Percent Composition

% Composition = mass of element     * 100%      
                         mass of compound

Percent Composition is how much an element's or molecule's mass takes up in a compound. In other words, calculating how much of a compound is made up of a certain molecule or atom.

Now for the fun part...counting the load of cra— I mean the mass of each element in a compound

Eg. C₆H₁₂O₆

Step 1 Calculate the mass of each element (the C, H, and O)

C = 12.0 X 6(as you can see from the compound there are 6 Carbons) = 72.0 grams of Carbon
H = 1.0 X 12 = 12.0 grams of Hydrogen
O = 16.0 X 6 = 96.0 grams of Oxygen

Now add them up together and you should get 180.0 grams.

Step 2 PLUG IT!

Take the weight of each element and stick it into the formula...

% composition = 72.0/180.0 = 40% Carbon

%c composition = 12.0/180.0 = 6.7% Hydrogen

% composition = 96.0/180.0 = 53.3% Oxygen

THAT IS ALL.......JOKES ON YOU. Now I will guide you with using the crap from the top to apply it onto another pile of said crap called Empirical Formula

Step 1 always assume its 100.0 grams of material (Just do it).
Step 2 Convert those percentages we derived several seconds ago into grams.
Step 3 Now convert your crap into moles.......for the visual viewers....

                              =>    =>

Step 4 divide the brown spot (MOLES) by the smallest value

Eg. C = 48.0%
      H = 9.33%
      O = 42.7%  (remember these should add up to 100%)

48.0 X 1 mole/12.0g= 4.0.....
9.33 X 1 mole/1.0g = 9.333.....
42.7 X 1 mole/16.0g= 2.66875...
 
WE'RE ALMOST DONE THE CRAP.
Now take all 3 of those measurements and divide each one by the smallest measurement
(In this case: 2.66875)
you should get C = 1.5
                     H = 3.5
                     O = 1
EEW look at those ugly numbers.....lets make them prettier by scaling all of them, whatever you multiply to 1 you have to do to all 3...... (hint hint multiply everything by 2)

Your attention please, the moment you have been waiting for....(drum roll)  THE ANSWER!! C3H7O2

That's all folks!!!

Let the kids (all of you) decide: would you rather prefer to learn Alchemy or Chemistry?

To reduce, or not to reduce, that is the question...

Hello dear readers, thank you for visiting again to read our newest posts on Chemistry.

First off, you might think that today is one of those math-brain-attack days where your brain automatically  freezes after viewing our blog post. It's alright, we get that kind of feedback all the time, so you're not alone on this. Today's lesson will definitely involve less math in comparison to mole conversion (as far as I know, anyway...)

On first glance, it would seem that the following topics are focused on math, as opposed to chemistry. This is only partially true. Let's say in math, if you don't reduce a fraction on a test, you will lose marks unless your teacher is an incompetent nincompoop. However, on a chemistry test, it could be a completely different scenario. You may possibly even earn marks for not reducing the fraction...in a sense.

Empirical formula
All compounds have an empirical formula. What this means is that the compound is written with the lowest ratio of atoms in the compound. For example, we know that Glucose is C₆H₁₂O₆.
The ratio of its atoms are 6:12:6. In lowest terms, this would be 1:2:1, therefore the empirical formula of Glucose is CH₂O.



On the other hand, Molecular Formula is then the "not-in-lowest-terms" form of formula writing for compounds. And this form of writing states the exact number of each atoms contained in a certain molecule.

Let's try having a brain freeze now,


For example:
Given the molecular formula for butane is C4H10
What is the empirical formula of such?

[First, you look for the greatest common factor/divisor of 4 and 10, which is 2.]
[Then , divide both numbers by 2, you will get 2 C's and 5 H's]
[Ta Da]
[You answer should be: C2H5]

These formulas below you may want to must keep them closely in mind when a test date comes along,

Molecular Formula = Empirical Formula  x a Whole Number 
Molecular Formula Mass = Empirical Formula Mass x a Whole Number
Mass of one mole = Empirical formula Mass (g) x a Whole Number

Moving on ... don't worry you will have a 'brain melt' sooner than you think.

So,  how do you determine the empirical formula given the masses of each atoms???

The Million Dollar Question here today is: Determine the empirical formula of Ca, O, and H given 13.5g of Ca, 10.8g of O, and 0.675g of H. {Now, get out some scrap paper, folks and give this question a try. Stay tune for full-solution.}

1) Make friends with mole (unless you want your arm to be bitten off...the decision is entirely up to you)
                      1 mol Ca       
13.5g Ca x     40.1 g Ca         = 0.337 mol Ca

                      1 mol O          
10.8g O  x     16.0 g O          = 0.675 mol O

                      1 mol H       
0.667g H x     1.0 g H           = 0.667 mol H

2) Divide each value of mole by the smallest number of moles calculated when friend-making with moles.

 0.337     mol Ca  => 1.00
 0.337

 0.675     mol  O  => 2.00
 0.337

 0.667     mol H   => 1.98 ≐ 2.00
 0.337

3) Brain Melt time...

The final answer to the 'Million Dollars Question' is CaO2H2 and you can also present your answer in this form: Ca(OH)2 and the million dollars will still be awarded to you!
Are you the ultimate winner of the million dollars tonight?

Mole Conversion, Part Deux

The Hall of the Mole King
It is your third night in the Mole King's prison. Three days ago, you unintentionally stumbled and intruded upon his lair. It was but the moles' duty to put you through the tests.

On the first day, they taught you of their sacred number, 6.022*10²³. On the second day, they built upon your known knowledge of chemistry, telling you of the subtle yet absolute connection between the elements, the mole, and their weight.

Earlier today, His Blindliness the Mole King presented you with the ultimate challenge of their race, conversions between the number of atoms, to weight, and back. He handed to you two objects. The first, a faint green emerald, clear as crystal, smooth as silk; the second, a round blue sapphire, as rich and dark as the ocean. These two objects, as the Mole King told you, were called Grues and Bleggs, respectively.

Grues, the Mole King told you, are a pure substance. Even at the atomic level, they are Grues, and every single Grue atom weighed 27.2u. Bleggs were similar in that they were also a pure substance, yet each Blegg atom weighed 19.3u.
The Mole King challenged you to find, by sunset, the approximate number of Grue atoms that resided within the stone, and the weight of the Blegg. He told you a mere two pieces of information in addition to what you learned the last two nights: that the Grue weighed 21.3 grams, and that 4.05*10²³ Blegg atoms forms the dark sapphire in your hand. If, by sundown, you have not yet come up with the answer, the mole kingdom will devour you. Literally. But of course, you already know how to derive the answer, and silently smirk as the Mole King leaves.

You know that you must first convert to moles, and then the desired quantity. You decide to find the answer to the Grue first. As Grue atoms weigh 27.2u, that means a mole of Grue atoms weighs 27.2g. Through this, you discover that the chunk of Grue is (21.3/27.2)=0.78 mol. And, by multiplying that with the moles' holy number, you obtain the answer, 4.71*10²³ atoms.

By similar reasoning, you know that a mole of Blegg weighs 19.3g. By dividing the number of atoms with the moles' glorious constant, you are aware that you hold, in your hand, [(4.05*10²³)/(6.022*10²³)] = 0.673 moles. Again, by simply multiplying this by the weight of one mole of Blegg, you find that the Blegg stone weighs 13.0g.

As the Mole King returned, you confidently tell him the answer. A look of shock registers, surprised that you could beat the challenge. Yet as he promised, he released you, back to where you came. But you yourself know, that never will you forget the lesson of the moles.

Mole Conversion, Part UNO!!!

Hi everyone! Today, we are going to talk about an interesting, exciting, fascinating, amazing, wonderful, fantastic topic; I'm sure you've got it! It's MOLE CONVERSION!!! 

It is very important that you make good friends with the "moles" and be able to convert them correctly without being bitten in the arm!

Anyways, enough of chatting. Let's get started and I promise you that you'll be excellent friends with the "moles" at the end of this blog.



STEPS TO BEING FRIENDS WITH MOLES

【Particle/Atom/Formula unit → Mole】

Whenever we are given a bunch of atoms and we need to find the number of moles,
we just need to simply divide the number of atoms by Avogadro's number.

Example: How many moles of Oxygen atoms are in 4.56 x 10¹² O atoms?
First: we write down the information we know
       4.56 x 10¹² O atoms

Secondly: we divide the # of atoms by Avogadro's number (6.022 x 10²³)
       4.56 x 10¹² O atoms  x      1 mol O    
                                6.022 x 10²³

Thirdly: we punch in the numbers into our best friend calculator
       4.56 x 10¹² O atoms  x      1 mol O      = 7.572235138....x10^-12
                                 6.022 x 10²³

Lastly: we record the answer with correct number of sig figs
       since "4.56" has the least # of sig figs, our final answer is 7.57 x 10^-12mol O

You are getting the hang of it right?! That's good, because an amazing time-travelling trip has been scheduled for ALL of you. Wait, we're on time...please a take a seat in the time-travelling shuttle...

【Mole → Particle/Atom/Formula unit】

Now, let's do a reversed version of what we just did above. When we are given a number of moles to find how many atoms are present, we'll need to multiply the # of moles by Avogadro's number.

Example: How many molecules are present in 3.2 mol of HCl?

3.2 mol HCl x  6.022 x ²³     = 1.9 x 10²⁴ molecules HCl
               1 mol HCl

However, if the question wants us to find a specific TYPE of atoms in a compound, we'll need to do one more step.

Example: How many molecules of Oxygen are present in 5.6 mol of SO₄?

Since the question wants us to only find the number of oxygen molecules, but not including the sulphurs, so we'll have to multiply the total number of mols by the number of atoms per molecule.

                                  ↓There are 4 oxygen atoms per 1 SO₄ molecule
5.6 mol SO₄ x 6.022 x 10²³ x    4 atoms O     = 1.3 x 10²⁵ atoms O
             1 mol SO₄        1 molecular SO₄

We are almost there. Let's just take a quick break from this...FUN FACT TIME: did you know that mole does not only can refer to the chemistry unit mole, animal mole (digging those ugly holes in the ground making people to fall inside their "traps"), it can  also refer to as a type of sauce(eatable ones, of course)!!! Mole sauce apparently is a sauce used in the Mexican cuisine and is essentially just chili peppers gravy. Nothing too fancy.

Okay, enough laughters, getting back to the real deal...

【Mole → Grams】

Last blog post, we've learnt about molar mass, molecular mass and formula mass. It's time for us to put those numbers in use! When we are ask to convert mol into grams, we can multiply the molar mass of the elements by # of moles.
                             
Example: What is the mass in grams of 4.78 moles of Iron?

Atomic mass of Fe = 55.8 u
molar mass of Fe = 55.8 g/mol

4.78 mol Fe x 55.8 g/mol  = 267 g Fe
                1 mol Fe


【Grams → Mole】


After we've learnt how to convert mole to grams, it'll be easy for you to do the reverse. To convert grams to mole, you can divide the # of moles by the molar mass of elements.

Example: How many moles are in 18.66 grams of Carbon?

Atomic mass of C = 12.0 u
Molar mass of C = 12.0 g/mol

18.66 g C x 1 mol  C  = 1.56 mol C
            12.0 g C

Awesome! That's pretty much wraps up mole conversion......part one!!! But anyways, here's some jokes about moles and Avogadro to light up you day....as usual :D

Q: What did Avogadro teach his students in math class?
A: Moletiplication (multiplication)

Q: What kind of fruit did Avogadro eat in the summer?
A: Watermolens

Q: What is Avogadro's favorite kind of music?
A: Rock 'N' Mole
Q: What did Avogadro invent for his wife to use as a night cream?
A: Oil of Molay


At last, here's a present from Pinch of KCN to everybody, it's a game about mole conversions!!! *clap clap clap*
It is a very short RPG-styled game which contain 10 unique questions about mole conversions.The file might seems a bit large, so please allow 3-5 mins to download.
※This game is completely made by us, so don't worry, it will NOT spread scary viruses to your computer※


Now please download this game and allow an exceptionally adorable mole to enhance your chemistry learning experience. Look how cute he looks with those fake glasses I accessorized him with:)

Full solution to questions in the game:

https://docs.google.com/document/d/1IQArFYrIubcUVZ3JZ_hzM2athoYRSj4vf7HN4Kt4w3k/edit

Game Download:

[WinRar is required to open the game (Compressed file)]

http://www.mediafire.com/?ay6f6cv08jvdc9i

OR

(Click "download now"→"Regular download"→Wait for 20 secs)

http://www.4shared.com/file/rBDYXz3n/Mole_game_-_by_pinch_of_kcn.html

Simple steps to open this game:
1. Download the file from one of the links provided above
2. Extract the file with WinRar
2. Click this icon:

5. Open the game:

The Mole, Mass, Amedeo Avogadro

Recap

For most of our high school life, we have been using typical units of measurement such as grams, and occasionally touching on atomic weight. In both chemistry and physics, all atoms have a certain weight to them. The units to these values are called "Atomic Measurement Units", also referred to as "amu"s, "u"s, and "daltons".

Amedeo Avogadro

Amedeo Avacado Avogadro was an 18th century, Italian scientist. He is most famous for his contributions to molecular theory, especially for what is known today as "Avogadro's Law" and "Avogadro's Constant". 

Avogadro's Law, Constant, and the Mole

Avogadro's Law states, that under the same conditions of temperature, volume, and pressure, two different samples of gas will contain the same number of particles. In essence, this means that the number of molecules or atoms of a gas is independent of the size and mass of the gas. 

Avogadro's Constant is currently defined by the International Bureau of Weights and Measures as "the number of Carbon-12 atoms in 12 grams of mass". Currently, this number is approximately 6.022*10²³. To put this into perspective, it is a number about a thirty thousand times larger than today's average computer can hold in a single variable. In base 2, it is approximately 2⁷⁹. The average computer today can store 2⁶⁴bits in a single variable. This number is 100 times the number of stars in the observable universe. If this many pieces of typical paper were stacked on top of each other, it would be tall enough to reach the sun and back (assuming of course the sun doesn't instant incinerate them all), 250 billion times. This astronomically large number, Avogadro's Constant, is the mole unit. 

Mass

In chemistry, there are three kinds of mass describing pure substances. The formula mass, and molecular mass are the mass of ionic and covalent compounds, respectively. Molar mass, on the other hand is the mass of one mole of a pure substance.

For example: Potassium Cyanide, or KCN has 3 elements in it. They are Potassium, Carbon, and Nitrogen.

Their atomic weights are:

1 x Potassium: 39.1u

1 x Carbon: 12.0u

1 x Nitrogen: 14.0u

The sum of this is the formula mass of Potassium Cyanide, 65.9u. This translates to 65.9grams/mol for Potassium Cyanide.

Another example: Water or H₂O, containing Hydrogen and Oxygen

Atomic weights are:

2 x Hydrogen: 2 x 1.0u

1 x Oxygen: 16u

The sum of this is the molecular mass of water, 18.0u. This translates to 18.0grams/mol for water.

Conclusion

The Mole is important for many branches of science, not just for chemistry. This constant number allows for scientists to determine number of atoms and molecules in a given sample with relative certainty.

For us students, it just means that unit conversions are going to become a lot messier.

Here's an amusing video to lighten up your day. Though it has nothing to do with today's topic, except the KEY word: Mole! Enjoy.

Lab Day and ALL the Fun Stuff

Once again, today is a Lab Day....on aluminum foil. Today's task for the lab is to measure the thickness of aluminum foil. It may sound utterly simple to you at the moment, but when you think about it....how are you suppose to measure the thickness of a sheet of aluminum foil using just a RULER? Well you can....follow along since you must know the steps before the quiz next class.

So first, we take 3 square-looking sheets of aluminum foil one at a time  that were said to have a dimension of 15cm x 15cm (we later found out they don't.) and measure the length and the width of each of these sheet using a ruler. Then record the data (to the corect number of sig figs of course, you must know that by now!) into the chart that we copied out from the lab book.

Next, weigh the three pieces separately on a centigram and jot that down as well.

The end...we have completely all the hands-on work for this lab. You might want to ask "That's it???" Well, there's several numerical values that we need to find out still.
Here are some of the basic formulas that you might want to shove into your brain,

Volume (rectangular prism) = Length x Width x Height (V=LWH) –⑴
Density = Mass ÷ Volume –⑵

And the next two are derived from the density one above for those of you who are not proficient at math like we are.

Mass = Densiy x Volume –⑶
Volume = Mass ÷ Density –⑷

And density of aluminum is given to be 2.70g/cm^3.

One more fun fact:
You can ACTUALLY calcualate the thickness of a sheet of any material without knowing it's density. Just a ruler can do your job. Stay tuned...we'll do it the proper way for now. And our little trick can come along later

Okay, since the mass of of the aluminum foil is known, and the density is given, we can find out the volume using formula ⑷. And since the length, the width, and the volume of that piece of aluminum foil is known, simply use formula to determine the height, thus the thickness of that sheet of aluminum foil.

E.g. The measured values for the length and width of the aluminum sheet are 15.76cm and 15.15cm respectively. The mass is 0.96g. (And the density of aluminum is 2.70g/cm^3)

From  ⑷, V= 0.96g ÷ 2.70g/cm^3 = 0.355556cm^3 (keep more decimals that you actually need to guarantee the accuracy of your final answer OR use the answer in your calculator in the next step without any rounding!)

***1cm^3 = 1mL***

From ⑴, V =3.55556cm^3 = 15.76cm x 15.15cm x H
               H = 3.55556 ÷ (15.16 x 15.15) =1.5x10^-3 cm

∴ Thickness = Height = 1.5 x 10^-3 cm

Simple huh!!!


One other thing we need to calculate is the experimental error (the value is to be expressed in percentage):

The known value, aka the accepted value, in this lab is 1.55x10^-3...don't worry about this number, the omniscient Ms.Chen will give you that if you need to calculate such on a quiz/test. Just plug in the numbers, nothing fancy.

E.g. using the previously calculated thickness (1.5 x 10^-3 cm)

Experimental error =  |1.5 x 10^-3 cm - 1.55x10^-3|   x 100% = 3.2%
                                                1.55x10^-3     

Revealing a dirty trick:
Now, I can sum up the process in just three words: FOLD (multiple times), MEASURE, then DIVIDE (the number of times you've folded)! Quite simple huh! One thing you must remember: NEVER use this method on a test. NEVER!! Not because this procedure is incredibly flawed for high schoolers, rather it's perfectly fine practicing it in your everyday life. It's just...we won't actually have a piece of that with us for tests.

Alright then, I think you're pretty prepared for the quiz next class. Just review the materials from some of our recent posts on density and sig fig then I'm sure you will ace it!

Hurray!!! Video game time...
 

Graphing In Excel

Excel is a beautiful piece of software. Since 1982, it has risen in popularity to be one of the most widely used spreadsheet applications in the world. One of its many functions is creating graphs. It can create any type of graph one can imagine, given the right data.

For our assignment, we were asked to create the graph of Volume vs. Temperature graph of water. To avoid being verbose, here's a video about Microsoft Excel 2007's best fit line or trend line function.