Homework is the Percent Yield worksheet which was handed out in class. Here it is if you happened to have misplaced it: http://www.csun.edu/~jte35633/worksheets/Chemistry/12-4PercentYield.pdf
I got up at 4am and finished it at 8am, so I know it is tough. If you dig in for an hour and a half of concentrated attention, that would satisfy me, and we can finish next week. If you get on a roll, keep going. However, it's hard to make progress if we expect to only "get it" in class, so give it a good shot.
To make it a little easier, here is an app for calculating the molar mass: molar mass calculator.
And the answer sheet has been done by me, so if you find my mistakes, please send me a note.
By the time I had finished, I developed a system: (Here is a form to do the homework out- print out 5 pages- don't worry, it takes very little ink. Percent Yield workpage and Answer sheet
- Write the balanced equation.
- Figure out the molar masses that will be needed; write in a box on the side. Under that write the actual yield, which is given in the problem.
- For Questions 2, 3, 4,9 and 10, you need to find the limiting reactant. Maybe you should do those last. Example 9.8 + the self-check 9.7 is the best example to follow (on page 275 and A20 in your text.) (oops, I had better check my work. I have some fixing to do on those.)
- Write the equation: on the left , the given in grams, and on the right, what you are trying to find:
- 3.74 g Na x ___________x____________x__________=[ ] g Na2O2
- Fill in all the molar masses and ratios. [grams>moles>>mole ratio>grams]
- Calculate to find the theoretical yield.
- Using the actual yield (given) and the theoretical yield (worked), calculate the percent yield.
- Think: Does this seem right? Be alert for calculator problems.
Lastly, here is the solution page, again:
Please check your work as you go.
***already 1 mistake, caught by Lily.
#9 -the limiting reactant is O2
the theoretical yield is 18.14gZnO
the percent yield is 80.4%
Also, Avery caught some mistakes (copying and calculator) in problems 2 and 7, listed in the comments below. This is why we show our work!
In question two, you wrote that the molar mass of K2O is 92.4, therefor giving a theoretical yield of 8.18g of K2O, and a percent yield of 90.0%. However, the molar mas of K2O is not 92.4, it is 94.2, giving a theoretical yield of 8.34g of K2O, and a percent yield of 88.3%.
ReplyDeleteIn question seven, you wrote that 15.8*1/17.1*4/4*30.01= 21.8 It actually equals 27.7, giving a percent yield of 78.7%
ReplyDeleteWhat's a good precision of measurement? I mean, should we do molar mass to four significant digits? Or two? Because it makes quite a large difference in the percent yields.
ReplyDeleteAlso, in #9, isn't Sulfur Dioxide SO2, not just S?
ReplyDeleteThe number of significant figures in molar mass can vary. In calculations, include one more significant figure than the data given in the problem so that they do not limit the number of significant figures in the answer.
ReplyDelete