¿How Concentrated are Molars?

I understand that the long-awaited winter break is just around the corner, and many of you, even us, the dedicated chemistry bloggers, are starting to feel a little unwilling to blog. jkjk. Anyhow, today we will be presenting you a new but yet old idea.

Molar concentration (aka. molarity) may sound like it has something to do with your molar teeth. Put it this way: No. And please, spare us a minute to explain this simple concept to you. If you have been following tightly to our blog, you must have encountered a post about two months ago on Density. The idea of molar concentration is very similar to the concept of density except of its units...sort of...

Molar Concentration can be defined as the concentration measured by the number of moles of solute per litre of solution.

Forgot what a solute is? Let us tell you. A solute is the chemical that's being dissolved, thus having a smaller quantity.

What about a solution? It is the larger quantity which is also a homogeneous mixture.

The must-know formula is:


                               moles of solute (mol)                                          mol  
Molarity =            volume of solution (L)             OR    M=      L                ---- (1)


From here, you may derive 2 more formulae to assist you when finding moles of solute or volume of solution given that the other two terms are known.


mol = M x L  ---- (2)      and       L= mol/ M    ------ (3)

E.g. How many moles of CaCl2 are dissolved in 225 mL of 0.0350 mole/L CaCl2?

Step 1: Since, you are given the the values of the molarity and the volume of CaCl2, and are finding the value of moles; your eyes should be aiming at formula (2).

Step 2: Substitute the numerical values into the formula, be sure that all the units are correct. In this case 225mL should convert to L before substitution.

 mol  = M x L
        = (0.0350 mole/L) x (0.225L)
        = 0.007875mol CaCl2

"Mr. PinchofKCN, I'm finished. The question you gave me is so easy-peasy-lemon-squeezy. Give something at least a little bit more challenging, please."

The answer should be rounded to the correct number of sig fig...in this case 3 sig fig. The correct answer should be 0.00788!

I know all of you are thinking about the same thing as SpongeBob. And Mr.PinchofKCN's generous answer to you is 'you may'!!! Since today is the last class before 2012 (and I'm sure you all know what's going to happen in 2012). So, most likely, you can be excused for the rest of your life. No more chemistry!!! Yay to you, nay to us!


Do you believe in eschatology? (Well, we shall wait and let the event reveals itself:)

Lab 4C: Naming a Hydrate

What is a Hydrate?
A "hydrate" is a name given to a compound, formed by the addition of water to the molecule. When writing the formula of a hydrate, first write the original molecule, and attach a "· nH₂O", where "n" is the number of water molecules present in the hydrate. For example, Cobalt(II) chloride hexahydrate is CoCl₂·6H₂O. 


Lab Summary
In our lab over the last two classes, we wasted productively spent most of the class heating and waiting for a crucible to cool. In our first class, we did this process three (for some groups, four) times. The first time, we heated it, using Bunsen burner, ring stand, and pipestem triangle, to eliminate any water already present in the crucible. Between heatings, we also measured, using a centigram, the weight of the crucible. At the first weighing, we discovered the weight of our crucibles. Afterwards, we used another weighing and the first weighing to derive the weight of the hydrate. Finally, after heating the crucible to evaporate the water, we weighed it again, and derived the amount of water that was evaporated through heating.

Results

After being generously given the weight of one mole of the anhydrous compound (the compound to which water was attached in the hydrate) and the formula, we could correctly produce the empirical formula of the original hydrate we were given.

Afterwards, we added a drop of water to the anhydrous salt, and observed it bond, forming a hydrate once again, thus wasting all of our original patience hard work.

To end this sarcastic blog post, below is a picture of some crazy dude burning a hydrate (that contains methane) in his hands. Compliments of Google Image Search. Until next time, stay safe, don't eat in the classroom, and wash your hands. No, seriously.

Empirical Formula of Organic Compounds

∴

Good evening to our fellow blog readers! Does your brain hurts after reading our last five posts? Or have you had a nightmare about the Mole King? Well, here's the "good" news for you! Today's lesson will be involving with even more math! HURRAY!

First of all, seems like it has been over half a year since we had chemistry in Science 10, so I don't expect anyone of you would still remember what an organic compound is. But that's totally fine, because I'm going to give you a brief recap right now:

An organic compound is any molecular compounds that contains carbon, usually with the support of oxygen and hydrogen. Two common examples are methane(CH₄) and glucose (C₆H₁₂O₆). However, some carbon containing compounds such as CO₂ and carbonate does not considered as organic compound.

← Caffeine (C₈H₁₀N₄O₂) is a nitrogenous organic compound



Empirical Formula of Organic Compounds

Since we have been receiving positive feedback on the Million Dollar Question from last time, we decided to hold another brain melt section called the "Double Dare"!!!

Our thousand points question for today is:

What is the empirical formula of a compound that when a 0.952 gram sample is burned and produces 1.35g of CO₂ and 0.826g of H₂O?


Step 1. Find out the moles of CO₂ and H₂O
(Tips: grams to mole → divide the # of moles by the molar mass of elements)

mol CO2 = 1.35g CO₂  x  1 mol CO₂   = 0.0307 mol
                                                          44.0g CO₂

mol H2O = 0.826g H₂O   x  1 mol H₂O   = 0.04589 mol
                                            18.0gH₂O

∴During the burning, 0.0307mol carbon dioxide and 0.04589mol water were produced.


Step 2. Find out the moles of Carbon and Hydrogen
(Tips: mole of compound to mole of element → multiply the total number of mols by the number of atoms per molecule)

C = 0.0307mol  x  1C   = 0.0307 molC
                   
H = 0.04589mol  x 2H   = 0.09178 molH
               

∴Now, we know that the empirical formula of the original substance is
C0.0307H0.09178, however, we DO NOT like decimals like that. So we'll need to convert it into whole numbers.


Step 3. Convert to whole numbers (We all have a pet-peeve for ugly things, now fix it and satisfy our needs.)
(Tips: Divide each value of mole by the smallest number of moles)

C = 0.0307÷0.0307 = 1
H = 0.09178÷0.0307 = 2.9896

∴Since 2.9896 is closed to 3, the answer is CH₃.

Now, don't put down your pens yet. We still have to do the last step: Check Mass.


Step 4. Check Masses (Like we all do everyday to ensure our beautiful public images...JOKES)
(Tips: To check if the mass balance, add the mass of carbon and hydrogen together, then compare to the sample's original mass)


0.0307 molC  x 12 g  = 0.3684g of C
                        1mol

0.04589 molH x 1  g  = 0.04589g of H
                         1mol

Total Mass = 0.3684g + 0.04589g = 0.41429g
Substance's original mass = 0.952g

0.41429g ≠ 0.952g

OMG!! The total mass does NOT equals the original mass! No way! Does it mean that I should go back to step one and do this all over again?

Now, calm down.

If the mass of carbon and hydrogen does not equal the mass of compound, it means that there is another element involved, which is oxygen.


Step 5. Find the mass of Oxygen (In short, Find MoO.)
(Tips: mass of O = mass of compound - mass of C + H)

Mass of O = 0.952g - 0.41429g = 0.53771g


Step 6. Convert mass of Oxygen to moles (We love to interact with moles very much!)
0.53771g O  x  1  mol  = 0.0336 mol O
                        16.0 g


Step 7. Convert to whole numbers (again...feeling tired already?)
C = 0.0307÷0.0307 = 1
H = 0.09178÷0.0307 = 2.9896
O = 0.0336 ÷0.0307 = 1.0945

∴The FINAL CORRECT answer for this question is CH₃O.


Phew! This is pretty much for today! Here's two jokes about chemistry for you. Though they are not related to today's topic.


Q: What do you get when you mix iron, bromine, uranium, argon, and yttrium?
A: FeBrUArY (iron=Fe,bromine=Br,uranium=U,argon=Ar,yttrium=Y)

Q: What did Donald Duck say in his chemistry class?
A: Quark, quark, quark


Not laughing hard enough? Well, try something organic:

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.

Have You Ever Been "Floatified"? - Density

Hello fellow Pinch Of KCN blog readers, today we'll be touching on the topic DENSITY. Just to start off, and the Question of the Post today is: have you ever visited or been in/on the Dead Sea? Let's assume you haven't. And you all know that we, the wonderful Pinches of KCN have certainly...not been there. Anyhow, the Dead Sea is obviously not a sea that had been ill. On the contrary,this Dead Sea, located somwhere in Jordan, is a world class phenomenon. And how everything seems to float on the water surface in the Dead Sea appear to be rather unimaginable but yet factual. And the concept behind how even the fattest man on earth will possibly float in the Dead Sea is the idea of DENSITY!!!!



So DENSITY, this word sounds pretty familar...so what exactly is density again?

Density is a property of matter, it is used to measure the heaviness of an object with a constant volume. The closer the particles are together, the denser the object is.

To find Density you must have Mass divided by Volume!!!   *D = M/V*

Most common units for Density,

For Solids it should be grams per centimeter cubed (g/cm^3).
For Liquids it should be grams per milliLitre or kilograms per Litre (g/mL, kg/L).

The density of water should be common knowledge:
*1cm^3 water = 1mL of water
*Density of water = 1000 g/L
                            =1.0 g/mL

If the density of an object is greater than the density of the liquid it is submerged in, it will sink, if the density of an object is lower than the liquid, it will float. This is what people mean by saying "it's so salty you could float a rock in it." Salt makes water denser!And this is how the Dead Sea, aka the Sea of Salt in Hebrew (יָם הַ‏‏מֶּ‏‏לַ‏ח) seems to "floatify" every object.

Apart from just floating, density also determines the layers in mixtures and suspensions. Denser liquids will sink to the bottom, and lighter liquids float to the top when left still.

An example of the density of aluminum, 135g  = 2.7 g/mL
                                                                 50mL

Since it's getting late already and we all wanted to get a good night sleep, we'll be wrapping up now...

Let us welcome Mr.Lego and his...strange presentation on density.

**Curtain unfolds**     "Enjoy the show" - a 60-year-old man's voice pierced through the mist of darkness!
          

Measurement and Uncertainty

Every measurement is not exact, they are approximate estimations which we try to get as close as possible to the "exact value".
The two types of uncertainties are "Absolute Uncertainty" and "Relative Uncertainty".


Absolute Uncertainty
Absolute uncertainty is the inaccuracy in a measurement quantity.

Method 1 to express uncertainty:

1. First, collect at least three pieces of data from your measurements
2. Then, we will cancel out unreasonable pieces of data (if present)

3. Add all values together, then divide by the trials number to get an average
4. Subtract the average from the highest and lowest values
5. Record answer

Example:
Lengths of buffalo hair

Trial              Length
1                    22.3cm
2                    10.4cm
3                    22.4cm
4                    22.1cm (Lowest measurement)
5                    22.6cm (Highest measurement)

Since the second trial is clearly erroneous data, we will ignore out this measurement.
Average of lengths: (22.3+22.4+22.1+22.5) ÷ 4 = 22.3cm (Do not forget the UNITS!!!)
Difference between highest measurement: 22.6-22.3 = 0.3cm
Difference between lowest measurement: 22.3-22.1 = 0.2cm
(We use 0.3 as the uncertainty because it is the greater number)

Answer: 22.3 ± 0.3cm is the average length of buffalo hair.


Method 2 to express uncertainty:

We can figure out the uncertainty scale of each measuring tool, and then when we are measuring, we'll make the estimation as precise as possible. This way, we can use the uncertainty scale of the instrument to determine the answer.



Relative Uncertainty

The relative uncertainty is the percentage ratio between the uncertainty of an estimate to the real value of a measurement.

The formula for calculating relative uncertainty is the following:

Absolute    uncertainty 
estimated measurement

We can express relative uncertainty in two ways:
~ in percentage (%)
~ in significant figures

In percentage

1. First, we use the formula above to calculate an answer
2. Then, we multiply the answer by 100%

Example:Measuring the length of part of a rotten hot dog
The absolute uncertainty is ±0.5cm
The estimated measurement is 10cm
Thus the relative uncertainty is 0.5cm /10cm * 100%=5%


In significant figures
The last digit of a measurement is always uncertain, where the number of sig. figs. represents relative uncertainty.


Want some more practice on this topic?
Download the worksheet made by us! :D
(Topic included: calculating absolute uncertainty, reading measurements, review of sig. figs.)
http://www.mediafire.com/?hrvvwzloc4oyc1d
The link above is not working? Try clicking HERE to download!

And get the answer to the worksheet here:
http://www.mediafire.com/?4fabc5rq07r9i97
The link above is not working? Try clicking HERE to download!

Precision and Accuracy, Significant Figures, Rounding Rules

Accuracy vs. Precision

Hi there again. You have probably heard these two words "Accuracy" and "Precision" at least once in your life time. And you have possibly encountered at least 10 people who have been using these two words interchangeably; possibly you're even one of them yourself, which makes it at least 11 people so far. And maybe up until this very moment, you still have completely no idea what I'm trying to point out here. Okay, let me keep this real simple: Accuracy ≠ Precision. So you've been using BAD English all along. (Haha, we're not trying to critize our enthusiastic supporters here, just keep in mind that, we still love you very much, even if you use bad English ,dear!) Now, let us take a deeper look at these two words by CORRECTLY defining them.

Definitions:

Accuracy is how close the measurement is to true value of the measured quantity.

Precision is how close the measurement is to other ones.

As can be seen in the chart below showing the dart boards of drunk/sober men with good/bad aim, accuracy is being sober, and precision is having a steady hand.

Significant Figures

• more precise digits means there are going to be more significant figures
• the last digit of a value/measurement is usually estimated or uncertain (eg. 2938.234, 4 would be the uncertain number)
• to calculate you must include all of the certain digits and only ONE uncertain digits
• leading zeros are not counted as significant figures(eg. 0.0000001, the number of significant figures is 1)
• trailing zeros without a decimal point do not count. (eg. 10, 1000, 10,000, there is only one significant figure)
• trailing zeros AFTER a decimal point are significant digits. (eg, 32.000 has 5 significant figures) 

Rounding Rules

Calculations With Significant Figures

Addition and Subtraction
Problem: Kofia Dicted adds 1.3 grams more sugar in her morning coffee than the average person, which is 12.544 grams. How much sugar does she add in her coffee?

      12.544 g

+      1.3     g

=     13.844 g

As Kofia Dicted is aware of significant figures, she knows that her 1.3 added grams are imprecise, and that the total amount of sugar she adds cannot be more precise than 1.3grams. Thus she rounds off the sum to the first uncertain digit.

13.844g →13.8g
Answer: Kofia Dicted puts approximately 13.8grams of sugar in her coffee.

Multiplication and Division (Programming techniques mentioned are all real)
Problem: Terri Buljo Oaks is making a computer program that multiples with significant figures in mind to avoid actually thinking while doing chemistry homework. The first question is 12.544g x 1.3g

Below is her program's source written in pseudocode.

String.amountOfSigFigs(factor1)=sigfig1;
String.amountOfSigFigs(factor2)=sigfig2; //Finds the number of sig figs each number has
Variable Short Sigfigs=Math.Min(sigfig1,sigfig2); //Finds the lesser number of sig figs
Variable Double Float: product=factor1*factor2; //Finds the product of the two numbers
Print(Math.Round(product,sigfigs)); //Rounds to the number of lesser sig figs and displays the result

Variables used:
factor1: 12.544
factor2: 1.3
sigfig1: 5
sigfig2: 2
SigFigs: 2
product: 16.3072
Printed result: 16
     12.544 g

x      1.3     g 
        3.7632g²
+   12.544  g²

=   16.3072g²

16.3072g² →16g²

Unfortunately Terri's program does not account for non-decimal trailing zeros or units, so she got all of those wrong and naturally failed the course for life. 

Acids Formation

Naming "non-acids"

Naming ionic "non-acid" names are similar to naming other ionic compounds.

Step 1: First name the positive ion (typically a metal).
Step 2: Now name the negative ion (typically a non-metal).
Step 3: Change the ending of the negative ion to "-ide"

Note that the total charge on a compound must always be zero.

Some examples are: NaCl, K₂S; Sodium Chloride, Potassium Sulphide.

Naming simple acids

Simple acids are solutions of hydrogen bonded with non-metals from group 16 and 17 on the Periodic Table.

Step 1: Place the prefix "hydro" at the beginning of the name

Step 2: The name of the non-metal follows and its ending is to be substituted  to "ic"
            - E.g. S- sulphur will be become sulphuric.

Step 3: Lastly, the word acid is added to the very end.

Some examples of simple acids are listed below,

Formula                       Non-metal Names                        Acid Names

HBr                             hydrogen bromide                     hydrobromic acid

HI                               hydrogen iodide                         hyrdroiodic acid

HF                              hydrogen fluoride                       hydrofluoric acid

Naming Complex Acids

Complex acids are defined as solutions of hydrogen bonded with a poly-atomic ion that's negatively charged.

Step 1: The word or prefix of "hydrogen" or "hydro" is taken away in this type of naming.

Step 2: If the negatively charged poly-atomic ion name ends in the following suffix, it must be changed accordingly as shown below,

            "-ate"  →  "-ic"             E.g.  Chromate →Chromic

            "-ite"   →  "-ous"          E.g.  Nitrite →Nitrous

Step 3: Now, the word acid is placed at the end.

Some of the examples of complex acids are listed below,

Formula Name                        Acid Name

H₃PO₄                                 phosphoric acid

HNO₃                                     nitric acid

H₂SO₄                                sulphurous acid

Law of Definite Composition (Proust's Law)

• Every chemical compound always contain a fixed proportion of its composite elements by mass
• For instance, CO₂(carbon dioxide) has 1 atom of C(carbon) and 2 atoms of  O(oxygen). The total mass of CO₂ is 44g (C=12g, and 2xO=2x16g=32g) which would apply anywhere in the universe.

Law of Multiple Proportion (Dalton's Law)

• Two or elements can be bonded differently to form multiple compounds where the ratios of the composite masses are different. 
• For example, the elements of C(carbon) and O(oxygen) may bond to form CO compound (C:O=3:4) and CO₂ compound (C:O=3:8)

Lab 3B: Paper Chromatography

Objective and Conclusion
In Lab 3B, we performed paper chromatography tests with green, a primary colour, and an unknown colour. In this lab, we performed R_f value calculations. Through this process we learned that many components correspond to a certain R_f value. This value is the quotient of the distance from the starting line to the sample spot, and from the starting line to the solvent front, being the dividend and divisor, respectively.








Results From Actual Experiment
On our chromatography paper testing green food colouring, it split into two colours: yellow and blue. This confirms that green is a mixture of the components, yellow and blue. In art, this is well known: that the secondary colour "green" is a combination of the primary colours "yellow" and "blue". Below is a video of a very similar experiment.

 

Paper Chromatography in the Real World

Apart from purely scientific uses, paper chromatogrphy is used for many applications. This is because it is cheap, relatively quick, and can be used to separate virtually any mixture. Often they are used to test for drugs in one's blood. Even CSI uses it to sequence DNA and RNA.

Heating and Cooling Curves, Seperation Techniques

Changes in Matter

Heating Curve

A: A solid, closely packed particles in an ordered manner, slow vibrating in fixed positions.
A→B: Gain of heat, KE¹ ↑², temperature ↑.
B: Particles still packed, similar to state A.
B→C: Changes from solid→liquid. Temperature does not change as heat energy is used to overcome the force of attraction (latent heat of fusion). Known as the melting point.
C: Liquid state.
C→D: Temperature ↑ KE ↑
D: Still a liquid, molecules begin to move more freely, starts to become a gas.
D→E: Changes from liquid → gas. Temperature does not change as heat is used to overcome the force of attraction. Known as the boiling point.
E: Gas state
F: As temperature continues to ↑, molecules vibrate faster and KE ↑.

Cooling Curve

P: Gas state. High levels of vibration and KE.
P→Q: Temperature ↓³KE ↓.
Q: Still a gas, molecules begin to form bonds to create liquid.
Q→R: Condensation continues, changes from gas→liquid. Same temperature (latent heat of vaporization).Known as the boiling point.
 R: Liquid state
 R→S: KE↓ Temperature↓ Molecules lose energy and move closer.
S: Liquid state, beginning to become a solid.
S→T: Changes from liquid→solid. Molecules become ordered into a neat pattern. Known as the freezing point. Temperature remains the same (latent heat of fusion)
T: Solid state.
T→U: Temperature↓
U: Room temperature

Separation Techniques
The basis of separation is founded on different components have different properties, or the magnitude of these properties. Examples of properties include: density, reactivity, volatility, magnetism, solubility, polarity, etc.

Strategies of separation use processes that discriminate between these properties. Strategies include: filtration, flotation, crystallization, extraction, distillation, chromotography, and so on. Below is a chart classifying common methods to separate components.

Type of Matter	Solid	Liquid	Gas
Solid	Gravity Separation	Crystallization	Adsoprtion
Liquid	Filtration	Chromatography Distillation	Centrifugation
Gas	Absorption	Demister (tool)	Adsorption

 

---

1: Kinetic Energy
2: Increases
3: Decreases

Physical and Chemical Changes

The Three States of Matter


The three states of matter are: solid, liquid, and gas. Particles behave differently depending on what kind of matter it exists in. Solids are usually stable, with little less vibrations and energy than liquid and gas. Liquids are free-moving, and typically don't stay a certain shape unless within a container. Matter in the form of gas is pretty much everywhere, moving quickly at all times. These are the three main types of matter that makes up our world (well there's plasma but I guess we're not there yet).

Chemical Changes

Chemical changes occur at the molecular level. These changes always involve creating a new compound or substance from the original materials. These changes are difficult to reverse, or even impossible. Combustion is a type of chemical change, and you'd be hard pressed to "unburn" wood logs.

Physical Changes

Physical changes, in contrast with chemical changes, do NOT produce a new substance or compound. Typically, physical changes are changes in state (melting, freezing, etc) or shape (a bent spoon, for example). These changes are much easier to reverse than chemical changes.

What's the Matter?

Matter is everything that has MASS and takes up VOLUME (or space).
It is made up of atoms and molecules which contain mass.

The three units of mass:
gram                 g        
kilogram           kg
milligram          mg       

Example:
A small penny coin has a mass of 2.5g



However, the mass of the M87 black hole can be as much as 3.580056 × 10³¹ kg!



For a grain of sand, it is estimated to have a mass of 2.5mg.




Matter can be seperated into two categories: "Mixture" and "Pure Substance".

Mixture
- Made out of substances that CANNOT be united during any chemical processes
- Have more than one set of properties
- Example: salt water, alcohol
- Can be classified as "Homogeneous" or "Heterogeneous"
               Homogeneous
                  
              
                ↑ Sugar added in a glass of water
              ~Different components are NOT visible to naked eye after being mixed
              ~ The components are spreaded uniformly throughout mixing
             ~ They do not undergo any reactions
             ~ Examples: sugar in coffee, milk in tea

             Heterogeneous

           
             ↑  Oil floating on top of a glass of water
            ~ Different components can be visibly seen after mixed together
           ~ Components seem to have very different properties so they don't blend
           ~ Physical appearence of components will not be changed after mixed
           ~ Examples: oil with water, noodles with pork, sand in water

Pure Substance
- Cannot be physcially or chemically seperated
- Have the same taste, color, composition, and texture between the substances
- Have only one set of properties
- Examples: Gold, oxygen gas
- Can be classified into an "Element" or a "Compound"

          Element
          ~ Composed of atoms
          ~ The simplest form of matter
          ~ Cannot be decomposed or break down
          ~ There are 115 known elements
          ~ Example: hydrogen, nitrogen, silver
          ~ Can classified into "Metal", "Metalloid" and "Non-metal"

                 Metals

                
                ↑ A testube with mercury
              ‧Metal elements are great conductors of heat and electricity
              ‧High melting points
              ‧Solid at room temperature
              ‧Example: iron, copper, silver

                 Metalloid

               
                ↑ Boron, a metalloid
              ‧They have both metal and non-metal properties
              ‧They are like non-metals when they "meet with" metals
              ‧They are like metals when the "meet with" metals
              ‧Example: boron, silicon

                Non-metal

                
                 ↑ Neon is an example of a non-metal
              ‧Heat and electricity isulators
              ‧Gain electrons easily
              ‧Example: neon, chlorine

          Compound
          ~ They are made up of two or more elements
          ~ Compounds are combined chemically
          ~ Molecules such as oxygen gas should composed of 2 atoms
          ~ Example: Chlorine gas, sulfur dioxide       
 
               Ionic-acid
              ‧Has a pH <7
              ‧Corrosive
              ‧Taste sour
              ‧Always have a hydrogen atom
              ‧Example: acetic acid, hydrochloric acid

               Ionic-Base
              ‧Has a pH >7
              ‧Corrosive
              ‧Taste bitter
              ‧Always have a hydroxide atom
              ‧Example: ammonia, sodium hydroxide

              Ionic-Salt
              ‧The mixture of an acid and a base (neutralization)
              ‧Always have a salt product + H2O
              ‧Example: HCl + NaOH → NaCl + H2O

              Covalent-Organic compound

              
                ↑  The structure formula of Melamine
              ‧A compound with a carbon atom
              ‧Example: Aspartic acid, Hexafluoropropylene

The diagram of Matter and it's sub-categories: