LAB 6D - Limiting Reactant and Percent Yield

interesting fact:
Zemmiphobia: Fear of the great mole rat.

Scary mole/cat is scary.

ANYWAYS

Today, fellow followers/readers, we did a lab once again! But this time, it was on stoichiometry - of course to summon our knowledge together to conduct a clear analysis of this chapter through an effective experiment.(Right before the test!)
I hope you know all your safety rules and no horseplay! (I'm watching you silently.)

Now I warn you, this lab is very tedious. Very. So I suggest you bring a book out just in case... (jokes of course! pay attention!)

To summarize what we did, we just obtained 25mL of sodium carbonate and calcium chloride solution in seperate graduated cylinders and poured them into one 250mL beaker. Now what do we see? It turned into a thick homogeneous white solution. We had to let it sit there for about five minutes in order to let the contents seperate into layers and suspend. Ahh you see it is a double replacement reaction - if you didn't already know.

MEANWHILE... (I better not catch you reading your book), we had to set up our apparatus! It consisted of a ring stand with a paper filter that we carefully folded into the shape of a cone. BUT before that, we had to put our names on it (IN PENCIL.. NOT PEN) and weigh the paper filter.

Now after the wait, we got to do the filtering of the precipitate and the solution! We had another beaker underneath the funnel, and we swirled the 250mL beaker with the contents before cautiously and slowly poured it into the funnel. It is important to do it in stages, and not all at once or it will be even more tedious!
Once that is completed, we took out our filter with the precipitate and placed it on a paper towel in a safe location to dry. When it dried (AKA next class), we weighed the contents with the filter paper and recorded the data.

NOW here comes the calculations/analysis:
1 Na2CO3(aq) + 1 CaCl2(aq) --> 2 NaCl(aq) + 1 CaCO3(s)

With this balanced equation, we calculated which reactant was the limited and the excess, and found the theoretical mass of the precipitate (CaCO3). With this information and the actual product (which we weighed), we used the percent yield formula (don't remember? refer to the previous posts!) to find the percent yield of the product.

.... And that concludes our lab! It was a pretty chill one :)

YAY LET'S TALK ABOUT PERCENTS! (And no not your grade percentages)

It's all about the ratios and percentages today! We'll be looking at percent YIELD and percent PURITY.


You're probably wondering, "WHAT IS THE DIFFERENCE? I DON'T EVEN-"
It's okay. Calm your horses. We will explain, don't worry.


PERCENT YIELD

• measures amount of product actually produced compared to the amount that would theoretically be produced by using stoichiometry
• often, there will be an excess of reactants (because in the real world, we seldom have ideal reactions) so the products resulted will always be less than what is expected; so percent yield is needed
• determines what percentage of yield is expected in a reaction 
• ratio of actual yield (amount of product obtained in grams) and theoretical yield (amount of product expected in grams)

NOTE: Remember to always round to one decimal place when dealing with percentages!

Example: Fe₂O_3(s) + 3CO_(g) → 2Fe_(l) + 3CO_2(g)

If 2.2g of Iron (Fe) is produced when 6.5g of Fe₂O₃ is reacted with excess CO, what is the percent yield?

Step 1: Find the expected (theoretical) amount of Fe to be produced by 15.0g Fe₂O_3.

(Using stoichiometry - remember to use the ratios of the elements/compounds!)

6.5g Fe₂O₃ x 1mol Fe₂O₃ /159.6g Fe₂O₃ x 2mol Fe / 1mol Fe₂O₃ x 55.8g Fe/ 1mol Fe

= 4.545g Fe is expected to be produced

Step 2: Calculate the % yield using the ratio formula.

We know that 2.2g of Fe is actually produced, and 4.545g Fe is expected

Therefore....

2.2g Fe/4.545g Fe x 100% = 48.4% of Fe is actually produced for every 100% predicted

If, for example, the percent yield is given and the amount of a reactant that reacts is also given, how would you find the number of grams a certain product will be produced?

Let's use the same equation: Fe₂O_3(s) + 3CO_(g) → 2Fe_(l) + 3CO_2(g)

The percent yield is 70.0% and 3.10g of CO is reacted with excess Fe₂O₃. Find the number of grams CO₂ produced.

Step 1: Calculate amount of the product expected.

3.10g CO x 1mol CO/ 28.0g CO x 3mol CO₂/3mol CO x 44.0g CO₂/1mol CO₂

_{= 4.871g CO2 is expected}

Step 2: Turn the percentage into a decimal and calculate the amount of product produced.

0.700 = amount of product produced / 4.871g

amount of product produced = 0.700 x 4.871g

= 3.41g CO<sub>2 produced 


ET VOILA! Ergo, you have your grams of the actual product produced! Easy, right?</sub>

_{OMG, if this doesn't make you laugh... you need a sense of humour. (Can I get an "awww")}

_{PERCENT YIELD}

• used to calculate the amount of purity in a reactant (since not all reactants are pure)
• this is needed to find the amount of reactant that is actually available (pure) in order to react and form products

Percent Purity=   Mass of pure substance    x 100

                     Mass of impure sample

Example: 1Cu + 1Cl2 > 1CuCl2

A 7.4g copper ore contains 3.2g of pure copper. What's the percent purity?

% purity = 3.2g / 7.4g x 100%

= 43.2% pure copper

By finding the grams of pure copper (which is already stated here), you can find how much of a certain product is produced. That's why % purity is important! :)

EXCESS AND LIMITING REACTANTS

Have you ever camped outside in the forest and sat around the fire munching on your smores? Chances are, you have made your own smores before! If you didn't already know what a smore is (what a sad life you've been living in), it consists of marshmallows, chocolate, and crackers. This is the "formula" for each smore you make:


2 crackers + 1 chocolate + 1 marshmallow --> 1 smore

Now let's say you have 10 crackers, 5 chocolates, and 3 marshmallows. How many smores can you make?

You can only make THREE smores. That's not enough to feed everyone around the campfire now is it?

SOLUTION:

You have 10 crackers, so 10 divided by 2 is 5. 5 smores can be created with 10 crackers.

For every one smore you make, one chocolate is needed so 5 chocolates can create 5 smores.

BUT WAIT- there are only 3 marshmallows (the most important ingredient!!) so only three smores can be made! Therefore, the marshmallow is the limiting reactant because it is completely used up and limits how many smores (the product) can be made. The other ingredients (the reactants) are in excess, so they are the excess reactants.

Now let's get started with the lesson!

In an ideal world, where everything is perfect, balanced equations determines exactly what should happen in a chemical reaction. However, we all know that reality is a pain in the butt, and that not every reactant will be used up. Usually, there will be an excess of a certain reactant and another reactant that will be completely used up to create a product.

As stated above, the limiting reactant is one that will be completely used up and the excess quantity will be the one that will have some left over.

2Na + 1Cl₂ -->2NaCl

In this example, there are 5 chlorine (Cl2) molecules and 6 Na atoms. The equation states that two Na atoms would react with one chlorine molecule. Thus, only three molecules of chlorine is needed for 6 Na atoms. The rest of the chlorine molecules are in excess and will remain unreacted.

Example: 50.0g of Na reacts with 53.5g of Cl2. Which one is in excess and by how many grams?

Step 1: Write a balanced equation.

2Na + 1Cl₂ -->2NaCl

Step 2: Convert each reactant to the product. The one who produces the smaller amount of product will be the actual amount produced.

50.0g Na x 1mol Na/23.0g Na x 2mol NaCl/2mol Na x 58.5g NaCl/1mol NaCl

= 127.17g NaCl

53.5g Cl₂  x 1mol  Cl₂/71.0g Cl₂ x 2mol NaCl/1mol Cl₂ x 58.5g NaCl/1mol NaCl

= 88.16g NaCl <---- the actual product

(Cl2 is limited reactant because it limits the product produced)

Step 3: Convert the limiting reactant to the excess reactant to find out how much of the excess would be sufficient to react with 53.5g of Cl2.

53.5g Cl₂  x 1mol  Cl₂/71.0g Cl₂ x 2mol Na/1mol Cl₂ x 23.0g Na/1mol Na

= 34.7g Na is needed in order to create the actual product

Step 4: Subtract the amount of the excess reactant that is reacted with (or needed) from the amount that it started with before the reaction has taken place to get the excess amount.

50.0g Na - 34.7g Na = 15.3g Na is in excess.

Think you got the hang of it? Here are some practice problems for you: (have fun *evil smile*)

1) How many grams of CaCl2 will be formed when 24.3g of Ca reacts wieh 32.6g of Cl2? Which is the limiting and excess, by how much?
2) Given the balnced equation: Pb + F --> PbF2. How many grams of lead(II) flouride will be produced? Which is in excess and by how much?
3) If 20g of Beryllium is reacted with 35g of Sodium Oxide...
  a) Write the balanced equation.
  b) How much beryllium oxide can be formed?
  c) What is the limiting reactant.

STOICHIOMETRY

STOICHIOMETRY!!!! 

(Like seriously, I could not pronounce this word at the beginning...)



Anywayyyyy WHAT does this weird word actaully mean???
Well, we all know that "metry"means measurement right?? (if not, well now you know)
And "stoichio" means element in Greek! 
So PUT IT ALL TOGETHER, you get....the measurments of elements!!!! That's what Stoichiometry means. :)

- Stoichiometry is all about the quantities of the substance that are produced by chemical reactions.
- It shows the quantitive relationship with the products and the reactants in a chemical reaction.

For example: 

4NH3 + 5O2 ---> 6H2O + 4NO

= 4 molecules of NH3/moles react with 5 molecules of 02/mole of 02 to produce 6 molecules of H2O/moles of H20 and 4 molecules of NO/moles of NO

SOOO...
The first step in solving stoichiometry problems is to BALANCE THE CHEMICAL EQUATION!!
- That is very important beause the coefficients tells us the ratio of the moles and molecules in the equation.
- And remember to always include the units! Otherwise you might get confused later on in your calculations. It's only for your own good. :)

Example!!!

P4 + 5O2 ---> 2P2O5

How many moles P2O will be formed when there is 7.40 moles of P4?

7.40 P4 X 2mol P2O5/mol P4 = 14.8 mol P2O5

There will be 14.8 mol of P2O5!

Now that we got the basics, let's add in some MOLARITY!!! 


haha it's a mole!

So let's review theee formulas,

Molarity = mol/L
Litres = mol/M
mol = L X M



Example 1:

2 FeCl3(aq) + 3MgSCN2(aq) ---> 2Fe(SCN)3(s) + 3 MgCl2(aq)

How many moles of MgCl2 would be formed if 50.0 mL of 0.200 M FeCl3 is reacted with sufficient MG(SCN)2?

convert 50.0 mL into L which = 0.05 L

mol = (0.05 L)(0.200 M)
      = 0.01 mol FeCl3 X 3 mol MgCl2/2mol FeCl3 
      = 0.015 mol MgCl2

There will be 0.015 mol MgCl2.

Example 2:

2 FeCl3(aq) + 3MgSCN2(aq) ---> 2Fe(SCN)3(s) + 3 MgCl2(aq)

100.0 mL of Mg(SCN)2 is reacted with FeCl3 to produce 11.5 grams of Fe(SCN)3. What is the molarity of the Mg(SCN)2 solution??

11.5g Fe(SCN)3 X 1 mol Fe(SCN)3/230.1g Fe(SCN)3 X 3 mol Mg(SCN)2/2 mol Fe(SCN)3 x 1 mol/0.100L = 0.750 M

There is 0.750 M.

Example 3:

1Zn(s) + 2HCl(aq) ---> 1ZnCl2(aq) + 1H2(g)

At STP, what colume of H2 gas would be produced from 21.2g of HCl?

OMGG STP!!!! lets refress our memories, STP is 22.4L!

21.2 of HCl X 1 mol/36.5g HCl = 0.581 mol HCl X 1H2/2 HCl = 0.290 mol H2 X 22.4 L/1mol = 6.51 L H2

There will be 6.51 L H2




And that's it for STOICHIOMETRY! :D If you want more practice, here is a good website to check out!http://www.sciencebugz.com/chemistry/chprbstoich.html

THE RETURN OF THE MOLE.. (dun dun dun)

Hello all! Today we will be talking about ENERGY and how to calculate it! This means that we will be bringing the mole concept back...

Yes I know I know... we've all missed the beloved mole... Who didn't?!


Without further ado, let's get back to business. When we talk about energy in a chemical equation, we are talking about the change in energy, expressed in kilojoules (kJ) per mole, that is either absorbed or released in the chemical reaction. 


As you know already from the last post:
energy being released - exothermic
energy being absorbed - endothermic


In exothermic reactions, since energy is being released, the change in enthalpy will be negative and the energy term will be in the products' side.
Endothermic reactions will have a positive change in enthalpy because energy is being absorbed (increased) rather than having a decrease amount of it. The energy term will be on the reactants' side.
eg. The freezing of water shows that heat/energy is being released.
       H₂O (s) → H₂O (l) + 6.01 kJ          
In this case, the reactant has a higher energy than the compound in the product.
**Note: Even though ΔH is negative, it is written just as what is included in the result of the product so the sign is neglected.  **


When we calculate the energy change of a reaction, the ΔH is expressed in kJ per mole of one of the chemicals in the equation. The coefficients of the compounds in the balanced equation is used for the ratio of kJ/mol.


If the exothermic equation was: 2H₂O (s) → 2H₂O (l) + 12.02 kJ
The ΔH of one mole for H2O would be: -12.02 kJ/ 2 mol H2O     OR     -6.01 kJ/1 mol H2O


3 things to keep in mind:

1. ΔH changes with different reactions (the energy term is different for each)
2. ΔH is different for each compound, like molar mass, ΔH depends on the chemical being referred as the coefficients may be different for each chemical reaction.
3. ΔH is NOT a constant like Avogadro`s number.

When given the amount of moles of a compound produced, refer to the mole chart to calculate the energy! However, energy is now used so be sure to add that into your original chart!

Here are some questions you can try! The answers are posted below. :)

Remember to use sig figs!!

1. How many moles of CH4 are needed to produce 1500kJ of energy?

CH4 + 2O2 → CO2 + 2H2O + 812kJ

2. Calculate the amount of energy required to produce 23.0g of NO2.

N2 + 2 O2 + 67.6 kJ → 2 NO2

3. How many atoms of Oxygen are needed to produce 1000. kJ of energy?

4NH3 + 5 O2 → 6 H2O + 4 NO + 905 kJ

Answers:

1. 1.8 mol CH4

2. 16.9 kJ

3. 6.654 x 10²⁴ atoms of O

GOODLUCK!

EXOTHERMIC AND ENDOTHERMIC REACTIONS! (going into thermodynamics)

Woot! Today's blog is about..... EXO AND ENDOTHERMIC REACTIONS!

Exothermic and endothermic reactions are always there in a chemical reaction! They have to do with the absorption and the releasing of heat or in other words energy.


Exothermic means the release of energy to the suroundings.


Endothermic means the absorbtion of energy from the surroundings.

For example:
The formation of snow in clouds releases energy and are exothermic reactions.
Melting ice absorbs energy and are endothermic reactions.

Furthermore:
Chemical bonds are what holds the molecules together
- If energy is added, it breaks the bonds because there is an increase in kinetic energy.
- If it gives off energy, it joins together and the molecules have little movement.
- more energy to break bonds than it gives off to form bonds = endothermic!
- less energy to break bonds than it gives off to form bonds = exothermic!
- Enthalpy (H) is the heat in the system.

ENERGY DIAGRAMS!!! (shows the change from start to end)
- Reactants start with a certain amount of energy, energy is added to start reaction, and is then released OR absorbed in process.
- Relative amount of energy determines if the reaction is endothermic or exothermic 

Things to know about the potential energy graph...
1) Energy of reactants is the total potential energy of all reactants.
2) Energy of products is the total potential energy of all reactants.
3) Energy of activated complex is the potential energy of "transition state" between reactants and products.
4) Activation energy  is the energy that must be added to get the reaction to progress (from reactants to activated complex).


Perhaps an analogy may help some understand potential energy diagrams better!
Below, A is the reactant, and the pushing up the hill is the activation energy needed in order for the rock to progress. When it reaches the top of the hill, it is at the potential maximum energy of the transition state before the energy is changed. As A (the rock) rolls down the hill and comes to a stop, it becomes the product (B). Since the end position of the rock is further down from it's origin, then the change in energy is negative. This is an example of an exothermic reaction because the energy in the end is less than at the start, so energy is released.

5) Change in Enthalpy is the change in potential energy (energy of products - energy of reactants).

okay that is all...!!

- Melody