Download Significant Figures

Significant Figures
Zeros appearing between nonzero digits are significant, for example:
* 60.8 has three significant figures
* 39008 has five significant figures
Zeros appearing in front of nonzero digits are not significant, for
example:
* 0.093827 has five significant figures
* 0.0008 has one significant figure
* 0.012 has two significant figures
Significant Figures
Zeros at the end of a number and to the right of a decimal are
significant, for example:
* 35.00 has four significant figures
* 8,000.000000 has ten significant figures
Zeros at the end of a number without a decimal point may or may not
be significant, and are therefore ambiguous, for example:
* 1,000 could have between one and four significant figures.
This ambiguity could be resolved by placing a decimal after the
number, e.g. writing "1,000." to indicate specifically that four
significant figures are meant, but this is a non-standard usage.
To specify unambiguously how many significant figures are implied,
scientific notation can be employed:
* 1×103 has one significant figure, while
* 1.000×103 has four.
Significant Figures
Multiplication and Division
When multiplying or dividing numbers, the result is rounded
to the number of significant figures in the factor with the least
significant figures. Here, the quantity of significant figures in
each of the factors is important—not the position of the
significant figures. For instance, using significance arithmetic
rules:
*
*
*
*
*
*
*
8 × 8 = 6 × 101
8 × 8.0 = 6 × 101
8.0 × 8.0 = 64
8.02 × 8.02 = 64.3
8 / 2.0 = 4
8.6 /2.0012 = 4.3
2 × 0.8 = 2
Significant Figures
Addition and Subtraction
When adding or subtracting using significant figures rules, results
are rounded to the position of the least significant digit in the
most uncertain of the numbers being summed (or subtracted). That
is, the result is rounded to the last digit that is significant in each
of the numbers being summed. Here the position of the significant
figures is important, but the quantity of significant figures is
irrelevant. Some examples using these rules:
* 1 + 1.1 = 2
- 1 is significant up to the ones place, 1.1 is significant up to
the tenths place. Of the two, the least accurate is the ones place.
The answer cannot have any significant figures past the ones
place.
* 1.0 + 1.1 = 2.1
- 1.0, 1.1 are significant up to the tenths place. So will the
answer.
Significant Figures
Addition and Subtraction
* 100 + 110 = 210
- 100, 110 are significant up to the ones place, even though
these digits are zeroes. So will the answer.
* 1×102 + 1.1×102 = 2×102
- 100 is significant up to the hundreds place, while 110 is up
to the tens place. Of the two, the least accurate is the hundreds
place. The answer should not have significant digits past the
hundreds place.
Significant Figures
Exact Numbers
Some numbers are considered to have an infinite number of
significant figures and are called exact numbers. An example
would be a whole number from a balanced equation:
6.021 x 4/2 = 12.04
where 4/2 is perhaps the stoichiometric ratio of product to
reactant. Since both 4 and 2 are exact numbers they can be
considered to each have more significant figures than 6.021 which
has only four significant figures. Thus, the rules for multiplication
and division indicate that the result should have four significant
figures.
Significant Figures
Both Multiplication/Division and
Addition/Subtraction with Significant Figures
●
when doing different kinds of operations
with measurements with significant figures,
do whatever is in parentheses first,
evaluate the significant figures in the
intermediate answer, then do the remaining
steps
3.489
×
(5.67
–
2.3) =
2 dp
1 dp
3.489
×
3.37
=
12
4 sf
1 dp & 2 sf
2 sf
Significant Figures
Calculations
In a long calculation involving mixed operations, carry as
many digits as possible through the entire set of calculations
and then round the final result appropriately. For example,
(5.00 / 1.235) + 3.000 + (6.35 / 4.0)=4.04858... + 3.000 +
1.5875=8.630829...
The first division should result in 3 significant figures. The
last division should result in 2 significant figures. The three
numbers added together should result in a number that is
rounded off to the last common significant digit occurring
furthest to the right; in this case, the final result should be
rounded with 1 digit after the decimal. Thus, the correct
rounded final result should be 8.6. This final result has been
limited by the accuracy in the last division.
Significant Figures
Example Problems
5.692 m + 0.6 m + 4.33 m
= 10.6 m
4.51 cm x 3.6666 cm
= 16.5 cm2
7.310 km / 5.70 km
= 1.28
3.00 x 10-5 – 1.5 x 10-2
= 0.0000300 - 0.015
= -0.01497
= -1.5 x 102
Dimensional Analysis
●
●
●
●
●
Many problems in chemistry involve using
relationships to convert one unit of measurement to
another
Conversion factors are relationships between two units
– May be exact or measured
Conversion factors generated from equivalence
statements
2.54cm
1 in
– e.g., 1 inch = 2.54 cm can give
or
1 in
2.54 cm
Arrange conversion factors so given unit cancels
– Arrange conversion factor so given unit is on the
bottom of the conversion factor
May string conversion factors
– So we do not need to know every relationship, as
long as we can find something else the given and
desired units are related to
Dimensional Analysis
Example Problems
1 lb = 453.6 g
convert 0.0833 lb to milligrams
0.0833 lb x 453.6 g/lb x 1000 mg/g
= 3.78 x 105 mg
1 mi = 1609 m
convert 11.0 x 103 km/s to mi/hr
km
m
1 mi
s
11.0×10
×1000
×
×3600
s
km 1609m
hr
3
=2.46×107
mi
hr
The Atomic Theory
1. Elements are composed of atoms.
- tiny, hard, unbreakable spheres.
2. All atoms of a given element are identical
- all carbon atoms have the same chemical and physical
properties
3. Atoms of a given element are different from those of another
element.
- carbon atoms have different chemical and physical properties
than sulfur atoms.
4. Atoms of one element can combine with atoms of another element to
form compounds.
- Law of Definite Proportions
- Chemical formulas
5. Atoms are indivisible in chemical processes (Lavoisier).
- all atoms present at the beginning of a chemical reaction are
present at the end of the reaction
- atoms of one element cannot change into atoms of another
element
The Atomic Theory
electron
neutron
proton
The Atomic Theory
Isotopes
●
●
●
●
●
all isotopes of an element are chemically identical
– undergo the exact same chemical reactions
all isotopes of an element have the same number of
protons
isotopes of an element have different masses
isotopes of an element have different numbers of
neutrons
isotopes are identified by their mass numbers
– protons + neutrons
The Atomic Theory
Isotopes
• Atomic Number
 Number of protons
example: Lithium – 3 protons
Z
-Z=3
• Mass Number
 Protons + Neutrons
example: Nitrogen-15 15
7N
 Whole number
• Abundance = relative amount found in a sample
Atomic Mass Units
Instead of using hydrogen, the lightest element,
for historical reasons, we arbitrarily assign the
carbon-12 atom a mass of exactly 12 atomic mass
units (12 amu or simply 12 u) and then scale the
mass of every other atom to that. Thus, atomic
masses are actually relative masses.
Note that while mass numbers are whole numbers
representing the sum of protons and neutrons in
the nucleus, atomic masses are not generally
whole numbers.
The Atomic Theory
Isotopes
The weighted average:
Let's say that we have two sets of data with 20 in the
first and 30 in the second:
1: 62, 67, 71, 74, 76, 77, 78, 79, 79, 80, 80, 81, 81, 82,
83, 84, 86, 89, 93, 98
2: 81, 82, 83, 84, 85, 86, 87, 87, 88, 88, 89, 89, 89, 90,
90, 90, 90, 91, 91, 91, 92, 92, 93, 93, 94, 95, 96, 97,
98, 99
average 1: 80
average 2: 90
What is the overall average?
The Atomic Theory
Isotopes
Two ways to calculate overall average.
1. sum all numbers and divide by total:
4600/50 = 86
2. calculate a weighted average:
20×8030×90
=86
2030
or
20
30
×80
×90
2030
2030
The Atomic Theory
Isotopes
The average atomic mass is a weighted average of atomic
masses of isotopes.
avg. mass= fraction A×mass A fraction B×mass B
The Atomic Theory
Isotopes
Example Problems
How many protons, neutrons and electrons in:
a)
17
8
b)
199
80
c)
63
29
O
8 protons, 8 electrons, 9 neutrons
Hg
80 protons, 80 electrons, 119 neutrons
Cu
2+
29 protons, 27 electrons, 34 neutrons
The Atomic Theory
Isotopes
Example Problems
Potassium has two major isotopes, potassium-39 and
potassium-41. The atomic mass of potassium-39 is 38.964 amu
and its percent abundance is 93.1%. If the average atomic
mass of potassium is 39.098 amu what is the atomic mass of
potassium-41?
41
39.098=0.931×38.9640.069×Mass K
41
Mass K =
39.098−0.931×38.964
0.069
= 41
Chemical Compounds
Molar Mass
The mass of 1 mol of a substance in grams is defined
as it’s molar mass. Since a mol of a substance is an
Avogadro’s number of particles we can simply use the
numerical value of the atomic mass of the element from
the periodic table in grams units to find the molar
mass. Thus:
The molar mass of H atoms is 1.008 g/mol
The molar mass of H2 molecules is 2.016 g/mol
The molar mass of H2O molecules is 2 X 1.008 + 16.00 =
18.02 g/mol
Thus, for compounds we can find the molar mass (or the
formula mass) by using the molecular formula or the
formula unit and simply adding up the atomic masses of
each atom in the formula.
The Mole
We generally use the SI counting unit, the mole, to
represent an Avogadro’s number of atoms or
molecules.
The numerical value of the mole is defined as being
equal to the number of atoms in exactly 12 grams of
pure carbon-12. This is an Avogadro's number of
carbon-12 atoms or 6.022 x 1023 atoms.
The mole is nothing more than a counting unit like
“dozen” or “pair”. It's definition is purely for
convenience to make the atomic masses of the
elements in atomic mass units (which are hard to
weigh in the lab) coincide with molar masses of the
elements.
Chemical Compounds
Calculate the molar mass of V2O4 and Na2CO3.
One mole of V2O4 contains two mols of V and four mols of O and
the masses of these would be:
2 x 50.942 g/mol + 4 x 15.999 g/mol = 165.88 g/mol
One mole of Na2CO3 contains two mols of Na, one mol of C and
three mols of O. The molar mass is:
2 x 22.990 g/mol + 12.011 g/mol + 3 x 15.999 g/mol
= 105.99 g/mol
The Mole
How many moles of ions are in 27.5 g of MgCl2?
MgCl2
Mg2+ + 2Cl-
mols MgCl2 = 27.5 g / 95.21 g/mol
= 0.289 mol
mols ions = 3 x mols MgCl2
= 3 x 0.289 mol
= 0.867 mol
The Mole
What is the mass in grams of 7.70 x 1020 molecules of caffeine,
C8H10N4O2?
Mass of one mol of caffeine = 194.19 g/mol
This is the mass of an Avogadro's number of caffeine molecules,
6.022 x 1023.
mass of 7.70 x 1020 molecules of caffeine =
20
7.70×10 molecules
×194.19g /mol
23
6.022×10 molecules/mol
=0.248 g
The Mole
A sample of glucose, C6H12O6, contains 1.25 x 1021 atoms of
carbon. How many atoms of hydrogen does it contain? How
many mols of glucose does it contain?
From the formula for glucose there are twice as many
hydrogens as carbons so there must be 2 x 1.25 x 1021 = 2.50 x
1021 atoms of hydrogen.
Since each molecule has 6 carbons the number of glucose
molecules will be 1.25 x 1021 / 6 = 2.08 x 1020. Division by
Avogadro's number will give the number of mols of glucose:
20
2.08×10 molecules
−4
=3.46×10
mol
23
6.022×10 molecules/mol
Chemical Compounds
Empirical Formula
Summary of empirical formula determination steps:
1. Use the given % composition, or determine what the
%composition is and then find the mass of each
element in a 100 g sample of the substance.
2. For each elemental mass, determine the number of
mols of the element in the 100 g sample using the
corresponding atomic mass from the periodic table.
3. If necessary, divide each mol quantity by the
smallest of the numbers of mols to get whole numbers.
If any fractions result, multiply all results by the
appropriate number to obtain all whole numbers.
4. Use the whole numbers from step 3 to generate an
empirical formula.
Chemical Compounds
Empirical Formula
When a 0.350 g sample of cinnabar containing Hg and S is
heated it decomposes to give 0.302 g Hg. What is the
empirical formula of cinnabar?
Mols Hg = 0.302 g / 200.59 g/mol
= 1.51 x 10-3 mol
mass S = 0.350 g – 0.302 g = 0.048 g
mols S = 0.048 g / 32.065 g/mol
= 1.5 x 10-3 mols
mol ratio Hg:S
1.51 x 10-3:1.5 x 10-3 or 1:1
empirical formula HgS
Chemical Compounds
Empirical Formula
A sample of a chromium compound has a molar mass of
151.99 g/mol. Elemental analysis shows that it is composed of
68.43% Cr and 31.57% O. What are the empirical and
molecular formulas for this compound?
In a hypothetical 100 g sample there will be 68.43 g Cr and
31.57 g O.
mols Cr = 68.43 g / 51.996 g/mol = 1.316 mol
mols O = 31.57 g / 15 999 g/mol = 1.973 mol
mol ratio V:O 1.316:1.973 or 1:1.5 or 2:3
empirical formula: V2O3 empirical formula mass: 149.88 g/mol
ratio molar mass to efm 151.99:149.88 or 1:1
molecular formula is V2O3
Oxidation States
Rules for assigning oxidation numbers:
1. The oxidation state of an atom in it's elemental
form is zero.
2. For ions composed of only one atom the oxidation
number is equal to the charge on the ion.
3. The oxidation state of oxygen in most compounds is
-2. A notable exception is in peroxides such as
hydrogen peroxide in which the oxidation state of
oxygen is -1.
4. The oxidation state of hydrogen is generally +1
except when it is bonded to metals such as sodium
(NaH) in which case it's oxidation number is -1.
Oxidation States
5. Fluorine has an oxidation number of -1 in its
compounds ... always.
6. The sum of the oxidation numbers of all of the
atoms in a molecule or ion must equal the net charge
on the molecule or ion.
7. Oxidation numbers do not necessarily have to be
integers. ie. In Fe3O4, iron has an apparent oxidation
number of +2 2/3.
Oxidation States
What is the oxidation number of V in V2O4?
4 x O = -8
2xV= ?
----Total charge 0
Oxidation number on each V must be +4
Nomenclature of Compounds
Metal-containing Binary Compounds
Compounds containing only two different elements are
binary compounds and are named by using the element name
of the first element followed by the name of the second
element modified with the suffix 'ide'.
exceptions:
NaCl
Al2O3
sodium chloride
aluminum oxide
NaOH
KCN
sodium hydroxide
potassium cyanide
Nomenclature of Compounds
Many metals can exist in more than one oxidation state
and form can form two or more compounds with a given
anion.
Fe+2
Fe+3
ferrous ion
ferric ion
FeCl2
FeCl3
ferrous chloride or iron(II) chloride
ferric chloride or iron(III) chloride
Binary Acids
When binary compounds with acidic properties are
discussed in the context of their acidity we alter
their names by using the prefix 'hydro' and suffix
'ic':
HCl
HF
HCN
hydrochloric acid
hydrofluoric acid
hydrocyanic acid
Nomenclature of Compounds
Most polyatomic ions are anions rather than cations and
the most common modifiers used as suffixes are 'ate' and
'ite', depending primarily on oxidation numbers. Also,
it is most common for polyatomic anions to have oxygen
in them and are called oxoanions.
NH4+
CO3-2
NO3NO2-
ammonium
carbonate
nitrate
nitrite
The 'ate' suffix is applied to the anion which has the
primary atom in the higher oxidation state and 'ite' to
the anion with the primary atom in the lower oxidation
state.
See Table 3.5 in the text for an extensive list of
anions and their names. You will need to know these!
Nomenclature of Compounds
Oxo Acids (acids of polyatomic anions)
H2CO3
H3BO3
carbonic acid
boric acid
H3PO4
H3PO3
phosphoric acid
phosphorous acid
H2SO4
H2SO3
sulfuric acid
sulfurous acid
HClO
HClO2
HClO3
HClO4
hypochlorous acid
chlorous acid
chloric acid
perchloric acid
Nomenclature of Compounds
Binary Compounds of Nonmetals
The nomenclature of these compounds is similar to the
binary compounds mentioned earlier except for those in
which the primary element can exist in multiple
oxidation states.
HCl
hydrogen chloride
PCl5
PCl3
CO2
CO
P4O6
phosphorous pentachloride
phosphorous trichloride
carbon dioxide
carbon monoxide
tetraphosphorus hexaoxide
Nomenclature of Compounds
Name these compounds: NiO, N2O5, HgCl2
Nickel (II) oxide
Dinitrogen pentoxide
Mercury (II) chloride
Stoichiometry
Chlorine dioxide is a bleaching agent used in the paper
industry and can be prepared by the following reaction:
NaClO2 + Cl2
ClO2 + NaCl
What mass of Cl2 is needed for complete conversion of 30.5 g of
NaClO2 ?
Balance the equation:
2NaClO2 + Cl2
2ClO2 + 2NaCl
Mols NaClO2 = 30.5 g / 90.442 g/mol
= 0.337 mol
mols Cl2 = 0.337 mol NaClO2 x 1 mol Cl2 / 2 mol NaClO2
= 0.169 mol
Stoichiometry
Chlorine dioxide is a bleaching agent used in the paper
industry and can be prepared by the following reaction:
NaClO2 + Cl2
ClO2 + NaCl
What mass of Cl2 is needed for complete conversion of 30.5 g of
NaClO2 ?
Mass Cl2 = 0.169 mol x 70.91 g/mol
= 12.0 g
Stoichiometry
The fertilizer ammonium sulfate is prepared by the reaction
between ammonia and sulfuric acid:
2NH3 + H2SO4
(NH4)2SO4 + 2H2O
What mass of NH3 in kg is needed to make 3.5 metric tons of
ammonium sulfate?
Mols (NH4)2SO4 = 3.5 tons x 1000 kg/ton x 1000 g/kg / 132.14
g/mol
= 2.6 x 104 mols
mols NH3 = 2.6 x 104 mols (NH4)2SO4 x 1 mol NH3/ 1 mol
(NH4)2SO4
= 2.6 x 104 mols
mass NH3 = 2.6 x 104 x 17.03 g/mol = 4.5 x 105g or 4.5 x 102 kg
Limiting Reagents
For the reaction between Zn and HCl:
Zn + 2HCl
H2 + ZnCl2
if 12.5 g HCl and 7.3 g Zn are reacted how much ZnCl2
can be expected?
Assume HCl is limiting:
mols HCl = 12.5 g / 36.46 g/mol
= 0.343 mol
mols ZnCl2 = 0.343 mol HCl x 1 mol ZnCl2 / 2 mol HCl
= 0.171 mol
Limiting Reagents
For the reaction between Zn and HCl:
Zn + 2HCl
H2 + ZnCl2
if 12.5 g HCl and 7.3 g Zn are reacted how much ZnCl2
can be expected (in grams)?
Assume Zn is limiting:
mols Zn = 7.3 g / 65.409 g/mol
= 0.11 mol
mols ZnCl2 = 0.11 mol Zn x 1 mol ZnCl2 / 1 mol Zn
= 0.11 mol
This is the smaller of the two so Zn is the limiting
reagent.
Limiting Reagents
For the reaction between Zn and HCl:
Zn + 2HCl
H2 + ZnCl2
if 12.5 g HCl and 7.3 g Zn are reacted how much ZnCl2
can be expected (in grams)?
Mass ZnCl2 = 0.11 mol x 136.296 g/mol
= 15 g
Reaction Yield
The theoretical yield of a reaction is the maximum
amount of product that can be obtained for a given
amount of limiting reagent. In practice, the
theoretical yield is rarely obtained ... most often
the actual yield is less than the theoretical yield.
The efficiency of the reaction is described by the
%yield:
% yield = (actual yield/theoretical yield) x 100
Reaction Yield
Determine the % yield for the reaction:
KClO3
KCl
+ O2
if 2.41 g KClO3 produces 0.67 g O2.
2KClO3
2KCl
+ 3O2
First, calculate the theoretical yield:
mols KClO3 = 2.41 g / 122.55 g/mol
= 0.0197 mol
mols O2 = 0.0197 mol KClO3 x 3 mol O2 / 2 mol KClO3
= 0.0295 mol
mass O2 = 0.0295 x 31.999 g/mol
= 0.944 g
Reaction Yield
Determine the % yield for the reaction:
KClO3
KCl
+ O2
if 2.41 g KClO3 produces 0.67 g O2.
Now the percent yield:
% yield = (0.67 g/ 0.944 g) x 100
= 71 %
Solution Stoichometry
Molarity = mols solute/liters of solution
or, in symbolic form:
M = n/V
or, rearranging the equation:
n = M x V
Thus, 0.346 mol of HCl dissolved to make 550 mL of
solution has a molarity of:
MHCl = 0.346 mol/0.550 L
= 0.629 mol/L
Solution Stoichometry
Sometimes, a concentrated solution must be diluted for
use in an experiment and the concentration of the
diluted solution is needed. The key to this
calculation is the fact that if a sample of a
concentrated solution is diluted the number of mols of
solute does not change, just the volume of the
solution changes. This can be represented
symbolically:
or
Mi
ni = nf
x Vi = Mf x Vf
where i is initial and f is final
Solution Stoichometry
What will be the final volume of a solution if 10 mL of a
1.5 M solution is to be diluted to give a 0.30 M
solution?
Mi x Vi = Mf x Vf
Vf = Mi x Vi / Mf
= 1.5 M x 10 mL / 0.30 M
= 50 mL
Solution Stoichometry
How many grams of NaCl are required to precipitate most of
the Ag+ ions from 2.50 x 102 mL of 0.0113 M AgNO3 solution?
Cl- + Ag+
AgCl
mols Ag+ = 0.0113 mol/L x 0.250 L
= 0.00283 mol
mols Cl- = 0.00283 mol Ag+ x 1 mol Cl- / 1 mol Ag+
= 0.00283 mol
mass NaCl = 0.00283 mol x 58.44 g/mol
= 0.165 g
Solution Stoichometry
A 3.664 g sample of a monoprotic acid was dissolved in water. It
took 20.27 mL of a 0.1578 M NaOH solution to neutralize the
acid. What is the molar mass of the acid?
HA + NaOH
NaA
+ H2O
mols NaOH = 0.1578 mol/L x 0.02027 L
= 0.003199 mol
mols HA = 0.003199 mol NaOH x 1mol HA / 1 mol NaOH
= 0.003199 mol
molar mass = mass/mols
= 3.664/0.003199
= 1145 g/mol
Reaction Types
Precipitation Reactions
Some ionic compounds are very soluble such as NaCl
while others are only soluble to an extremely small
extent .. so much so that we think of them as insoluble.
If a solution is prepared with the cations an anions of
an insoluble ionic compound in it, the ions will combine
to form the solid form of the compound in the solvent
and will precipitate out of solution.
Reaction Types
Precipitation Reactions
In order to represent these types of reactions we use a
shortened version of the balanced equation called the
net ionic equation. Suppose we mix a solution of
Pb(NO3)2 and NaI. In each of these solutions there is
dissolved a strong electrolyte, however, when mixed
there will be Pb+2 ions and I- ions in solution and PbI2
is an insoluble compound .. a precipitate of PbI2 will
form:
Pb+2 + 2NO3- + 2Na+ + 2IThe net ionic equation is:
Pb+2 + 2I-
PbI2
2Na+ + 2NO3- + PbI2
Reaction Types
Soluble
Insoluble
Compounds that contain an
alkali metal cation or
ammonium ion
Compounds that contain
the nitrate ion
Most acetate derived
compounds
except
Most chlorides, bromides, except
iodides
Most sulfate compounds
except
silver acetate
Ag+, Hg22+, Pb2+
with cations Ba2+,
Pb2+, Ag+, Ca2+
Reaction Types
Soluble
Insoluble
Most carbonate and
But not
phosphate compounds with with
alkali metal ions or
ammonium ions
other metal cations
Hydroxides and sulfides
But not
with alkali metal ions or with
ammonium ion and Ca2+,
Sr2+, Ba2+
other metal cations
Reaction Types
Indicate whether or not the following compounds are soluble:
1. Ca3(PO4)2
insoluble
2. Mn(OH)2
insoluble
3. K2S
soluble
4. ZnSO4
soluble
5. Hg(NO3)2
soluble
Reaction Types
Acid-Base Reactions
A definition:
Acid: a substance that dissociates in water to
produce hydrogen ions (and an anion):
HA
H+
+
A-
where A- stands for an anion of some type such as Clor NO3-.
Base: a substance that dissociates in water to
produce hydroxide ions (and a cation):
MOH
M+
+
OH-
Reaction Types
Acid-Base Reactions
The Bronsted-Lowry definition:
Acid: a substance that donates a proton to a base:
HA
+
B
HB+
+
A-
Base: a substance that accepts a proton from an acid:
HA
+
B
HB+
+
A-
acid
conjugate base
base
conjugate acid
Reaction Types
Acid-Base Reactions
H2O +
H2O
H3O+
+
OH-
H3O+ and H2O are a conjugate acid-base pair
as are OH- and H2O.
Reaction Types
Oxidation-Reduction (Redox) Reactions
These reactions are ones in which there is a transfer
of electrons from one molecule to another resulting in a
change of oxidation number on some of the atoms
involved:
2Na
+
Cl2
2NaCl
In this case the oxidation number on Na and Cl2 are zero
(remember the rules for assigning oxidation numbers?)
and the oxidation number of Na in NaCl is +1 and on Cl
in NaCl is -1. The oxidation number of Na has increased
and we say that it has been oxidized ... the oxidation
number of Cl has decreased and we say it has been
reduced.
So, if an atom's oxidation number increases or it
formally loses electrons it has been oxidized and if an
atom's oxidation number decreases or it formally gains
electrons it has been reduced.
Reaction Types
Half Reactions
In order to explicitly show the gain and loss of
electrons we can split the redox reaction into half
reactions:
Na
2e- +
Na+
Cl2
+
e2Cl-
(shows oxidation reaction)
(shows reduction reaction)
In order for the overall reaction to be balanced the
two half reactions must add together so that there are
no electrons in the equation. This means that for the
above example we would have to multiply the first
reaction by 2 in order to cancel out the numbers of
electrons.
Half reactions are useful for balancing redox
equations, especially the more complex ones.
Reaction Types
MnO4-(aq) + Br-(aq)
acid solution)
--->
Mn+2(aq)
+
Br2(aq)
(in
1. Separate the reactants and products into the appropriate
oxidation and reduction half reactions.
2. Balance the numbers of atoms of the oxidized or reduced
species and write the number of electrons involved.
3. Balance the number oxygens (if any) by adding H2O to the
appropriate side of the equation.
4. Balance the number of hydrogens by adding H+ to the
appropriate side of the equation.
5. If necessary, multiply the equations by the appropriate
factors so that the number of electrons is the same in each
half reaction.
6. Add the half reactions.
Reaction Types
1. MnO4- + 5eMn2+
Br2 + 2e2Br2. 8H+ +MnO4- + 5e2BrBr2 + 2e3. 16H+ + 2MnO4- + 10e10Br5Br2 + 10e4. 16H+ + 10Br- + 2MnO4-
Mn2+ + 4H2O
2Mn2+ + 8H2O
2Mn2+ + 5Br2 + 8H2O
The Combined or Ideal Gas Law
The three proportionality laws can be combined into
one all-encompassing equation:
Boyle's Law:
rate
rate
gasA
gasB
=

M wB
M wA
Charles' Law: V ∝T or V =k ' T
Avogadro's Law: V ∝n or V =k ' ' T
Combined:
nT
V∝
P
or: V =
RnT
P
More commonly expressed as:
PV = nRT
where R is the universal gas constant:
R = PV/nT
Standard Conditions
●
●
●
since the volume of a gas varies with pressure
and temperature, chemists have agreed on a set
of conditions to report our measurements so
that comparison is easy – we call these
standard conditions
– STP
standard pressure = 1 atm
standard temperature = 273 K
– 0°C
Molar Mass and Density
mols = mass/molar mass
or: n=
m
M
substituting into PV = nRT
PV = mRT/M
solving for M:
M=
mRT
PV
Thus, we can find the molar mass of a substance in the
gas phase, knowing the mass, temperature, pressure and
volume. This allows us to find the molecular formula
of a substance if we first determine the empirical
formula from mass percent composition data.
Molar Mass and Density
Similarly: PV = mRT
M
mRT
PM=
V
or:
PM = dRT
(density, d = m/V)
M = dRT/P
Mixtures of Gases ... Dalton's Law
John Dalton studied the effect of gases in a
mixture. He observed that the total pressure of a gas
mixture was the sum of the partial pressure of each
gas.
Ptotal = P1 + P2 + P3 + .......Pn
The partial pressure is defined as the pressure of
a single gas in the mixture as if that gas alone
occupied the container. In other words, Dalton
maintained that since there was an enormous amount of
space between the gas molecules within the mixture
that the gas molecules did not have any influence on
the motion of other gas molecules, therefore the
pressure of a gas sample would be the same whether it
was the only gas in the container or if it were among
other gases. This assumption that molecules act
independently of one another works fine as long as
there is a lot of space between gas molecules in the
mixture and the temperature is not too low.
A useful concept when dealing with gas mixtures is
mol fraction. Consider, for a mixture of gases at
constant V and T:
The fraction of mols of gas A is XA = nA/nT. Since
V and T are constant XA is also equal to PA/PT. XA is
the mol fraction of gas A in the mixture.
Collecting Gases
●
●
●
●
gases are often collected by having them
displace water from a container
the problem is that since water
evaporates, there is also water vapor in
the collected gas
the partial pressure of the water vapor,
called the vapor pressure, depends only on
the temperature
– so you can use a table to find out the
partial pressure of the water vapor in
the gas you collect
if you collect a gas sample with a total
pressure of 758.2 mmHg* at 25°C, the
partial pressure of the water vapor will
be 23.78 mmHg – so the partial pressure of
the dry gas will be 734.4 mmHg
– Table 5.4*
Example Problems
In alcohol fermentation, yeast converts glucose to ethanol and
carbon dioxide:
C6H12O6
2C2H5OH + 2CO2
If 5.97 g of glucose are reacted and 1.44 L of CO2 gas are
collected at 293K and 0.984 atm, what is the percent yield?
Mols glucose = 5.97 g / 180.156 g/mol
= 0.0331 mol
mols CO2 = 0.0331 mol glucose x 2 mol CO2 / 1 mol glucose
= 0.0663 mol (theoretical yield)
mols CO2 (actual yield) = PV/RT
= (0.984 atm x 1.44 L)/( 0.0821 x 293 K)
= 0.0589 mol
% yield = (0.0589 mol / 0.0663 mol) x 100 = 88.8%
Example Problems
At 741 Torr and 44 oC, 7.10 g of a gas occupy a volue of 5.40 L.
What is the molar mass of the gas?
M = mRT/PV
= 7.10 g x 0.0821 Latm/molK x 317 K / (741/760) atm x 5.40L
= 35.1 g/mol
Example Problems
A mixture of gases contains 0.75 mol N2, 0.30 mol O2 and 0.15
mol CO2. If the total pressure of of the mixture is 1.56 atm
what are the partial pressures of each gas?
The mole fraction of each gas may be used here:
total number of mols = 0.75 + 0.30 + 0.15 = 1.2 mols
XN2 = nN2/nT = PN2/PT
PN2 = (nN2/nT) x PT = (0.75/1.2) x 1.56 atm = 0.98 atm
similarly:
PO2 = (0.30/1.2) x 1.56 atm = 0.39 atm
PCO2 = (0.15/1.2) x 1.56 atm = 0.20 atm
Example Problems
A gaseous mixture of O2 and Kr has a density of 1.104 g/L at
435 torr and 300K. What is the mol fraction of O2 in the
mixture?
D pure O2 = MP/RT
= 31.999 g/mol x (435/760) atm / 0.0821 x 300
= 0.744 g/L
D pure Kr
= 83.8 g/mol x (435/760) atm / 0.0821 x 300
= 1.95 g/L
Dmixture = XO2 x DO2 + XKr x DKr
= XO2 x DO2 + (1-XO2) x DKr
XO2 = (Dmix – DKr)/(DO2 – DKr) = (1.104 – 1.95)/(0.744 – 1.95)
= 0.70
Molecular Velocities
• all the gas molecules in a sample can travel at
different speeds
• however, the distribution of speeds follows a
pattern called a Boltzman distribution
• we talk about the “average velocity” of the
molecules, but there are different ways to take
this kind of average
• the method of choice for our average velocity
is called the root-mean-square method, where
the rms average velocity, urms, is the square
root of the average of the sum of the squares
of all the molecule velocities

r a tgea s A M w B
=
r a tgea s B M w A
Effusion
Graham’s Law of Effusion
• for two different gases at the same
temperature, the ratio of their rates of
effusion is given by the following equation:

rate gasA
M wB
=
rate gasB
M wA
Temperature and Molecular Velocities
• KEavg = ½NAmu2 (for one mol of gas particles)
 NA is Avogadro’s number
• KEavg = 1.5RT
 R is the gas constant in energy units, 8.314
J/mol∙K
1 J = 1 kg∙m2/s2
• equating and solving we get:
 NA∙mass = molar mass in kg/mol
rate
rate
•
gasA
gasB
=

M wB
M wA
as temperature increases, the average
velocity increases
Ideal vs. Real Gases
•
•
•
Real gases often do not behave like ideal
gases at high pressure or low temperature
Ideal gas laws assume
1) no attractions between gas molecules
2) gas molecules do not take up space
 based on the kinetic-molecular theory
at low temperatures and high pressures
these assumptions are not valid
The Effect of Molecular Volume
●
●
●
at high pressure, the amount of space occupied
by the molecules is a significant amount of
the total volume
the molecular volume makes the real volume
larger than the ideal gas law would predict
van der Waals modified the ideal gas equation
to account for the molecular volume
– b is called a van der Waals constant and is
different for every gas because their
molecules are different sizes
nRT
V=
n b
P
or
(V −n b )=
nRT
P
The Effect of Intermolecular Attractions
●
●
●
at low temperature, the attractions between
the molecules is significant
the intermolecular attractions makes the real
pressure less than the ideal gas law would
predict
van der Waals modified the ideal gas equation
to account for the intermolecular attractions
– a is called a van der Waals constant and is
different for every gas because their
molecules are different sizes
nRT
n 2
P=
−a( )
V
V
or
n 2 nRT
Pa( ) =
V
V
Van der Waals’
Equation
●
combining the equations to
account for molecular
volume and intermolecular
attractions we get the
following equation
– used for real gases
– a and b are called van
der Waal constants and
are different for each
gas
n 2
( Pa( ) ×(V −nb)=nRT
V