Who's Worth More: Your Lil Brother or Lil Wayne?

We began class today by learning about percent composition in our notes packet. A percent composition compares the mass of each element present in 1 mole to the total mass of the compound.

Sample Problem:

Penicillin F has the formula C₁₄H₂₀N₂SO_{4. What is the mass percent of each element?}

To begin this problem, we calculated the total mass of the compound:

14 x 12 + 20 x 1 + 2 x 14 + 32 + 4 x 16 = 312 grams/mole

To determine the mass percent of Carbon, we multiplied the amount of carbon in the compound (which we used to determine the total mass of the compound) by 100 and divided by the total mass of the compound. We repeated this process with hydrogen, nitrogen, sulfur, and oxygen to find the rest of our results. We found the following:

Carbon had a mass percent of 53.8 %

Hydrogen had a mass percent of 6.4 %

Nitrogen had a mass percent of 8.9%

Sulfur had a mass percent of 10.3%

Oxygen had a mass percent of 20.5 %

After going over the notes, Mr. Lieberman introduced us to the 'How Much are You Worth?' project. The project asks us to calculate the cost of ourselves, a friend, family member, or favorite actor, although, of course, we are all priceless! We are to show all work in a factor-lable style, and are required to include a picture of our person. A maximum of 5 extra credit points can be given for creativity in your presentation style. Mr. Lieberman showed us some examples of cool projects that have earned students extra credit in the past, but remember that going above and beyond is not required.

Then we took our review quiz which we will have some time to finish in class tomorrow.

The % Composition Bubble Gum Lab is due tomorrow.

Web Assign is due Monday.

'How Much are You Worth?' project due Monday.

The Unit Exam is Thursday 10/28.

The next scribe is...Korri H!

'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.

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

Wholly GuacaMOLE!!!

Today, we began by getting back our tests from last week. After going through some of our mistakes and questions, Mr. Lieberman suggested that if you still have troubles with naming compounds, you should practice some more problems. Then, we moved on to notes about the mole.
The mole is a counting unit that means 602 billion trillion (similar to a dozen, which is 12) or 602,000,000,000,000,000,000,000, or in scientific notation 6.02x10^23. It is also known as Avogadro's number in honor of Avogadro. A mole is really large in terms of things that we can see, such as soft drink cans, as Avogadro's number of these can cover the surface of the earth to a depth of 200 miles.
The mole is used in the same way a dozen is used:

• 1 dozen cookies=12 cookies
• 1 mole of cookies=6.02x10^23
• 1 dozen cars=12 cars
• 1 mole of cars=6.02x10^23

*The number is always the same, but the mass is different* (Mole is abbreviated mol.)

Moles of particles:

• 1 mole of carbon = 6.02x10^23 C atoms
• 1 mole of H2O = 6.02x10^23 H2O molecules (2 moles of hydrogen and one mole of oxygen)
• 1 mole of NaCl = 6.02x10^23 NaCl "molecules" (6.02x10^23 Na^+ ions and 6.02x10^23 Cl^- ions)

Molar Mass: the mass of 1 mole (in grams) is equal to the numberical value of the average atomic mass (from the periodic table)

• 1 mole of C atoms = 12.0 g
• 1 mole of Mg atoms = 24.3 g
• 1 mole of Cu atoms = 63.5 g

To find the molar mass of molecules and compounds, you add the molar masses together:

1 mole of CaCl2 = (1 mole of Ca x 40.1 g/mol) + (2 moles of Cl x 35.5 g/mol) = 111.1 g/mol CaCl2

Here are some notes for calculations with moles:

• 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

The artificial sweetener aspartame has the formula C14H18N2O5. To find out how many moles of aspartame are in 225g of aspartame: 225g of C14H18N2O5 x (1 mol/294 g/mol) = .765 moles of aspartame. (the 294 is from C14 (14x2) + H18 (18x1) +N2 (2x14) + O5 (5x16) = 294 g/mol)

Sorry about the bad formatting. I still couldn't insert the exponents and other such things. For more information about the mole, visit http://www.visionlearning.com/library/module_viewer.php?mid=53 and for more information about Avogadro, visit http://chemistry.about.com/od/famouschemists/a/avogadro.htm.

Don't forget about the worksheet from last week as well as the one from today. Also, there is a quiz on Wednesday!

~Kaitlyn Y.

The next scribe is Alex K.