general chemistry i, unit i: study guide - cracking...
TRANSCRIPT
General Chemistry I | Unit I | 1
General Chemistry I, Unit I: Study Guide
CDS Chapter 1: Atomic Molecular Theory Law of Conservation of Mass – the total mass of all products of a chemical reaction is equal to the total mass of
all reactants of the reaction
The total mass of the elements that make up the compound must equal the mass of the starting compound
Example: copper carbonate – made of copper, carbon and oxygen
o A 100.0g sample of copper carbonate contains 51.5g copper, 38.8g oxygen and 9.7g carbon
o The sum of these constituent masses is 100g, the same as the mass of copper carbonate
o Regardless of the specific sample of copper carbonate, or the mass of the sample, the fraction of the
mass of the sample which is copper (or oxygen or carbon) is always the same (51.5%, 38.8% and 9.7%
respectively)
Law of Definite Proportions – When two or more elements combine to form a compound, their masses in that
compound are in a fixed and definite ratio
The Law of Definite proportions does not prove anything about atoms – especially not how they combine to
make compounds
The mass relationships between a fixed amount of nitrogen and oxygen in three different oxides shows that
there are a few specific masses of oxygen which will combined with the fixed nitrogen, and those specific
masses are integer multiples of each other
o This reveals that oxygen exists as fixed units of mass known as an “atom”
Law of Multiple Proportions – When two elements combine to form more than one compound, the masses of
one element that combine with a fixed mass of the other element are in simple integer ratio
o If we fix the mass of one element, the masses of the other element in the three compounds are always
in simple integer ratio
Atomic Molecular Theory
o Each element is composed of very small, identical particles called atoms
o All atoms of a single element have the same characteristic mass
o The number and masses of these atoms do not change during a chemical transformation
o Each compound consists of identical molecules, which are small, identical particles formed of atoms
combined in simple whole number ratios
CDS Chapter 2: Atomic Masses and Molecular Formulas Assumptions: knowledge of the Atomic Molecular Theory, Law of Conservation of Mass, Law of Definite
Proportions, Law of Multiple Proportions
1 L oxygen + 2 L hydrogen gas = 2 L water vapor
o The volume of gases is not conserved here – the combined volume of the reactants is 3 L, but the
product is 2 L
o However, the volumes of the gases are in simple integer ratios
o Arbitrary amounts of hydrogen and oxygen cannot be combined because they combine in a fixed ratio
by mass
Law of Combining Volumes – When gases combine during a chemical reaction at fixed temperature and
pressure, the volumes of the reacting gases and products are in simple integer ratios
From Atomic Molecular Theory: atoms and molecules react in simple integer ratios
o A possible explanation for the integer volume ratios is that these are the same integer ratios as the
particles react in
o Equal volumes of the two gases must contain equal numbers of particles, regardless of whether they are
hydrogen or oxygen
o Basis for this conclusion: uniqueness of integers – Occam’s Razor
General Chemistry I | Unit I | 2
Avogadro’s Law – Equal volumes of gas contain equal numbers of particles, if the volumes are measured at the
same temperature and pressure
Problem: 1 L of hydrogen + 1 L of chlorine = 2 L of hydrogen chloride
o This is the same as saying 1 hydrogen atom + 1 chlorine atom = 2 hydrogen chloride atoms
o This violates AMT – unless hydrogen is diatomic
The rest is observation-based – read over to review
CDS Chapter 3: Structure of an Atom Assumptions: Atomic Molecular Theory, measurements of relative atomic masses, properties of an element and
its atoms
o Electricity consists of individual charged particles called electrons – assigned a negative charge
o Atoms in elemental form are neutral (without charge) – so atoms must contain a positive charge that
equals the charge of the electrons – there has to be an integer number of negative and ositive charges
Rutherford’s Gold Foil Experiment
o Alpha particles – positively charged, much more massive than electron
o Beam of alpha particles will be fired at the gold foil
o Observations and conclusions:
Observation Conclusion
The greatest number of the alpha particles pass directly through the gold foil without any deflection.
The gold foil consists mostly of empty space.
A much smaller number of alpha particles experience small deflections in paths.
Since these particles are deflected, whatever they came close to must have been more massive than they are – they came close to the nucleus.
A very small fraction of alpha particles are deflected back in the direction of the beam they came from.
A tiny fraction of the alpha particles must encounter something other than empty space – in order to rebound, an alpha particle must hit something much more massive than itself.
o Simple atom model: each gold atom must be mostly empty space and most of the mass of the gold
atom must be concentrated into a very small fraction of the volume of the atom – this concentrated
mass is the “nucleus.”
o If the positive charges of the atom are concentrated in the nucleus, then the negative charges in the
atom (the electrons) must be in the much larger space outside the nucleus.
X-ray emissions
o Since the positive charge is an integer, there must be particles of positive charge in the nucleus known
as protons
o Use spectroscopy to identify that a sample contains a particular compound or element just by looking at
frequencies of light emitted
o There is a parabolic relationship between atomic number and x-ray frequency
o This shows that the atomic number is a physical property of the atom – because it is an integer, it must
be counting something within the atom – the only thing it could be describing is the number of protons
CDS Chapter 4: Electron Shell Model of an Atom Elements with similar atomic numbers are very different both physically and chemically
Assumptions: AMT, relative atomic masses, structure of an atom, atomic numbers
General Chemistry I | Unit I | 3 o Coulomb’s Law (edit 8/17/15: you will need to know this for the rest of general chemistry and possibly
for the rest of your god damn life) – algebraic expression which relates the strength of the interaction
between two charged particles to the sizes of the charges on the particles and the distance between
them.
The strength of the interaction between particles is either the force that one particles exerts on
the other or the potential energy that exists when two particles interact. General Chemistry likes
to focus on the potential energy, symbolized by V: V =(q1)(q2)
r
When V is a large negative number, the potential energy is very low and the charges are
strongly attracted to one another – it requires a lot of work to pull them apart.
V will be a large negative number when…
The charges are large
r is small
Elements with very different atomic numbers can have similar chemical properties
o Fluorine and chlorine are both gases and exist as diatomic molecules in nature – both are highly
reactive, combine with hydrogen to form acids, combine with metals to form solid salts – but atomic
numbers are different
o Alkali metals – lithium, sodium, potassium – soft metals with low melting points – form salts with
chlorine w/ similar molecular formulas
Immediately before each of these elements in the list are three elements that are also very
similar to each other – noble gases
Immediately after each of these elements are three elements w/ similar properties – alkaline
earth metals
Halogens
o Groups of elements appear “periodically” in the rankings of the elements by atomic number.
Periodic Law – The chemical and physical properties of the elements are periodic functions of the atomic
number.
The attraction of an electron at distance r from the nucleus is given by the potential energy in Coulomb’s Law:
𝑉(𝑟) =(+𝑍𝑒)(−𝑒)
𝑟
o An electron close to the nucleus is more strongly attracted to the nucleus (more negative potential
energy)
o Electron in an atom with a large atomic number (Z) would be more strongly attracted to its nucleus
than an electron in another atom with a smaller atomic number
o The ionization energy is the amount of energy required to remove an electron from an atom.
Large negative V(r) would require a large ionization energy
General Chemistry I | Unit I | 4
Observations of ionization energy (IE) graph:
o Ionization energy increases and then abruptly decreases over and over with increasing atomic number.
o The top of each peak is a noble gas.
o The bottom of each valley is an alkali metal.
o Between each alkali metal and valley, ionization energy increases almost linearly.
o In each period, the IE increases as one moves from alkali metal to noble gas.
o The sharp drops in ionization energy after each noble gas are due to a large increase in r (the distance
between the valence electrons and the nucleus).
Conclusions:
o In each period of the Periodic Table, from an alkali metal to a noble gas, the outermost electrons are all
about the same distance from the nucleus.
o Each additional electron is added to this consistent “layer” – “electron shell”
o At the end of the period, the shell is full and unable to accommodate more electrons – the next
electron is in a new shell, much farther from nucleus.
o Alkali metals have a single electron in the outermost shell – that’s why they’re similar.
o Because this outermost shell is so important, it is the “valence shell” and contains “valence electrons.”
Successive electrons can be removed – ionization energies for each one
o The second ionization energy is the energy needed to remove an electron from the positive ion to form
an ion with a +2 charge
o Observations from table 4.2:
The first electron in Na (sodium) is easy to remove, but the second one is harder.
The first electron is in a valence shell, far from nucleus. Therefore, the r value in
Coulomb’s Law is large, so the electron’s potential energy is small.
General Chemistry I | Unit I | 5
The second electron is closer in and more strongly attracted; its r value is small, so its
potential energy has a large negative value.
Conclusion: Sodium (Na) only has one valence electron.
The first and second electrons in Mg are relatively easy to remove, third one is harder.
Mg has two valence electrons.
General Chemistry I | Unit I | 6
What the CDS doesn’t cover
Main-group cations and
anions (2.12)
Common Polyatomic Ions
(2.12) Formula Name Formula Name
Cation Singly charged anions (cont.)
NH4+ Ammonium NO2
- Nitrite
Singly charged anions NO3- Nitrate
CH3CO2- Acetate Doubly charged anions
CN- Cyanide CO32- Carbonate
ClO- Hypochlorite CrO42- Chromate
ClO2- Chlorite CrO7
2- Dichromate
ClO3- Chlorate O2
2- Peroxide
ClO4- Perchlorate HPO4
2- Hydrogen phosphate
H2PO4- Dihydrogen phosphate SO3
2- Sulfite
HCO3- Hydrogen carbonate SO4
2- Sulfate
HSO4- Hydrogen sulfate S2O3
2- Thiosulfate
OH- Hydroxide Triply charged anion
MnO4- Permanganate PO4
3- Phosphate
Percent yield Percent yield =
Actual yield of product
Theoretical yield of product× 100%
Limiting reactants / excess
reactant
1. Convert the masses of reactant to moles.
2. Compare these mole values to the balanced chemical equation. For example, if
your reactants were 2 mol H2O and 1 mol CO2, and you found that your samples
contained 0.5 mol H2O and 0.3 mol CO2, you should know that H2O is the limiting
reactant because the chemical equation says the amount of H2O has to be two
times that of CO2 for a full reaction to take place.
General Chemistry I | Unit I | 7
3. Use the limiting reactant and basic stoichiometry to find the amount of product in
grams.
4. The mass of excess reactant can be found by converting only the amount of excess
reactant required by the chemical equation to grams and subtracting that from the
initial mass of the excess reactant. (In the example above, convert only 0.25 mol of
CO2 to grams, since the amount of H2O reacting must be twice that of the amount
of CO2.)
Molarity Molarity =
Moles of solute
Liters of solution=
mol
L
Diluting solution Moles of solute (constant) = Molarity × Volume = Mi × 𝑉i = Mf × 𝑉f
Solution stoichiometry The same as normal stoichiometry, just including molarity.
Titration (3.9) Titration is a procedure for determining the concentration of a solution by allowing a
measured volume of that solution to react with a second solution of another substance
(the standard solution) whose concentration is known.
Percent composition /
empirical formulas (3.10)
Percent composition can be used to find the empirical formula of a compound.
1. Convert given masses to percentages of the mass of the initial sample.
2. Use these percent masses to make a hypothetical sample of 100 grams – the
percent composition values now become the masses of the constituent elements in
this 100g sample.
3. Convert the grams to moles.
4. Divide each molar value by the smallest molar value to find the mole ratio and
empirical formula.
Combustion analysis For combustion involving only carbon, hydrogen and oxygen:
1. Find the molar amounts of carbon and hydrogen from the provided masses of CO2
and H2O. Example:
1.023 g CO2 ×1 mol CO2
44.01 g CO2×
1 mol C
1 mol CO2= 0.02324 mol C
2. Find the number of grams of carbon and hydrogen in the sample using the molar
masses you just found.
General Chemistry I | Unit I | 8
3. Subtract the masses of carbon and hydrogen from the mass of the starting sample.
This will give you the “missing” mass of oxygen in the compound.
4. Convert the mass of oxygen to moles.
Now that you have mole values for all three of the constituent elements, you can divide
each of the mole values by the smallest of the three mole values to find the C:H:O ratio
and the empirical formula.
For combustion involving carbon, hydrogen, oxygen and another element:
1. Determine the mass of each element.
2. Convert each of the masses to composition percentages.
3. Determine the mass of each element present in a 100g sample of the molecule you
are combusting.
4. Find the molar ratios to form the empirical formula.
5. Find the molecular formula using molar mass information provided.
Net ionic equations An ionic equation explicitly shows all of the ions involved in a chemical reaction. It
can be derived from the molecular equation. A net ionic equation is an ionic equation
without the “spectator ions,” which are ions that do not undergo any change during the
course of the reaction (they are simply there to balance the charge).
Ionic equation: Pb2+ (aq) + 2 NO3
- (aq) + 2 K+ (aq) + 2 I- (aq) 2 K+ (aq) + 2 NO3- (aq) + PbI2 (s)
Net ionic equation: Pb2+ (aq) + 2 I- (aq) PbI2 (s)
Oxidation numbers (4.6) An atom in its elemental state has an oxidation number of 0.
An atom in a monatomic ion has an oxidation number identical to its charge.
An atom in a polyatomic ion or in a molecular compound usually has the same
oxidation number it would have it were a monatomic ion.
o In general, the farther left an element is in the periodic table, the more
probable that it will be “cationlike.” Metals usually have positive
oxidation numbers. The opposite is true for nonmetals.
o Hydrogen can be either +1 or -1.
o Oxygen is usually -2.
o Halogens are usually -1.
The sum of the oxidation numbers is 0 for a neutral compound and is equal to the
net charge for a polyatomic ion.
Redox reactions (4.6-4.7) Oxidation is the loss of one or more electrons by a substance (element, compound, or
ion), and a reduction is the gain of one or more electrons by a substance. An
oxidation-reduction (redox) reaction is any process in which electrons are transferred
from one substance to another.
Identifying redox reactions:
Reducing agents
o Causes reduction
o Loses one or more electrons
o Undergoes oxidation
o Oxidation number of atom increases
Oxidizing agent
o Causes oxidation
o Gains one or more electrons
o Undergoes reduction
o Oxidation number of atom decreases
General Chemistry I | Unit I | 9
Half-reaction method for
balancing redox reactions
Redox stoichiometry (4.10) It’s the same as titration. Really, check for yourself if you don’t believe me. Bitch.
Atomic radii (5.14) An atom’s radius is half the distance between the nuclei of two identical atoms
when they are bonded together.
A comparison of atomic radius versus atomic number shows a periodic rise-and-
fall pattern.
o Atomic radii increase going down a group of the periodic table but
decrease going across a row from left to right.
o The increase going down a group occurs because successively larger
valence-shell orbitals are occupied.
o The decrease in radius from left to right across the table occurs because of
an increase in effective nuclear charge caused by the increasing number of
protons in the nucleus (increasing nuclear charge).
Valence shell electrons are
o Strongly shielded by electrons in the inner shells, which are closer to the
nucleus
o Less strongly shielded by other electrons in the same shell
o Only weakly shielded by other electrons in the same subshell, which are
the same distance from the nucleus
Because electrons in the same shell are at approximately the same distance from
the nucleus, they are relatively ineffective at shielding one another. At the same
time, the nuclear charge Z increases across a row. Therefore, the effective nuclear
charge for the valence shell electrons increases across the period, drawing all
the valence shell electrons closer to the nucleus and shrinking the atomic radii.