Download Cumulative Review, entire quarter

Chem 139 Cummulative Review
A] Science
Observation, Identify patterns
form testable Hypothesis (educateed guess)
Experimentally test hypothesis
Hypoth. fails: revise and retest
Hypoth. true: independently confirm
Proven hypothesis a relation only: becomes a law
Proven hypothesis includes explanatory model: becomes a theory
B] Numbers
Scientific Notation :
0.0032 = 3.2 x 10-3
0.3200 = 3.200 x10-1
32000 = 3.2 x 104
Sig Figs
Leading zeros don’t count, trailing zeros right of the decimal do count as significant.
Multiply or divide :
sig figs in answer = smallest number of sig. figs. in the numbers multiplied.
Add or subtract:
smallest sig fig in answer is the same place as the largest least significant digit
in the numbers added
3240
+ 45.3
3285.
21.583
- 6.30
15.28
C] Changes
Physical changes do not result in a change in chemical identity of the substances
involved. They are usually reversible by another physical process.
Chemical changes do change the chemical identity of the substances involved.
A chemical change almost always involves a physical change as well since different
chemical compounds have different physical properties.
indications of chemical change:
production of heat or cooling
color change
precipitation of solid
evolution of gas
D] Unit Conversions or dimensional analysis
Set up factors so units cancel leaving the correct unit for the answerr.
28 mg | 10-3g | 1 ml = 7 x 10-3 ml
| 1 mg | 4 g
Conversion factors:
1 mg = 10-3 g
density = 4g/ml
4g
1 ml
1 mg
10-3g
or
or
10-3 g
1 mg
1 ml
4g
E] Nuclei and atomic anatomy
Atoms made from negatively charged electrons (e-),
positive protons (p) and
electrically neutral neutrons (n).
The protons and neutrons have almost all the mass and are in the nucleus.
The atomic symbol for an element contains the symbol ,
the atomic number and
the atomic mass number,
possibly the electric charge as well:
During any nuclear change,

The sum of all the mass numbers of the initial nuclei and particles = the sum of the
mass numbers of all the product particles.

The total electric charge is the same for products and initial particles.

If atomic numbers are included on β particles, then the sum of these numbers is
the same for products and initial particles.
Example:
131
53

0 
I  131
54 Xe  1 
The third rule is often neglected and is not, in fact a basic law of nuclear science.
The conservation of electric charge and conservation of mass number are
fundamental physical laws.
The atomic number is redundant with the element symbol:
The number of protons in the nucleus
= atomic number and defines the atom’s identity.
If an atom has 54 protons in the nucleus, it is xenon, regardless of the number of
neutrons or electrons in the atom.
Isotopes are atoms of the same element with different masses due to different
numbers of neutrons in the nucleus.
Nuclear radiation: There are four main types of radiation which come from nuclear
decay:
α2+ = 4He2+
β- = e-
β+ = e+
γ = high energy light.
The atomic numbers for these are 4, -1, +1 and 0 , respectively. We will not deal
with γ emissions here.
Exam 1 ends here
F] Elements, compounds
In the formula for a chemical compoud, the element symbols and subscripts give the
atoms combined to make that particular compound
(NH4)2 C2O4 is the formula for ammonium oxalate,
There are 2 nitrogens, 8 hydrogens, 2 carbons and 4 oxygens in one molecule.
The nitrogen and hydrogen occur in two groups of 4 hydrogens and one nitrogen
The carbons and oxygens occur as one group of 2 carbons and 4 oxygens.
Chemical compounds are electrically neutral; if composed of charged ions, the number
of ions must give a total charge of zero.
For (NH4)2 C2O4 , NH4 is an ion with charge +1 : NH4+ ammonium
C2O4 is an ion with charge -2 : C2O4-2 oxalate
Most ionic compounds are formed from metals or hydrogen and nonmetals combined.
NaCl FeS PbF2 SnI4 Ag2O BaSO4
Other compounds are not formed from ions and are usually made of nonmetals.
N2O5, SO3, ICl3, XeF4 .
G] Nomenclature
See note.
Table of ions to memorize.
No prefixes in ionic nomenclature except with hydrogen (dihydrogen phosphate)
H] Balancing reactions
First make sure the compounds have valid formulas (to start with, we will only use
correct formulas, which are given).
1) No changes in the formulas of the compounds can be made. This would result in a
different chemical reaction, if the new formulas are valid compounds. Do not change the
subscripts!
2) Coefficients in front of formulae give how many molecules of that formula are
needed to balance the equation.
3) Adjust coefficients so that there are the same number of atoms of each type on
each side of the equation. See example BE1 below.
In counting atoms, multiply the coefficient by the subscript to get the total.
For example, in 5 (NH4)2C2O2, there are 10 N, 40 H, 10 C and 20 O.
That is the whole idea behind balancing chemical equations.
I] Types of reactions
Synthesis: reactants combine to make one compound:
NH3 (g) + HCl (g)  NH4Cl (s)
2Cu (s) + O2 (g)  2CuO (s)
Decomposition: a compound breaks into two or more products:
CaCO3 (s)  CaO (s) + CO2 (g)
Single replacement: the positive ion or negative ion is replaced by another: (SR)
Cu (s) + 2AgNO3 (aq)  2Ag (s) + Cu(NO3)2 (aq)
2KI (s) + Cl2 (g)  2KCl (s) + I2 (s)
Double replacement: switch positive ions: (DR)
NaCl (aq) + AgNO3 (aq)  AgCl (s) + NaNO3 (aq)
NH4I (aq) + KOH (aq)  NH3 (g) + H2O (l) + KI (aq)
2NaBr (aq) + Ba(Cl)2 (aq)  BaBr2 (aq) + 2NaCl (aq) = No Reaction
The key to knowing whether the products are solids or gases is to memorize the
solubility rules and the small number of typical gas forming reactions.
J] Predicting products
This will be restricted to the four types of reaction given above.
Guiding questions:
Can I put the two reactants together to make a valid compound? (synthesis)
Can I break the compound down into two or more valid compounds? (decomp)
Look for gas-forming reactions or removal of water here.
Is one reactant an element which can replace an element in the other compound?
(SR)
If you switch the positive ions in the reactants, can you produce a solid or gas?
(DR) Solubility rules and gas reactions.
Don’t forget the diatomic gases when they appear as products:
K] Moles and chemical equations
When calculating with the number of atoms or molecules, use Avogadro’s
number to convert:
1 mole of things = 6.022 x 1023 of them.
Avocado’s number: the number of Avocados a mole can eat.
You will not need it other than finding the number of tiny things in a certain number of
grams of compound.
[L] Stoichiometry: calculating with balanced chemical equations.
The following diagram summarizes the sorts of calculations which can be done with the
balanced chemical equation.
From the balanced chemical equation, we get several possible conversion factors. For
example, given the reaction
2 Al(s) + 3H2SO4(aq)  Al2(SO4)3(aq) + 3H2(g)
The following conversion factors can be obtained:
2mol Al
3mol H 2 SO4
3mol H 2 SO4 3mol H 2
3mol H 2 SO4 1mol Al2 ( SO4 )3
2mol Al
1mol Al2 ( SO4 )3 2mol Al 1mol Al2 ( SO4 )3 3mol H 2
3mol H 2
These simply use the coefficients from the balanced reaction to relate quantities of the various
reactants and products in the reaction. The formulas of the compounds also give us conversion
factors:
2mol H
1mol SO4
3mols SO4
2mols Al
Suppose we start with 135. g of Al. How many grams of aluminum sulfate can be produced?
The road map for this problem is :
So,
135 g Al 1mol Al 1mol Al2 (SO4 )3 342.17 g Al2 (SO4 )3
 856 gAl2 (SO4 )3
26.98 g Al
2mol Al
1mol Al2 (SO4 )3
You may be given starting quantities for both reactants. Then you need to determine which
runs out first and how much product can be made.
Suppose we start with 13.5g Al and 30g of H2SO4 . We can determine which reactant runs out
first (limiting reactant) by calculating how much product can be made by completely reacting
each reactant.
13.5 g Al 1mol Al 1mol Al2 (SO4 )3 342.17 g Al2 ( SO4 )3
 85.6 gAl2 ( SO4 )3
26.98 g Al
2mol Al
1mol Al2 ( SO4 )3
30 g H 2 SO4 1mol H 2 SO4 1mol Al2 (SO4 )3 342.17 g Al2 (SO4 )3
 34.9 g Al2 (SO4 )3
98.09 g H 2 SO4 3mol H 2 SO4
1mol Al2 (SO4 )3
Since the sulfuric acid produces less product, it is the one which runs out first; it is the limiting
reactant. This also tells us that 34.9g of aluminum sulfate are produced. We can also
determine the amount of aluminum left.
34.9 g Al2 ( SO4 )3
1mol Al2 (SO4 )3
2mol Al
26.98 g Al
 5.50 g Al used;
342.17 g Al2 ( SO4 )3 1mol Al2 ( SO4 )3 1mol Al
13.5g Al – 5.50g Al = 8.0g Al remaining.
Now suppose you are told that when the reaction was run, 30.0g of aluminum sulfate was
obtained. We can determine the percent yield. From the above, the theoretical yield, which
could be obtained if all the limiting reactant went to producing product, is 34.9g of aluminum
sulfate. The actual yield of 30.0g is some percentage of the theoretical yield:
30.0 g
100%  86.0% yield
34.9 g
Net ionic equations.
Starting from a balanced equation,
1) if there is no water (or (aq) state labels), there is no ionic equation.
2) If an ionic compound is soluble, separate it into individual ions.
CaCl2(aq) --> Ca2+ (aq) + 2Cl- (aq)
3) If an acid is one of the strong acids, one H+ dissociates.
HCl(aq) --> H+ (aq) + Cl- (aq)
H2SO4(aq) --> H+ (aq) + HSO4- (aq)
4) All gases, insoluble solids, weak acids and covalent molecules are written as
the full molecule (or formula unit).
PbS(s), CO2 (g), CH3CO2H, CH3COCH3 (a polar, soluble molecule)
5) Cancel all ions which remain in solution on both sides of the equation.
M] Electrons in atoms, configurations
Electrons in atoms are located in orbitals, regions of space where electrons form standing
waves. The basic shapes are
s orbitals, spherical cloud centered on nucleus
p: two-lobed orbitals, 3 per set, oriented along the xy&z axes; from period 2 and up.
d set of 5 four-lobed orbitals, beginning with the third period and up.
f set of seven orbitals beginning with the fourth period and up.
Electrons are placed in the orbitals from the lowest energy up. For the sets of multiple orbitals
of equal energy, electrons are placed one in each orbital, then
paired up. Two ways to remember the filling order are the filling
diagram below and the periodic table labeled with the orbital set
being filled.
Examples of electron configurations:
N : 1s2 2s2 2p3 _↑↓_ _↑↓_
1s
2s
_↑_ _↑_ _↑_
2p
Mn: 1s2 2s2 2p6 3s2 3p6 4s2 3d5 _↑↓ ↑↓_ ↑↓ ↑↓_ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑_ ↑_ ↑_ ↑_ ↑_
1s 2s
2p
3s 3p
4s
3d
short form: [Ar] 4s2 3d5
O: 1s2 2s2 2p4 or, as a box diagram, _↑↓_ _↑↓_ _↑↓ _↑_ _↑_
1s
2s
2p
Atoms with unpaired electrons are called paramagnetic because they are attracted to a
magnetic field, but not as strongly as ferromagnetic materials like iron. If all electrons are
paired, the atom is diamagnetic.
When the ion or element is near the middle of the d block, the s and d orbitals are so close in
energy that it is better to spread the electrons out, 1 per orbital, than to fill the s orbital.
Cr: [Ar] 4s1 3d5
instead of
[Ar] 4s2 3d4 .
Ions.
For negative ions, just add the extra electrons in the same filling order.
For positive ions, remove electrons from the outer shell or period first.
On the filling diagram , this would be row by row, from the top down.
N] Molecular structure: Lewis Structures.
Chemical bonds are pairs of electrons.
1 Add up the electrons on the neutral atoms and subtract the charge
on the ion, if there is one.
2. H andF are always on the outside, forming one bond each.
3. Place the least electronegative element in the center and make
bonds to the other atoms.
4. Distribute the rest of the electrons on the more electronegative atoms.
5. Redistribute the electrons so that every atom larger than boron is surrounded by an octet of
electrons. For this purpose, the bonds around the atoms count for both atoms since the
bonding pair is around both of the atoms.
If necessary, atoms below the second period can accomodate more than eight
electrons.
Electron pairs on atoms can be moved to make double bonds to make octets.
6. If a choice of where to put the double bond (as in the nitrate ion) resonance structures must
be included with each structure showing a different choice of double bond location.
7. For ions, place the structure in brackets and write the charge outside the upper right corner of
the bracket.
8. If needed, the formal charge on an atom is calculated as the number of valence (outer s & p)
electrons in the neutral atom minus the number of electrons the atom has in the structure,
counting one electron for each bond.
The formal charges add to give the total charge on the structure.
The most stable structure will have the smallest formal charges.
Example BE1:
(NH4)3PO4 + Ca(OH)2  Ca3(PO4)2 + NH3 (g) + H2O
First things to notice: phosphate ions stay together on the product side, so we can count
phosphates instead of just P, O, etc.
Let’s count atoms on each side to see where we are:
species
N
H
PO4
Ca
O
Reactant side (left)
3
3*4 + 2 = 14
1
1
2
Product side (right)
1
5
2
3
1
Note that you multiply the number
outside parentheses by the
subscript inside to get the total
number.
Clearly we are not balanced. I would start with calcium phosphate on the right and put trial
coefficients on the left to balance Ca and phosphate.
2(NH4)3PO4 + 3Ca(OH)2  Ca3(PO4)2 + NH3 (g) + H2O
This gives
species
N
H
PO4
Ca
O
Reactant side (left)
6
30
2
3
6
Product side (right)
1
5
2
3
1
Now look at the nitrogens. We have 6 on the left, try a 6 in front of NH3 on the right.
While we are at it, try 6 waters on the right to balance those extra oxygens on the left.
2(NH4)3PO4 + 3Ca(OH)2  Ca3(PO4)2 + 6NH3 (g) + 6H2O
Now we have:
species
N
H
PO4
Ca
O
Reactant side (left)
6
30
2
3
6
Product side (right)
6
18 + 12 = 30
2
3
6
And we are done!