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.

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!!!

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

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

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.

Mole Conversions --- By Tina Zhao

Mole Conversions

One- Step mole conversions

Conversions Particle/Atom/Formula Unit into Mole

Ex.1:How many moles of Carbon atoms are there in 3.01 X10^24C atoms?

             3.01 X 10^24   atoms 1 mol C/ 6.022ｘ10²³

Drawing and Interesting Graphs Using MS EXCEL --- by Ria Park

Since a graph represents the relationship between variables, sometimes a graph helps us to solve the density problem.

Density = Mass/Volume

Because a slope is equal to y/x(Rise-y over Run-x), the volume should be on x-axis. On the other hand, the mass should be on y-aixs.

We did the graphing with MS Excel. It was easy to make a graph and we can get the ideas easily.

Steps
- Open the MS Excel program and complete the table
- Click 'Insert', select 'scatter', then the graph will come out.
- Click anyone of the points and choose "Add Trendline" -> "Linear" or "Polinomial"
- Display Equation on the Chart
- Make it pretty!


And here are three graphs that we did in the class.

This is a video that demonstrates how to make a graph on excel.

http://www.youtube.com/watch?v=8B8kFVNzlQ8