversion of X to Y

A balanced equation shows that the same amount of atoms are present in the reactants and products. The reaction is \( MnO_2 \rightarrow Mn_3O_4 + O_2 \). Balance the oxygen atoms first by placing a 4 in front of \( MnO_2 \) and a 2 in front of \( Mn_3O_4 \). Then balance manganese by placing a 3 in front of \( MnO_2 \). Therefore, the balanced equation is \( 3MnO_2 \rightarrow 2Mn_3O_4 + O_2 \)

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Most popular questions from this chapter

Each copper(II) sulfate unit is associated with five water molecules in crystalline copper(II) sulfate pentahydrate \(\left(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\right)\). When this compound is heated in air above \(100^{\circ} \mathrm{C},\) it loses the water molecules and also its blue color: $$ \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{CuSO}_{4}+5 \mathrm{H}_{2} \mathrm{O} $$ If \(9.60 \mathrm{~g}\) of \(\mathrm{CuSO}_{4}\) are left after heating \(15.01 \mathrm{~g}\) of the blue compound, calculate the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) originally present in the compound.

What is the mass (in amu) of a carbon-12 atom? Why is the atomic mass of carbon listed as 12.01 amu in the table on the inside front cover of this book?

A sample of iron weighing \(15.0 \mathrm{~g}\) was heated with potassium chlorate \(\left(\mathrm{K} \mathrm{ClO}_{3}\right)\) 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.

Silicon tetrachloride \(\left(\mathrm{SiCl}_{4}\right)\) can be prepared by heating Si in chlorine gas: $$ \mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(l) $$ In one reaction, 0.507 mole of \(\mathrm{SiCl}_{4}\) is produced. How many moles of molecular chlorine were used in the reaction?

What is the difference between a chemical reaction and a chemical equation?
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