Lab 6D

A stoichiometry based lab that takes 2 days, a lab day and a results day.

Day 1- On this day it involved getting 25 mL of Na2CO3 solution and 25 mL of CaCl2 solution poured into one beaker. We observed. It looked like a thick milky solution.

We let the solution set aside and moved on to set up the ring stand, filtering apparatus, and beaker. We gingerly poured the solution bit by bit through the filter into to the beaker waiting below. There`s not much to do so we waited.

Once all of the solution drained we took the filter paper out and let it dry overnight until next class.

Day 2-  We weighed the mass of our filter paper and what we got from mixing the two reactants was:

1Na2CO3 + 1CaCl2 - 2NaCl + CaCO3

Basically salt and chalk

Percent Yield and Purity

First, let's look at Percent Yield.

Sometimes not all the product is recovered or all the reactants are not used up, so we calculate the Percent Yield. The basic formula is: amount of product obtained / amount of product expected X 100 to get percent.
Basically, when dividing the two numbers, if you get something over 1, something, somewhere, has gone terribly wrong...

Now let's use this info in an example:
15.0 g of CH4 reacts with excess Cl2 according to the following equation:
CH4 + Cl2 = CH3Cl + HCl
a total of 29.7 grams of CH3Cl is produced. What is the percentage yield?

First, we have to find the mass of CH3Cl that is expected:
15.0 g CH4 X 1 mol CH3Cl / 16.0g CH4 X 1 mol CH3Cl / 1 mol CH4 X 50.5 g CH3Cl / 1 mol CH3Cl = 47.34 g

Now let's figure out the percentage yield:
29.7 g / 47.34 g X 100 = 62.7 %
*Remember not to round until the end and use sig. figs.*

Simple enough, right? Let's do a backwards one to see if you really get it:
What mass of CuO is required to make 10.0 g of Cu according to the reaction:
2NH3 + 3CuO = N2 + 3Cu + 3H2O
if the reaction has a 58.0% yield?

We have to find the mass of CuO:
10.0 g Cu X 1 mol / 63.5 g Cu X 3 mol CuO / 3 mol Cu X 79.5 g CuO / 1 mol CuO = 12.52 g

Now, we have to divide this amount by 0.580 to get the LARGER number that will allow for the LOSS that will occur during the reaction:
12.52 g / 0.580 = 21.6 g


One more example:
What mass of K2CO3 is produced when 1.50 g of KO2 is reacted with an excess of CO2 according to the reaction:
4KO2 + 2CO2 = 2K2CO3 + 3O2
if the reaction has a 76.0% yield?

Calculate the mass of K2CO3:
1.50 g KO2 X 1 mol KO2 / 71.1 g KO2 X 2 mol K2CO3 / 4 mol KO2 X 138.2 g K2CO3 / 1 mol K2CO3 = 1.458 g

Now multiply that by the percent yield:
1.458 g X 0.760 = 1.11 g


Phew! Any more questions? Watch this video:

If you have this down, the next part will be easy.

What's more pure than a sparkly diamond?

Percentage Purity Time!

Only the pure part of a substance will actually react so sometimes we need to calculate the percentage purity of a substance.
The formula is basically the same as percentage yield: mass of pure reactant / mass of impure reactant X 100

Example:
If 100.0 g of FeO produces 12.9 g of pure Fe according to the reaction:
2FeO + 2C + O2 = 2Fe + 2CO2
what is the percentage purity of FeO used?

Same as before, find the mass of FeOl:
12.9 g Fe X 1 mol Fe / 55.8 g Fe X 2 mol FeO / 2 mol Fe X 71.8 g FeO / 1 mol FeO = 16.6 g

Calculate the percentage purity:
16.6 g / 100.0 g X 100 = 16.6 %


As you can see, everything is basically the same as percentage yield. Got it? Good. Have fun!

Excess and Limiting Reactants

This entry focuses more on the reactant side of a chemical equation
*aside: after all there's no reaction without them!!

A balanced equation assumes that there is a perfect amount of reactants in order for the reaction to occur; but it's theoretical. Say what?

In real life however, this is of course, impossible! In order for a reaction to occur there must be too much of one reactant to ensure that enough of the reactants will combine.

Therefore, one reactant will be fully used and be the limiting reactant (because it limits how much of the other reactant is reacted) and the other reactant will have some molecules left over and be the excess reactant.

The Four Steps to Success! OR The Four Steps to finding the excess reactant and its amount...

1. Write a balanced equation! Simple eh? If an equation is given but is unbalanced - balance it. And even if there is an equation that looks suspiciously balanced - check it anyway.

2. Convert each mass of the reactants to the mass of the same product.

3. Take the limiting reactant (the mass of the reactant that results in a smaller mass value of the product) and convert that into the excess reactant (the mass of the other reactant).

4. Subtract your accurately approximated and brilliantly solved for value from step 3 from the theoretical mass  given.

OR

Watch this video, if you prefer to have someone teach it vocally to you!

Voila!

YOU  CAN NOW FIND THE AMOUNT OF EXCESS REACTANT!!

 

I suppose you may want to test this theory for yourself,

Very well... Example 1!:

Okay seriously awkwardly dancing and cheering people cut it out!

A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen.  Find which reactant is in excess and by how much. You may begin:


4 NH_3(g) + 5 O_2(g)4 NO_(g) + 6 H₂O_(g)











If your awesomeness is as above average as your IQ (or not) you may want to try a more advanced method.
This is of course assuming that you are a chemistry wiz (or not) and are more concerned about how much time you spend doing chemistry, then doing a slower and more certain process (or not). Well I can certainly see that you all fit the bill.

LET US PROCEED!


Example 2: 90.0 g of FeCl₃ reacts with 52.0 g of H₂S.  What is the limiting reactant?

2FeCl_{3 +} 3H₂S ---> 6HCl + 1Fe2S3


52.0 g H₂S x 1mole H₂S  x 2 moles FeCl_{3  x 162.3 g} FeCl_{3  =  165 g} FeCl₃
                            34.1 g H₂S     3 moles H₂S      1 mole FeCl₃ 


The limiting reactant is FeCl_3. 


This website, will help you to master your stoichiometry skills.  It has stoichiometry questions that involve every day situation too!

http://www.chemteam.info/Stoichiometry/Limiting-Reagent.htm

Stoichiometry Calculations

As previously mentioned... stoichiometry is calculations which involve chemical reactions!

Just in case you have forgotten what chemical reactions may look like (which i doubt ;) this will help!

From our studies with moles, we have learned quite a few different conversions which involve them.

This chart sums up mole conversions, before our virgin ears ever heard the word stoichiometry:




Stoichiometry simply adds one more step to this simple conversion chart.

In a balanced chemical reaction, once given the mass, particles or volume of either the reactant or product; a chemistry 11 student can then find the mass, particles or volume of any other reactant or product!

How is this possible?!

Well since we know how many moles of a certain substance per so many moles of another substance, we can use this as a conversion.

For example in the balanced chemical equation:

2 Fe + 3 Cl₂ = 2 FeCl_{3; 
where we know there are 123.4 grams of iron (Fe), we can find the amount of product, (FeCl3):}


123.4 g Fe x 1 mol Fe     x  2 mol FeCl3   x  162.3 g FeCl3  = 717.8 g FeCl3


                    55.8 g Fe        1 mol Fe              2 mol FeCl3

Remember this concept can be applied to any sort of calculation involving a reaction; given you remember some important facts!
                                          Such as: Volume at STP (22.4L), M = moles / litres


Here`s a website with in depth information, for our blog`s eager learners!

  http://www.shodor.org/unchem/basic/stoic/

What is Stoichiometry?

Stiochiometry:
-"Stoichio" comes from Stoichion, the english equivalent is "element"
-""metry" comes from Metron, the english equivalent is "measurement"

Both from the Greek language. Basiccally measuring elements.

Stoichiometry is about measuring the substances within a reaction, this includes the reactant itself and the product. 

When finding the measurements always remember to balance the equation.
CaCO3 -> CaCO + CO2
Identify the ratio = 1:3 By finding the ratio you are now able to do some calculations.

Energy Calculations

I'm back and better than ever!

It is now time for the mole to resurface and assist us in some calculations...

Energy in Equation:
- Energy absorption or release can be placed directly in an equation
Eg. CH4 + 2O2 = CO2 + 2H2O + 812 kJ
- Exothermic reactions have energy term on the RHS and negative Enthalpy, or energy released
- Endothermic reactions have energy term on the LHS and positive Enthalpy, or energy absorbed

Energy Calculations:
- Enthalpy is energy change of a reaction expressed in kJ per mole of one of the chemicals
- Using the above reaction as an example, we use coefficients of balanced equations for this exothermic reaction
-812 kJ/1 mol CH4 or -812 kJ/2 mol O2 or -406 kJ/1 mol O2
For products:
-812 kJ/1 mol CO2 or -812 kJ/2 mol H2O or -406 kJ/1 mol H2O

Now that we have the moles, we can use it for tons of different calculations!
Don't forget your sig figs because they have returned too!!!
Using the above reaction, calculate the amount of energy released when 0.35 moles of H2O is produced?

0.35 moles X -812 kJ/2 mol H2O = -140 kJ or 140 kJ released

Now calculate the moles of CH4 needed to produce 2100 kJ of energy.

2100 kJ X 1 mol CH4/-812 kJ = 2.6 moles CH4


Calculate how many grams of O2 would be needed to produce 1500 kJ of energy.

-1500 kJ X 1 mol O2/-406 kJ X 32 grams/1 mol = 120 grams of O2


Now, as you can probably tell, the return of the mole means that we are now able to do tons of Energy Calculations. STP, Particles, Formula Units, Grams, Atoms, etc. So, this is a good thing, right...? Of course!

Need a bit more help? Here's a video:

Energy Diagrams

Energy Diagrams chart potential energy of chemicals as they become products.
- Reactants start with a certain amount of energy that is added to start the reaction which is then either released or absorbed as the reaction proceeds.
- The amounts of energy determine whether the reaction will be exothermic or endothermic

Here is a Exothermic Reaction Diagram:

1) Energy of Activated Complex - Potential energy of "transition state" between reactants and products
2) Activation Energy - Energy that must be added to get the reaction to progress
3) DeltaH - Change in potential energy during reaction = Energy of Products - Energy of Reactants
~ positive change in enthalpy (H)  is an endothermic reaction, while a negative change is exothermic


Here is a Endothermic Reaction Diagram:

Here's a video for a bit more clarity:

Endothermic and Exothermic Reactions

Endothermic and Exothermic Reactions


Every single chemical reaction involves some kind of change in energy which fall into one of two categories Endothermic or Exothermic.

An endothermic reaction is one that absorbs energy. For example: instant ice packs absorb energy and are therefore endothermic.

An exothermic reaction is one that releases energy.  For example explosions release energy and are therefore exothermic.

At an atomic level... molecules are held together by chemical bonds.  These bonds can break if energy is added and they can form together if energy is given off to form bonds.

In an endothermic reaction it takes more energy to break the bonds then the reaction gives off.  But in an exothermic reaction it takes less energy to break the bonds then the reaction gives off to form bonds.

For an in-classroom demonstration of types of endothermic and exothermic reactions watch this video!