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Chapter 3: Problem 79

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 \(0.6498 \mathrm{~g}\) of one of the compounds leaves a residue of \(0.6018 \mathrm{~g}\). Heating \(0.4172 \mathrm{~g}\) of the other compound results in a mass loss of \(0.016 \mathrm{~g}\). Determine the empirical formula of each compound.

Short Answer

Expert verified
The empirical formulas of the two compounds are HgO for Compound 1 and Hg₂O for Compound 2.

Step by step solution

01

Compound 1: Calculate moles of Mercury

For Compound 1, the initial mass is 0.6498 g and the residue mass of Hg is 0.6018 g. To find the moles of Hg, divide the residue mass by the molar mass of Hg: moles of Hg = (0.6018 g) / (200.59 g/mol) = 0.003 mol
02

Compound 1: Calculate moles of Oxygen

For Compound 1, determine the mass of oxygen (O) by subtracting the residue mass of Hg from the initial mass: mass of O = (0.6498 g - 0.6018 g) = 0.048 g Next, calculate the moles of O by dividing the mass of O by the molar mass of O: moles of O = (0.048 g) / (16.00 g/mol) = 0.003 mol
03

Compound 1: Determine empirical formula

For Compound 1, the mole ratio of Hg to O is 1:1, since both have equal moles (0.003 mol). Therefore, the empirical formula of Compound 1 is HgO.
04

Compound 2: Calculate moles of Oxygen

For Compound 2, the mass loss after heating is 0.016 g. This mass loss corresponds to the mass of O. Determine the moles of O by dividing the mass of O by the molar mass of O: moles of O = (0.016 g) / (16.00 g/mol) = 0.001 mol
05

Compound 2: Calculate initial mass of Mercury

For Compound 2, determine the initial mass of Hg by subtracting the mass of O from the initial mass of the compound: Initial mass of Hg = (0.4172 g - 0.016 g) = 0.4012 g
06

Compound 2: Calculate moles of Mercury

To find the moles of Hg, divide the initial mass of Hg by the molar mass of Hg: moles of Hg = (0.4012 g) / (200.59 g/mol) = 0.002 mol
07

Compound 2: Determine empirical formula

For Compound 2, the mole ratio of Hg to O is 2:1, since the moles of Hg are twice the moles of O (0.002 mol : 0.001 mol). Therefore, the empirical formula of Compound 2 is Hg₂O. In conclusion, the empirical formulas of the two compounds are HgO for Compound 1 and Hg₂O for Compound 2.

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Most popular questions from this chapter

One of relatively few reactions that takes place directly between two solids at room temperature is $$ \mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}(s)+\mathrm{NH}_{4} \mathrm{SCN}(s) \longrightarrow $$ In this equation, the \(\cdot 8 \mathrm{H}_{2} \mathrm{O}\) in \(\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}\) indicates the pres- ence of eight water molecules. This compound is called barium hydroxide octahydrate. a. Balance the equation. b. What mass of ammonium thiocyanate \(\left(\mathrm{NH}_{4} \mathrm{SCN}\right)\) must be used if it is to react completely with \(6.5 \mathrm{~g}\) barium hydroxide octahydrate?

Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane: $$ 2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g)+2 \mathrm{CH}_{4}(g) \longrightarrow 2 \mathrm{HCN}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ If \(5.00 \times 10^{3} \mathrm{~kg}\) each of \(\mathrm{NH}_{3}, \mathrm{O}_{2}\), and \(\mathrm{CH}_{4}\) are reacted, what mass of \(\mathrm{HCN}\) and of \(\mathrm{H}_{2} \mathrm{O}\) will be produced, assuming \(100 \%\) yield?

Methane \(\left(\mathrm{CH}_{4}\right)\) is the main component of marsh gas. Heating methane in the presence of sulfur produces carbon disulfide and hydrogen sulfide as the only products. a. Write the balanced chemical equation for the reaction of methane and sulfur. b. Calculate the theoretical yield of carbon disulfide when \(120 . \mathrm{g}\) of methane is reacted with an equal mass of sulfur.

Anabolic steroids are performance enhancement drugs whose use has been banned from most major sporting activities. One anabolic steroid is fluoxymesterone \(\left(\mathrm{C}_{20} \mathrm{H}_{29} \mathrm{FO}_{3}\right) .\) Calculate the percent composition by mass of fluoxymesterone.

A compound contains \(47.08 \%\) carbon, \(6.59 \%\) hydrogen, and \(46.33 \%\) chlorine by mass; the molar mass of the compound is \(153 \mathrm{~g} / \mathrm{mol} .\) What are the empirical and molecular formulas of the compound?
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