Cooper (II) Chloride Lab

Day 1

To start the lab we took the mass of a baby food jar then added 4.00grams of Copper (II) Chloride to the jar and 50.0grams of distilled water. We stirred the solution until it was completely dissolved. After polishing off a nail with steel wool we placed it in the container and let it sit.

Day 2

We pulled of the remainder of the nail out of the jar. Then, we slowly poured out solution careful to not lose any copper product. We did the same two more times after washing it with HCl and more distilled water. We set the jar with copper out to dry. To finish off we took the ending mass of the nail.

Day 3

We recorded the mass of the baby food jar with the copper and then subtracted the previous mass of the baby food jar with no solution to find out the mass of product we had. Then it was time to start calculations. A single replacement reaction took place place but the charge of iron was unknown so we had to create a balanced chemical equation for each possible charge of iron (2+ or 3+) and find theoretical yield of each then use our percent yield formula to calculate which had the highest percent yield; therefore, telling us the charge of the iron. After calculations were done, it was found that iron with a 2+ charge created a percent yield above 100% which is impossible, so the charge on iron in or reaction was 3+.

Here are some pictures of the lab:

Percent Yield

The basic formula for percent yield is:

Use your knowledge of stoichiometry and limiting reagent (with the help of my two previous blog posts) to solve for theoretical yield. Actual yield comes from the results of your lab.
The following practice problems  can help solve for percent yield:  http://tmlimitingreagents.weebly.com/percentage-yield-and-actual-yield-practice-problems.html

Limiting Reagents

We learned two approaches to solve limiting reagent problems:

Approach 1:
1. Balanced chemical equation
2. Convert all info to moles
3. Calculate the mole ratio from the given info. Compare calculated ration to actual ratio.
4. Use the amount of limiting reagent to calculate the amount of product produced.
5. If necessary, calculate how much is left in excess of non-limiting reagent

Approach 2:
1. Balanced chemical equation
2. Convert info to moles
3. Use stoichiometry for each individual reactant to find mass of product produced
4. The reactant that produces lesser amount is the limiting reagent
5. The reactant that produces larger amount is excess reagent
6. To find amount of remaining excess reactant, subtract the mass of excess reagent consumed from total mass of excess reagent given.
**This approach is the one I prefer to use

The following video and link can be used for extra practice:
https://www.khanacademy.org/science/chemistry/chemical-reactions-stoichiome/limiting-reagent-stoichiometry/v/stoichiometry-limiting-reagent
http://www.chemteam.info/Stoichiometry/Limiting-Reagent.html

Stoichiometry

Solving problems using stoichiometry uses a balanced chemical equation. Then you start with what is given. Convert grams of the given substance and convert to moles using the periodic table, then multiply by the mole ration from the balanced chemical reaction, and finally back to grams using the periodic table again. Here's a chart to map it out a bit better:

The following video can help explain more: https://www.youtube.com/watch?v=SjQG3rKSZUQ

Here are some practice problems:  http://www.chemistry.wustl.edu/~coursedev/Online%20tutorials/Plink/Stoichiometry/stoichset.htm

Reactivity Series Lab

In this lab we watched different redox reactions occur and then solved for their net ionic equations. We were able to use aqueous solutions of water, hydrogen chloride, copper sulfate, and river nitrate. With those we reacted each with solid forms of lead, copper, calcium, magnesium, zinc, and tin and watched to see if any sort of chemical reaction took place. Using our results we had to create our own reactivity series.
In order to predict their solution we had to use a reactivity series table that showed us whether the single replacement reaction would occur. Here is a link to a site that describes this table: http://www.bbc.co.uk/education/guides/zqjsgk7/revision
Here are some images showing some of the reactions that took place and others that had no reaction at all:

Redox Reactions

Redox reactions consist of electrons being transferred from the metal to the nonmetal. If a species loses electrons, it is said to be oxidized, and this is considered the reducing agent. If a species gains electrons, it is said to be reduced, and this is considered the oxidizing agent. An easy way to remember this is from the acronym OIL RIG. It stands for: Oxidation is loss; reduction is gain. Here's a visual for it:

http://www.ict4us.com/r.kuijt/images/en_oxidation_reduction.jpg

The first type of redox reaction we learned of was redox single-replacement reactions. It is set up as a single element reacting with a compound.. For this type of reaction, it it good t remember that "like attacks like". In this way, metal attacks a metal, while a nonmetal attacks a nonmetal.

The second type we learned of was synthesis. These reactions happens when two or more reactants come together and form one product. So A+B creates AB. Decomposition is the exact opposite of this, so there is one reactant breaking down to two or more products. This would then be AB creates A+B. 

The final type we talked about was combustion reactions. In this type of reaction, when a hydrocarbon reacts with water, the products are always water and carbon dioxide. Here is an example:

Here are some video's to further explain these concepts:
https://www.youtube.com/watch?v=RX6rh-eeflM
https://www.khanacademy.org/science/chemistry/oxidation-reduction/redox-oxidation-reduction/v/oxidizing-and-reducing-agents-1

Acid Base Reactions

These types of reactions will produce a water and a salt, and the water produced is the driving force of this reaction.
Here's a small example of what it would look like:

http://lrs.ed.uiuc.edu/students/mihyewon/images/HClNaOH.gif

Within the acid-base reaction, there is a possibility for both strong acids and bases, as well as a weak version of both. Strong acids produce H+, protonate completely, HCl, HBr, HI, and are the strongest if the oxygens outnumber the hydrogens by 2 or more. Strong bases contain an -OH- anion, disassociate completely, all group 1 and 2 metals plus the -OH anion are the strongest. Weak acids do not protonate completely, are not on our memorized list Weak bases do not disassociate completely, are not on our memorized list

A good thing to remember is when looking at the molecular diagram, to always look for the parents. If there are more parents, this means it is weak, or if there are less, it is strong. 

http://mgh-images.s3.amazonaws.com/9780073402680/5120-4-3IRC1.png

Here's a link to walk through the solving of these problems: http://science.widener.edu/svb/pset/acidbase.html

Here's a video for further explanation: https://www.youtube.com/watch?v=ANi709MYnWg

Double Replacement Reactions

In a double replacement reaction the driving force is the formation of a solid. The reactants which must be aqueous and ionic undergo disassociation, the ionic compounds break apart into cations and anions. The cations then replace eachother and create two new compounds, one being a solid. 

To know whether a solid is produced or not, you will need to know the solubility rules.

Some extra practice can be found at: http://www.chemteam.info/Equations/DoubleReplacement.html

Describing Chemical Reactions

Clues a chemical reaction has occurred are color change, formation of a solid, bubbles, and heat is produced or absorbed. The anatomy of a chemical equation is:

Subscripts tell us number of atoms of each element in a molecule. Coefficients tell us the number of molecules. In a combustion reaction, reaction with oxygen that produces a flame, the product is always CO2 + H2O. To balance these equations you must follow the CHO method. Balance in order starting with Carbon then Hydrogen then Oxygen. The following link can help describe the CHO method a bit more http://www.onlinemathlearning.com/balance-chemical-equation.html