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Chapter 3: Problem 102

When \(2.750 \mathrm{g}\) of the oxide of lead \(\mathrm{Pb}_{3} \mathrm{O}_{4}\) is strongly heated, it decomposes and produces \(0.0640 \mathrm{g}\) of oxygen gas and \(2.686 \mathrm{g}\) of a second oxide of lead. What is the empirical formula of this second oxide?

Short Answer

Expert verified
The empirical formula of the second oxide of lead is \(Pb\).

Step by step solution

01

Identify the Mass of Lead in the Original Oxide

We start the solution by calculating the mass of lead (\(Pb\)) in the original oxide (\(Pb_3O_4\)), which is the difference between the total mass of the compound and the amount of oxygen produced. This can be done by subtracting the mass of the oxygen gas produced (\(0.0640 \, g\)) from the total mass of the oxide (\(2.750 \, g\)), giving us \(2.750 \, g - 0.0640 \, g = 2.686 \, g\) of lead.
02

Identify the Mass of Lead in the New Oxide

From the question, we know the mass of the new oxide compound is \(2.686 \, g\). However, the amount of lead (\(Pb\)) is the same as what was present in the original compound, since no lead was lost or gained during the reaction. Therefore, the mass of the lead in the new oxide is also \(2.686 \, g\).
03

Appraise Mass of Oxygen in the New Oxide

The mass of the oxygen in the new compound can be obtained by subtracting the mass of lead from the total mass of the new compound. Thus, \(2.686 \, g - 2.686 \, g = 0\)g of oxygen is present in the new oxide.
04

Establish Empirical Formula

From steps 2 and 3, we know that the total amounts of lead and oxygen in the new oxide are \(2.686 \, g\) and \(0 \, g\) respectively. Since we're requested to find the empirical formula and not the molecular formula, there's no need to convert these masses to moles. The empirical formula of the new oxide is lead(\(Pb\)) as there was no oxygen left in the new compound.

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