Tuesday, October 12, 2010

Review for Chapters 4-6

The Exam is a week from Thursday and I suggest you start reviewing now. I am in the process of checking with some other faculty about posting the pdf of complete solutions for the problems in the back of each chapter. Someone brought it to my attention that the solutions manual in the library only works out the solutions to the odd numbered problems. I would like to post a solutions manual (as pdf) that works out both even and odd numbered problems. Please be patient as I verify that I am allowed to do this.

The most important thing to realize from chapter 4 is that percent composition and empirical formula are two ways of calculating the same thing. Both essentially embody the law of definite proportions. The definition of a compound is that it will always have the same mass percentage of elements. ALWAYS.  If I tell you a certain compound is always 58% iron by mass then you can assume that 58% of any given mass will be iron. Likewise, if I tell you the percentage composition of multiple elements within a compound you can use the molar masses (from the periodic table) to convert into the lowest whole-number ratio of elements per formula unit. So then the big question is this: How does the empirical formula (lowest whole-number ratio)  relate to the molecular formula? The two are related by a whole number ratio of elements or a whole number ratio of molar mass values. If I tell you the molar mass of a compound is 44 g/mole and I tell you that the empirical molar mass is 11 grams per mole then I would multiple the empirical formula unit (subscripts) by 4 to obtain the molecular formula unit. (molecular formula molar mass/empirical formula molar mass= small whole number). As I stated in class, you usually do not modify subscripts on formula units (true for balancing equations), however, in this case the empirical formula unit is not necessary a real compound. It merely represents a relationship or ratio between elements. There are cases where the empirical formula is actually the molecular formula. (CH4 is an example) In this case the molecular formula molar mass= empirical formula molar mass.

Chapter 5 explores the nature of chemical reactions. We look at different types of reactions and different ways to analyze if two things will actually react. The biggest indicator is whether or not the products will form something that is more stable than the starting materials. Key indicators that a reaction has occurred can be observed by the following: driving off a gas (like carbon dioxide), forming a solid in solution (precipitate), a permanent change in color, production of heat or light. (Note here that something will change color slightly when dissolved in aqueous solution (color will become fainter). This is a permanent color change. Why is it not a chemical reaction? Do you know?)
We studied many different kinds of reactions: combination reactions (two compounds coming together to form one compound), decomposition reactions (a compound breaking apart into two or more compounds), single-displacement reactions (elemental metal like aluminum displacing another metal within a compound), double displacement reactions (precipitation or acid/base reactions), combustion reactions (metal or nonmetal combining with oxygen to form an oxide or carbon dioxide/water). It is important to be able to identify the type of reaction you are analyzing. It is only after you have identified the type of reaction that you can begin to pinpoint the products.

There are certain rules of thumb you can use at this level to assess reaction type: Is a metal present in elemental form (ie- it is not combined with anything else in a compound)? In this case you probably have a single displacement reaction (if the elemental metal is higher on the activity series than the metal bound with the compound). Can you identify hydrogen ions or the hydroxide (OH-) ion in solution? If so- you probably have an acid/base double-displacement reaction. Is something combining with oxygen gas? You probably have combustion. Do you only have one product? You probably have some kind of combination reaction. Does a solid form in the presence of ions in solution (do you have ions that are identified on the solubility rules as insoluble with each other?) You have a precipitation reaction. Work examples to learn how to identify these key characteristics.

In chapter 6 we expand upon the notion of a balanced chemical equation into converting quantities of reactant into quantities of product. This introduces the ubiquitous stoichiometric relationship in chemistry. Yes, as the title of my blog states it, everything in introductory chemistry boils down to stoichiometry (pun intended). You must understand stoichiometric relationships to understand how to figure out your amount of product and to understand the transfer of heat in a reaction. Both depend on moles of reactant compared to moles of product. If you don't have a properly balanced equation you haven't properly denoted the conservation of mass in your chemical equation and you will fail to define these critical relationships.
What is the easiest way to figure out limiting reagent given grams of two reactants? What is the first step in any problem involving grams? Convert to moles- ALWAYS. Use your molar mass from the PT to figure out moles of each reactant. Then, use your balanced chemical equation to figure out how much product you could make from the amount of moles of reactant that you have. The reactant that gives you the least amount of product is your limiting reactant. The amount of product you calculated from this reactant is your theoretical yield. Then, you go in the lab and measure the amount you actually got from the reaction. Actual/theoreticalX 100 is your percent yield.
How did we describe energy as factoring into these molar relationships? The amount of energy emitted or absorbed in any chemical reaction is directly affected by the stoichiometric relationship of reactants/products. If you know your reaction absorbs 50 kg/mole of starting material then how much heat to you absorb for 0.5 mole of starting material? Half of 50 or 25 kj/mole. You use the coefficients in the balanced equation to calculate energy used/created per amount available.
What is calorimetry? Calorimetry is how we measure the transfer of heat. We talked about a very simple demonstration you can do to illustrate this. Take a coffee cup and fill it with water. Take the temperature of the water. Then, add a piece of ice. Measure the temp again (let all the ice melt). What is the final temp? By using this exercise we can measure the heat absorbed by the water in the styrofoam cup. What is the heat absorbed? It is the amount of heat required to melt the given amount of ice in the cup. In this case the surroundings in the water and the system is the ice. Heat flows from water to the ice to melt the ice. (Heat always flows from hot to cold). Use q=mass X specific heat X change in temperature to calculate the amount of energy required to melt the ice. Of course, don't forget to negate the sign of the heat value you calculate based on a loss of energy from the water. The water is losing energy (-q). So the ice must be gaining heat/energy (+q). Review these kinds of concepts and do many practice problems.