Functional Groups

  Today we leanred about fuctional groups.  They are called hydrocarbon derivatives.  It is because the carbon is bonded other than hydrogen.   Each specific functional group gives rise to a family of organic compounds.

Example incude: Halocarbons, alcohols, organic acids, aldehydes, ketones, ethers, esters, amilines, and amides. 

The main ones we will focus on are:
 Halides & Nitro Compound

 Alcohols
Aldehydes & Ketones

1) Halides and Nitro Compound
-Halocarbons: R-X where X=F,Cl,Br,I
-A halocarbon is an organic compound with a halogen atom(F,Cl,Br,I) bonded to its structure.
-Nitro compound is same as Halides except only No2 is attached to the Carbon
 
Alcohols
- Is formed from a hydrocarbon that is convalently bonded to a hydroxyl group.
-expressed as R-OH where R represents a hydrocarbon chain or ring and OH represents at least one hydroxly group consisting of an exygen atom and a hydrogen atom.
Naming alcohols
-find parent chain containing the hydroxy group(OH)
-change ending to ol
-OH gets lowest possible locant
-when more than one OH include Greek prefixes.
Eg: Two OH groups are labelled diol


Ethanol



Aldehydes

-Gerneral formula is R-COH

-A Carbon in a chain is double bonded to an oxygen and single bonded to a hydrogen

Naming

-add al at the end

2-methylbutanal

Kentones
-Almost same as Alhydes
-The double bonded oxygen to the carbon chain is somewhere in the middle.  Unlike Alhydes it is always in the beginning or end
-stucture is R-Co-R
Naming
-Add one for the ending
-locate the ketones with lowest subscripts in the chain


3-methy-2-pentanone

That concludes what we had lernt today.  Now review again with these cool interesting videos!!!!!!!!

Alkenes and Alkynes

It's another part of the organic chemistry we have learned in chapter 23. Alkenes and Alkynes are double bonds and triple bonds respectively. The carbon can form bouble and triple bonds with Carbon atoms. When multiple bonds form fewer hydrogens are attached to the Carbon atom.

The naming rules are almost the same as the Alkanes: the postion of the double or triple bonds always has the lowest number and is put in front of the parent chain.

ALKENES
Hydrocarbons with one or more double bonds located between carbon atoms leading to an "unsaturated" hydrocarbon. The ending of a formula is changed from -ane to -ene (for alkenes).

GEOMETRIC ISOMERS : we give the double bonds formula each a different name based on their geometry.



Example:

Cis-2-butene

trans-2-butene







ALKYNES: hydrocarbons with one or more triple bonds located between carbon atoms leading to an "unsatrated" hydrocarbon.

Naming: the ending is changed fron -ane to -yne for alkynes.

Example: 

Organic Chemistry: The Chemistry of Carbon Compounds

One of the most important element to sustain life is to have the element of carbon (C) in substances like sugar, plants, and us, human. We cannot live without carbon. Organic chemistry is responsible for many of the every-day products that are used around the world.

Properties of Organic Compounds:

• low melting point
• weak or non-electrolytes
• can for chains of carbon atoms that are links in a straight-line, circular pattern and branched pattern.

• Can like with other atoms in...
• Single Bonds

• Double Bonds

• Triple Bonds





Then, here come a newer stuff, Alkanes, a unbranched/straight chain. For example, a hydrocarbon is a compound that contains only hydrogen and carbon. There are different types of hydrocarbons and there are different ways to represent them. Non-polar nolecules are immiscrible with water, which means they are non-dissolvable.

Here is a 3D modle of a tetrahdron:

Alkanes are saturated hydeocarbon which have all carbon atoms bonded by single bonds. Saturated means that it is not possible for another atom to bond to the structure. Naming of alkanes: the names of all hydrocarbons end in '-ane' because they are 'alkanes'!!!

Examples would be:

Easy to memorize, huh? Ofcourse it is!

Then we had learned about the branched hydrocarbons. Hydrocarbons can have "side branches" with are also hydrocarbon chains: These hydrocarbons are call substitured hydrocarbons or brached hydrocarbons.

Examples would be like:

Naming: the ames of all alkyl groups end in '-yl' bcause they are alkyl

Pentane C5H12       Pentyl C5H12
Butane C4H10        Butyl C4H10
Propane C3H8        Propyl C3H8
Ethane C2H6           Ethyl C2H6
Methane CH4          Methyl CH4

Rules of alkane nomenclature:

1. Find and name the longest continuous carbon chain
and place at the end of the name
2. Identify and name groups attached to this chain.
3. Number the chain consecutively, starting at the end
nearest a side group. (i.e. the lowest numbered carbon)
4. Designate the location of each side group by an
appropriate number and name.
5. Assemble the name, listing groups in alphabetical
order.

Electron Dot and Lewis Diagrams

It's very easy to draw a electron dot diagram and a Lewis Diagram. They are both about the same. Just for the Lewis Diagram, the electrons that are bonded between elements are a straight line, instead of a paired electron dot.

The nucleus is represented by the atomic symbol. For individual elements determine the number of Valence electrons. Electrons are represented by dos around the symbol. Four orbitals (one of each side of the nucleus) each holding a maximum of 2 e- (recall: maximum of 2 electrons perorbital). Each orbial gets 1 e- before they pair up.

The dots are placed in 4 groups of one or two electrons, with 8 electrons representing a closed shell or noble gas configuration. The dots are placed on the 4 sides in pairs as a reminder that electrons are paired in the orbitals.

Here are some practise problems from the internet:

Periodic Table Trend

              Today in class, we learned about the Periodic Trends.  The trends tells us the ralationship between elements on the table.  For example there are eight important trends we need to now.  There is the Metallic Properites trend, Atomic Radius trend, Ionization energy trend, electronegativity trend, reactivity trend, ion charge trend, melting/boiling point trend and the density trend.

Metallic properties trend
-elements tend to be more metallic in nature toward the left side of the periodic table
-metallic properties like malleability are result of the little valance electrons
-usally low ionization energies
-elements more metallic to the left side of the stair case
-elements more metallic when down a group

Atomic Radius Trend
-radius decreases when reading down from a group
-radius increases when reading from left to right of a row
-elements have higher enegy level at the bottom.  The atom is more sprend apart making the radius bigger
-elements atomic radius decreases from left to right of the row because the atomic number increases making the nucleaus more attractive to the other electrons.  Thus the atom is more packed together and not wide apart

Ionization of energy trend
-is the process of removing one or more elctrons from an atom to produce an ion
-opposite trend of atomic radius
-energy becomes bigger from left to right of the row because the atom is more tightly packed since the   electrons are attracted to the positve nucleuas with protons.  Thus harder to remove an electron from the atom
-energy decreases as reading down the group because electrons more far apart because of higher energy level. Thus electrons are easier to be remove and requir less energy

Electronegativity trend
-is the ability to attract electons from a neighbouring atom
-same trend as ionization energy
-elements to the right of the row has high ionization energy.  Thus they have higher attraction to their neighbors
-elements down a group has lower ionization energy since their atom is not tightly packed together.  Thus they are loose and will not have enough energy to attract their neighbors

Reactivity Trend
-is how readily an element will react with other elements
-the reacticity trend for metals is different from non metals.  For example when you read down the alkaline metal group the ionization energy decreases.  Therefore it is easier for the element to loos its valace electron and that will make the element more reactive
-for non metals, they tend to want to again electrons.  As you read upward the halogen group the element becomes more reactive since its got higher ionization level.  This means that elements have a lower tendency to lose electrons and a greated tendency to gain them.  Thus elemts become more reactive as you proceed up the group for non metals

Ion Charge
-Elements ion charges depend on thier group

Melting point/boiling point trend
-is the temperature at which it changes from solid to a liquid
-melting point of a substance depends the stregh of their bond.  If the bond is weak the melting point iwll be lower
-melting point for metals usually decreases as read down a group
-melting point for metals usually increases when moving to the right of the row until it reach the middle
-for the halogen group and the noble gas group, the melting point increases as read down the group
-element at the center of the circle has the highest melting point





Density

-The density usually increase by the atomic number

Predicting Valence Electrons(kevin)

..........

Electronic Structure of the Atom

In this Chapter, we have learned the energy level of the electrons surrounding the nucleus. Different elements have different electronic energy level. An Energy Level is the amount of energy, which an electron in an atom can possess. ("n" is the number of the energy level). The energy difference between two particular enegy level is called the quantum of energy.

                                                                What is energy level?

Quantum of energy

In atomic physics and quantum chemistry, electron configuration is the arrangement of electrons of an atom, a molecule, or other physical structure. It concerns the way electrons can be distributed in the orbitals of the given system (atomic or molecular for instance).

For example:  H hydrogen is 1s¹
                              C carbon is 1s² 2s² 2p²
                      Mg magnesium is 1s² 2s² 2p⁶ 3s²
                            

Ground State：
when all the electrons of an atom are in their lowest possible energy levels

Excited State:

when one or more of an atom's electrons are in energy levels other than the lowest available level.

Types of Orbitals:

An atomic orbital is a mathematical function that describes the wave-like behavior of either one electron or a pair of electrons in an atom. This function can be used to calculate the probability of finding any electron of an atom in any specific region around the atom's nucleus. The term may also refer to the physical region defined by the function where the electron is likely to be.

Predicting the Number of Valence Electrons:

In chemistry, valence electrons are the electrons of an atom that can participate in the formation of chemical bonds with other atoms. Valence electrons are the "own" electrons, present in the free neutral atom, that combine with valence electrons of other atoms to form chemical bonds. In a single covalent bond both atoms contribute one valence electron to form a shared pair. For main group elements, only the outermost electrons are valence electrons. In transition metals, some inner-shell electrons are also valence electrons.

The videos below are examples of how to find valence electrons from the electron configuration:

.

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 ! 




Stochiometry Involving Moles

        

Today in Class we learnt how to calculate from moles to grams and vice versa.  The  balance equation shows the relationship between moles of the substances nicely.  However in real experiments we deal with the mass of the substances which is in grams.  Therefore we need to know how to convert one way to the other.

Review

How to find the mass of substance B when you are given the mass of substance A?

Solution:  A(grams)--> moles of A-->moles of b--->mass of B(grams)






Road map for all calculation involving Stoichiometry

 



Practice Web#1


 Practice Web#2



Road map focusing on grams to moles

 

Example:

WHAT IS THE MASS OF O2 IS REQUIRED TO REACT WITH 80.0 GRAMS OF CH4?

CH4 + 2O2--> CO2 + 2H2O



Step 1:  Check if the equation is balance.

 Solution:  Each side has 4H, 4O, and 1C.  Thus the equation is balanced.

Step 2:  Convert 80g of CH4 to moles of  CH4

Solution: 80gCH4*1molCH4/16.0gCH4= 5molCH4

Step3:  Convert 5molCH4 to molO2

Solution: 5molCH4*2molO2/ 1molCH4= 10mol of O2

Step 4:  Convert 10molO2 to O2g

Solution: 2.5mol02*32gO2/1molO2= 320g02

Step5:  Check your Sig Figs

Solution: Number use in the quetion was 80.0 so that is three sig figs.  Thus 320g0--->3.20*10^3g of O.