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Chapter 5: Problem 85

There are two binary compounds of mercury and oxygen. Heating either of them results in the decomposition of the compound, with oxygen gas escaping into the atmosphere while leaving a residue of pure mercury. Heating 0.6498 g of one of the compounds leaves a residue of 0.6018 g. Heating 0.4172 g of the other compound results in a mass loss of 0.016 g. Determine the empirical formula of each compound.

Short Answer

Expert verified
The empirical formulas of the two binary compounds of mercury and oxygen are HgO and Hg2O.

Step by step solution

01

Determine the mass of oxygen in each compound

For the first compound: Mass of mercury = 0.6018 g Since the initial mass is 0.6498 g, the mass of oxygen is: Mass of oxygen = 0.6498 g - 0.6018 g = 0.048 g For the second compound: Mass loss due to heating = 0.016 g (this is the mass of oxygen that escaped) So, the mass of mercury = 0.4172 g - 0.016 g = 0.4012 g
02

Convert mass to moles

Using the molar mass of mercury (Hg) = 200.59 g/mol and oxygen (O) = 16.00 g/mol to convert masses to moles: For the first compound: Moles of Hg = 0.6018 g Hg * (1 mol Hg / 200.59 g Hg) = 0.00300044 mol Hg Moles of O = 0.048 g O * (1 mol O / 16.00 g O) = 0.003 mol O For the second compound: Moles of Hg = 0.4012 g Hg * (1 mol Hg / 200.59 g Hg) = 0.00199932 mol Hg Moles of O = 0.016 g O * (1 mol O / 16.00 g O) = 0.001 mol O
03

Find the simplest ratio of moles

For the first compound: Divide the moles of both elements by the smallest value (0.003) to find the simplest ratio: Ratio_Hg = 0.00300044 mol Hg / 0.003 mol = 1.000133 Ratio_O = 0.003 mol O / 0.003 mol = 1 So, the empirical formula for the first compound is HgO. For the second compound: Divide the moles of both elements by the smallest value (0.001) to find the simplest ratio: Ratio_Hg = 0.00199932 mol Hg / 0.001 mol = 1.99932 Ratio_O = 0.001 mol O / 0.001 mol = 1 The ratio of Hg and O is approximately 2:1. Therefore, the empirical formula for the second compound is Hg2O.

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Most popular questions from this chapter

A 0.755-g sample of hydrated copper(II) sulfate $$\mathrm{CuSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O}$$ was heated carefully until it had changed completely to anhydrous copper(II) sulfate (CuSO_) with a mass of 0.483 g. Determine the value of \(x .\) [This number is called the number of waters of hydration of copper(II) sulfate. It specifies the number of water molecules per formula unit of \(\mathrm{CuSO}_{4}\) in the hydrated crystal.]

When the supply of oxygen is limited, iron metal reacts with oxygen to produce a mixture of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\). In a certain experiment, \(20.00 \mathrm{g}\) iron metal was reacted with \(11.20 \mathrm{g}\) oxygen gas. After the experiment, the iron was totally consumed, and 3.24 g oxygen gas remained. Calculate the amounts of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) formed in this experiment.

In using a mass spectrometer, a chemist sees a peak at a mass of \(30.0106 .\) Of the choices \(^{12} \mathrm{C}_{2}\) \(^{1} \mathrm{H}_{6},\) \(^{12} \mathrm{C}\) \(^{1} \mathrm{H}_{2}\) \(^{16} \mathrm{O},\) and \(^{14} \mathrm{N}^{16} \mathrm{O},\) which is responsible for this peak? Pertinent masses are \(^{1} \mathrm{H}\), \(1.007825 ;^{16} \mathrm{O}, 15.994915 ;\) and \(^{14} \mathrm{N}, 14.003074.\)

Hydrogen peroxide is used as a cleansing agent in the treatment of cuts and abrasions for several reasons. It is an oxidizing agent that can directly kill many microorganisms; it decomposes on contact with blood, releasing elemental oxygen gas (which inhibits the growth of anaerobic microorganisms); and it foams on contact with blood, which provides a cleansing action. In the laboratory, small quantities of hydrogen peroxide can be prepared by the action of an acid on an alkaline earth metal peroxide, such as barium peroxide: $$\mathrm{BaO}_{2}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{BaCl}_{2}(a q)$$ What mass of hydrogen peroxide should result when \(1.50 \mathrm{g}\) barium peroxide is treated with \(25.0 \mathrm{mL}\) hydrochloric acid solution containing \(0.0272 \mathrm{g}\) HCl per mL? What mass of which reagent is left unreacted?

An ionic compound \(\mathrm{MX}_{3}\) is prepared according to the following unbalanced chemical equation. $$\mathbf{M}+\mathbf{X}_{2} \longrightarrow \mathbf{M X}_{3}$$ A \(0.105-\mathrm{g}\) sample of \(\mathrm{X}_{2}\) contains \(8.92 \times 10^{20}\) molecules. The compound \(\mathrm{MX}_{3}\) consists of \(54.47 \%\) X by mass. What are the identities of \(\mathrm{M}\) and \(\mathrm{X}\), and what is the correct name for \(\mathrm{MX}_{3} ?\) Starting with 1.00 g each of \(M\) and \(X_{2}\), what mass of \(M X_{3}\) can be prepared?
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