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Chapter 3: Problem 102

A sample of a compound of \(\mathrm{Cl}\) and \(\mathrm{O}\) reacts with an excess of \(\mathrm{H}_{2}\) to give \(0.233 \mathrm{~g}\) of \(\mathrm{HCl}\) and \(0.403 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) Determine the empirical formula of the \(\mathrm{com}-\) pound.

Short Answer

Expert verified
The empirical formula of the compound is \(\mathrm{ClO}_7\).

Step by step solution

01

Identifying Given Information

The information we know is that \(\mathrm{HCl}\) is produced by the amount of \(0.233 \mathrm{g}\) and \(\mathrm{H}_{2} \mathrm{O}\) is produced by the amount of \(0.403 \mathrm{g}\).
02

Calculating Moles of Compounds

The molar mass of \(\mathrm{HCl}\) is \(35.5+1=36.5 \mathrm{g/mol}\) and the molar mass of water is \(18 \mathrm{g/mol}\). So, the number of moles of \(\mathrm{HCl}\) is \(\frac{0.233}{36.5}=0.00639 \mathrm{mol}\) and the number of moles of \(\mathrm{H_2O}\) is \(\frac{0.403}{18}=0.0224 \mathrm{mol}\). This is important to note because the moles of H2O will give us the moles of O in the compound.
03

Getting Moles of Elements

The number of moles of \(\mathrm{Cl}\) would be equal to the number of moles of \(\mathrm{HCl}\), which is \(0.00639 \mathrm{mol}\). The number of moles of \(\mathrm{O}\) would be equal to the number of moles of \(\mathrm{H_2O}\), which is \(0.0224 \mathrm{mol}\).
04

Determining Empirical Formula

The empirical formula is obtained by dividing the moles of each element by the smallest number of moles calculated, which in this case is the number of moles of \(\mathrm{Cl}\): \(\frac{0.00639}{0.00639}=1\) for Cl and \(\frac{0.0224}{0.00639}\)=3.5 for O. But an empirical formula should consist of whole numbers, multiply both numbers by 2 to give Cl1O7. Hence the empirical formula of the compound is \(\mathrm{ClO}_7\).

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Most popular questions from this chapter

The annual production of sulfur dioxide from burning coal and fossil fuels, auto exhaust, and other sources is about 26 million tons. The equation for the reaction is$$\mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)$$How much sulfur (in tons), present in the original materials, would result in that quantity of \(\mathrm{SO}_{2} ?\)

Without doing any detailed calculations, estimate which element has the highest percent composition by mass in each of the following compounds: (a) \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\) (b) \(\mathrm{NF}_{3}\) (c) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (d) \(\mathrm{C}_{2952} \mathrm{H}_{4664} \mathrm{~N}_{812} \mathrm{O}_{832} \mathrm{~S}_{8} \mathrm{Fe}_{4}\)

Why is the actual yield of a reaction almost always smaller than the theoretical yield?

What is wrong or ambiguous with each of the statements here? (a) \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) is the limiting reactant in the reaction$$\mathrm{NH}_{4} \mathrm{NO}_{2}(s) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ (b) The limiting reactants for the reaction shown here are \(\mathrm{NH}_{3}\) and \(\mathrm{NaCl}\). $$\begin{aligned}\mathrm{NH}_{3}(a q)+\mathrm{NaCl}(a q) &+\mathrm{H}_{2} \mathrm{CO}_{3}(a q) & \longrightarrow \\ & \mathrm{NaHCO}_{3}(a q)+\mathrm{NH}_{4} \mathrm{Cl}(a q) \end{aligned}$$

Consider the reaction \(3 \mathrm{~A}+2 \mathrm{~B} \rightarrow 3 \mathrm{C}\). A student mixed 4.0 moles of \(\mathrm{A}\) with 4.0 moles of \(\mathrm{B}\) and obtained 2.8 moles of \(\mathrm{C}\). What is the percent yield of the reaction?
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