¿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: