Percent Composition

It is a percentage that you can calculate the mass of one of the elements in a chemical formula.

The Questions of the Fives:

1. If a compund contains 394.0g of Au, 96.3g of S and 192.0g of O, calculate the total malar mass and % compostion.

2. What percent of Rutin, which is found in buckwheat and in the fruit of the Fava D'anta tree etc, is C₂₇H₃₀O_x , has a total molar mass of 610g/mol. Find the number of oxygens there are in this compound and the percentage composition.

3. Calculate the percentage of the bold species in each of the following:
       a) CaCl2 + 2H2O

4. If a compund contains 394.0g of Au, 96.3g of S and 192.0g of O, calculate the total malar mass and % compostion.

5. What percent of Rutin, which is found in buckwheat and in the fruit of the Fava D'anta tree etc, is C₂₇H₃₀O_x , has a total molar mass of 610g/mol. Find the number of oxygens there are in this compound and the percentage composition.

5.Determine if the percent compostion of ammonium oxalate (NH4)2C2O4) is accurate
%N=22.6%
%H=6.5%
%C=19.4%
%O=51.6%

Solutions:

3. Total MM= 682.3g/mol

     % of Au= (394.0g/mol)/(682.3g/mol)=57.74%

     % of S= (96.3g/mol)/(682.3g/mol)=14.1%

     % of O=(192.0g/mol)/(682.3g/mol)=28.14%

4. MM oxygen= 610.0-324.0-30.0=256.0g/mol

          (256.0g/mol)/16=16 oxygen atoms

     % of C27=(324.0g/mol)/(610.0g/mol)=52.11%

     % of H=(30.0g/mol)/(610.0g/mol)=4.92%

     % of O=(256.0g/mol)/(610.0g/mol)=41.97%

5.MM ammonuium oxalate=124g/mol
% of N: 28.0g per mol/ 124g/mol x 100= 22.6%
% of H: 8.0 g per mol / 124g/mol x 100 = 6.5%
% of C: 24,0g per mol / 124g/mol x 100 = 19.4%
% of O: 24.0g per mol/ 124g/mol x 100= 51.6%
Thus the Percent Compostion is correct!!

Empirical and Moliecular Formula

Today In Class we learnt about two diffrent formulas.  They were very interesting because they are somewhat related.  The two formula we leart are:

Empirical Formula
-It is the simple formulae expressing the compostion of a compound.  It gives the simples whole number ratio of atoms or ions.

eg. The empirical in this case will be Fe3O4 as they are calculated by dividing the number of grams they have
by the total number of mass of the combining compound which is the molar mass.  After Fe and O is divided by the MM than we can use the lowest ratio in this case 1.29 to divide by the other rations like 1.72.  At the end it resulted 3 Fre and 4 Oxygens.  This shows the lowest number formula of the compound.

Molecular Formula
-Is a multiple of the empirical formula.  If the lowest common denominator of the subscripts is 1, the empirical and molecular foulae for the compoud are the same.

eg. Molecular Formula

An organic compound is analyzed and found to be 69.5% carbon, 7.3% hydrogen, and 23.2% oxygen
by mass.  The molar mass of this compound is 276.0g/mol. If the EF is found to be C4H5O than what is its Molecular Formula

Easy!
If we know the the most simple formula of the compound and we were given the molar mass of the original compound which is 276.0g/mol than we can find the MF.

1. We first Find the MM of C4H5O the EF of the compound=69.0g/mol
2. Than we will devid the MM of the EF by the MM of MF which will be 276.0g/mol/69.0g/mol
3. You will than receive the number 4
4. Than you will need to multiply the number 4 to the original EF which will be 4(C4H50)
5. You will than get the MF C16H20O4

Mole Conversion Quiz!

                      Today we had a easy quiz on Mole Conversions.  It was easy because the quetions are based on the following:

1. find the molar mass of a compound
2. convert molecules, atoms, to grams
3. convert grams, atoms molecules to mole

Map that helped me prepare for the quiz

                   After the quiz we spend 10min working on another work sheet on Mole Convertions again.  How fun!!

Harder Mole Conversion

At the previous class of the mole conversion, we've learnt:
-From particles → moles.
-From moles → particles/molecules/formula units/atoms
-From moles → grams
-From grams → moles

The previous sections showed how to perform single-step conversions between moles and any of mass, volume, or number of particles.  This section shows how to convert between mass and volume, number of particles and mass, and so on.  The box table below summarizes the conversion factors needed.


                            Conversion                    |              Conversion Factor       
          
                                                                        6.022x10^23 particles
                 -mole → particles                     |               1 mole
                ______________________________________________________       
               
                  -mole → mass                        |      molar mass g                                  
                                                                             1mol
                ______________________________________________________

                -mole → volume                       |           22.4 L
                                                                           1 mol
                ______________________________________________________

               -molecules → atoms                |           # atoms      
                                                                         1 molecule

Mole Conversions

There are two main types of mole conversions:

1. Gram to Mole Conversion
2. Mole to Gram Conversion
3. Particles to moles Conversion
4. Moles to Particles Conversion

First of all,
Let's see some Grams to Moles Conversion

Moles to Grams Conversions:



Conversions between particles and moles:

More information about Mole Conversions:



Chapter 4 Moles

Chapter 4 Moles



Avogradro and his mole number!

 

                                    What is a mole?

         A mole is a number, just like a dozen is the number 12.  However the mole number is called the Advagadro's number which is about 6.022*10^23.

Why do is mole so important in Chemistry?
-Acording to Avogadro's Hypotheses, Gas with equal volume at same temperature and pressure will have the same number of particles.


Equal volumes of gas, at the same temperature and pressure, contain the same number of molecules.-This means each of these gas will contain 1 mole and 1 mole more=6.022*10^23  partciles,unites or molecules



Review: Chapter 3

Significant Figures:
-measured or meaningful digits.
-can be precisedif there are many of them.
e.g 24.37  24.6

Accuracy:
-It is how close the measurement (or average measurement) comes to the accepted or real value.

Precision:
-It is how reproducible a measurement is compare to other similar measurements.

Absolute Uncertainty:
-the uncertainty expressed in the units of measurement, not as a ratio.

Relative Uncertainty:


-Relative Uncertainty =         Absolute Uncertainty
                                       ———————————
                                        Estimated Measurement


Density:
-defined as its mass per unit volume.

-formula of density:             Mass
                         D    =     ————
                                       Volume

Lab 2E Quiz and Graphing

Today we have just done a quiz on Lab 2E.

The quiz was just all about calculating the thickness of a certain object, such as alunminum foil.

Equation:

Density=Mass/Volume

If you want to calculation the thickness based on the information of an object:
Density=2.70g/cm^3
Mass=5.0g
The object Area=15cm by 16cm
And calculation the thickness:
It goes,
Volume=(2.70g/cm^3)/5.0g=0.54m^3
Thickness=0.54cm^3/15cm/16cm=2.25*10^-3cm

Next, we went to lab room and learnt how to graph base on the information of mass and volume.

 

Lab 2 E

               Today in class, we had an exciting lab on determining the thickness(height) of the aluminum foil.The lab was to help us get more familiar with sig figs, accuracy, precision, uncertainties and the formula of density.  Thus we all enjoy this super amazing lab!!!

Purpose: Find accurate and precise measurement of the thickness of the aluminnum foil expressed using correct number of sig figs.

Procedure

1. We start off by measureing all measuring the length and width of foil that is at least 15 cm on each side

2. Than we use the centigram to measure the mass of each foil.

3. After we calculated the Mass we than use the density formula to calculate the volume of the foil.

4. When we finished calculating the volume of the aluminum foil, we than use the lengh and width to calculate

the height because LxWxH=V

eg.V=0.36cm^3

L=15.60cm

W=15.35cm

Hight=0.36cm^3

--------------

15.60cm*15.35cm

=1.5*10^-3(Remember  round you sig figs till the very end!!)

5.  The final step is for us to calculate the experimental error when it comes to the accepted value for thickness of the aluminum foil which Mr.Chen gave us was 1.55*10^-3.

Formula Expremental error:

Experimental error=Your measurement-Accepted value

---------------------------------------------

accepted value *100

6.At the end of the experiment we wrote a conclusion on how accurate and precise

our messurement was.  Our measurement was precise because our experimental error

was only 3.2%. It was also precise because we got almost the same value each

time when we calcute the thickness of the foil for 3 trials.

Need more help go to:   http://www.mefeedia.com/watch/29045208

Note:Quiz next class on everything we done today on the lab !!!!

Density

Density:
-defined as its mass per unit volume.
-or, its weight per unit volume.
-Osmium is the most densest known substance at standard conditions for temperature and pressure.
-less dense fluids float on more dense fluids if THE FLUIDS DO NOT MIX.

 Here is an example of density between fluids.















-formula of density:             Mass
                         D    =     ————
                                       Volume


Example: An iron bar has a mass of 12.00 g and a volume of 1.25 L.  What is the iron bars density?

    As we know D=M/G, we could simply plug in the number, D=12.00g/1.25L which equals to 960 but don't forget the units! Its 960 g/L.

Leif Tung