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.