FAQ's Unit2 (4 questions)

1. Is there an "easy way" to balance a chemical equation?In Chapter 4, the "easy" or simple method for balancing chemicalequations is a "change and test" strategy and balancing equationsby "inspection". These ideas involve systematically changingone stoichiometric coefficient of either a reactant or productand checking the resulting modified chemical equation to see if it is balanced.A simple example of the strategy will be applied to the reaction of hydrogengas and oxygen gas to yield water.

H2 (g) + O2 (g) = H2O (l)

In this case,checking the first reactant H2, you will note there are2 mols of H atoms in H2 (g) and there are 2 mols of H atoms in the product,H2O, so the number of hydrogen atoms moving from reactants to products,is balanced. Checking the "balance" in terms of oxygen atoms,you discover 2 mols of oxygen atoms as a reactant but only 1 mole of oxygenin the product, H2O. The oxygen part of the equation is not balanced.In order to balance it, place a 2 as the stoichiometric coefficient forH2O as follows

H2 (g) + O2 (g) = 2 H2O (l)

Now "check " the new equation for mass balance. This time theoxygen atoms are balanced with 2 mols of oxygen atoms as reactants and productsbut now the hydrogen is not balanced. This can be corrected by placinga 2 as a stoichiometric coefficient on the reactant, H2 (g).

2 H2 (g) + O2 (g) = 2 H2O (l)

Check this new equation. It is a balanced chemical equation, 4 mols ofhydrogen atoms and 2 mols of oxygen atoms as both reactants and products.

It is important that you check all chemical reactions for mass balancebefore you begin any of the stoichiometric calculations presented in Units3 and 4. Generally, it is assumed you are working with a balanced equationunless you are instructed to balance, however texts sometimes have "misprints".

2. How do I use the "Solubility Rules" presented in Figure4.7, page 167 ? The solubility guidelines presented on page 167 providea "rule of thumb" guide for you to predict the solubility of ioniccompounds in water. The dissociation of ionic compounds often referred to"salts" for historic reasons, was observed by Svante Arrheniusat the end of the nineteenth century. His work explained why ionic saltsdissolved in water conducted electricity because the resulting hydratedcations and anions could act as charge carriers in the aqueous solution.

Figure 4.7 is divided into two categories. The "top" framerepresents essentially soluble ionic compounds in water with a fewexceptions noted to the right of the frame. For example, all chloridesare soluble, that is form solvated cations and anions in water, exceptsalts of silver (Ag+) divalent mercury (Hg(2)2+) and lead (Pb2+). Thismeans magnesium chloride, MgCl2 (s) dissolves in water to give Mg2+ (aq),meaning one, "solvated" magnesium 2+ cation and two, solvatedchloride ions, Cl- (aq).

The 'bottom" frame of Fig, 4.7 represents insoluble ionicsalts, generally In this frame, the entries to the right are exceptionsthat are soluble. A quick read of the bottom frame, suggests salts of variousanions including carbonate, CO3(2-) and hydroxide, OH- , are insolublein water except for ammonium carbonate or ammonium hydroxide or theseanions as compounds with the Group 1A metal ions, such as sodium, potassiumetc. In this case you would say calcium carbonate, CaCO3 (s), is insolublein water meaning it does not dissolve and does not resultin the formation of solvated ions in solution. However sodium carbonate,Na2CO3 (s) is soluble because it is an exception and does result in theformation of three solvated ions in solution. Can you give the threeions in solution that result from dissolving sodium carbonate in water ?Ans: Two sodium ions, Na+ (aq) and one carbonate ion, CO3(2-)(aq).

The solubility guidelines are used to predict whether or not an ionicsalt is soluble in water. From Unit 1, you should be aware that an ionicsalt is a compound formed between a metal cation or ammoniumcation and any of the nonmetal anions or any of the polyatomicanions listed in Table 3.2, page 118, is soluble in water. If it issoluble then you must be able to write down how many and what types of ionsare in the resulting solution. For example, see Example 4.2, page 163 andother exercises which are part of Objective 3.

3. Can you show me how the "oxidation number rules"on page 191 work ? At this point in the development of the course, oxidationnumber is a "mechanically" discovered number without any theoreticalsupport. The "guidelines" offer a recipe to determine them. Youdon't need to memorize these guidelines but you should know how to use themand where they are located in the text. Exercise 4.12 has you determinethe oxidation numbers for the elements in iron (III) oxide, Fe2O3a "neutral" compound and carbonate ion, CO3(2-), I willuse these as examples to answer this question.

Fe2O3 First, Rule 6 says the sum of the assigned oxidation numbersmust be zero for a neutral compound, so whatever oxidation numbersyou give to iron and oxygen in this neutral compound, their sum must = zero.Second, Rule 5 says oxygen, O, in compounds or ions is "always"assigned an oxidation number -2 and hydrogen, H, is "always"+1. (Note while the text discusses exceptions to rule 5, you will not maketoo many serious errors if you just thin of oxygen as -2 and hydrogen as+1 ). Using these two rules, you can assign oxidation number in the Fe2O3case. Oxygen is -2 and there are three oxygens in the compound, so the "total"contributionof oxygen is 3 x (-2) = -6. Now, Rule 6 says the sum numbers assigned toFe and O must equal zero or (?) + (-6) = 0. Therefore the "number"assigned to Fe must be +3 because there are two "irons" in thecompound and 2 x (+3) = +6 . Checking: the total assigned to all iron presentis +6 and the total assigned to all oxygen present is -6 and (+6) + (-6)= 0.

CO3(2-) You can use Rule 5 to assign -2 each O in the structure.Similar to the previous case, there are three "oxygens" presentfor a total contribution of -6. In the case of ions, Rule 6 indicatesthe sum of assigned numbers must equal the charge on the ion, herethat is -2. So your thinking about the possible oxidation number on carbonhas the "equation", (?) + (-6) = -2. Solving this equation resultsin +4 and since there is only one "carbon" in the ion, CO3(2-),the oxidation number of carbon must be +4. Checking: the total assignedto carbon, +4 and the total assigned to the three "oxygens", -6and (+4) + (-6) =-2 . For more discussion and examples, see Example 4.8,page 191 and other activities which are part of Objectives 5 and 6.

4. How do I know if a reaction is an oxidation-reduction reaction? An oxidation -reduction reaction requires the identification of someelement or species "gaining" or "going up" in valueof its assigned oxidation number as you move from reactants to products.That specie so identified is said to be "oxidized" and is the"reducing agent" Also, you must identify some element orspecies whose relative oxidation number "decreases" or "goesto a lower value" as you move from reactants to products. This specieis said to be "reduced" and is the "oxidizing agent".Seethe discussion beginning on page 193,"Recognizing Oxidation-ReductionReactions"