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					                CHEMISTRY SOL REVIEW NOTES

Naming Compounds
Before naming a compound, you have to figure out what kind of compound it is. We will
consider three types:

   1. Ionic Compounds Without a Transition Metal.

       Basically if the compound contains a metal, it is ionic. But there are different sets
   of rules for transition metals.
       a.      So for a compound with any other metal, apply these rules:

               The metal ion's name does not change regardless of charge
               The non-metal's name ends in ide.

       For example: AlCl3 = aluminum chloride
                    Na2S = sodium sulfide

       b.      In going backwards (from name to formula), we will have more fun. In
       such a case the total charge of the (+) and (-) ions in the compound has to be
       ZERO.

       Example: What is the correct formula for calcium phosphide?

       Here, we have to consider the common charges for calcium and phosphorus,
       which are +2 and -3, respectively.                Ca+2 and P-3

       So the answer is Ca3P2.

       Other examples:        potassium oxide :      K+1 and O-2 give K2O
                              aluminum bromide       Al+3 and Br-1 yield AlBr3.

       c.       Polyatomic Ions

       When metals are bonded to polyatomic ions, which consist of two or more atoms
       with one overall charge, the same rules apply, but you have to learn the names
       and charges of common polyatomic ions.
           Polyatomic Ion                                Name
                OH-1                                   hydroxide
                SO4-2                                    sulfate
                PO4-3                                  phosphate
               NO3-1                                     nitrate
               CO3-2                                   carbonate
               HCO3-1                       hydrogen carbonate or bicarbonate
               ClO3-1                                   chlorate
               NH4+1                                  ammonium


2. Ionic Compounds With a Transition Metal.

The only difference here is that we have to specify the charge of the transition
metal ion by using a Roman numeral, and keep in mind that a transition metal is
an element with an atomic number of 21 to 30, 39 to 48 or 57 to 80.


    Roman numeral                      Charge
           I                             +1
          II                             +2
         III                             +3
         IV                              +4
          V                              +5
         VI                              +6

Example:      manganese(II) oxide contains Mn+2 and O-2. So we just need one of
              each and the formula becomes MnO.

              Copper(I) oxide is Cu2O.


Example: What is the correct name of CrCl3 ?

The charge of Cr is unknown = x . But chloride = (-1). The sum of the charges
              has top be zero, so:

                     x +3(-1) = 0.
                     x = 3.
Answer:       CrCl3 = chromium (III) chloride.

3. Covalent Compounds. These are formed from non-metals that share electrons.
Because there are many sharing possibilities between two non-metals, the formula
cannot be guessed unless we have a naming system that reveals the number of
atoms involved.
       For this we use a set of prefixes:


                          Prefix                  Number of atoms
                          mono                          1
                             di                         2
                            tri                         3
                           tetra                        4
                          penta                         5
                           hexa                         6

       The only time we drop a prefix is if the mono is to appear at the beginning of the
       name.

       Examples:      CO = carbon monoxide ( note we don't say monocarbon monoxide)
                      CO2 = carbon dioxide

                      dinitrogen pentoxide = N2O5.
                      phosphorus trichloride PCl3.

       Note that none of the above compounds contain a metal. Metals do not form
       covalent compounds, so we generally don't use prefixes for compounds
       containing metals.

                                   Significant Figures

There are two kinds of numbers in the world:

      exact:
           o  example: There are exactly 12 eggs in a dozen.
           o  example: Most people have exactly 10 fingers and 10 toes.
      inexact numbers:
          o example: any measurement.
              If I quickly measure the width of a piece of notebook paper, I might get
              220 mm (2 significant figures). If I am more precise, I might get 216 mm
              (3 significant figures). An even more precise measurement would be 215.6
              mm (4 significant figures).


PRECISION VERSUS ACCURACY

Accuracy refers to how closely a measured value agrees with the correct value.
Precision refers to how closely individual measurements agree with each other.
                      accurate                                 accurate
                                              precise
              (the average is accurate)                           and
                                            not accurate
                     not precise                                precise




Rules for Working with Significant Figures:

   1. Leading zeros are never significant.
      Imbedded zeros are always significant.
      Trailing zeros are significant only if the decimal point is specified.
      Hint: Change the number to scientific notation. It is easier to see.
   2. Addition or Subtraction:
      The last digit retained is set by the first doubtful digit.
   3. Multiplication or Division:
      The answer contains no more significant figures than the least accurately known
      number.

EXAMPLES:

Example     Number of         Scientific
            Significant       Notation
              Figures
 0.00682         3            6.82 x 10-3            Leading zeros are not significant.
  1.072          4           1.072 (x 100)        Imbedded zeros are always significant.
   300           1              3 x 102           Trailing zeros are significant only if the
                                                         decimal point is specified.
   300.          3             3.00 x 102
  300.0          4            3.000 x 102

EXAMPLES
Addition                              Even though your calculator gives you the answer
                                      8.0372, you must round off to 8.04. Your answer
                                      must only contain 1 doubtful number. Note that the
                                      doubtful digits are underlined.



Subtraction                           Subtraction is interesting when concerned with
                                      significant figures. Even though both numbers
                                      involved in the subtraction have 5 significant figures,
                                      the answer only has 3 significant figures when
                                      rounded correctly. Remember, the answer must only
                                      have 1 doubtful digit.
Multiplication                        The answer must be rounded off to 2 significant
                                      figures, since 1.6 only has 2 significant figures.




Division                              The answer must be rounded off to 3 significant
                                      figures, since 45.2 has only 3 significant figures.




Notes on Rounding

      When rounding off numbers to a certain number of significant figures, do so to
       the nearest value.
                                                                         4
           o example: Round to 3 significant figures: 2.3467 x 10 (Answer: 2.35 x
              104)
                                                                       3              3
           o example: Round to 2 significant figures: 1.612 x 10 (Answer: 1.6 x 10 )
      What happens if there is a 5? There is an arbitrary rule:
           o If the number before the 5 is odd, round up.
           o If the number before the 5 is even, let it be.
              The justification for this is that in the course of a series of many
              calculations, any rounding errors will be averaged out.
                                                                     2              2
           o example: Round to 2 significant figures: 2.35 x 10 (Answer: 2.4 x 10 )
                                                                     2              2
           o example: Round to 2 significant figures: 2.45 x 10 (Answer: 2.4 x 10 )
                                                                             2
           o Of course, if we round to 2 significant figures: 2.451 x 10 , the answer is
                                 2                    2                     2
              definitely 2.5 x 10 since 2.451 x 10 is closer to 2.5 x 10 than 2.4 x 102.
EMPIRICAL FORMULA

Example: Calculate the empirical formula for a compound that has 43.7 g P (phosphorus)
and 56.3 grams of oxygen. First we convert to moles:

                         43.7 grams P       1 mol
                                                        = 1.41
                                        x
                                                        moles
                               1        30.97 grams
                         56.3 grams O      1 mol
                                      x             = 3.52 moles
                               1        16.00 grams



Next we divide the moles to try to get an even ratio.

                                                1.41
                                  Phosphorus:          = 1.00
                                                1.41
                                                3.52
                                  Oxygen:              = 2.50
                                                1.41

When we divide, we did not get whole numbers so we must multiply by two (2). The
answer = P2O5

The Discovery of the Atom
Thompson’s “Plum Pudding” Model
The first major discovery that set off modern atomic theory was that atoms aren’t in fact the
smallest things that exist. J. J. Thompson discovered the electron in 1897, which led him to
posit a “plum pudding” model (a.k.a. the “raisin pudding” model) for the atom. Electrons
are small negative charges, and Thompson suggested that these negative charges are
distributed about a positively charged medium like plums in a plum pudding.
Rutherford’s Gold Foil Experiment
In a series of experiments from 1909 to 1911, Ernest Rutherford established that atoms have
nuclei. His discovery came by accident and as a total surprise. His experiment consisted of
firing alpha particles, which we will examine in more detail shortly, at a very thin sheet of
gold foil.




This unexpected result shows that the mass of an atom is not as evenly distributed as
Thompson and others had formerly assumed. Rutherford’s conclusion, known as the
Rutherford nuclear model, was that the mass of an atom is mostly concentrated in a
nucleus made up of tightly bonded protons and neutrons, which are then orbited by
electrons.




Concept of mole/molar ratio

   1) How many moles of sodium atoms correspond to 1.56x1021 atoms of sodium?


   2) How many moles of Al are in 2.16 mol of Al2O3?


   3) Aluminum sulfate, Al2(SO4)3, is a compound used in sewage treatment plants.
         a. a. Construct a pair of conversion factors that relate moles of aluminum to
             moles of sulfur for this compound
         b. b. Construct a pair of conversion factors that relate moles of sulfur to moles of
            Al2(SO4)3
         c. c. How many moles of Al are in a sample of this compound if the sample also
            contains 0.900 mol S?
         d. d. How many moles of S are in 1.16 mol Al2(SO4)3?


  4) What is the total number of atoms in 0.260 mol of glucose, C6H12O6?




Percent composition and empirical formulas

  5) Calculate the percentage composition by mass of each element in the following
     compounds:
         a. a. NaH2PO4
         b. b. NH4H2PO4
         c. c. (CH3)2CO


  6) Quantitative analysis of a sample of sodium pertechnetate with a mass of 0.896g found
     0.111g Na and 0.477g technetium (Tc). The remainder was oxygen. Calculate the
     empirical formula of sodium pertechnetate, NaxTcyOz.


  7) A substance was found to be composed of 22.9% Na, 21.5% B, and 55.7% O. What is the
     empirical formula of this compound?


  8) When 0.684 g of an organic compound containing only C, H, and O was burned in
     oxygen 1.312g CO2 and 0.805g H2O were obtained. What is the empirical formula of the
     compound?



Balancing equations
  9) Balance the following reactions:
         a. a. Ca(OH)2 + HCl  CaCl2 + H2O
         b. b. AgNO3 + CaCl2  Ca(NO3)2 +AgCl
         c. c. Fe2O3 + C  Fe + CO3
         d. d. NaHCO3 + H2SO4  Na2SO4 + H2O + CO2
         e. e. C4H10 + O2  CO2 +H2O
         f. f. Mg(OH)2 + HBr  MgBr2 + H2O
         g. g. Al2O3 + H2SO4  Al2(SO4)3 + H2O
          h. h. KHCO3 + H3PO4  K2HPO4 + H2O + CO2
          i. i. C9H10O + O2  CO2 + H2O




Stoichiometry/limiting reactants

  10) Chlorine is used by textile manufacturers to bleach cloth. Excess chlorine is destroyed
      by its reaction with sodium thiosulfate, Na2S2O3:
               Na2S2O3(aq) + 4Cl2(g) + 5H2O(aq)  2NaHSO4(aq) + 8HCl(aq)

          a.   a. How many moles of Na2S2O3 are needed to react with 0.12mol of Cl2?
          b.   b. How many moles of HCl can form from 0.12mol of Cl2?
          c.   c. How many moles of H2O are required for the reaction of 0.12mol of Cl2?
          d.   d. How many moles of H2O react if 0.24mol HCl is formed?


  11) The incandescent white of a fireworks display is caused by the reaction of phosphorous
      with O2 to give P4O10.
          a. a. Write the balanced chemical equation for the reaction.
          b. b. How many grams of O2 are needed to combine with 6.85g of P?
          c. c. How many grams of P4O10 can be made from 8.00g of O2?
          d. d. How many grams of P are needed to make 7.46g P4O10?


  12) In dilute nitric acid, HNO3, copper metal dissolves according to the following equation:
              3Cu(s) + 8HNO3(aq)  3Cu(NO3)2(aq) + 2NO(g) + 4H2O(aq)

      How many grams of HNO3 are needed to dissolve 11.45g of Cu?



  13) The reaction of powdered aluminum and iron(II)oxide,
             2Al(s) + Fe2O3(s)  Al2O3(s) + 2Fe(l)

      produces so much heat the iron that forms is molten. Because of this, railroads use the
      reaction to provide molten steel to weld steel rails together when laying track. Suppose
      that in one batch of reactants 4.20mol Al was mixed with 1.75mol Fe2O3.

          a. a. Which reactant, if either, was the limiting reactant?
          b. b. Calculate the mass of iron (in grams) that can be formed from this mixture
             of reactants.


  14) Silver nitrate, AgNO3, reacts with iron(III) chloride, FeCl3, to give silver chloride, AgCl,
      and iron(III) nitrate, Fe(NO3)3. A solution containing 18.0g AgNO3 was mixed with a
      solution containing 32.4g FeCl3. How many grams of which reactant remains after the
      reaction is over?



Theoretical and percent yield

  15) Barium sulfate, BaSO4, is made by the following reaction:
             Ba(NO3)2(aq) + Na2SO4(aq)  BaSO4(s) + 2NaNO3(aq)

      An experiment was begun with 75.00g of Ba(NO3)2 and an excess of Na2SO4. After
      collecting and drying the product, 63.45g BaSO4 was obtained. Calculate the theoretical
      yield and percent yield of BaSO4.



  16) Aluminum sulfate can be made by the following reaction:
             2AlCl3(aq) + 3H2SO4(aq)  Al2(SO4)3(aq) + 6HCl(aq)

      It is quite soluble in water, so to isolate it the solution has to be evaporated to dryness.
      This drives off the volatile HCl, but the residual solid has to be treated to a little over
      200C to drive off all the water. In one experiment, 25.0g of AlCl3 was mixed with 30.0g
      H2SO4. Eventually, 28.46g of pure Al2(SO4)3 was isolated. Calculate the percent yield.


Answers


  1) 2.59x103mol Na atoms

  2) 4.32mol Al
  3) a. 2mol Al/3mol S      b. 3mol S/1mol Al2(SO4)3             c. 0.600mol Al    d. 3.48mol S
  4) 3.76x1024 atoms
  5) a. 0.215mol    b. 0.0916mol           c. 0.0794mol                  d. 4.31x108mol
  6) a. 19.2% Na, 1.68% H, 25.8% P, 53.3% O
     b. 12.2% N, 5.26% H, 26.9% P, 55.6%O

      c. 62.0% C, 10.4% H, 27.6% O

  7) Theoretical data (83.89% C, 10.35% H, 5.76% N) are consistent with experimental
      results.
  8) 0.474g O
  9) NaTcO4
  10) Na2B4O7
  11) C2H6O
  12)
           a. a. Ca(OH)2 + 2HCl  CaCl2 + 2H2O
           b. b. 2AgNO3 + CaCl2  Ca(NO3)2 + 2AgCl
           c. c. 2Fe2O3 + 3C  4Fe + 3CO3
           d. d. 2NaHCO3 + H2SO4  Na2SO4 + 2H2O + 2CO2
           e. e. 2C4H10 + 13O2  8CO2 + 10H2O
           f. f. Mg(OH)2 + 2HBr  MgBr2 + 2H2O
           g. g. Al2O3 + 3H2SO4  Al2(SO4)3 + 3H2O
           h. h. 2KHCO3 + H3PO4  K2HPO4 + 2H2O + 2CO2
           i. i. C9H10O + 14O2  9CO2 + 10H2O
   13) a. 0.030mol Na2S2O3 b. 0.24mol HCl c. 0.15mol H2O
       d. 0.15mol H2O

   14)   a. 4P + 5O2  P4O10 b. 8.85g O2         c. 14.2g P4O10              d. 3.26g P
   15)   30.31g HNO3
   16)   a. limiting reactant is Fe2O3  b. 195g Fe is formed
   17)   26.7g of FeCl3 are left over
   18)   theoretical yield = 66.98g BaSO4, % yield = 94.73%
   19)   % yield = 88.74%


MOLARITY

 The concentration of a solution is typically given in molarity. Molarity is defined as the
number of moles of solute (what is actually dissolved in the solution) divided by the
volume in liters of solution (the total volume of what is dissolved and what it has been
dissolved in).

                                               moles of solute
                                  Molarity =
                                               liters of solution


Molarity is probably the most commonly used term because measuring a volume of
liquid is a fairly easy thing to do.

Example: If 5.00 g of NaOH are dissolved in 5000 mL of water, what is the molarity of
the solution?


One of our first steps is to convert the amount of NaOH given in grams into moles:

                     5.00g NaOH                1 mole
                                   x                                = 0.125 moles
                           1           (22.9 + 16.00 + 1.008)g
Now we simply use the definition of molarity: moles/liters to get the answer

                                        0.125 moles
                           Molarity =                    = 0.025 mol/L
                                        5.00 L of soln


So the molarity (M) of the solution is 0.025 mol/L.




Periodic Trends in Radii




All of the following trends in atomic properties are directly related to the trends in atomic
radii.
Periodic Trends in Ionization Energy




Periodic Trends in Electron Affinity
Periodic Trends of Electronegativity




BONDING

How to Build a Lewis Structure?
For example, oxygen has 6 electrons in the outer shell, which are the pattern of two lone
pairs and two singles.




                        Incorrect Structure      Correct Structure


One good example is the water molecule. Water has the chemical formula of H2O, which
means there
Example: Write the Lewis structure for carbon dioxide (CO2).
Answer: Carbon is the lesser electronegative atom and should be the central atom.




Example: Write the Lewis structure for methane (CH4).
Answer: Hydrogen atoms are always placed on the outside of the molecule, so carbon
should be the central atom.




GAS LAWS

An equation that chemists call the Ideal Gas Law, shown below, relates the volume,
temperature, and pressure of a gas, considering the amount of gas present.

                                      PV = nRT
Where:

P=pressure in atm
T=temperature in Kelvins
R is the molar gas constant, where R=0.082058 L atm mol-1 K-1.

The Ideal Gas Law assumes several factors about the molecules of gas. The volume of
the molecules is considered negligible compared to the volume of the container in which
they are held. We also assume that gas molecules move randomly, and collide in
completely elastic collisions. Attractive and repulsive forces between the molecules are
therefore considered negligible.

Example Problem: A gas exerts a pressure of 0.892 atm in a 5.00 L container at 15°C.
The density of the gas is 1.22 g/L. What is the molecular mass of the gas?

Answer:
PV = nRT
T = 273 + 15 = 228
(0.892)(5.00) = n(.0821)(288)
n = 0.189 mol
 .189 mol       x grams
            x             = 1.22 g/L
  5.00L         1 mol
x = Molecular Weight = 32.3 g/mol




BOYLES LAW
P1V1= P2V2

Example:

Let's try an example with our new equation. If the volume of an gas is 0.312 liters at 822
kPa, how would we find a new volume at 948 kPa?

Formula: PV = P1V1

0.312 L x 822 kPa = 948 kPa x V1
(0.312 L x 822 kPa) / 948 kPa = V1
V1 = (0.312 L x 822 kPa) / 948 kPa
V1 = 0.2705 L
V1 = 0.271 L (to the correct number of significant digits)




CHARLES LAW
Example: In the experiment above the initial volume and temperature of the gas is 0.5L, 5
0C. Assuming the pressure and moles of gas is constant, what is the volume of the gas if
the temperature is increased to 80 0C? Let T1 and V1 be the initial temperature and
volume and let T2, V2 be the final temperature and volume. Then according to Charles
law




COMBINATION GAS LAW

In the combined gas law, the volume of gas is directly proportional to the absolute
temperature and inversely proportional to the pressure.

This can be written as PV / T = constant. Since for a given amount of gas there is a
constant then we can write P1V1 / T1 = P2V2 / T2.

      P1 is the initial pressure
      V1 is the initial volume
      T1 is the initial temperature (in Kelvin)
      P2 is the final pressure
      V2 is the final volume
      T2 is the final temperature (in Kelvin)

This equation is useful if you have the current volume, temperature, and pressure of a
gas, and if you have two of the three final values of the gas.
For example if you have 4.0 liters of gas at STP, and you want to know the volume of the
gas at 2.0 atm of pressure and 30o C, the equation can be setup as follows:

(1.0)(4.0) / 273 = (2.0)(V2) / 303
(V2)(2)(273) = (1)(4)(303)
V2 = 2.2
Therefore the new volume is 2.2 liters.




                                          STP
STP is Standard Temperature and Pressure. STP is Oo Celcius and 1 atmosphere of
pressure. Gases properties can be compared using STP as a reference.

To calculate the pressure of the gas the partial pressure of the water must be subtracted
from the pressure in the container. The partial pressure of the water can be obtained from
the table below.

Temperature (oC) Pressure (mmHg)
0                4.6
5                   6.5
10                  9.2
11                  9.8
12                  10.5
13                  11.2
14                  12.0
15                  12.8
16                  13.6
17                  14.5
18                  15.5
19                  16.5
20                  17.5
21                  18.7
22                  19.8
23                  21.1
24                  18.7
25                  23.8
26                  25.2
27                  26.7
28                  28.3
29                  30.0
30                  31.8
35                  42.2
40                  55.3
45                  71.9
50                  92.5
55                  118.0
60                  149.4
65                  187.5
70                  233.7
75                  289.1
80                  355.1
85                  433.6
90                  525.8
95                  633.9
100                 760.0
105                 906.1

For example if a gas is collected over water at 22oC and 1 atm of total pressure, the
pressure of the gas would be calculated as follows:

1 atm = 760 mmHg therefore 760 mm - 19.8 mm = 740.2 mmHg would be the pressure
of the gas.


HEAT                                     Where: q = mc∆T

C = specific heat in cal/g-°C
q = heat added in calories,
m = mass in grams
ΔT = rise in temperature of the material in °C.

The value of C for water is 1.00 cal/g-°C.

Example Problem: If a 2.34 g substance at 22°C with a specific heat of 3.88 cal/g-°C is
heated with 124 cal of energy, what is the new temperature of the substance?
Answer:

        q
ΔT =
       MC

            (124)
ΔT =                  = 13.7°C
       (2.34)(3.88)

new T = 22 + 13.7 = 35.7°C

Nuclear Chemistry

   1. Alpha decay follows the form:




   2. Beta negative decay follows the form:




   3. Gamma decay follows the form:
ELECTROCHEMICAL



The following is a diagram of an electrochemical cell with zinc and copper acting as the
electrodes.




REDOX REACTIONS
OXIDATION

                                     Cu (s) ----> Cu2+ + 2 e-

REDUCTION

                                2 Ag+ (aq) + 2 e- ------> 2 Ag (s)

				
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