Tuesday, November 30, 2010

Percent Composition

Today we learned about Percent Composition.
Percent composition is the percentage by mass of a "species" in a chemical formula.

Before I tell you how to calculate percent composition, check out this video! It should help out a bit.
Percent Composition 

To calculate the percent composition of a compound
  • Calculate the molecular mass (molecular weight, formula mass, formula weight), MM, of the compound
  • Calculate the total mass of each element present in the formula of the compound
  • Calculate the percent compositon (percentage composition): % by weight (mass) of element
        = (total mass of element present ÷ molecular mass) x 100 
The percent composition of a compound tells you which elements are in the compound, and how much of each there is.

After you calculate the percentage of each compound, all of the percentages should add up to 100.

 In this picture, it shows that there is 13% gravel and cobble, 30% salt and clay, and 57% sand in a sediment. This is an example of percent composition. Someone has already found out how much sand, salt, clay, gravel and cobble is in the sediment, and as you can see, they all add up to 100%.

1. Calculate the composition of C2H5OH.
        Step 1:
  • Calculate the Molar Mass of this object.
  • Carbon has a mass of 12.0g. There are 2 atoms of Carbon in this substance, therefore the mass of carbon in this substance is 24.0g/mol.
  • Hydrogen has a mass of 1.0g. There are 6 atoms of Hydrogen in this substance, therefore the mass of hydrogen in this substance is 6.0g/mol.
  • Oxygen has a mass of 16.0g. There is one atom of Oxygen in this substance, therefore the mass of oxygen in this substance is 16.0g/mol. 
  • Add it all together to calculate the Molar Mass: 46g/mol.
       Step 2:
  • % of Carbon: 24.0g / 46.0g. This is the mass of carbon divided by the total Molar Mass. Then multiply it by 100. = 52.2%.
  • % of Hydrogen: 6.0g / 46.0g. Multiplied by 100 = 13%.
  • % of Oxygen: 16.0g / 46.0g. Multiplied by 100 = 34.8%.
As you can see, if you add these three numbers together, it turns out to be 100%.

2. If a compound contains 8.5g of F, 19.0g of Mg, and some amount of Be, and has a total mass of 41, calculate the % composition.

This question is harder because we have to figure out what the amount of Be is.
  • The total MM is 41g. So add 8.5g and 19.0g to get 27.5. 41-27.5 = 13.5g of Be.
  • MM of Fluoride: 8.5g
  • MM of Magnesium: 19.0g
  • MM of Beryllium: 13.5g
  • % of F: 8.5g / 41g x 100 = 20.7%
  • % of Mg: 19.0g / 41g x 100 = 46.3%
  • % of Be: 13.5g / 41g x 100 = 33%
Add it all up and its 100% !
Pretty simple, right?

Just in case it still doesn't make sense, here is a step-by-step tutorial of an "Easy" example of Percent Composition.
Easy Example

If you're just breezing through this, here is a step-by-step tutorial of a "Harder" example of Percent Composition - made by the same guy!
Hard Example

Wednesday, November 24, 2010

The Mole

Sorry for the extremely late post, apparently when I thought I pressed publish post I just pressed preview, and never actually published it... -_-

Anyways, here's the lesson.
Today we learnt about moles!
And no, we aren't talking about the moles that are cute furry little creatures that live in the ground.
And no...we aren't talking about the little bumps on your face (that sometimes sprout hair. Yum).
A mole is a unit, just like m, km, g, L, etc. Pretty simple, right?

Here's a great video on the mole being a unit. Skip to about 0:30 to get the actual song!
A Mole is a Unit
It's pretty catchy if I do say so myself...that's going on my Ipod.

Now who came up with the mole? A man named Avogadro came up with it, and he also came up with a hypothesis and a number.
This man right here is Amedeo Avogadro.
His hypothesis was that two equal volumes of gas, at the same temperature and pressure, contain the same number of molecules.
If they have the same number of particles, the mass ratio is due to the mass of the particles.

Avogadro's Number:
The number of particles in 1 mole of any amount of substance is...

Now you may be thinking. WHAT? How am I supposed to memorize such a big number!?
It's actually a lot easier then you may think!
The mole is extremely important to chemists because it allows them to count atoms.

Here's another video on Mole Day!
Both of these songs I've posted are actually really helpful! Hahah I really like this one..

Too bad we already missed mole day. :'(

Tuesday, November 23, 2010

Even more mole conversions!!!

Last class we worked on the simpler mole conversions
- only going from mole to particle and reverse, and gram to mole and reverse

But!!! today we did harder mole conversions
- we went from particles to mass and reversed
(that means we had to convert through moles) yikes!!

The very first thing you need to know about harder mole conversions is the MOLE MAP!!
it's very helpful!!!

MM= molar mass
in each box all the top amounts are divided by the bottom amounts
With this tool you can go through any conversions using moles!!

Here is a video simplifying mole conversions
Mole conversions made easy :)

Whenever your doing Mole conversions NEVER forget AVOGADRO'S number
(6.022x 10 ^23)

This is what we learned today and next class we have a quiz!!
If you want to try some practice questions here is a link to ms. Chens website
extra worksheets on ms chens website

Saturday, November 20, 2010

Mole Conversions

So, we have now started our new unit on the mole! yay!

Before we start, make sure you are comfortable calculating molar mass.

The first type of conversion is particles --> moles
Remember: 6.022 x10^23 particles/mole

Say you want to convert 3.01 x 10 ^24 particles of carbon to moles.

As with unit conversionis, you need the same units in the beginning of the equation, and on the bottom, so they can cancel eachother.

3.01 x 10^24 particles of carbon x 1 mole
                                                      6.022 x 10^23 particles

If you do the math, that means that there are 5.00 moles of carbon! Make sure you round off to the correct number of significant digits. The least number of sig figs in the equation is 3, therefore the answer should also have 3 sig figs.

Now, converting moles to molecules.

Lets try converting 0.75 moles of CO2 to atoms of oxygen.
Follow the exact same rules as given in the example above. Always follow those rules, these are just different units we are using.

0.75 moles x 6.022 x10^23 molecules
                     1 molecule
Once you multiply, you will calculate that the answer is 4.5 x 10^23 molecules CO2

We can also do conversions between moles and grams.
Lets calculate the grams of 2.04 moles of carbon
Use molar mass! The molar mass of Carbon is 12.0 grams/mol
Now lets do the math!
2.04 moles x 12.0 g
                    1 mole = 24.5 grams of carbon
For compounds, to get the molar mass, just add up all the masses of each atom of each element.

Last but not least, grams to moles!

How about we convert 3.45 grams of carbon back to moles

Remember everything we have done so far! Find the mass of Carbon and plug it in. Then just follow the steps and do the math:

3.45 g x 1 mole
              12 grams                      and all this math = 0.288 moles of carbon

There you have it! How to convert from anything to everything! Hope this helps.

Wednesday, November 17, 2010

Quiz Day

This class we had a quiz/test on experimental error and scientific notation.
It was pretty long.
Hopefully everyone did okay on it!

...Well I don't really have much to say.

Here's a review video on accuracy and precision.
Accuracy and precision

Since we didn't learn anything this class, here's a cute video of some guinea pigs eating corn.

K I'm done with being a loser now..

Saturday, November 6, 2010

Accuracy/Precision and Measurement/Uncertainty

Accuracy and Precision

The textbook definitions (which are always useful) for accuracy and precision are:

-Precision is how reproducible a measurement is compared to other similar measurements. So how hard or easy is it to reproduce that measurement?

-Accuracy is how close the measurement (or average measurement) comes to the accepted or real value.

A good example of this would be a game we played in elementary school. We would play frisbee-golf where you have to throw the frisbee into the hulahoop (like a golf hole) and play the entire playground exactly like it is a golf course. So if you think about the hulahoop being the accepted or real value, accuracy is how close you can come to it when throwing the frisbee.

Measurement and Uncertainty

Every measure you will take or see is only an estimate, which has some degree of uncertainty.

Absolute uncertainty:

The uncertainty expressed in the units of measurement. There are 2 methods of doing this. Make sure to try both!

Method 1: Make at least 3 measurements. Calculate the difference between the average and the lowest or highest reasonable measurement. Be sure to include a +- sign at the beginning of the answer (I don't have that symbol on my computer)

Method 2: When making a measurement, always measure to the best precision you can (remember, precision is reproducibility!) Therefore, you should estimate to a fraction 0.1 of the smallest segment on the instrument scale.

Now, time for exactions! Here is how to calculate relative uncertainty:

Relative Uncertainty = absolute uncertainty / estimated measurement

Relative Uncertainty can be expressed in percentages or using significant figures.

Here is a video for your viewing pleasure, the guy really seems to know his chemistry stuff!

Super Cool Video

Thursday, November 4, 2010

Lab Day!

Today we did a Lab on the thickness of aluminum foil!!

I know it doesn't sound very interesting but it was actually pretty interesting how to do it!

  • We took 3 almost square pieces of foil.
  • We measured all for sides and came up with an average lengh and width of 1 piece of foil.
  • We got to use a centigram balance to find the mass of each piece of foil

centigram balance ( ours were much older)

With this data we collected we had to use formulas for density, and volume

Volume = (length)(width)(height)

*height is our thickness that we were originally trying to find

Here is a video on how to find the thickness of aluminum foil using these formulas=

Next class we have a quiz on this lab!!