Tuesday, October 26th

Today, Chemistry class started off just like any other day by reviewing the blog that was written the night before. Shidan had written about finding the empirical formula when giving the molar mass of a compound and percents of specific elements in a compound. Mr. H reminded us of what homework was due upcoming. That included a Webassign that was due on Thursday and the fact that we had a upcoming test on Thursday.

We began the day with work in our packet with problem 3 on page 13. This problem was a two part problem asking to name two different compounds. One compound was an oxide with 50.0% Sulfur and the other was another oxide with 40.0% Sulfur. The percents were by mass. First, we began with the compound that contained 50.0% Sulfur. To begin the problem Mr. H told us that we could assume that there were 100g of the compound. So with that information were able to conclude that there were 50g of of Sulfur in the substance. Next, we had to find the number of moles in Sulfur and to do that the number of grams of Sulfur in the substance was divided by the molar mass of Sulfur. This equaled 1.559 mols. Then, we had to find the number of moles of oxygen in 50g of Oxygen. It was 50g because there was 100g of the substance and 50g of sulfur, so this was the left over amount. The total amount of grams of Oxygen in the substance was then divided by the molar mass of Oxygen which was 16.00 rounded. This equation equaled 3.125 mols. The two equations for calculation mols are below on the blog and in the picture.

Mols of Sulfur: 50g x (1 mol/32.07g)=1.559 mols.
Mols of Oxygen: 50g x (1 mol/16.00g)=3.125 mols.

The next step was finding the ratio of oxygen atoms to sulfur atoms which was found by the equation y/x. Y being oxygen and X being sulfur. The ratio in this compound approximately when rounded 2:1. For every two oxygen atoms there were one sulfur atom. The empirical formula was SO2. The name of the compound was sulfur dioxide, which was a topic that we covered in Unit 2.

After completing the first part of the problem, Mr. H told us to the second part on our own. Following the same steps as above it was assumed that there were 100g of the substance, and because 40% was Sulfur it there were 40g of Sulfur and 60g of Oxygen. Next the number of mols was calculated just like above by the following equations.

Mols of Sulfur: 40.0g x (1 mol/32.07g)=1.247 mols.
Mols of Oxygen: 60.0g x (1 mol/16.00g)=3.75 mols.

Next, the same formula for finding the ratio was used as previously to conclude that the ratio of Oxygen atoms to Sulfur atoms was approximately 3.00 when rounded. We were able to conclude that the empirical formula was SO3. Again, we had to name the molecular compund and it was Sulfur Trioxide.

Picture of the problem that we did in class on packet page 13.
After the problem, we switched to a second lab that was an extension of the lab that we had done the previous day. It was focused around making specific calculations with the information that we are given. Since this lab was very dangerous, Mr. H did it himself as the class watched. Mr. H performed a reaction where before the reaction there were 91.46g of Sucrose with the beaker and after there were 67.28g of of Carbon with the beaker. The third piece of data that we had was the mass of the beaker which was 50.66g. The three calculations that we had to make where the mass of Sucrose, Carbon and H20. The mass of Sucrose was found by taking the subtracting the mass of the sucrose with the beaker and subtracting the mass of the beaker. There were 40.80 g. The mass of carbon was the mass of carbon after the reaction which was 67.28g was subtracted by the mass of the beaker which meant that there were 16.62g of Carbon in Sucrose. The mass of H20 was the mass of sucrose subtracted by the mass of carbon. The mass of H20 was 24.18g. Then, the percents of Carbon and H20 in sucrose was found. The equations follows.

%C: 16.62g/40.80 x 100=40.7%
%H20: 100%-40.7%=59.3%

This extension lab that we completed perfectly complimented the lab that we had worked on the previous day which was Lab MR3. With the help of learning how to do the new calculations we were able to successfully complete our individual labs. The demonstration for the lab follows and the password to gain access to the video is gbs.
http://www.dropshots.com/chemistryclassroom#date/2010-10-25/20:33:04

Working on the labs in our groups brought us to the end of class. Before class ended, Mr. H repeated what the homework was for the night.

Tuesday, October 19

Converting to Atoms, Moles, and other measurements
After going over last nights blog and talking to us about national mole week, Mr. Henderson began to focus in on the big idea for the day. Just as we have been doing for the past week or so, we started off by converting one unit to another. However, today we started with slightly more difficult conversion problems. Of course, I'm talking about converting moles to atoms, atoms to moles, atoms to grams, grams to moles and so on. Before I begin to go over pages 6 and 7 (the pages we worked on today), it is important to have a basic understand for some of the quantities we will be addressing.

First off, the one you will be hearing most about in this blog is the mole. 1 mole is 6.022x10^23 of something. This number was created to help us convert one atom to a useful unit, like the gram. This leads into my next point, the mass of atoms. On our periodic table, the mass of an element is known as XX amu's. This number can also be used to obtain the elements grams per mole. How you do this is actually very very very very simple. Take the atomic mass of the element and drop it into the g/mol unit! So, for example, oxygen's atomic mass is about 16 amu's. Therefore, oxygen is also 16g/mol. This is used so that we can speak of the units of each element in more relative terms, such as grams or grams per mole.

Now lets begin going over the packet pages starting with pg.6. The top of page 6 asks how many donuts or in 2 dozen donuts. This question is there to help us understand that a mole is nothing more than a way to help us speak more easily about atoms, like a baker uses the dozen to talk about the number of donuts he is baking. The next question asks how many atoms are in 2 moles of water. Since we know that 1 mole of atoms is 6.022x10^23, we automatically know that 2 moles of water would be 12.044^23 atoms, or 2 x 1 mole of atoms. Questions 4-6 are very similar and are used to help us understand just how large a mole is. The first question asks how much mass would 1 mole of .62kg basketballs have? So, as we would start of all conversions, we begin with the unit we are given which is 1 mole of basketballs, in this case. Now, our next step would be to put 6.022x10^23 basketballs / 1 mole of basketballs. Our final calculation would be to multiply the mass of a basketball / 1 basketball. We are now left with (6.022x10^23)x.62 kg. After doing these calculations, we get the answer......... 3.7kg x 10^23! If you still don't understand just how massive this is, the mass of the of the moon is also given. It is only 7.4x10^22. 1 mole of basketballs is 5 times as massive as the very moon that orbits our Earth!

(Yup, that moon)

Next, on Questions 7-9, we began to use the mole for what it is truly meant for; Converting atoms to more useful units of measurement. Our first question is How many atoms of lead(Pb) are in 5.0x10^-4 grams of lead? Now lets begin our conversion with 5.0x10^-4 g of Pb. Because we are looking to obtain a number of atoms as our answer, we will start with the next corresponding conversion factor, the mole. 1mole/207.2 g of Pb should be your next step because this is the number of grams that would result from 1 mole of lead atoms. Also, as we discussed earlier, it is the same amount as the atomic mass of lead on the periodic table. Finally, we will multiply 6.022x10^23 atoms/ 1 mole to cancel out our moles and get an answer. Although 0.0005 grams of lead may seem like a small amount relative to our everyday lives, It is still 1.45 x 10^18 atoms of lead! A massive number of atoms for a minuscule amount of grams.


The answer for pages 6 are as follows:

#1)24 donuts
#2)12.044x10^23 molecules
#3)10 (every power added to 10 multiplies the given amount by 10, in this case, moles)
#4)3.7 x 10^23kg
#5)3.8 x 10^ 14 years
#6)3.6 x 10^26 cubic centimeters
#7)1,45 x 10^18 atoms
#8)9.0 x 10^18 atoms
#9) 2.1 x 10^-22 grams (that's 0.000000000000000000000021 grams!)

As for pg 7, we just began to discuss the molar mass of compounds and the way to do this is probably the way you would have guessed. Just add together the mass of each elements in the compound. To obtain the molar mass of Barium Chloride (BaCl2), add the mass of 1 mole of barium atoms(137.33 g/mol) to the mass of 2 moles of chloride atoms (2 x 35.34 g/mol). Your answer should add up to 208.23 g/mol. We will continue on to finish this sheet tomorrow but all the answers we have gotten in class so far are as follows.

#1) 32 g/mol
#2) 58.45 g/mol
#2.5) H20= 18 g/mol (IMPORTANT FOR LAB MR1.)
#4) 208.23 g/mol

With the final 10 minutes of class, Mr. Henderson let us get back to our lab groups and finish up the lab with the knowledge that we had gained from question #2.5 on pg.7. Now we know that water is about 18 g/mol. Next, each group would have to find the amount of moles that was assigned to your group. This could be done by dividing the amount of molecules assigned to each group and dividing it by 6.022x 10^23 Luckily, all our group were assigned a number of molecules that could easily be divided by a mole. Finally, all that had to be done to solve this seemingly impossible conversion was to multiply the number of moles your group had by the grams per mole (g/mol) of water, which we know is 18 g/mol. Now you would be left with xx grams of water, and since we know water's density is 1 gram per milliliter, the last step was to change grams into mL and present your findings to Mr.Henderson.

I hope this blog helped clear up any confusion about molar mass and atomic conversions. Today was definitely a jam packed day. Good Luck!