Balancing Equations and Chemical Reactions

Today we began our discussion on the balancing of equations. One of the most important things to remember as you balance a chemical equation is the Law of Conservation of Mass, or in other words, what you start with you have to end up with. When dealing with an equation, you may only change the coefficients, NOT the subscripts. Two things you can do when balancing an equation is first: to start simplifying the atoms that only appear once in the formula, and second: to always balance and oxygen last.

There are five types of chemical reactions. The two we began discussing in class today were:

1. Synthesis reactions. These types of reactions occur when two substances (generally elements) combine and form a compound. Reactant + Reactant --> 1 product or A + B --> AB. An example of a synthesis reaction is 2 H2 + O2 --> 2 H2O

2. Decomposition reactions. This kind of reaction occurs when a compound breaks up into the elements or in a few to simpler compounds. 1 reactant --> product + product or AB --> A + B. An example of this type of reaction is 2 H2O --> 2 H2 + O2

In order to further explain these two reactions, Mr. Lieberman showed us two demonstrations.
http://www.youtube.com/watch?v=8ncN7AdROrY

The first video demonstrates a synthesis reaction. Liebs used a balloon filled with hydrogen gas and a candle as the two reactants. When the open flame touched the candle, it reacted violently with the hydrogen inside the balloon, causing the balloon to pop (although it was more of an explosion) and the product of the reaction was water.

The second video shows an example of a decomposition reaction. A large graduated cylinder was filled with hydrogen peroxide and some dish soap. When Liebs added the catalyst, the hydrogen peroxide reacted and foamed over the top of the graduated cylinder. Liebs calls this demonstration "elephant toothpaste."
http://www.youtube.com/watch?v=f_ftXxCwvDw

So that pretty much sums up what we learned in class today. Sorry everybody, my computer won't let me upload/embed the videos, so I just posted the web addresses to each of the videos. Hopefully they work.
Also, the next scribe is................. Nirali P.

Magnesium Oxide Lab

First we went over the pre-lab. For number....
1. Subtract the massof iron (85.65 g) from the mass of the iron oxide produced (118.37)
2. Using molar mass, (O= 16 G) Set up the conversion, which should be the number of grams you found in number one, (Oxygen) multiplied by one mole of oxygen over 16 g
3. Using the same method, you should be able to find the moles of iron (molar mass= 55.9)
4. Next set up the ratio of moles, which should be number of moles of iron over moles of oxygen. You should end up with 3/4, so the empirical formula should then be Fe3O4

In the lab, here is what we did.
Using a crucible, we heated the magnesium oxide with the Bunsen burner.
Every three minutes, you take the lid off and allow oxygen to get in the crucible, sometimes you could even see the metal glow when this would happen. After heating for a while, and cooling, we massed out the magnesium oxide product. The mass should go up, since we were adding oxygen through the experiment.

Empirical Formula

Today in class we went over some of the notes and discussed the empirical formula.
Sample Problem: 71.65% Cl, 24.27% C, 4.07% H
1. assume 100 grams
a. 71.65 g.*(1 mol./35.5 g.)
b. 24.27 g.*(1 mol./12 g.)
c. 4.07 g. *(1 mol./1g.)

2. convert grams to moles
a. 2.02 mol.
b. 2.02 mol.
c. 4.07 mol.

3. calculate ratio
a. 2.02/2.02=1
b. 2.02/2.02=1
c. 4.07/2.02=2.01=about 2

4. empirical formula is CClH2

5. relate empirical formula to molecular formula (molar mass)
(49.5 g.)*2=
98.96 g/mol.

The next scribe will be Bailey Monte

Hydrate Lab!

Today in class...

We started off by going over the quiz on Mole Conversions. For the rest of the class though, we did the Formula of a Hydrate Lab.

We started off by finding the mass of the empty test tube (to subtract it from the other masses later) and then the mass of the hydrate inside of the test tube

Next, we placed the test tube in the stand to heat it.
Then we began to heat the hydrate by using the Bunsen burner...

The blue hydrate eventually turned white, and it lost all of the water within it, the water left over towards the end of the tube can be seen below...


This new anhydrous salt was left, and we then had to find it's mass.


In the post lab, you should be able to use these masses to find moles of water lost during heating, the percent of water in the hydrate, the moles of anhydrous salt, and the ratio of moles of water lost per moles of anhydrous salt.

Percent Composition

Today in class we learned about percent compositions and took our quiz on molar mass and unit conversions.

To start class today we talked about percent composition. Percent composition is used to describe the make up of a compound by mass percentage. It compares the mass of each element present in 1 mole to the total mass of the compound.

In our notes we did this sample problem:

Find the mass percent of each element in C14H20N2SO4.

14(12)+20(1)+2(14)+32+4(16)=312 (The numbers in the parenthesis are rounded versions of the atomic mass)

This format is repeated for the rest of the components. The decimal answer you get is your mass percent.

% Composition of Bubble Gum

Yesterday in class we we began by briefly reviewing for our quiz which was pushed back until today. After our review of unit conversions and molar mass problems we began the Percent Composition of Bubble Gum Lab.

In this lab the purpose was to find out the percent sugar in bubble gum and write an empirical formula for bubble gum. The notes on empirical formulas will be displayed in a later post.

Here is the procedure for this lab:

1. Obtain a piece of bubble gum and its wrapper. Record the mass of the gum and its wrapper.

2. Unwrap the gum (save the wrapper). Chew the bubble gum. While Chewing the gum record the mass of the wrapper.

3. When the gum has lost its sweetness, place the chewed gum in the wrapper and record the mass of the gum and its wrapper.

Note: In this lab our class used Double Bubble chewing gum because as Mr. Lieberman said, "Double Bubble is the worlds greatest gum, for 15 seconds."

The calculations for this lab are due on Friday.

More Unit Conversions

Today we learned about conversions.

We started today looking at some more problems and explanations of molar mass and how to use it. Again molar mass is just the mass of 1 mole in grams. This is found on the periodic table and it is the atomic mass.

Look on the last post for examples of what the equations look like.

One of the most important things we learned today was that EVERYTHING must go through moles.

If it is Grams to Particles,it has to go through moles and if it is Particles to grams then it also has to got through moles.

We then did some learning checks problems like the Nutra-sweet problem. For problems like those to calculate the molar mass of something you take each element and treat it like its own part. For the Nutra- Sweet problem it would look like this

C14H18N2O5

(12x14) + (1x18) +(14x2)+(5x16). this is how you find the molar mass of a compound, then you complete the problem like a simple one.

After that we then had the rest of the class period to work with our partners on the Mole Workshop. Those problems are tricky they just need to be thought out and make sure to show your work. Remember that only one person in your pair needs to turn it in tomorrow but it is best if both partners do the work. I am sorry that I can not post the notes on here, I can not seem to find them on slideshare but if you have any questions just post it here or ask a teacher. This website http://www.chem1.com/acad/webtext/intro/MOL.html could also assist you.

Homework tonight: Finish the Mole Workshop and the other Mole worksheet

Study for the quiz tomorrow over dimensional analysis and unit conversions.

Test is coming up the 28th.

Alright well thats all I have have fun studying.

next scribe is Trevor. have fun

Bobby