Atomic Structure

Today we have learn the atomic structure of an atom. There are three parts to consist an atom: Proton, Neutron and Electron.

Atomic Number: The number of protons found in the nucleus of an atom that atoms have no overall electrical charge. Atomic Number= number of protons= number of electrons

Subatomic Particles

Ions:
Most atoms are capable of either gaining or losing electrons. A few elements, like hydrogen, are able to do both. They can do this by accepting electrons from , or giving electrons to, other atoms. Atoms that have gains or lost electrons are called ions.

Mass Number:
The mass number (A), also called atomic mass number or nucleon number, is the total number of protons and neutrons (together known as nucleons) in an atomic nucleus.

Atomic Mass:
The atomic mass (m_a) is the mass of a specific isotope, most often expressed in unified atomic mass units.The atomic mass is the total mass of protons, neutrons and electrons in a single atom (when the atom is motionless).

Isotopes:
Isotopes are variants of atoms of a particular chemical element, which have differing numbers of neutrons. Atoms of a particular element by definition must contain the same number of protons but may have a distinct number of neutrons which differs from atom to atom, without changing the designation of the atom as a particular element.

Atomic Theory

Greek philosophers believed that atomos that were the smallest pieces of matter.
- Aristotle who believed in four element of earth, air, fire and water
- Alchemist who desired to turn common metals into gold
Their activities marked the beginning of our understanding of matter

But it is not a scientific theory because it could be tested through  observation.

There are a Earliest theory about atomic theory who is Democritus, a 300 b.c. greek philosopher who said that atoms are invisible particles.

Later in the late 1700s came Lavoisier after Democritus, he stated the first version of the law of conservation of mass and Law of difineite proportions

After that, Proust in 1799 proved that Lavoisier's Laws by experiments.
Follow up is Dalton in early 1800s who defined atoms as solid and indestructible spheres base on the Law of Conservation of Mass

Later, J.J. Thomson (1850s)  rose the first theory that have positive and negative charges in atoms and demonstrated the existance of electrons using a cathrode ray tube.

Rutherford in 1905 shwoed atoms have a positve, dense center with electrons outside it and ecplains why electrons spin around nucleus and suggested atoms are mostly empty space.

Atomic Theory IV
Neils Bohr (1885-1962), studied gaseous smaple of atoms, which were made to glow by passin gan electronic current through them.

That's about all of the history of chemists in the earlier eras. These are the ones of the greatest chemists in human history. We learn new things from them. Even the scientists today apply these theories into their experiement and exploration of chemistry. So it is worthwhile to learn about them. I hope you have learned something from this particular blog.

Percent Yield & Percent Purity

Percent Yield is the calculation of the amount of product produced to the amount of product one expected.



FORMULA
                         grams of actual product recovered
Percent Yield= -----------------------------------------------  x 100%
                         grams of product expected from stoichiometry

On the other hand, Percent Purity is the percentage of the mass of a substance to the mass of the impure substance. This means there are usually some impure substances in a chemical reaction. But we need to calculate the amount of impure substance in an equation for example.

 FORMULA:
                 mass of Pure Substance
% purity=----------------------------- x 100%
                 mass of Impure Substance


For example,

>If a 156.0 g of Cu ore contains 60.00 g of pure Cu metal. What is the percent purity?
                 60.00g of Cu
% purity=------------------ x 100% = 38.46 %
                156.0g of Cu ore
>There are 6.5 grams of impure Zn, the percent purity is 87%, what is the mass of the pure Zn?
                       x
87%= -----------------------
              6.5 g of Impure Zn

so x= 0.87 * 6.5g Impure Zn
     x= 5.6g of pure Zn

Lab 6D

Today we did a super cool expriment on Limiting Reactant and Percent Yield in a precipitation reaction.  Our objective is ofcourse to first observe the reaction in the double replace ment reaction
Na2Co39(aq)+CaCl2(aq)------->NaCl(aq) + CaCO3(S).  Then we determine the limiting reactant and the excess reactant.  At the end will will than classify the Percent Yield by comparing our product produce in this experiment and the theoritical product.

Procedure:
1.  In the begining of the lab we put on our safety equipments.
2. We obtain Na2Co3 solutions and CaCl2 solution and record our measurements.
3. Than we poor both sultions into a beaker and leave it sit for 5min
4. Than we write our name on a filter paper and set up the filter apparatus like this

5. We then place grducated cylinder underneath the funnel to start filtering. 
6.  We will use a wash bottle to rinse any NaCl left on the filter paper so that only the precipatate remains.
7. At the end we remove the filter paper with the precipitate CaCO3 into a dry location so that we can measure the percent yiled next class.

End of the Lab
With the double replacement balanced equation  1Na2CO3(aq)+1CaCl2(aq)------->1NaCl(aq) + 1CaCO3(S) we can find the limiting reactant. To find the limiting reactant, we simply just convert one reactant to another to see if they exceed more they are presented or the Presented is more than it is needed

E.g  we use (25ml) of 0.70M Na2CO3 and (25ml) of 0.50M of CaCl2.  In the end we converted them into moles.  0.018mol Na2CO3 and 0.012mol CaCl2.

0.0125molCaCl2* molNaCO3/molCaCl2=0.0125mol CaCl2.  Thus CaCl2 is the limiting since it needs more than it is given.

Conclusion
To find the percent yield of this experiment, we will need to weigh the mass of the filter product which is next class.  The formula to find the percent yied is

Percent Yield=   actual mass produced (grams)    x 100

           theoretical mass produced (grams)

Stoichiometry: Excess and Limiting Reactants Percent Yield

In this section, we learned that for a balanced equation, there is some conditions that the reaction may not present ( may be caused by pressure, temperature, concentration, etc.). Sometimes it is necessary to add more reactant into the chemical reaction in order to react.

One reactant is the EXCESS QUANTITY that would have some left overs at the end. The second reactant is used up is called the LIMITING QUANTITY.

The videos belows tells you how to calculate excess quantities and limiting quantities based on the equation.
ENJOY!

Excess and Limiting Reactants

Today in class we learned the reactants in a chemical reaction is not always exact which means some reactant may have more amount than others.  Thus there will be excess of reactants during the reaction.  We also learned about the limiting reactant; a reactant that is not present in alarge enough quantity to fully react with another reactant.  The limiting reactant is significant because it helps us determine how much product can be formed.  This concludes what we learned today.

Examples





In this cass the car bodies are the litmiting reactant becaucause not matter how many tires there are only 8 car bodies are availible to make a car.  Thus only 8 cars can be formed and there is 16 tires excess.



A sample of 111.6g of Fe is mixed with 96.3g of  S.

a) Which reactant is in excess

b) Which is the limiting reactant?
c) When this reaction is carried out, what mass of FeS will actually be produced?

Solution
1) Write out the balance equation.
1Fe+1S----->1FeS

2)Convert Fe and S into moles
Mol Fe= 111.6g Fe *1molFe/55.8gFe =2.00 mol Fe
Mol S= 96.3g S * 1molS/32.1gS= 3.00mol S

3) Use mol ration to determined how much S is need to react with Fe
2.00mol Fe* 1molS/1molFe= 2.00mol S needed.  Thus S is the excess reactant because 3.00 moles of S is presented in the beginning. Since S is the excess than Fe will be the litming reactant.

4)Use the mass of the limiting reactant to find the mass of FeS that actually produced because we cannot start with the mass of the excess reactant S.  We need to use Fe the limiting reactant or we will get more mass of the product than we actually can.
111.6g Fe* 1molFe/55.8g Fe*1molFeS/1molFe*87.9gFeS/1molFeS=176gFeS

Molarity and Stoichiometry

So, today we've taken stoichiometry one step further, which is including the molarity.

Remember the stuff we did last day on stoichiometry was basically to take gram stuff into the mole stuff and such and such, but today, we involved molarity.

As we know, molarity is a concentration stuff, so for example:

-How many mL of 0.124M NaOH contain enough NaOH to react with 15.4 mL of 0.108 M H2SO4 ?

-First of all, we have to wrtie the chemical reaction out and balance it.

    #mL       15.4mL
2 NaOH + H2SO4 ---> 2 H2O + Na2SO4
0.124 mol   0.108 mol

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

     1L               1L

and the question has given out a lot of information, as you can see, it is asking the milliliter of NaOH (in red). And we all know the molarity can be written as moles per L, so the question said 0.124 M, which is equivelent to 0.124 mole per L (in yellow). and another piece of information is 15.4 mL (in green) of 0.108M, or 0.108 mole per L (in aqua).

And now let's start doing some conversion!!


 15.4 mL           1 L          0.108 mol H2SO4     2 mol NaOH     
-------------- x --------------- x ----------------------------- x ---------------------
   1 L           1000 mL                   1 L              1 mol H2SO4   
                  1 L             1000mL
    x ------------------------- x ------------ = 26.82580645 mL NaOH   <---- WRONG!
       0.124 mol NaOH        1 L

Be careful !! sig fig counts for mark!!

26.82580645 mL NaOH, 3 sig figs = 26.8 mL NaOH

+

 

YAY !