Showing posts with label Moles. Show all posts
Showing posts with label Moles. Show all posts

Wednesday, October 20, 2010

'MOLE'ten Lava


Honors chemistry period six is now on round two for these scribe posts because we had to skip Elim.

Mr. Lieberman began the class by letting us correct and ask questions about the worksheet "Mole Problems 2." The worksheet was more practice, that we all needed, on converting particles, moles, and mass. For those that missed it, 6f should be scratched out.

We also corrected and asked questions about the second converting sheet, the one with no name.
Like Kaitlyn stated in her post, the conversions are:

  • For grams to moles, divide by molar mass
  • For moles to grams, multiply by molar mass
  • For particles to moles, divide by Avagadro's number
  • For moles to particles, multiply by Avagadro's number



    We, then, moved on to the lab, "% Composition of Bubble Gum," which we did not do a pre-lab for. Mr. Lieberman told us to fit all of the lab on one page, because it is a relatively short one. The procedure was as follows:
    1. Take the gum and mass it with the wrapper (record)
    2. Unwrap and chew gum
    3. During the chewing process, mass the wrapper (record)
    4. Place chewed gum on wrapper and mass it (record)
    Amazingly, the mass of the gum dropped drastically, but that's because the sugar in the gum was consumed by the chewer of the gum. Our lab group's gum had a mass of 6.13 grams and the sugar within the gum had a mass of 3.97 grams. From here we had to answer 5 post-lab questions:
    1. Calculate the mass of sugar in the bubble.
    2. Calculate the % composition of the sugar (by mass of the bubble gum).
    3. Calculate the moles of the sugar and gum using the molar masses given the procedure.
    4. Using your answer from question number 3, determine a possible empirical formula for the bubble gum.
    5. Researchers have found that the ideal formula for the gum is GS2, where G is a fictional elemental symbol for gum and S for sugar. How does your gum compare to the ideal? What might be some sources of error?
    An empirical formula can be made by the following procedure from http://chemistry.about.com/od/workedchemistryproblems/a/empirical.htm
    1. Assume you have 100 g of the substance (makes the math easier because everything is a straight percent).
    2. Consider the amounts you are given as being in units of grams.
    3. Convert the grams to moles for each element.
    4. Find the smallest whole number ratio of moles for each element.
    There is a mole quiz tomorrow, and this lab is due friday. Don't fall behind on all this work, it is alot to do. Also, there is a "mole day" party on saturday at 6:02 AM, but Mr. Lieberman did not give us alot of information on it.

    -Emilio I
    The scribe for tomorrow's class is: Zoe S.

    Tuesday, October 19, 2010

    Insert Mole Title Here

    So first off, I'd like to congratulate Kaitlyn on a wondeful posting job.
    Now for class, we had kind of a work day in store for us. We got more practice with moles so we could better understand what it is and also we got more practice with unit conversion.
    First off, we went to our notes and finished two "Learning Check!". First problem we had was:
    Question: How many atoms of K are present in 78.4 g of K?
    Work: Take the amount (78.4 g), multiply it by (1 mole of K over 39 (molar mass) grams of K), then multiply that by (Avogadro's number (6.02 times 10^23) over 1 mole of K).
    Answer: 1.21 times 10^24.

    Our second Learning Check! was:
    Question: What is the mass (in grams) of 1.20 X 10^24 molecules of glucose (C6H12O6)?
    So this question is actually working the opposite way of our other problem. We've got to find the mass (instead of molecules) and we start with the molecules (instead of the mass).
    Work: 1.20 X 10^24 times (1 mole of glucose divided by Avogadro's number) times (the Molar mass divided by 1 mole of glucose).
    To find the molar mass for this problem, we must take all the subscripts, multiply them by the mass of their individual atomic masses, then add all of those together. In this problem, the molar mass would be:
    6 (subscript of Carbon) X 12 (atomic mass of Carbon) + 12 (hydrogen) X 1 + 6(oxygen) X 16 = 180 grams
    Once we plug in the molar mass, our answer should be 359 grams per mole.

    So, in order to really understand how we got this, you must already know about moles and molar mass. If you do not understand these fully, please either look back at your notes (you can find extras in moodle), or just look below at Kaitlyn's post.

    After we finished those two problems, we had the rest of the period to work on our mole workshop. This compromised of 5 questions all of which we should have done by tomorrow with our partner. An extra copy of this worksheet can also be found in moodle under unit three worksheets (unfortunately the answers are not there).

    Homework: Finish the Mole Workshop as well as the second mole worksheet. Both were given out before class.
    Important dates: 10/20 (tomorrow!) review quiz on unit conversion and moles.
    10/28 Unit exam, remember to have your homework problems done

    Random fact of the day: Black whales are born white.
    Also, I'd like to point out we had our first visitor from Japan.

    Next Scribe will be Elim J