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Chapter 3: Problem 63

Write a formula for (a) the chloride of titanium having Ti in the O.S. \(+4 ;\) (b) the sulfate of iron having Fe in the O.S. \(+3 ;(c)\) an oxide of chlorine with Cl in the O.S. \(+7 ;\) (d) an oxoanion of sulfur in which the apparent O.S. of \(S\) is +7 and the ionic charge is \(2-\).

Short Answer

Expert verified
The formulas for the compounds are: (a) \(TiCl_4\), (b) \(Fe_2(SO_4)_3\), (c) \(Cl_2O_7\), and (d) \(S_2O_7^{2-}\).

Step by step solution

01

Part (a) - Chloride of Titanium in O.S. +4

The chloride of titanium will be a binary compound consisting of titanium (Ti) and chloride (Cl). Titanium is in oxidation state +4, meaning it has lost 4 electrons, and chloride always has a charge of -1 (gained 1 electron). The formula for the compound will be written by equaling the total positive charge to the total negative charge. Here, \(Ti = +4\) and \(Cl = -1\). So, to equalize the charges, we need 4 chloride ions for each titanium ion, which gives the formula \(TiCl_4\).
02

Part (b) - Sulfate of Iron in O.S. +3

The sulfate of iron includes iron (Fe) and sulfate (SO4) ions. Here, iron is in oxidation state +3 and sulfate ion always carries a charge of -2. Thus, two iron(III) ions will be needed to balance the charges of three sulfate ions. This gives us the iron(III) sulfate formula as \(Fe_2(SO_4)_3\).
03

Part (c) - Oxide of Chlorine in O.S. +7

The oxide of chlorine will contain chlorine(Cl) and oxygen(O). Chlorine in O.S. of +7 means it has lost 7 electrons, and oxygen always has a charge of -2 (gained 2 electrons). To balance the charges, we need seven half oxygen ion for each chlorine ion, which results in the formula \(Cl_2O_7\).
04

Part (d) - Oxoanion of Sulfur in O.S. +7 and ionic charge 2-

An oxoanion of sulfur will contain sulfur(S), oxygen(O) and carry an overall ionic charge of -2. Here, sulfur (S) is in an oxidation state of +7 and oxygen (O) is in an oxidation state of -2. Oxygen is more electronegative than sulfur and will pull electrons towards itself. Sulfur will have its own oxidation state and oxygen will have -2 (-II in traditional notations). In order to balance out the total charge of -2 for the ion, 4 oxygen atoms (-8) and one sulfur atom (+7) will be needed, as \(+7 - 8 = -1\). To balance out the total charge assigned to the ion of -2, results in the formula \(S_2O_7^{2-}\).

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Most popular questions from this chapter

Some substances that are only very slightly soluble in water will spread over the surface of water to produce a film that is called a monolayer because it is only one molecule thick. A practical use of this phenomenon is to cover ponds to reduce the loss of water by evaporation. Stearic acid forms a monolayer on water. The molecules are arranged upright and in contact with one another, rather like pencils tightly packed and standing upright in a coffee mug. The model below represents an individual stearic acid molecule in the monolayer. (a) How many square meters of water surface would be covered by a monolayer made from \(10.0 \mathrm{g}\) of stearic acid? [Hint: What is the formula of stearic acid?] (b) If stearic acid has a density of \(0.85 \mathrm{g} / \mathrm{cm}^{3}\), estimate the length (in nanometers) of a stearic acid molecule. [Hint: What is the thickness of the monolayer described in part a?] (c) A very dilute solution of oleic acid in liquid pentane is prepared in the following way: $$\begin{aligned} &1.00 \mathrm{mL} \text { oleic acid }+9.00 \mathrm{mL} \text { pentane } \rightarrow \text { solution }(1)\\\ &1.00 \mathrm{mL} \text { solution }(1)+9.00 \mathrm{mL} \text { pentane } \rightarrow \text { solution }(2)\\\ &1.00 \mathrm{mL} \text { solution }(2)+9.00 \mathrm{mL} \text { pentane } \rightarrow \text { solution }(3)\\\ &1.00 \mathrm{mL} \text { solution }(3)+9.00 \mathrm{mL} \text { pentane } \rightarrow \text { solution }(4) \end{aligned}$$ A 0.10 mL sample of solution (4) is spread in a monolayer on water. The area covered by the monolayer is \(85 \mathrm{cm}^{2} .\) Assume that oleic acid molecules are arranged in the same way as described for stearic acid, and that the cross-sectional area of the molecule is \(4.6 \times 10^{-15} \mathrm{cm}^{2}\). The density of oleic acid is \(0.895 \mathrm{g} / \mathrm{mL} .\) Use these data to obtain an approximate value of Avogadro's number.

The oxidation state of I in the ion \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}\) is (a) -1 (b) \(+1 ;(c)+7 ;(d)+8\)

A hydrate of copper(II) sulfate, when heated, goes through the succession of changes suggested by the photograph. In this photograph, (a) is the original fully hydrated copper(II) sulfate; (b) is the product obtained by heating the original hydrate to \(140^{\circ} \mathrm{C}\) (c) is the product obtained by further heating to \(400^{\circ} \mathrm{C}\) and (d) is the product obtained at \(1000^{\circ} \mathrm{C}\) A \(2.574 \mathrm{g}\) sample of \(\mathrm{CuSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O}\) was heated to \(140^{\circ} \mathrm{C},\) cooled, and reweighed. The resulting solid was reheated to \(400^{\circ} \mathrm{C},\) cooled, and reweighed. Finally, this solid was heated to \(1000^{\circ} \mathrm{C},\) cooled, and reweighed for the last time. $$ \text {Original sample } \quad \quad\quad\quad\quad \text {\(2.574 \mathrm{g}\) } $$ $$ \text {After heating to \(140^{\circ} \mathrm{C}\) } \quad \quad\quad\quad\quad \text {\(1.833 \mathrm{g}\) } $$ $$ \text {After reheating to \(400^{\circ} \mathrm{C}\)} \quad \quad\quad\quad\quad \text {\(1.647 \mathrm{g}\) } $$ $$ \text {After reheating to \(1000^{\circ} \mathrm{C}\)} \quad \quad\quad\quad\quad \text {\(0.812 \mathrm{g}\)} $$ (a) Assuming that all the water of hydration is driven off at \(400^{\circ} \mathrm{C},\) what is the formula of the original hydrate? (b) What is the formula of the hydrate obtained when the original hydrate is heated to only \(140^{\circ} \mathrm{C} ?\) (c) The black residue obtained at \(1000^{\circ} \mathrm{C}\) is an oxide of copper. What is its percent composition and empirical formula?

A \(1.562 \mathrm{g}\) sample of the alcohol \(\mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{CH}_{3}\) is burned in an excess of oxygen. What masses of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) should be obtained?

A 0.732 g mixture of methane, \(\mathrm{CH}_{4}\), and ethane, \(\mathrm{C}_{2} \mathrm{H}_{6},\) is burned, yielding \(2.064 \mathrm{g} \mathrm{CO}_{2} .\) What is the percent composition of this mixture (a) by mass; (b) on a mole basis?
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