1. Mole Theory Percent Composition Yield Problems

2. Let’s Review… Remember that we can determine the percentage of a defined part by dividing the amount of that part by the total of all of the parts in the system. (and then multiplying by 100) In equation form, it is: Percentage = Amount of part Total of all parts x100

3. An Example: What is the mass percentage of oxygen in the compound sodium sulfate? Using our strategy for writing correct chemical nomenclature gives us the formula Na 2 SO 4 The calculation of the formula weight will give the result: fw = 2(23) + 1(32.1) + 4(16) = 142.1 g/mole

4. Continuing… What is the mass percentage of oxygen in the compound sodium sulfate? The mass percentage of oxygen will be calculated as: % Oxygen = 4(16) 142.1 = 64 142.1 x100 =45.04 % The formula weight The four oxygens in the formula

5. Now for the “New Stuff” Because the chemical formula of a compound is set by nature (and non- variable) And…because the atomic weights of the elements in that compound are set by nature (and essentially non-variable)… The ratio of mass of part to formula weight of a compound is a non-variable quantity and may be used as a factor in other problems.

6. What does this mean??? If we look at the sample problem that was presented earlier: Because sodium sulfate has the formula Na 2 SO 4, and because the atomic weights of the elements Na, S, and O will not change… We are free to say that there will always be 64 grams of Oxygen in 142.1 grams of Na 2 SO 4.

7. Therefore… I am free to establish a factor that looks like this: 64 grams O 142.1 grams Na 2 SO 4 () and use it in what are called “yield problems”

8. An example… How many grams of oxygen are contained in 300 grams of sodium sulfate? Using the factor created in the last slide, we can set up the following calculation: 300 grams Na 2 SO 4 x 64 grams O 142.1 grams Na 2 SO 4 () =135.12 grams Oxygen Notice how the factor will allow you to cancel.

9. An important note: You should remember from you time in Chem I that it is possible to invert a factor. You would do this in order to be sure that you could cancel out the units given in the problem. The following example illustrates this…

10. Example #2 How many grams of sodium sulfate would be needed to have 100 grams of oxygen? The logic: –This problem seems to be “backward” – we are given the yield (the 100 grams of oxygen) and are asked to determine from where it came. –Since the problem is “backward”, we will have to invert the factor…

11. Working this out: We will set up the calculation like this: 100 grams of Oxygenx 142.1 grams Na 2 SO 4 64 grams of Oxygen =222.03 grams Na 2 SO 4 Notice here how we invert the factor so that we are able to cancel out the grams of Oxygen in this case. You are always free to invert a factor as needed.

12. A final note… Calculations like the ones shown in this presentation will appear throughout the AP exam. It is a fair estimation that up to 15% of the entire exam will involve calculations using factors that you will have to derive and set up. You will need to be good at this….