Saturday, February 11, 2012

Lab 5B - By Sally Chen

Objectives
1.  To observe a variety of chemical reactions.
2.  To interpret and explain observations with balanced chemical equations.
3.  To classify each reaction as one of the four main types (Synthesis, Decomposition, Double Replacement, and Single Replacement).
Decomposition

 Synthesis
Single Replacement
Double Replacement
Supplies
- Equipment:  lab burner, 6 test tubes (one will be flame heated), test tube clamp, medicine dropper, wood splints, crucible tongs, steel wool, safety goggles, lab aprons.
- Chemical Reagents:  copper wire (bare), iron nail, 0.5M sodium carbonate solution, mossy zinc, 2M hydrochloric acid solution (6%), manganese (IV) oxide.

Procedure
1.  Lab aprons + safety goggles.
2.  Observations before, during and after each reaction.  Record.
       Reaction 1
       1. Burner (high heat).
       2. Hold copper in the hottest part for a few minutes (crucible tongs).
       Reaction 2
       1. Clean iron nail till shiny with steel wool.
       2. Tube with CuSO4, half of nail covered.
       3. After 15 min, remove nail, note changes of solution and nail.
       Reaction 3
       1. 1/3 of tube filled with solid Copper (II) sulphate pentahydrate.
       2. Heat the tube, move gently over flame.
       3. Heat till no changes observed.
       4. Save content for Reaction 4.
       Reaction 4
       1. Wait till the tube from 3 cool.
       2. Medicine dropper:  2 to 3 drops of water.
       Reaction 5
       1. Test tube 1:  1/4 full of CaCl2 solution, test tube 2:  1/4 full of Na2CO3 (aq).
       2. Pour the content in test tube 1 into 2.
       Reaction 6
       1. Test tube:  mossy zinc
       2. Add HCl (aq) till zinc is fully covered by HCl
       Reaction 7
       1. Test tube:  1/2 hydrogen peroxide (aq)
       2. Small amount of MnO2 (catalyst)
       3. Put a glowing but not burning splint into mouth of the tube to test the gas.
       4. Wash hands thoroughly before leaving.
Place all liquid and solid waste into the designated waste containers.

Observations
Reaction 1
Before:  copper: orange
During:  the flame turns from blue to green almost instantly; copper gradually turns black.
After:  copper:  black
Reaction 2
Before:  nail: silver; CuSO4 (aq): sky blue
During:  nail covered by CuSO4 gradually turns orange; CuSO4 turns from blue to dark green.
After:  nail:  thicker with orange powder-like copper sticking onto the nail; liquid:  dark green
Reaction 3
Before:  CuSO4·5H2O:  sky blue solid in powder form.
During:  CuSO4·5H2O gradually turns from blue to white, from outside to inside, water drops form at the mouth of the test tube; there was a little white steam; some white powder gradually turns black or even orange/yellowish in the end.
After:  CuSO4:  white, partially yellow/orange-ish solid (one big piece, no more powder)
Reaction 4
Before:  CuSO4:  white, orange-ish solid (one big piece)
During:  instantly turns blue
After:  CuSO4·5H2O:  sky blue solid (one big piece)
Reaction 5
Before:  Na2CO3:  clear liquid; CaCl2:  clear liquid.
During: instantly forms white solid and clear liquid.
After:  white solid (CaCO3) in transparent liquid (NaCl)
Reaction 6
Before:  zinc:  silver metal; HCl:  clear liquid
During:  instantly starts to form small bubbles; the amount of zinc lessens.
After:  clear liquid (ZnCl2) and a little mossy zinc.
Reaction 7
Before:  H2O2:  clear liquid; MnO2:  black powder
During:  instantly starts to form a lot of big bubbles (really active); glowing stick starts to burn; test tube gets really hot; white steam
After:  black liquid (black powder in clear liquid)


Questions That Might Help You to Understand the Lab Better
1. Q:  What types of reactions are reaction 1 to 7?
A:  1 Synthesis, 2 Single Replacement, 3 Decomposition, 4 Synthesis, 5 Double Replacement, 6 Single Replacement, 7 Decomposition
2. Q:  Chemical equations for reaction 1 to 7?
A:  1)  2Cu (s) + 1O2 (g) ---> 2CuO (s)
2)  1CuSO4 (aq) + 1Fe (s) ---> 1Cu (s) + 1FeSO4 (aq)
3)  1CuSO4·5H2O (s) ---> 1CuSO4 (s) + 5H2O (l)
4)  1CuSO4 (s) + 5H2O (l) ---> 1CuSO4·5H2O (s)
5)  1Na2CO3 (aq) + 1CaCl2 (aq) ---> 1CaCO3 (s) + 2NaCl (aq)
6)  1Zn (s) + 2HCl (aq) ---> 1ZnCl2 (aq) + H2 (g)
7)  2H2O2 (aq)   MnO2  >  2H2O (l) + 1O2 (g)

2. Q:  What are some errors students may make during the lab?
A:  R1) Burner is not turned to high heat so Cu and O2 do not react.
R2) Iron nail is not cleaned with steel wool; there might be FeO on the surface of the nail; therefore there might be changes in observation.
R7) A burning splint is put into the mouth of the test tube, therefore no obvious changes are noted.
Or, a glowing splint is put at the mouth of the test tube, therefore no obvious changes are noted.

3.  Q:  What is copper reacting with in reaction 1?
A:  Oxygen in the air.

4.  Q:  What does the colour change in reaction 2 indicate?
A:  It indicates that Cu2+ released in water produces the blue colour, Fe2+ replaced Cu2+ to cause the colour change.

5.  Q:  How to test H2 and O2?
A:  H2:  place a BURNING stick at the mouth of the test tube.  If the gas starts burning or create a little "boom" sound, it's H2.
O2:  place a GLOWING stick at the mouth of the test tube.  If the stick starts burning, the gas is O2.

6.  Q:  What do you learn from this lab?
A:  There are 4 main different types of reactions, decomposition (with one reactant forming two or more products), synthesis (with two or more reactants forming one product), single replacement (with one single-element reactant and one double-element reactant forming another single-element product and another double-element product) and double replacement (with two reactants switching + ions to get two products).\

7.  Q:  What types of reactions are not included in this lab?
A:  Neutralization (acid + base = salt + water, a special double replacement reaction) and combustion (burning of organic material in oxygen).

Tuesday, January 10, 2012

Lab 4C Percent Composition of Unknown Hydrates - By Sally Chen

Objectives
1.  To determine the percentage of water in an unknown hydrate.
2.  To determine the moles of water present in each mole of this unknown hydrate when given the molar mass of the anhydrous salt.
3. To write an empirical formula of the hydrate.

Supplies
Equipment:  lab burner, crucible and lid, crucible tongs, pipe stem triangle, ring stand and ring, centigram or digital balance, lab apron, safety goggles.
Chemical reagents:  approximately 5 g of a hydrate, water.

Procedures
1. set up equipments, make sure flame is blue
2. heat empty crucible
3. cool empty crucible
4. measure the mass of crucible
5. fill 1/3 of crucible with hydrate, record mass
6. heat and then cool, record mass
7. reheat & cool, record mass
8. take the lower mass in 6 and 7
9. add water to the content, record changes.

Questions That Might Help You to Understand the Lab Better
1. What did you learn from the lab?
A:  1) we can decide some hydrates' percent composition by heating
2) we cannot decide some hydrates' percent composition because they would decompose before heating is done.
3) Hydrates might have different colors than their anhydrous salts.
4) Not all hydrates would go through a color change during heating.

2.The reason the answer is not accurate:
A:  1) the hydrate decomposes under high temperature (mass of anhydrous salt )
2) the flame is not blue so carbon would accumulate under the crucible (mass of anhydrous salt )
3) experiment errors
4)the crucible does not cool completely before weighing (mass of anhydrous salt )
5) water is not completely driven off (mass of anhydrous salt )

3. How to name the hydrates?
A:  Na2CO3 10H2O = sodium carbonate decahydrate
CuSO4 5H2O = copper sulphate pentahydrate
(1 mono, 2 di, 3 tri, 4 tetra, 5 penta, 6 hexa, 7 hepta, 8 octa, 9 nona, 10 deca, 11 hendeka, 12 dodeca)

4. Why heating the crucible first before starting the experiment?
A:  to drive off water in the crucible so the experiment would be more accurate.




Sunday, January 8, 2012

Standard Temperature and Pressure--By Tina Zhao

Standard Temperature and Pressure:
-Molar volume of a Gas at STP. Gases expand and contract (change volume) with changes in temperature and pressure. We have a Standard cordition to compare volume of gases called STP.

-In chemistry, IUPAC established standard temperature and pressure (informally abbreviated as STP) as a temperature of 273.15 K (0 °C, 32 °F) and an absolute pressure of 100 kPa (14.504 psi, 0.986 atm), An unofficial, but commonly used standard is standard ambient temperature and pressure (SATP) as a temperature of 298.15 K (25 °C, 77 °F) and an absolute pressure of 100 kPa (14.504 psi, 0.986 atm).

- At STP 1 mole of gas occupies 22.4L
Thus we can create the conversion factors:
      22.4L of gas/ 1 mole of gas  OR 1 mole of gas/ 22.4L of gas
 Example: calculate the volume occupied by 3.4 g of ammonia at STP.
    First step: Molar mass of ammonia(NH3)
     (1x14)+(3x1)=17g/mol
    Step two: Moles of ammonia
      3,4g of NH3/ 17g= 0.20 moles of NH4
    Step three: Molar volume
      0.2 moles x 22.4L= 4.5 L
(volume occupied by 3,4 g of ammonia at STP=4.5L)
http://www.gaston.k12.nc.us/schools/highland/faculty/blpadgett/Course%20Outline%20and%20Syllabus/CH%2011%20Molar%20Volume%20Worksheet.pdf if you want do more example!!!

Thursday, January 5, 2012

Molar Concentration -- by Ria Park

Before Start - Rewview
 - Solute is the chemical that has the smaller quantity.
 - Solvent is the chemical that has the larger quantity.


Molar Concentration(Molarity)
 - is the number of moles of solute in a specific mount of litre of a solution.
 - We use "M" or "mole/L" to denote molarity.

               Molarity = moles of solute(mol) / volume of solution(L)

   e.g. If there is 3.40 moles of NaCl in 1.2 litres of solution, what is the molar concentration?
           
               Molarity =  moles of solute(mol) / volume of solution(L)
                             = 3.40 moles of NaCl / 1.2L
                             = 2.8333333...
                             = 2.8 mol/L NaCl  or  2.8 M
  

   e.g. Calculate the molarity of a solution that has 0.738 moles NaOH in 2.40L of solution.
   
              Molarity = mol / L            M  of NaOH = 0.738 mol NaOH / 2.40 L
                                                                           = 0.3075
                                                                           = 0.308 M NaOH  or  0.308 mol/L NaOH

   e.g. How many moles of KCl are contained in 1.5 L of 2.00 mol/L KCl?

              moles KCl = molarity × volume
                               = 2.00 mol/L × 1.5L
                               = 3.0 moles KCl

   e.g. What volume of 0.34 mol/L NaCl do we must take to obtain 0.021 mole of NaCl?

            volume = moles / molarity
                       =  0.34 mole NaCl / 0.021 mol/L NaCl
                       = 16 L




And here is very interesting video for fun learnig.

 

http://www.youtube.com/watch?v=WsQk8C098zA

Tuesday, December 13, 2011

Empirical Formula of Organic Compounds By Sally Chen

- An organic compound is a covalent compound containing carbon.
- Process:
     -find CO2 and H2O (mole)
     -find C and H
     -C:H
     -CxHx
     -check the answer
     -if the totals don't balance, there is another element in the compound (usually O or N)
              -difference of totals = mass of the other element
     -CxHx or CxHxUx
- Example
     An organic compound weighed 99.99 g was burnt and in the product there is 191.29 g carbon dioxide and 117.36 g water.  (there IS the element O in the compound)
           - CO2:  191.29 g x 1 mol / 44.0 g = 4.3475 mol
             H2O:  117.36 g x 1 mol / 18.0 g = 6.52 mol
           - C:  4.3475 mol
             H:  13.04 mol
           - C:H = 1:2.999424957 = 1:3
           - Check:  C:  4.3475 mol x 12.0 g / 1 mol = 52.17 g
                       H:  13.04 mol x 1.0 g / 1 mol = 13.04 g
                       52.17 g + 13.04 g = 65.21 g < 99.99 g
           - O:  99.99 g - 65.21 g = 34.78 g
             34.78 g x 1 mol / 16.0 g = 2.17375 mol
           - C:H:O = 1:2.999424957:2.17375 = 1:3:2.17375 = 2:6:1
           - C2H6O


Monday, December 5, 2011

Chapter 4 Empirical & Molecular Formula & Percentage Composition -- By Nemo Jin


Chapter 4 
--Empirical & Molecular Formula & Percentage Composition
Percentage Composition

Definition:
--Percentage Composition is the percentage of the mass of a certain element in a compound 
Calculation:
--Calculate the molar mass of the compound
--Calculate the mass of each element in the compound
--The %composition of X element = mass of X element / total mass of the compound * 100%

Ex.
In ionic compound FeO
The percentage composition of Fe = (55.8 / 55.8 + 16.0)*100% = 77.7%
Applied skills:
(Connect Empirical Formula with Percentage Composition)


--Use percentage composition to find out the empirical formula of a certain compound.

Question: 
A compound is made of 72.8% Oxygen and 27.2% Carbon. Find out the empirical formula of that compound.

--Assume there is 100g compound 
--so that the mass of oxygen is 72.8g and the mass of carbon is 27.2g
--find out how many moles by doing “72.8g/16.0g=4.55 moles” and “27.2g/12.0g=2.27 moles” 
--the biggest number is divided by the smallest “4.55/2.27=2”
--write out the empirical formula, gives you the formula of Carbon Dioxide CO2


Empirical Formula And Molecular Formula 


--Empirical Formula of a compound gives you the lowest-term of ration of atoms in the formula, for example C2H3. 2 and 3 are both not divisible by any number that is larger than 1.


--Molecular Formula of a compound gives you all atoms exist in that compound. For example C4H6. 4 and 6 are both divisible by 2.


Conversion.
--Convert empirical formula to molecular formula you just need to multiply every atoms by n
--Convert molecular formula to empirical formula you just need to divide every number by the same number n.