Lab 6D fun: "Will I die if I touch them by accident?"

Hello you all! DUN DUN DUN the horrific lab based on the well-known subject, stoichiometry is on his way to here. Before his expected arrival, let's laugh it off to ease up the pain from you know, The Lab :D
(Please please read the caption, my friend)

HA HA HA. Coincidentally, hydrochloric acid is one of the most important chemical components in our lab today. So, my sincere advice to you: please DO what Professor Glickman is doing (IF AND ONLY IF you wish to get smack in the head by Professor Bingham, that is in the case where he can still feel his back and get send to the principal's office and later wait to die in the juvenile  detention centre.)


You get the point, folks. Goggles on, and let's rock and roll. I mean carefully perform the lab.

The procedure for this experiment is utterly simple. So I suggest you to prepare a chair beforehand because there won't be any left once everyone begins to notice how easy it gets. JOKES

Without further ado, let's get down to the business. The Day 1 portion of this experiment involves a 25mL Na2CO3 solution and a 25mL CaCl2 solution both contained in a separate graduated cylinder to be poured together into a 250mL beaker. Now observe.
You will find that the mixture becomes a milky white solution, appearing to be a homogeneous colloid. Sort of like milk but with a thicker consistency.

Then, with the previously set-up filtering apparatus (ring stand + funnel+filter paper +beaker/Erlenmeyer flask), pour the mixture slowly into the funnel through the filtering paper. Be sure the beaker is directly underneath the funnel's opening, or else, you will wet the poor table. Make sure you drain it slowly and slowly, or else the mixture will clog the hole. If you wish, you can even rest and wait until the partial mixture is completely 'down the drain.' (And that's what I said the chair is for!) Then do so repeatedly until most of the liquid component is inside the beaker. Now remove the filter, put it on a piece of paper towel then let it sit and dry until next day.

"Will I die if I touch them [the chemical substances] by accident?" - Speaker A
"You sure will. NOOOOOOO, it's just chalk. Some people even eat chalk!" - Speaker B

On Day 2, just weigh the mass of the dry filter paper, then you are ready to ummmm "talk N troll"!!!
Let's take a look at the chemical equation for this lab:

 1Na2CO3 + 1CaCl2 ---> 2NaCl + 1CaCO3


The two products produced are NaCl and CaCO3, or you can simply call them salt and chalk to de-frighten yourself.


Now that's everything for today. Hopefully you have calm yourself down from the sudden horror that you've created. Because... 
































(Can't seem to see what we prepared you for, WELL, you need to call your optometrist immediately to get your eyes properly examined...) 

STOP IN THE NAME OF STOICH?!?!?!?!

           Percent Yield
Percent Yield is used when you cannot recover the whole product or when the reactants haven't been used up yet.

The actual boring definition of Percent Yield - the ratio amount of product obtained to amount of product expected by calculation, expressed as a % blah blah blah.

Mrs. PinchOfKCN will enlighten you by giving you the ancient formula for percent yield.

%

Now an example, 

2 C(s) + 1O₂(g) --- > 2 CO(g)

What is the percent yield of CO2 if 44g was predicted to be formed, and 34g actually formed?

Step. 1 USE THE FORMULA.

Step.2  DO THE MATH.

Percent yield = 100 x ( 34 grams CO₂ actual / 44 grams CO₂ predicted ) = 77 %

Moving on.... Percent Purity.

Reactants are sometimes impure and we must calculate the pure parts.

Boring Definition - ratio of the mass of the pure substance to the mass of impure.

Of course Mrs. PinchOfKCN has a formula for everything(love does not count).

Easy Example, If there was 121.2 g of solid was obtained, but showed that only 109.2g of it was aspirin. Calculate the percent purity of the product.

Percent purity = 109.2 ÷ 121.2 × 100% = 90.0%

Harder Example, 

Chalk is almost pure calcium carbonate. We can work out its purity by measuring how much carbon dioxide is given off. 10 g of impure chalk was reacted with an excess of hydrochloric acid. 2.128 liters of carbon dioxide gas was collected at standard temperature and pressure (STP).

CaCO₃ (s) + 2HCl (aq) → CaCl₂ (aq) + H₂O (l) + CO₂ (g)

Step 1 Calculate the molar mass of CaCO3 = 100g.

Step 2 Calculate the amount of grams CaCO₃ from the volume of CO₂

(hint: convert to moles)

2.128L CO₂(22.4 L/mol CO₂)* 1 mol CaCO₃/1 mol CO₂*100gCaCO₃/1 mol CaCO₃=9.5g CaCO₃

Step.3 Use the formula and find the percent purity.

Percent purity = 9.5 ÷ 10 × 100% = 95%

If you did not look at any of my examples, I'm not going to be mad at you. Just watch the videos =.= ........

WOAH AP CHEM!!!

Excess = Garbage (aka: the hardest type of mole question)

¡Bienvenido al blog de PinchOfKCN! ¿Como estas? ¡Ahora, voy a escribir un pasaje para el tema "Excess and Limiting Reactants"! ¡Es muy muy interesante! ¿No?

*cough cough* Enough of speaking Spanish, now switching back to normal English. So how's everyone doing? I see many tired faces, have you guys been completely wiped out by the math midterm? Me too, and what's worse, I'll be having my Spanish midterm very soon, that's why I'm practicing some Spanish in the beginning of this blog. Hoho.
Now, back to our topic of the day:

Have you ever made pizza or birthday cakes? When you finished making, you probably have some left over ingredients, right? Those left over ingredients are exactly the same as the excess quantity after a chemical reaction occurs. On the other hand, a limiting reactant is the reactant which will be completely used up during the reaction.

Return of the Mole King
Now it is where the mathematical calculations take place. Consider the following question:

44.6g magnesium reacts with 56.3g oxygen gas produces magnesium oxide. Find:
a.) How many grams of MgO is produced
b.) The limited and excess reactant
c.) How much of the excess reactant is left over

1. First, we will have to write a balanced equation. If an equation is already given, remember to check if it is balanced or not before you move on to the next step!
2 Mg + 1O₂  →  2 MgO


2. Then, convert each reactant to MgO to find out which reactant is limited and which one is excess.
The route you may want to follow is: grams to mol, mol to mol (ratio), then mol to grams
Mg to MgO:  44.6g Mg  x   1 mol Mg  x  2 mol MgO  x  40.3g MgO  =  74.0g MgO
                         24.3g         2 mol Mg         1 mol MgO

3. Now, do the same thing with O2.
O2 to MgO: 56.3g O₂ x 1 mol O₂  x  2 mol MgO  x  40.3 MgO  = 142g MgO
                     32.0 g        1 mol O2      1 mol MgO

4. Next, compare the amount of product produced by each reactant, the one which produce MORE MgO will be excess reactant, and the one which produce LESS will be limited reactant.
In this case, oxygen is excess and magnesium is limited, where 74.0g MgO will be produced.

5. After that, we can calculate how much oxygen is used.
44.6g Mg  x  1 mol O₂  x  1 mol O2  x  32.0g       =  29.4g of O₂
             24.3g Mg      2 mol Mg    1 mol O2

6. Finally, subtract the grams of O2 used from how many O₂ you have at the beginning.
56.3g - 29.4g = 26.9g O₂ left in excess

These questions sure look complicated (or in fact they ARE complicated), but once you've mastered them, you'll score 100% on the next test.
Now, a quick update on the mole map:

Pretty scary, huh? It's alright, don't stress out too much, just practice more, and you'll be all prepared for next class' quiz.

Stoichiometry

[Don't panic, and no need to 'talk' in CAPS, WE WILL TEACH YOU HOW TO DO IT!]
From Wikipedia: "Stoichiometry is a branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions."

I personally think of it as just another way to test us on moles, but that's not the point. Moving on!

If the Wikipedian definition wasn't clear, Stoichiometry, for us, is simply finding a quantity of a product given some prior information. This information can come in the form of mass of a certain substance, molarity, moles, and even volume. Whatever the question may be, we first need a balanced equation. If you don't recall, click here for a refresher.

Let's use the combustion of methanol as an example.
2 CH₃OH + 3 O₂ → 2 CO₂ + 4 H₂O

As we can see, this equation is already balanced. This is important, because if we don't balance it, EVERYTHING is going to be buggered; proper ratios are essential for doing stoichiometry.

From the reaction above, we can see that one of the ratios, namely Oxygen to Carbon Dioxide, is 2:3. That means for every 2 Carbon Dioxide molecules (or moles), there are three Oxygen molecules (or moles). These ratios are called molar ratios, and are needed to solve questions in this unit.

Example: How many moles of Oxygen Gas are required to completely react with 6.00 moles of methanol?
Solution: 6.00molCH₄O*(3 mol O₂)/(2 mol CH₄O)
=9.00 mol O₂ (Don't forget your sig figs!)

Alright, that was child's play. Let's do a real question.
How many grams of Carbon Dioxide gas is created when sufficient Methanol is reacted with 10.4L of Oxygen gas at STP?

Solution: Recall that 1mol of any gas at STP is 22.4L, the ratio of Oxygen to CO₂ is 2:3.
10.4L O₂ * (1 mol O₂)/(22.4L O₂)*(2 mol CO₂)/(3 mol O₂)*(12.0+16.0*2)g CO₂/(1 mol CO₂)
=13.6gCO₂

In essence, it's just what we've been doing all year, conversions. Conversions with moles, units, volume, and mass. If you get the hang of it, no problem!

Below is a Javascript calculator that gives the amount of moles and grams of a substance relative to another substance in the reaction. This is a helpful checker, but it SHOULD NOT be used to do your homework for you. It doesn't do sig figs and only goes to 5 decimal digits, anyway.
Disclaimer: The creator of this calculator and the writers at PinchOfKCN are NOT responsible for the consequences that occur from using this program, including but not limited to low grades, suspension, or setting your computer on fire.

Molar mass of substance A:

Molar mass of substance B:


Molar ratio:

moles of A to

moles of B.
Grams of substance A:



Answer:
Moles of B:

Grams of B:

Calculate the Energy, We Must!

CALCULATING ENERGY 101

        What an honour to have Master Jedi, Yoda, teach us about calculating energy in endothermic and exothermic reactions. With his great wisdom and knowledge, lets " pitter patter lets get at her."

Young Padawans of PinchOfKCN, If you may not have noticed, absorbing(consumption) or release of energy can be found right in the endothermic/exothermic reaction. In exothermic reactions, energy is usually on the right side of the equation and has negative  ΔH. For endothermic reactions, energy is found on the left hand side and has positive  ΔH.

Young ones be CAREFUL, It's the "Return of The Moles", they bring back horrid memories and cookies...
 ΔH is usually expressed as kJ per moles of one of the chemicals in the equation. Basically it should look like this

-812kJ

1mol CH4

Eg.

CH4 + 2O2 --> CO2 + 2H20 + 812kJ 

Question: How many beauty marks(moles) of CH4 is needed to produce 2100kJ of energy?

Step. 1 - Choose an element, then write the equation.

-2100kJ(negative because energy is needed) x 1mol of CH4

                                                                           812kJ

And you should get your answer 2.6mol CH4

Just for the trainees, How many grams of CH4 is needed to produce 2100kJ of energy?

Take the answer from the question above and convert it to grams! 2.6mol x 16 grams

                                                                                                                      1mol 

You should get 41.6 grams of CH4 is needed to produce 2100kJ of energy.

You are all ready to become a Jedi after watching this video!

Master Yoda has 1 last thing to say...

"Teacher, Can You Have Less Empathy?'

Hello there, your long-awaited couple's therapist has finally and officially returned to town. Let's babble for a bit before we get started. So for the past few days, I was attending a very interesting seminar abroad, not too far, just in New York City. A couple hours flight. No big deal. But one thing at that seminar caught my full-attention: Did you know that releasing your feelings to your partner once and awhile retain a much more promising relationship than those who normally absorb partner's feeling without saying any words in return? That sounds kind of obvious isn't it? Well, that is certainly true. And that smoothly leads into our today's topic.

Ahem. Endothermic and Exothermic. They both have the root-word thermic. And presumably all of you have used a Thermos at some point of your life. Or anything kind that's similar. That will get you started with the definition. Anyhow, endothermic reaction has the tendency to absorb energy in the surrounding where exothermic reaction release energy. (And I'm sure you can tell which one has a more preferable behaviour. LOL)

So molecules are generally assembled together by chemical bonds.
If energy used to breaks bonds > energy gives off to forms bonds ->   endothermic reaction
If energy used to breaks bonds < energy gives off to forms bonds  ->   exothermic reaction

Empathy, I mean Enthalpy, H, is the heat contained in the system.

Endothermic and exothermic reactions can be made into a charted diagram where Potential Energy is the dependent variable and Reaction Progress is the independent variable. And an actual diagram like such can explain it all:

                       Exothermic reaction:

    Endothermic reaction:        

voilà and until next time!