Download Spring 2009 Final Exam Review – Part 2

2012-2013 Final Exam Review – Part 2
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Nomenclature (Names & Formulas)
o List of Polyatomic Ions
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Prefixes for Type III compounds
 Mono – 1
 Di – 2
 Tri – 3
 Tetra – 4
 Penta – 5
 Hexa – 6
 Hepta – 7
 Octa – 8
 Nona – 9
 Deca – 10
Type I – Ionic Compounds – have regular metals and nonmetals/polyatomic ions
in them. You MUST crisscross charges.
Type II – Ionic Compounds – have transition metals and nonmetals/polyatomic
ions in them. Their names have Roman Numerals representing the charge of the
metal cation in it. You MUST crisscross charges.
Type III – Covalent Compounds – consist of 2 nonmetals. No crisscrossing! You
use prefixes in the names.
Type Acids – Always have H+1 in the formula. There are binary acids which are
made of the hydrogen ion and another element or the polyatomic ion, CN1(cyanide). Oxyacids are made of the hydrogen ion and a polyatomic ion with
oxygen in it.
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Formula Writing Flow Chart
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Name Writing Flow Chart
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Practice Nomenclature
 calcium chloride
 dinitrogen pentoxide
 copper (II) nitrate
 hydrosulfuric acid
 nitrous acid
 potassium carbonate
 carbon dioxide
 gold (III) cyanide
 K2Cr2O7
 Mn(OH)2
 H3P
 HClO3
 IF7
 AgCl
 CaS
Moles
o A mole is a quantity, just like a dozen means 12, a mole is 6.022  1023, which is
a really large number.
o You can use the mole to convert between various different things like, mass,
molecules/ atoms, or the volume of a gas at STP.
MASS (g)
molar mass
1 mol
ATOMS/ MOLECULES
1 mol
6.022  1023
1 mol
molar mass
6.022  1023
1 mol
MOLES
1 mol
22.4 L
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22.4 L
1 mol
VOLUME OF GAS AT STP
Molar Mass: the mass of 1 mol of any substance
 For an element: it’s the atomic mass
 For a compound: it’s the sum of the atomic masses of all elements in the
substance
 Ex. Li2O Li = 2  6.94g = 13.88g
O = 1  16.00g = 16.00g
Li = 29.88g
% Composition: is the percentage of the mass of the element that is made up of
that particular element.
 % element = mass of element  100 =
molar mass
 Ex. % Li in Li2O
 % Li = (13.88g/29.88g)  100 = 46.35%
 % O = (16.00g/29.88g)  100 = 53.55%
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Empirical Formula: Is a chemical formula that is reduced to the lowest ratio of
one element to another. It can be the true formula for a substance but it does
NOT have to be.
 Ex. H2O, CaCl2, C12H22O11, Ca3(PO4)2
 Calculating an empirical formula from a % composition.
 1. Convert the % to grams.
 2. Convert grams to moles. (Divide by the molar mass).
 3. Pick the smallest answer from # 2 and divide each answer by
that number.
 4. If you get whole numbers or close to whole numbers, then you
have a subscripts for the empirical formula.
 If you end up with an answer somewhere between .3 and .8, you
need to multiply all the answers by whatever number will give
you whole numbers and then you will have the subscripts you
need.
 Ex. Find the empirical formula for a substance that is 11.21% H
and 88.79% O.
o 11.21g H
1 mol H = 11.09900mol H
1.01g H
o 88.79g O
1 mol O = 5.549 mol O
16.00g O
o 11.09900mol H/ 5.549 mol = 2
o 5.549mol O/5.549 mol = 1
o H2O
Molecular Formula: Is the true chemical formula for a substance.
 Ex. C6H12O6 , H2O2, NaNO3
 Calculating an empirical formula from % composition and knowing the
mass of the molecular formula.
 Calculate the empirical formula like above.
 Find the molar mass of the empirical formula
 Divide the molar mass of the molecular formula (which is given
to you) by the molar mass of the empirical formula which you
just calculated. You should get a whole number.
 Multiply each subscript in the empirical formula by the whole
number and you will have the molecular formula.
o Ex. A substance was found to be 49.23% C, 6.5% H,
and 43.83% O. The molar mass of the molecular formula
is 146g. Find the molecular formula.
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Practice problems to try
 1. Find the molar mass of the following problems:
 a. NaNO3
 b. Cu(OH)2
 c. H3PO4
 2. Find the % of each element in each substance in #1.
 3. How many molecules are there in 24 grams of FeF3?
 4. How many molecules are there in 450 grams of Na2SO4?
 5. How many grams are there in 2.3 x 1024 atoms of silver?
 6. How many grams are there in 7.4 x 1023 molecules of AgNO3?
 7. What is the volume of O3 in 27.00g at STP?
 8. How many molecules of CO gas are in 0.99L at STP?
 9. Calculate the % composition of each element in Na2SO4?
 10. Calculate the % composition of each element in FeF3?
 11. A compound has the following % composition: 58.84% Ba,
13.74% S, and 27.43% O. Determine the empirical formula.
 12. The empirical formula for a compound was determined to be
CH4O and the molar mass of the molecular formula is
approximately 192g. Determine the molecular formula.
Stoichiometry: Using the mole ratio of one substance to another in a balanced chemical
equation to determine how much product can be made from a specific amount of reactant.
Mass of Given
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1 mol GIVEN
Molar mass GIVEN
moles UNKNOWN from eqn
moles GIVEN from eqn
molar mass UNKNOWN
1 mol UNKNOWN
Ex. What mass of water is made from the combustion of 32.10g CH4 with excess
O2?
CH4 + 2O2  CO2 + 2H2O
32.10 g CH4
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30.00g O2
1mol CH4
16.05g CH4
2 mol H2O
1 mol CH4
18.02g H2O = 72.08g H2O
1 mol H2O
Limiting Reactant: the reactant that you run out of first when you do a reaction.
In stoichiometry, it is always the reactant the produces the LEAST amount of
product.
Ex. Determine the limiting reactant if 32.10g of CH4 reacts with 30.00g of O2.
1 mol O2
32.00g O2
2 mol H2O
2 mol O2
18.02g H2O = 16.89g H2O
1 mol H2O
Since 32.10g of CH4 produced 72.08g of H2O and 30.00g of O2 produced 16.89g of H2O, then the
limiting reactant is O2 and 16.9g of H2O was produced.
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Practice Problems to Try
 1. Ammonium nitrate has been used as a high explosive because it is
unstable and decomposes into several gaseous substances. The rapid
expansion of gaseous substances produces this explosive force.
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Calculate the mass of each product (yes, this means you are doing 3
separate problems) if 1.25g of ammonium nitrate (NH4NO3) is used?
2NH4NO3(s)  2 N2(g) + O2(g) + 4H2O(g)
2. In the reaction below, 13.0g of MgCO3 reacts with 26.0g of water
to produce Mg(OH)2 and H2CO3. Determine the limiting reactant
and the theoretical yield of Mg(OH)2.
MgCO3 (s) + 2H2O(l)  Mg(OH)2(aq) + H2CO3(aq)
Electrons
o The electrons in an atom reside within the sublevels of the orbitals in the atom.
Each orbital is associated with a specific amount of energy. When an electron
absorbs energy from an outside source, it jumps up energy levels to an excited
state. The electron then releases that energy in order to move back down into ins
ground state. Sometimes, that energy is released as visible light.
 Visible light is a small part of the electromagnetic spectrum. Every type
of radiation on the electromagnetic spectrum must obey certain rules.
The radiation moves in waves which have specific wavelength () and
frequency () that travel at the speed of light, c, which is 3.00  108 m/s.
 c=
 Electromagnetic radiation can also act a particles. When the energy is
released from the electron in the atom, it comes off in a packet, called a
photon. That photon has a specific amount of energy in it.
 E = h   (h = 6.626  10-34 Js)
 You can also calculate the mass of the photon by using the equation
made famous by Einstein, E = mc2.
o Practice problems
 1. If the frequency of the electromagnetic radiation increases, what
happens to the wavelength?
 2. If the energy of a photon decreases, what happens to the frequency
of the wave?
 3. Why do elements give off specific colors of light when heated?
(Think electrons and energy levels.)
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Each orbital in an atom has sublevels, s, p, d, and f. Electrons actually reside in
the sublevels of each orbital. We can write the ground state, lowest energy
electron configuration for all the electrons in an element.
o See the chart below, for the order that the sublevels are filled.
o Practice problems
 Write electron configurations for the elements below, both ground state
and abbreviated state.
 1. Na
 2. Cl
 3. Ag
 4. Sr
Bonding
o Ionic Bonds- occur between metals and nonmetals. The metal gives up its
valence electron to the nonmetal, creating the strong bond.
 This bond creates compounds that are usually solids at room
temperature, have high melting points and high boiling points.
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Covalent Bonds – a strong bond that occurs through the sharing of electrons
between 2 nonmetals.
 You can draw Lewis structures for covalent molecules.
 Determine the total # of valence electrons in the molecule.
 Divide that number by 2 to determine the number of pairs of
electrons available to bond
 Determine your center atom, it’s always the element that is
closest to the left side of the periodic table, except H. Why not
H?
 Draw your center atom and all other atoms around it, drawing a
line or bonding pair of electrons between the center atom and
each end atom.
 Subtract the number of bonding pairs from the total number of
electron pairs you started with. This is how many lone pairs of
electrons you have.
 Place lone pairs of electrons around each atom, always starting
with the end atoms, EXCEPT H. Why not H?
 If you run out of lone pairs and the center atom is still NOT
stable with 4 pairs of electrons around it, then you must create
double or triple bonds.
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 Practice Problems- Draw Lewis structures for the following
molecules.
 1. CBr4
 2. NH3
 3. CO2
Heat and Energy
o Define:
 Temperature
 Heat
 Heat Capacity
 Endothermic
 Exothermic
o How much heat given off or taken in by a process can be determine by the
equation, q = mCΔT.
o Problems to try
 1. If q ends up being a positive number, then the process is
endothermic or exothermic?
 2. If q ends up being a negative number, then the process is
endothermic or exothermic?
 3. A 35.2g sample of metal X requires 1251J of energy to heat the
sample by 25°C. Calculate the heat capacity of metal X.
 4.
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Solutions
o Factors that affect the rate at which a solute can dissolve in a solvent
 1.
 2.
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4. Why do each of these factors affect how fast the solute dissolves?
Molarity – a measure of concentration
 M = n/V
 n = moles of solute, V = volume of solution in L
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Dilution – when you dilute a solution, you add more solvent but the
amount of solute remains the same before and after dilution. Therefore,
the moles of solute before you dilute something equals the moles of
solute after the dilution.
 M1V1 = M2V2
Molality – another measure of concentration
 m = n/ kg
 n = moles of solute, kg = kg of solvent
Practice Problems to try.
 1. Calculate the molarity if you dissolve 15.0g of Mg(OH)2 in 400mL
of water.
 2. How many liters of water were used to create a 0.76M solution of
HCl using 5.00g of HCl?
 3. What mass of NH3 was used to create a 1.23M solution that has a
volume of 1.00L?
 4. What is the molality of a solution that was created by dissolving
19.0g of NaCl in 121kg of water?
 5. What is the molality of a solution that was created by dissolving
100.0g of CaCO3 in 140.0kg of water?
Colligative Properties – properties of a solution that depend only on the number
of particles in solution not what the solute is.
 Boiling Point Elevation – adding a solute to a solution raises the boiling
point of the solution because
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 Tb = kb i  m
Freezing Point Depression – adding a solute to a solution lowers the
freezing point of the solution because
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Tf = kf  i  m
Practice Problems to try
 1. Determine the new freezing point of water when 85.3g of O2 gas
are dissolved in 1500g of water. The Kf for water is 1.86°C/m. The
original freezing point of water is 0°C.
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Practice Problems
 1. What is the molality of a solution in which 0.32 moles AlCl3 has
been dissolved in 2,200 g water?
 2. What mass of water is needed to prepare a 1.20 molal solution
using 0.60 mol propyleneglycol?
 3. What is the molarity of 245.0 g of H2SO4 dissolved in 1.00 L of
solution?
 4. Sea water contains roughly 28.0 g of NaCl per liter. What is the
molarity of sodium chloride in sea water?
 5. How many grams of Ca(OH)2 are needed to make 100.0 mL of
0.250 M solution?
 6. If I add 45g of sodium chloride to 500g of water, what will be the
new melting point and boiling point of the solution. kb = 0.512C/m
and kf = 1.86C/m.
Acids and Bases
o According to Arrhenius, what is the definition of an acid? A base?
o According to Bronsted- Lowry, what is an acid? A base? A conjugate acid?
A conjugate base?
o What is the difference between a strong acid and a weak acid? Name 6
strong acids.
o Calculate the pH & pOH of the following acids and bases.
 1. A 4.04  10-5 M solution of HBr
 2. A 8.88  10-8 M solution of KOH
 3. 45.0g of HNO3 dissolved in 500. mL of solution.
 4. 45.0 g of NaOH dissolved in 500.mL of solution.
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